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a:
ctr
l
** x
dx
~~
d(^- log.r)
d(x-logx)
dx
Let us note that formula 2 of
7.3,
^
proved previously, could be proved in an analogous way, because ax
=
158
DIFFERENTIAL CALCULUS
HI.
In practice we sometimes apply Formula 1 several times; *
i
if
e.g. -
y
=
/(#),
=
z
x-
w=
g(y),
h(z),
we
calculate
i
the derivative
t
>
applying the equality:
(toe
dw
dw
dx
dz dy dx'
dz dy
As seen, the derivatives in the like ordinary fractions. 4.
dlogf(x)
^f(x)
dx
above formula behave
'
f(x)
This expression is called the logarithmic derivative of the function /. A knowledge of the logarithmic derivative gives the usual derivative at once. This is easy to show
by
considering the example of the function y
the logy
xlogx
is
=
aP (where
differentiated).
EXAMPLES. dx?
~ dx cos(,r
((3)
The
)2#
= 2x
derivatives of the hyperbolic functions.
hyperbolic sine and cosine (15)
2
~
sinh#
we understand the
and
cosher
= *
2.
To
find their derivatives let us note that
de~ x
___
dx
Hence we /^v
Now
-=
=
Mr
.
dx let
d(x)
d(~x)
dx
_
cosh^
, '
-- = r
dx
us write
v
(17)
___
find easily:
dsinha?
(16) v '
de~ x
cosh
,7?
.
.
smho?
.
By
the
functions
7.
DERIVATIVES OF THE FIRST ORDER
Since, as
it is
159
easily seen,
eosh 2 #
(18)
sinlAr
=1
,
we have .sinho;
d
dtanha? (19)
dx
The functions Namely,
let y
^1
0,
^ =- + //
sihha?
1
and cosh# have inverse functions.
=
.
Then e?~ e~ x
2y
x
+
=
2* 0, i.e. e
whence
I'^+l
i.e.
,
--
log (y
yy^+l
}
'y*
FIG. 16
FIG. 15
FIG. 14
(the A^alue y into account).
cosh2 #-- sinh 2 #
cosh# dx
being negative,
we do not
take.it
way we have
expressed x as a function of y. This is the required function inverse to sinh#. We denote this function by the symbol arsinh: arsinh a?
(20)
Similarly, (x ^ 0). This We find (21)
In this
we
the
find
function
arcosh x
= is
function
inverse
to
cosh a?
denoted by the symbol arcosh.
log (x + )/x
2
l)
,
x
^1
.
160
III.
DIFFERENTIAL CALCULUS
Analogously, a function inverse to cosh a? (x
X ^- \ J/ _LJ log \X the formulae Applying X"
(16)
-
,.. rt
is
and the formula on the
we
find easily that
d
darsinho? --
/rtftx
0)
}
derivative of the inverse function
(22)
<
dx
- --
darcosha? -
,
(23)
1 ,-^-r
dx
ic
This calculation
2
may
-
-
d
1
d^
1
/
log afi
I
.^
2
-
be performed also by
differ-
entiation of functions on the right sides of the formulae (22)
and
(23).
FIG. 19
FIG. 18
FIG. 17
We add that the following formula function inverse to tanho?: +
(24)
1
ar taring
= - log
is
found for the
i
3
Hence fartanh,T (25)
dx
a?
=
i-
dx\2
Bemark. The formulae on the differentiation of the fundamental functions with which we became acquainted in 7.2 and 7.3 together with the formula for the differentiation of composite functions make it possible for us to differentiate an arbitrary elementary function (that
7.
is
DERIVATIVES OF THE FIRST ORDER
101
a function obtained from the fundamental functions of an arbitrary number of compositions).
by applying
A
derivative of an elementary function mentary function.
Now
us
let
the
consider
following
is
an
itself
function
ele-
(non-
elementary): (26)
f(jr)
=
.T
2
-
for
sin
The general formulae
;/
/:
for the
/(O)
,
=
.
differentiation of the
elementary functions do not enable us to calculate /'(O). We must calculate this derivative directly from the definition of a derivative.
We
It
have
is
clear that- for
.r
^
the derivative
/'('')
may
be
by means
of general formulae concerning the of the differentiation elementary functions.
calculated
7.7.
Geometrical
interpretation
of the sign of a derivative
Lagrange's theorem enables us to establish the relation between the sign of the derivative and the increase or decrease of the function. 1.
to
an
>
holds for any x belonging // the inequality f'(r) interval ab, then the junction f is increasing in this
interval.
^1
f'(' r )
<
O everywhere, then the junction
is
decreasing.
> 0, According to the formula (tl) we have for j(x-\-li) > f(x) or /(.*-+ //) < /(.^), depending on whether we assume the derivative to be positive or negative everyli
where. In the first case the function the second one, decreases.
/
increases
and
in
Remark. Assuming f(x) > or /'(#) ^ everywhere, the function / is increasing in the wider sense or decreasing in the wider sense, respectively. 11
162
III.
DIFFERENTIAL CALCULUS
The following theorem
is
the converse to Theorem 1:
If a function f is differentiate at a point c and increases (or decreases) in a certain interval surrounding 2.
>
< 0). Namely, if / increases, then for h > we have f(c ^_ and so > o an d passing to f(c)>
this point, then f'(c)
(or f'(c)
the
fi
limit
we obtain
f'(c)
> 0.
Similar arguments can be applied in the case in which the function / decreases.
We
conclude from Theorem 1 that
3. // f'(c) > 0, then the function f is increasing in a certain neighbourhood of the point c (assuming the derivative to be continuous at the point c).
Similarly:
if
f'(c)
decreasing at the point
/'
<
?
then
the
function
is
locally
c.
Indeed, because of the the inequality f'(c) >
continuity implies
of
the
the function existence
of
a number 6>Q such that there holds f(c + h)>0 for \h\<6. This means that the derivative /'(#) is positive at any point of the interval c h<x
Bern ark. Theorem 3 may be expressed
in the follow-
ing way: if f'(c) ^ 0, then (assuming the continuity of the derivative) the function / is locally one-to-one at the point c, i.e. it is one-to-one in a certain interval
(d> Q). Thus, the function y = /(#) the inverse function %~g(y) in this interval. possesses As we already know, the derivative of this inverse function is equal to the inversion of the derivative of the
c~d
<x
function
/.
the above assumptions we see that for a given value of y the equation y = /(#) possesses one and only one solution for x belonging to the interval c <5, c + d.
By
DERIVATIVES OF THE FIRST ORDER
7.
ex
=
EXAMPLE. The function y
163
decreases for
x
<
1
JC
and
increases for x
>
1.
Indeed,
^L l^\
- x^ ~~ x
dx\x) This expression
Hence
it
x<
if
follows that lim
7)
3.5,
(cf.
<0
is
and
ex
x
1
-'-
and
=
>0
CXD.
so the function
and
increasing, tends to CXD as obtained this result in another
More
(xl)
2
generally,
we prove
if
>
1.
Indeed, lim
=
co
being unbounded
,
x tends to
way
x
in
CXD
4.6,
(we have ((3)).
in a completely analogous
that the function f(x) = exxa (x> 0) is decreasing a and, a (and a < 0) and increasing for x > for x < n a oo (cf. since lim e n 3.5, 7),
way
rc=co
Indeterminate expressions
7.8.
Gauchy's theorem in
makes
7.5
it
possible to cal-
culate the limit of the following form:
lim
(28)
?
W here
/(a)
=
-
Expressions of this kind are called "indeterminate ". expressions of the type |J
// the functions f and g are continuous in the closed b and are differentiate inside this interval x Interval a 1.
<
<
and
if
f(a)
(29)
assuming
~
= #(&), lim
then
=
i
im
the existence of the last limit.
104
III.
JIFFFKUKNTIAJj CALflTLUS
Let us write x
a
1
i
(30)
+ h. Then
it
has to be proved that
im
* --10
h
<
Now, the equations
HO
=
f(a)
//(a)
and the Cauchy
formula give f(a
|-
A)
^
-/0~
Since
lira <9A //-^
=-
+ /0_-
/(
:|
/0-f
+ 0, we
have
whence the formula (30) follows. A similar theorem concerns the left-side limit. When the derivatives /' and //' an* continuous at the points a and y'(a) ^ ^? formula (29) implies immediately the following de 1'Hospital
formula:
!
(
)
(31) -r-fl
A
y\+'l
<J
V*l
similar formula concerns the one-side limit
and the
one-side derivatives.
EXAMPLES, f(x)
(a)
To
find iim
= log(l +
and
,r)
-'--
<J(JT)
let
=
us write
x
.
Hence /\
and
so
(32)
X 2
(
of
)
the
De
1' Hospital, a French mathematician of the second half 17th century, known especially for popularizing the dif-
ferential calculus.
DERIVATIVES OF THE FIRST ORDER
7.
ft
Evaluate lim
(P)
_
= e?\ and -
Let us write f(x)
-.
jc=o
=^
1
.
Then we Hence
have
=
/(())
lim^-U OT = 1
(33)' V
.,_o
Remarks,
-
tj'(a)
-
formula (31) gives lmi^-|-
derivatives
but
/' (a)
But
if
0.
^
/'()
0,
then the
=
=
flf'(),
cannot be applied. In this case higher orders have to be applied.
then the formula the
.
'<
If
(a)
105
(3.1)
of
8.4.)
(Cf. (ft)
The formula
when a
=- oo.
may be
(21))
In other words,
if
applied also in the case ~ lim g(x) lim f(x)
9
r
^c---oo
ix)
then
assuming the existence of the
We
write
and
,r
define,
last limit.
an auxiliary function F(t)
t
by the conditions: Similarly, let
function ^ of
7
is
/''(O
G(t) --
=/(-)) for
^
for
g\\
t
and
continuous at any
and
two continuous functions:
and ^(
/-
(0)
=
(
>)
0.
=
0.
The
/-
as a composition
/.
it is also right-side
t
continuous at the point lim F(t) ^= +
0,
= \m\f t=
I
(cf.
=lim/(a?) \t/
Similarly, the function
to apply the formula
since
(21)),
4,
(18))
-0 =
1
* (0).
J--OG
(i
lim
is
continuous. To be able
W(t)
t=-\ o CT
-,-J-
(t)
has to be calculated.
166
III.
DIFFERENTIAL CALCULUS-
we have
'Now,
_
_ "
J^O.
for
"~"dF~"
Similarly,
#(<)
clx
-'7-
=
Thus
= lira
Besides indeterminate expressions of the form
determinate expressions of the form sidered.
By
lim /(a?)
Sometimes the calculation
form
Fix)
-7-77
may
also
= oo = Yimg(x)
in-
con-
and O(x)
=
if
lim/(#)
.
For, writing
we have
-
,
= lim ..
this last indeterminate expression
Similarly,
.
of indeterminate expressions
be reduced to the form -
r lim -~{
and
are
,
such an expression we understand
where
of this
00
--
=
and
ar=a
is
lim
of the
=
oo,
form -
.
then the
ar-a
indeterminate expression \imf(x)'f/(x) of the form O-oo x**a
may
be reduced to the form -, for
li
7.
lim
If
DERIVATIVES OF THE FIRST ORDER /(a?)
=
=
oo
we reduce the
lim g(x], then
x*=a
167 in-
x***a
g (x)] of the
determinate expression lim[f (x) x=a
also to the
form -
,
form oo
oo
writing
It remains to consider indeterminate expressions of the form 0, 00 and 1. They may be reduced to the above considered by means of the relations
EXAMPLES. -
,.
lim (costf)*- lim
(y)
-logcosrr
6*
^m
= e x ~ +QX---=
=1,
because (log cos x)' r ^^ +0
log cos a?
^
x~-fO
We
is
5.3).
an expression
y
when o?>
a^
1)
dx
lim wX ___ ajiBM-j-Q*^
= lim "t
*
form -
(29)
W "i
"^*^
1
=
1.
oo.
for lim
(cf .
=
of*
7.6, 3)
we obtain
,
-/
The proof that (cf.
4.6,
1
into account
is
/+(0)
~
oo
(8)).
.
an indeterminate form
this limit is equal to
be derived directly, without applying of the
ferential calculus
=
~\\ 1
-
The expression lim (xiogx)
of the type
,
and taking
whence (e)
0, /(O)
= rlim
'
= #*(logo? +1)
f*E(\C\CF 1 v
/r
iC^+O*^
7 v
the
of
Applying formula
the fact that -=- (#*
may
.
have
= rlim -
(
=
defined as follows: f(x)
/
This
=
1
Find the right-side derivative /^(O) of the func-
(S)
tion
tan x
,.
^=+0
,7?
dif-
108
DIFFERENTIAL CALCULUS
III.
The
7.9.
derivative of a limit
THEOREM. lim
(35)
//
f tl (jc)
tlie
equalities
- / (x)
Mm fa(x) = g (x)
and
??,=oo
?<
are satisfied in the interval a
-;
x
= oo
<, b, the
continuous and uniformly convergent (36)
f'(x)
- g(r)
functions then
f'tl
being
f lim /n (.r)
'i.e.
,
to
Let c be a given point of the interval ab. Let us estimate the difference
>
(37)
+
c
)-
*]-''<
Let
>
e
i ,(o)
=f.(c +
e*)-^)
be given. Since the sequence {f'n } there exists a 7c such that */,
is
.
uniformly
convergent to (38)
i/;c*)-ff(.*oi<*
n > k and for any x belonging The function g being continuous
for
convergent sequence 1), a number e >
formly
of
6.1,
(cf.
equality \x
c\
<
(5
(39)
to the interval ab.
as the limit of a unicontinuous functions
exists such that the
|sfW)-flf(<0|<e.
Thus,
assuming n
>
k
iand
\li\
<
6
we obtain the
following estimate:
i.e.,
in-
implies
by
(37),
.l^M'Urtc) Since
lim fn (c + A)
= /(c + h)
and
lim /n (o)
= /(c)
,
7.
DERIVATIVES OF THE FIRST ORDEK
169
we have *
A
M
i
6. Passing to the limit as A tends to if only |A| < therefore obtain equation (36).
Eemarks.
Even the uniform convergence
(a)
we
of a se-
of functions does not
imply the convergence of quence the sequence of derivatives, as is shown by the following
example:
(p)
The theorem remains true the equations (35)
if
we assume that the
satisfied at a
given point v instead of assuming it to be satisfied in the whole interval ab. For, if the sequence of functions is convergent at a point c and if the sequence of the derivatives is first of
is
uniformly convergent in the whole interval ab, then the sequence of functions is convergent in the whole interval ab.
e
Indeed, according to our assumptions, to a prescribed > there exists a number k such that the inequalities
(40)
|
f n (v)
- fk (c) <
and
K
|
|
are satisfied for n
Let us write
>
and a
k
F n (x)
=
<;
f n (x)
x
f'n (x)
- f'k (x) <
e
\
< b.
fk(x)
and .r=-}-h. Then
we have .F'
t)
)
Hence, by
(40),
The
We (41)
-
0/0
\Fn (v)\ <e + (b a)e (j?) is convergent.
/
rt
derivative of a
shall
\
we have
Thus the sequence 7.10.
(c
- fk (c) + h[fu (c
prove that
power
series
= 8 (1 -\-b
a).
170 for
DIFFERENTIAL CALCULUS
III.
any x lying within
=
i(x)
(42)
the interval of convergence of the series
Moreover, the series (42) and g(x)
(43)
=
a1
have the same radius of convergence.
We shall first prove the second part of the theorem. Let r be the radius of convergence of the series (42) and let < c < r. We shall prove the series g(c) to be convergent. Let c < C < r. Let us compare the terms of the series (44)
\a 1
\
+ 2\a
2
\c
+
+ n\an \cnn +
...
...
with the corresponding terms of the convergent 6.3,
(cf.
Theorem
(45)
|
1):
% |C + |a
Since lim \/n
series
1
2 2 |<7
(cf.
+ ...+
an \C
3.5, 8)
and
n
+
...
>
we have
1,
C <-
,-' I-
and thus
for large n.
Hence we conclude that the series (44) is convergent and so the series g(c) is absolutely convergent, too. Conversely, if the series (43) is absolutely convergent, then the series (42) is also convergent. Namely, \an cc^\
In this
= \x\- \an xn ~ < \x\- \nan xn ~ l
\
l . \
way we have proved
within the interval of
that any point lying convergence of the series (42)
belongs to the interval of convergence of the series (43)
DERIVATIVES OF THE FIRST ORDER
7.
and conversely: any point lying within the
171
interval of
convergence of the series (43) belongs to the interval of convergence of the series (42). Thus, both these intervals are equal.
Now, let us proceed to the proof of formula (41). Let x be an interior point of the interval of convergence of the series (42) and thus of the series (43), too. Let the interval ab be any closed interval surrounding this point and lying inside the interval of convergence. By Theorem 1 of 6.3 the series (43) is uniformly con-
vergent in the interval
ab.
This means that
when we
write
a sequence uniformly convergent to the function g in the interval ab. Applying the theorem of the previous section we obtain the forthe functions
fn
constitute
mula (41). As is easily series
seen, formula (41) states that a power can be differentiated, term by term, like a polynomial. oo
More
if
generally,
by
differentiating the series
u n (x) n-O
00
term by term we obtain the convergent
and
series
the
moreover,
if,
u'n (x)
functions
continuous, then 00
00
N u n (%)
-_-
(46)
71
EXAMPLE.
=
u'n (x]
Differentiating the series
_JL -1 ~
T^x we obtain
/ =
71
for \x
<
1:
_L_=14 (1-X)*
.
uniformly u'n (x)
are
172
JIT.
7.11.
DIFFERENTIAL CALCULUS
The expansion of the in
functions log(l
+ x)
and arc tan #
series
power
Let us consider the known formula (47) It is
obvious that the series appearing in this formula
the derivative of the series
x-
ti(x)
j$
+
jA *---
is
,.5
jA --
+
'--
...
r
Hence the functions iog(1-fj?) and 8(x) have the same and so they differ only by a constant (cf 7.5, 4). Let us write log(l + ,r) = H(x) + C. The constant C may
derivative
.
= 0; we obtain = C log(l + 0)-tf(0) - 0.
be calculated by substituting x
=
=
to
Hence
y
AS
(0).
log(L
+ 0)
So we have proved the functions log(l +x) and S(x) be identical in the interval !<#
otherwise:
log(l
(-18)
+ *) =x-
?+*-* +
J
(M <
...
1)
.
we deduce from the formula
Similarly,
d arc tan x
1
that arc tan x
x
Substituting
=
arctanO
0,
and
we obtain C
arc tan#
(49)
=
/v>3
jj
Thus, \x\
<
we have proved 1.
We
for
x
-~
-
5
taking into account 0. Consequently,
that
= >i5
+~
t.j
for
j^3
= C+x
-v7
~+
>
...
(
\x\
<
1)
.
<
that the formula (48) holds
show that it holds also for x 1 the series on the right side of
shall
Indeed, formula, i.e. the series $(1) == 1 and thus by the Abel theorem (
J+ J 6.3, 3)
=
1.
this
..., is convergent the function
7.
is
left-side
DERIVATIVES OF
continuous at the point
=
8(1)
lim
>S'(,r)
-,
J--1-0
ORDER
FIRST
T II K
Hence
1.
+ #)
lim log(l
173
=.
log 2
,
./--1-0
since the function log(l
+
is
,r)
continuous at the point
1.
Consequently,
Iog2- 1-i
(50)
Similarly,
i
1
i
+
1-...
we deduce from formula
the
Leibniz
right side of the equation (49)
by T(x]
(49)
formula: (si)
1'
4""
Denoting the
3
+ 5-
^
+J
"
we have '
lim 7
T(l)
whence formula
!
lirn
(.r)
arc tan, r
arc.
tan
.1
,
(51) follows.
Remarks, (a) Tn practice the Leibniz formula is not convenient for the calculation of the number TT; e.g. to calculate the first three decimals of the number rc/4 it would be necessary to take the sum of the first 500 terms
Much
of the Leibniz expansion.
following formula. Let a =- arc tan i,
=
,
i.
convenient
1
tana
e.
=
is
J,
the
tan/5
Since
i.
and
arc tan
(i
more
1
=
arc tan (it/4), so
Ti/4
= a+
ft
=-
arc tan
Hence, by formula (49), we obtain: W -:L_ ! i. JL ! - 1 " "* r 5 32 4 3 2 3 8 3-27 i r
-
-
i
J
-_ '"
l
-
+ arc tan J.
5 243 .
Other formulae giving even faster methods for the calculation of TT are known. (P)
(52)
The formula -
log(
1
(48) leads to the following inequality:
+ x) <
1
for
<x<1
.
174
DIFFERENTIAL CALCULUS
III.
For,
we have
+ a :)= 1 -| + f-...^T...
ilog( 1
(53)
<
For
#
<
1 this
is
an alternating
series satisfying
the conditions
x !>> 2
2
thus,
ofi
3
xn nse=0o n
_
and
>...
Inn
according to the theorem
series is less
than
its first
term,
3.3, i.
e.
=
the
:
'
sum
of
this
the inequality (52)
holds.
Similarly
+
(54)
we prove that
iog(l+0)>l
since adding to the
by
2,
series
for
(53)
0
the series (48) divided
we obtain
It is easily seen that by omitting the first term we obtain an alternating series satisfying the assumptions of Theorem 1 of 3.3. Hence we conclude that the sum of this series is positive.
Thus, formula (54) holds.
7.12*. Asymptotes
By applying the differential calculus it is possible to find the asymptotes of a given curve y /(a?). The straight line Y is called an asymptote of the curve aX-\-b
=
^
y =/(*), a
= lim f(x)
and
b
= lim \j(x)
ax\
;
hence it is seen that the direction of an asymptote is a limiting direction to which the direction of the tangent to the curve at the point [>, /(#)] tends, as x tends to oo;
DERIVATIVES OF THE FIRST ORDER
7.
175
moreover, the distance between this point and the asymptote tends to 0, because this distance is equal to [/(#) ax 6]cosa|, where a is the angle between the asymptote and the .X-axis. |
oo. Similarly we consider asymptotes for x tending to c the line to the (parallel Y-axis) Finally, straight
X=
an asymptote
called
is
lim x=c
f'(x)
==
co
=
lim
of the curve y f(x).
f(x),
we
Similarly
x
if
the
define
a;=c
o
asymptotes to the curve y =f(x),
x>
c.
FIG. 20
EXAMPLES. As
=
hyperbola y
it is
easily seen, the
1/x are the axes
To find the asymptotes - -.
calculate y'
a
=
and
tote.
=
1,
Hence lim y
and the
z=-0 7.13*.
y'
=
asymptotes of the
and Y.
= e~ lim y' = 0.
of the curve
=
1
and
straight line
A second asymptote is the
and lim
A
6
X
X
Y=
y
1
is
llx ,
we
Thus,
an asymp-
Y-axis, because lim y
-=
oo
#=-0'
oo.
The concept of a
differential
rigorous mathematical sense may be given to the concept of a differential in the following way. By the differential of a function y = f(x) at a point x with
176
DIFFERENTIAL CALCULUS
HI.
we understand the product
regard to the increment h f(x)h.
We df(x)
(55)
write
= f'(x)h
or,
more
briefly,
dyf(x)h.
The symbol
df(x) or dy, convenient in applications for takes into account neither the dependence simplicity, of the differential on the variable x with respect to which the differentiation is performed, nor the dependence on its
h. To satisfy these requirements we should have to write e.g. dx (y, h). Moreover, if we should like to point out that the differential is considered at a point # we should have to write [Y7 r (/(j?), h) Jr ^r l )V analogy with
the increment
,
the notation for the derivative at a point .r By formula (55), dx = A, for substituting f(x) .
we have
f'(x)
Thus we may substitute
We
in the formula (55) li Dividing both sides of
=
obtain
x
1.
f'(x)dx. dy equation by dx we have then
dx. this
where the right side is the quotient of two differentials (which may be considered as an explanation of the Leibniz notation for the derivatives). Speaking more strictly,
we have
- ^ / /? dx (x, (
f(x)
?J h)
for
every
ft
^
0.
The
formulae of the differential calculus may be written in the differential notation as is seen by the following examples:
dsinx
Formula 4
of
= co8#&r
,
dc
=
7.4 for the derivative of a
.
sum
leads
to an analogous formula for the differential of a sum:
d(y
(56)
Indeed,
let
+
ff)
y =/(#), z
= g(x). We
have
7.
DERIVATIVES OF THE FIRST ORDER
177
Analogously, we prove that d(yz)
(57)
Now we
= ydz + zdy
shall give the
formula for the differential
of a composition of functions. Let y g (y). f(oo) and z
=
=
Namely, dx z
=
dof(x)
-ij-dx
.
-
Then
g'(y)1'(x)dx
=
g'(y)dx y, ac-
cording to (55).
x+h
Let us note that by (55) we have also dy z = (j'(y)dv y. Comparing this equation with equation (58) we see that if z is a function of the variable y, then the formula for the differential z remains formally unchanged after representing y as a function of a new variable x (however, the index at d showing with respect to which variable one has to differentiate has to be obviously omitted).
The geometrical interpretation
of the
differential
is
immediately seen in Fig. 21.
The following theorem
holds:
the increment of the function
Ay
=
The f(x
difference between h) f(x) and the
+
12
178
III.
DIFFERENTIAL CALCULUS
dy, divided by the differential dx tends with this differential, i.e. together differential
(*& dx0 ^=-^
lim
(59)
=
to
.
Namely,
da?
and
In practice (e.g. in physical applications) the above theorem often makes it possible to replace the differential dy by the increment Ay. Exercises on 1.
7
Differentiate the following functions:
4) sec a?
5) co sec x
,
7) arc see a;
10) log sin a;
6) cotano?
,
8) arccoseco?,
,
11) log tan x
9
9)
arccotan^,
12) arccos(l
,
^y
1
13) arc tanJL
x)
,
8
2
,
14) logsinh^,
,
15) artanh|/iP.
*H7
Prove that
a function / is differentiate, then /' possesses the Darboux property (i.e. passes from one value to another through all the intermediate 2.
its
if
derivative
values).
<
<
(First prove that if f(a) /'(&), then a point c exists in the interval ab such that f(c) 0. The general
=
case
when
f'(a)
f'(b)
may be
easily reduced to the
previous one considering the function f(x)
Ax.)
7.
3.
DERIVATIVES OF THE FIRST ORDER
Given the
and the sum
series
f(x)
=
vi
2i w =i
X --%
Ixn
n+ l
\
rl
I
\
n-}-/
of the series of the derivatives.
the result obtained with the theorem in 4.
Prove that x, then
if
a
function
179
/
is
Compare
7.9.
differentiate
at
a point
in
v
f(x)
-
8how by an example that the converse relation does not hold: the above limit (i.e. the so called generalized, derivative)
may
exist although the function
is
not
dif-
ferentiable at the point x. 5. Give an example of a function having a generalized derivative at a point of discontinuity.
Give an example of an even function (i. e. such that f(x) - f(-x)) for which /'(O) = 0, although the function does not possess an extremum (even an improper) 6.
at the point 0. 7.
Prove that
maxima at a minimum 8.
a
<
x
if a differentiate function possesses the points a and ft, then it possesses also at a certain point between a and b.
Let a function
<
b
and
right-side limit
be continuous in an interval
/
differentiable inside this interval. If the
lim f(x) exists, then the right-side dex<=a+o
rivative of the function
equal: l+(a) 9.
lim
/ exists,
too,
and they are both
f'(x).
x=a+0 Prove that among
all
rectangles with the
same
perimeter the square has the greatest area. 10. Prove that among all triangles with a constant perimeter and with a constant base the isosceles triangle
has the greatest area. 11. Prove that among all triangles with a constant perimeter the equilateral triangle has the greatest area. 12. Prove that if the derivative /' is strictly monotone, then for any x the curve y = f(x) lies on one side of the tangent at this point.
180
HI.
DIFFERENTIAL CALCULUS
Hence conclude that
>I + x
s
eP
for
13.
Prove that (- + ^)log(l +x) \X O/
14.
Prove that
<
x
^
0.
<
1 for
<
x
1.
/Y
<:log(l
"
15. 16.
Expand
log 1 / W
----JL
in
+ #) <#
a power
x>
for
1.
series.
t)u
Given two points
P
and
(>
lying on the plane on the X-axis such
above the J-axis, find a point R sum of the segments PR and RQ is the shortest possible. Prove that the (acute) angles between these segments and the X-axis are equal.
that the
(Interpret this result in the theory of optics: the path of the ray, the ^Y-axis = the mirror). 17.
18.
-^
i
,
Evaluate
tana?
,.
,.
hm-
. '
Prove the existence
lira
|
..
\,
/TU
x )tan#,
s
'
hm
x i *
PRQ x.
of the limit (called the Euler
constant}: ...
&
o
-
n
(prove that the considered sequence is decreasing and bounded, applying the inequality from Exercise 11). 19.
Prove that
if
the differentiate functions
/
and g
satisfy the inequality f(x)g(x) ^ <)'(rff(%) for every x, then between any two roots of the equation f(oo) = there exists a root of the equation g(x) = 0. (Consider
the auxiliary function
8.
rO
DERIVATIVES OF HIGHER ORDERS
8.1. Definition
and examples
The derivative called
i(x] --;
of the derivative
of a function /
the second derivative of this function.
the third derivative
is
is
Similarly,
the derivative of the second de-
DERIVATIVES OF HIGHER ORDERS
8.
rivative. Generally, the n-th derivative
is
181
the derivative
We
denote the derivatives of l)-th derivative. of the kind: higher orders by expressions of the (n
r
i
9
,,
iff I ?
/
m Ol
AH) ?
/
?
?
dy
cTy >*> T~n> "
cfy
d*y
~' Tl* ax ax
,
T~3 ax
ax
So we have
let
E.g. /"'(/E)
^
fl C
and
6
k (r n \
_1^_J
(1)
dx
=
f(x)
We
a?.
/<*>(#)
=
have
for
/'(#)
>
w-
=
3^2
More
3.
,
/"(a?)
-
60?,
generally,
= w^-lJ-.-.-^-fc+l)^-*
for
kt^n,
n
If a is a real number (but not a positive integer), then we have
(2)
for
an arbitrary positive integer
fc
and
for
an arbitrary
x>~l. The function
ex
has the property that
its
derivatives
(^)
= sin
of all orders are identical with itself.
Taking /'(#)
(3)
/(#)
= sin
= coso?
a?
we have er
,
f
/'"(#)
Here the equation
sino?
(x)
n /<
=
>(#)
coso?
=
,
IV ,
w+4 /<
>(a?)
/
holds.
a?
.
The same
cos a?. equation is satisfied by the function f(x) Let us note by the way that it is convenient to consider the derivative of order as equal to the function />)(#)
n
=
= 0.
f(x).
The above given equation holds
The formula the
ft-th
also
for
(1) makes it possible to calculate easily derivative of a polynomial of degree n: /(#) =
182
III.
DIFFERENTIAL CALCULUS Substituting x
.
Jc
=
0, 1,
we
n)
...,
=
in
/
(fc
>(#)
(for
find:
Hence we obtain the formula s^n
/y*2
1(X)
(5)
-
which we shall generalize below. As an example of the calculation of derivatives of higher orders let us mention the formula for the second derivative of the inverse function.
dx
A we know, ^As
=
dy 1 -:
d*x
dy
2
~
^ T Hence
.
dx
dy
dx) dx
\
dx
dx2
^ ~
dx ?
IdyV dy
dy
i.e.
dx2
dy*
'
\dx,
In applications in physics the second derivative is often of importance. In particular, the acceleration of a particle as a derivative of the velocity with respect to the time is the second derivative of the distance covered
with respect to the time, the particle
mass
is
equal to
i.e.
m
-^ Hence the force .
-j-^
,
where
m
acting on
denotes
the
of the particle.
8.2*. Differentials of higher order
We
define the
second
a
function
/
as the differential of the differential of the function
/.
Speaking more
strictly,
we
differential
define
it
of
as the differential
8.
DERIVATIVES OF HIGHER ORDERS
183
with respect to the increment ft of the differential of / with respect to the increment ft; the variable with respect to which the differentiation is performed is x. Then, writing y = f(x) and denoting the second differential by d2 y we have
the function
-
d(dy)
= d(y'h) -
= y"W - y"(dx)*.
^(y'&)&
Hence we
is
see that the second derivative of a function / the quotient of its second differential over the square
of the
differential
dx (which ^explains the notation of
the second derivative). n
Generally, the n-ih differential d y
the differential, i.e. respect to the same increment ft). tt-th iteration of
is
=
defined as the l
d(d?~ y) (with
Then we have
tfy=y
(6)
We
n
(n
\dx)
.
prove this by induction. Formula
(6)
holds for
=
1 (and as we have seen for n 2, too). Assuming this formula for n, it has to be proved for w+1. Now,
n
dn+l y
- d(dn y) = d[y (n \dx) n ^^-[y
(
]
EXAMPLE. The derivatives
of higher orders of the the (cf. previous section) may be calmeans of differentials of higher orders as
inverse function
culated
by
follows. Let the function
The
x
g(y) be inverse to y
differentials dy, d*y, d*y etc. will 2
dy y, dly, dly
etc.
Since y'x'
1,
with respect to
^
(/7i/'
y,
Evidently, d y
we have dy we obtain
/7r '
Ty'
=
~
/(#).
be understood as
0, d*y
y'x'dy.
=
0,
...
Differentiating
184
DIFFERENTIAL CALCULUS
III.
Similarly,
= d*y =
[y'"(x')*
way we
In this
+ 3y"x'x" + y'x
obtain
equations
which
make
possible for us to calculate the derivatives x', x", #'",
it ...
?
successively: y'x'
"
y
-
1
f
(x')*
,
+ 3y"x'x" + y'x'" =
,
8.3. Arithmetical operations
Let y that
=
j(x), z
=
g(x).
We
verify easily
by induction
and
For the n-th derivative
of a product, the following
Leibniz formula holds: (9)
We
prove this formula by induction. For n 1, formula as the formula for the derivative of a product. Let us assume this formula to hold for n shall
we know and
let
this
us differentiate
it.
We
obtain
8.
DERIVATIVES OF HIGHER ORDERS
185
since (*\
+
\fc/
(cf.
1,
\-/ - n+1 *
/
\*-i]
\
\
;
(4)).
EXAMPLE. To
find the n-th derivative of the function
xe?, let
us write y
the two
first
=
e*
and
2?
,7?.
As
is
easily seen, only
terms do not vanish in the formula
(9).
Hence
n
d-
(xe?)
=
dx
e%T + w)
.
Taylor formula C)
8.4.
The Lagrange formula
the "first approximation" of the following Taylor formula: (11))
7,
(
is
Let us assume the function f to be n times differentiate b. Let us write h = b in an interval a x a. Then
< <
may
f(b)
be represented in the following form:
(10)
+
(
b
~a
n~ l )
f
n~l
\a
\
+ ft
where
where
and &' are suitably chosen numbers < < 1 and < &' < 1.
satisfying
the inequalities
Proof. According to equality
(10),
we have
Brook Taylor and Colin Maclaurin (cited in the following), respectively English and Scottish mathematicians of the first half x
(
)
of the 18th century.
186
III.
DIFFERENTIAL CALCULUS
Let us denote by gn (x) an auxiliary function obtained from Bn by replacing a by x, i.e. (13)
gn (x)
= /(&)_/(o>)-
Differentiating
Thus,
&)=0,
flf;(a?)
n ()
flf
-
we obtain
-TT-rr/^V)-
At the same time
-.Bn.
Applying the mean-value theorem,
we obtain
i.e.
(-!)!
whence
according to the second part of formula (11); the so called Cauchy-form of the remainder. To prove the first part of formula (11),
Lagrange formula for the remainder, we
shall
this
i.e.
is
the
apply to
DERIVATIVES OF HIGHER ORDERS
8.
n
=
the functions gn (x) and u n (x) rem ( 7.5, 5). We obtain
the Cauchy theo-
(b~x)
Taking into account the equations u n (b) = n~l and u'n (x) = we conclude that n(b x)
-Rn _ n
(
b
-a-0h)
~n
n- l . n)
u n (a)
0,
=
h
n
^
(n-l)l
Rn is obtained in
whence the remainder
Thus, the Taylor theorem Substituting b laurin formula:
=
x and a
i(x)
= /(O) + 11 /'(O) +
the Lagrange-forni.
proved completely.
is
=
0,
we
obtain the Mac-
/yfl~~l
/v
(14)
187
...
+
(
.
.
/'"-"(O)
+
,
jy,
where
Rn -
(15);
n
^i f \0x) w!
The above formula
is
= ^ ^""^ (n 1)! satisfied
n
f \0'x)
.
assuming the function
/
to be n times differentiable in the closed interval 0# or #0, according as to whether x > or x < 0. (As is easily seen from the proof, the assumption of the n times diffe-
be replaced by a weaker assume that the whole interval in the continuous
rentiability of the function it
assumption, namely, (n
l)-th derivative is
and that the
w-th.
is
may
sufficient to
derivative exists inside the interval; we have made for n = 1
these are the assumptions which in Kolle theorem).
EXAMPLES AND APPLICATIONS, (a) Let us apply the Maclaurin formula substituting f(x) = e^ and n = 2 in formula (14). Since f(x)
=e?^ f"(x)
we have e
x
=
and
/(O)
- 1 - /'(O)
,
188
III.
DIFFERENTIAL CALCULUS
Hence we conclude that
any x the inequality e*^l + x
(16)
holds, because
x2
>
for
e &x
and
>
0.
Inequality (16) possesses a clear geometrical content on the graphs of the functions y = ex and y = 1 + x. (P) The Taylor formula makes it possible to sharpen the de 1'Hospital formula (7, (31)) as follows: if the functions / and g possess continuous n-th derivatives and if
= g'(a)
/(a) (j(a)
-
,
f(a)
,
,
...
,
9
,
(n ' l)
...
(a)
,
/<-D(a)
-
-
but
,
,
^
g^(a)
,
then V (17) '
Indeed, according to our assumptions and the Taylor formula (10) (with b replaced by #), we obtain
and
Thus,
_ fM[a+6(x
a)]
/_(#[ ~~
~g(xj
'
ffi[a+S'(p-a)}
Passing to the limit for x tending to a and basing of the functions / (n and # (n) we obtain formula (17).
on the continuity
>
,
For example to calculate lim^ -, we write x~o xsmx # sin x and g(x) ^sino?. Thus we have -.
=
/(O)
=0-0(0),
r(a?)-l-cosa?, g'(x)
/'(O)
-
=
,
1(x)
=--
sin o?
na?)
=
,
=
2
.
+
a?
cos a?
,
=
DKKIVATIVES OF HIGHER ORDEKS
8.
Applying formula M Similarly
we
sina;
.T
..
lim s=o
*)
:
a?
=
n
(17) (for
=
-
we obtain
2)
=
.
2
sin a?
find
lim (cot a? ar=0
(19)
-)
= 0.
ay
\
,1 = xcmxsi&x and
_,
For, cot a? f(x)
180
,
writing
:
x
#sma?
= xcosxsmx
and
g(x)
=
a?
sin a;
we
obtain the required formula applying a calculation completely similar to the previous one.
Applying the Maclaurin formula to the function a?) and taking into account that
(Y)
= log(l +
f(x)
we
get
for
x
r
>
1.
In particular, we have, for x
n+l
-,
I
oo
^
Since the series
n-^l
i
convergent
(cf.
--
.
log--,/
/
--- log& T s = vhm /,1+ -1 + ...M,1,23 ,
n _>\
3, (14)),
We
is convergent, too. I lg ----) '^' n-l u& / of the last series by y. Thus we have
series
sum
^
-|
"i M
,
2
-
1 2
-
the
denote the
190
III.
DIFFERENTIAL CALCULUS
But lim [log(w+l) n
n*=oo
we obtain
quently,
-=
lim log -
logri]
oo
(cf.
7,
M
Exercise 18):
y
(20)
Approximately, the Euler constant y 8.5.
if
conse-
0;
Expansions
in
power
=
0,5772...
l (
).
series
We conclude immediately lim En = 0, then
from the formula
(12) that
(21)
w=0
In particular,
we have lim
if
Bn =
in the Maclaurin
n=oo
formula, then the function f(x) a power series:
be expanded in
may
00 rp
(22)
j(x)
,y
=/(0)'+ I j/'(0)
+|
/"(0)
+
...
=
i
X ^
2,
/v7l
J
Before proceeding to applications of the above theoderivatives / (n) are uniformly exists 0#, i.e. if a number such that the inequality > |/ (n) (@o?)| holds for each n
rem we note that if all bounded in the interval
M
M
and
<
any &
satisfying the condition has the expansion (22) in a (Maclaurin) for
For, \Rn
we have
(*)
Till
xn
= \
also lim
now
it
is
M, and 9
nl
En not
<
then
1,
power series. xn
since lim
r
=
0,'
0.
known whether
is
a rational
number
or an irrational number.
Leonhard Euler (1707-1783), a great Swiss mathematician; mathematics owes him, besides many new results, a systematic elaboration of the mathematical analysis of those times.
8.
DERIVATIVES OF HIGHER ORDERS
Eemarks.
The convergence
(a)
191
of the series appear-
ing on the right
side of formula (22) is not a sufficient equation (22), i.e. of the expansibility of
condition of
the function f(x) in a power
series. _-L
For example the function f(co) e x2 (for x ^ 0), = be in cannot a /(O) 0, expanded power series although its Maclaurin series is convergent; we have here /(w) (0) = for each
n.
(p)
power a
Formula series.
= /(O),
aa
=
may
(4)
if
For, f
/QQ\ & ij \
and
T- -/'(0),
a ^fi
I
be generalized
=
/ (x)
+ a^ x + a
a
a general then
to 2 oc?
-f
. . .
,
generally,
i
nl
"/ /W/fh \
/
In other words, if a function possesses an expansion power series, then it possesses only one such expansion, namely, the Maclaurin expansion (formula (22)).
in a
To prove
this statement,
we
shall first
prove by
in-
duction that the assumption
implies
Substituting in this formula x
APPLICATIONS.
e?
1.
0,
=1+
we obtain equation
+
+
To prove this formula, we note we have / (n>(#) = & and / (0) = (ri)
rivatives
< x,
of
all
orders
then jM(Qx)
In particular:
<
1'.
...
=
(23).
..
that writing
Moreover, the de-
1.
are uniformly bounded; for,
if
<
1.
e?
and
e=
1
+
if
>
x,
+^+
then
...
n /<
>(6te)
192
HI.
x
sin x
2
DIFFERENTIAL CALCULUS x*
-
--
1!
3!
a \-
5!
Applying formula
(3) of
8.1
we conclude immedia-
tely that the derivatives of all orders of the function sin are uniformly bounded (by the number 1) in the set of all real numbers. Moreover, /< n >(0) = sinO = for
n
=
and
0,2,4,...
Similarly, 3.
=1
cos#
=
/'(O)
we prove
1,
/' "(0)
=
HO) -
-1,
1,
...
that
^ + j-
.-
f
We for
shall
arbitrary
integers)
and
now
deduce real
for \x\
~
00
the
exponents
<
not
are
(which
positive
1:
/
/
K
v
.
-J
-J
n
2j
.
~~j "'
7l-0
We
Newton binomial formula
shall consider the
cases of x positive
and
of
x
negative, separately.
x
1
> 0.
Since
the remainder in the Lagrange-form following form: (cf.
(2)),
is
of the
n Since
(cf.
3.4,
6)
--
lim
for
|a?|
<
1,
to prove that lim
xn
=
0>
W>*
n=oo
Rn (%) =
it
suffices to
show
n=oo
that for a given x the sequence (1
+ Qn x)
a ~~
n ,
n=
,
1 2 ,
,
bounded. Now, according to the inequality x > a have 1< 1 + 9n x < 1 + x. Thus, 1 < (1 + Sn x} < (1
is
x (
)
We
depends on
add the index n n.
to
B
in
order to point out that
...,
we
8.
DERIVATIVES OF HIGHER ORDERS
a
+ #) < (1+ On x)a < 1 according < 0. Moreover, (1 + 6n x)~ n < 1.
or (l or a
Hence the considered sequence x < we have 2
is
as whether a
^
bounded.
Writing the remainder in the Cauchy-form,
0.
\n
_L;.
we have
Just as before,
to be proved that the sequence (1
it is
193
0'n )
n ~~ l
+ 9'n x) a
(I
_ Qn \nl (^+ *yn x/
~~
n
r
i
(l+0'n x) y
=
x.
a~ l
are
Then y
and
For this purpose we write and &'n > n y, whence 1 9'n <
bounded.
>
l
< 1 - 0'n y < On
)
1.
Thus,
(^j$~
the other hand, 1
<
y
according as whether a sequences are bounded.
1
and
&'n y <~L
1<(1
or
< 1-
>
so
#;_
< 0. Thus, lim Rn () = 0. or a
Hence
1
both
n==oo
EXAMPLES. Substituting a
.1
/^
For a
=
^
,
1
we obtain ,
1-3
.
we have
1
=1
1
_,
.
1/1
=
+i
1-3
1-3-5
2-4
2-4-6
,
x-\
2
Hence
~ i/flT^i
2^
2^1
2-4-6 13
194
III.
DIFFERENTIAL CALCULUS
The above formula makes
possible to apply to the expansion of the function arc sin a? in a power series the it
same method which we have applied expanding the functions logx and arctan#. Namely, as is easily seen, the series appearing in this formula is the derivative of the series
&
1-3 x5
1-3-5 x 1
hence
arc sin a?
But arc sin x
=
;
dx
At the same time i.
e.
arc sin a?
=
=
x2
[/l
S(x)
+C
.
and so (7 = 0, way we have the following
=
arc sin
8(x). In this
=
8(0)
formula:
1-3 x5
1 Xs
1-3-5 # 7
,
,,
_,
note that substituting x = i and taking into account the fact that sin n = i we obtain:
We
*
H 6
A
8.6.
-
~~
_
_L
1 2 '3
2
criterion for
1
ll^ _JL + 1-3-5 _1 *
-
23
'
2 4 5^~2 5 -
2 4 6 -
-
7~T
7
'"
extrema
We have proved in 7.4, 2 that if a ( differentiate) function possesses an extremum at a point x, then f'(x) 0. the theorem not the converse is true: However, equation = does not imply the existence of an extremum /'(#) at the point x; the
shows
example of the function
f(x)
=
oc?
The investigation
of the derivatives of higher orders leads to the following more precise criterion for this.
the existence of an extremum. Let us assume that f(c)
and
(n)
^ 0.
n
=
is
0, /"(c)
0,
...,
an even number,
/^-^(c)
=
the function f
/ (c) Then, if has a proper eoctremum at the point c, namely a maximum, n (n) if f ^(c) < 0, and a minimum, if f (c)> 0. But if n is
8.
DERIVATIVES OF HIGHER ORDERS
195
an odd number, then the function does not possess an extremum at the point c. (We assume the continuity of the n-ih derivative at the point e). Indeed, applying the Taylor formula
(24)
f(c
+ h) =
whence according to the assumptions
we
we have
of
our theorem
obtain J
1(c
(25)
+ h}-f()^ ~^\c+0k).
Let us assume n to be an even number and let f Because of the continuity of the function /
(li)
(/l)
point c
we know
equality \x
\h\
so
that
<
even
for
a
if
c\
<
/<">(
<
\h\
n).
maximum
> < 0.
there exists a d d implies f (n) (x)
+0/0 < 6,
Hence we
0.
This means
0.
the
at
such that the inSince \0h\ conclude
<
f(c)
by
<
d for
(25) n
(because h > that the function / possesses
then f(c + h)
at the point
<
(c)
c.
proved by analogous arguments, that if n is an even number and if / (n>(0) > 0, then f(c + h) f(c) > and thus the function has a minimum at the point c. Now, let us assume that n is an odd number. Let It is
/
(fl)
<
(c)
(in
the case
/
(7l)
(
>
the arguments are analo-
As before we choose a number 6 for \h\ < d. Then that /< w>(o+ 0h) < gous).
have the inequality
and
for
d
<
h
<
f(c
+ h)
f(c)
<
0,
>
in such a
for
<
h
<
e.
f(c)
>
f(c
i.
way we
d
we have the
inequality f(c /( c )>0 showing that f(c)
=
n
is
x possesses a minimum EXAMPLE. The function f(x) at the point 0, if n is an even number. If n is odd, then there is no extremum at this point.
196
DIFFERENTIAL CALCULUS
III
n Indeed, the first derivative of the function x not = x is the n-th derivative (cf. (1)) vanishing at the point
and
f
(n ^(
Kemark. The for all functions
previous theorem does not determine (arbitrarily many times differentiate)
whether an extremum exists at a given point
(e.g.
for
i
* e x
the function
considered in
8.4
derivatives of
all
which vanish at the point 0). However, it always permits us to answer this question for functions having a Taylor expansion (formula (21)) in a neighbourhood of a given point a. For, it follows from formula (21) that if the function / is not constant, then not all derivatives of this function vanish at the point a.
Geometrical
8.7.
interpretation
of
the
3,
if
second
derivative.
>
then the
Points of inflexion
As we have seen
in
7.7,
f'(c)
0,
function increases in a neighbourhood of the point c. Hence if f"(c) > 0, then the function /' increases; thus,
>
assuming f(c)
the function
0,
increases faster
/
and
faster. 1.
If e
/"(
the
>
0,
then in a certain neighbourhood of the f(x) lies above the tangent of this
curve y
=
point curve at the point [c,/(o)]
(thus,
it
is
directed with the
convexity downwards).
Analogously, if f"(c) below thi, tangent (thus,
< it
0, is
then this curve
lies
locally
directed with the convexity
upwards).
In
both cases
we assume
the continuity
of the
second
derivative.
us estimate the difference between the differences quotient and the derivative Indeed,
let
8.
DERIVATIVES OF HIGHER ORDERS
197
According to the Taylor formula, we have
+ h) - / (c) - hf(e) = \ f'(c + 0h) assuming f"(c) > 0, for sufficiently small 7,2
.
/ (c
Hence, have
also
+
j"(c+0h)>Q, from which
>
it
follows
h
we
that
Interpreting the differences quotangent of the angle between the secant this tangent and the JT-axis we conclude that for h >
h) f(c) f(c tient as the
hf'(c)
0.
f(c)
c+h FIG. 22
is
greater than
f'(c),
i.e.
greater
than the tangent of
the angle between the tangent and the X-axis. This means that the arc of the curve lies above the tangent. The proof of the second part of the theorem is completely analogous.
EXAMPLE. The parabola y = a?2 lies, in the neighbourhood of any point, above the tangent at this point. The second derivative is equal to 2. A curve y = f(oc) is said to possess a point of inflexion at the point c, if for sufficiently small increments h (]A| < d) the arc of the curve lies for positive increments
198
DIFFERENTIAL CALCULUS
III.
on the other side of the tangent to the curve at the point [e, /((?)] from what it does for negative increments. In other words:
if
number
a
d
>
exists
such that the
expression
is
<
positive for
or conversely:
Ji
<
d
negative
> h > - d.
EXAMPLE. The
and negative
<
for
h
<
d
>
for
and
h
>
d
for
positive
=
sin# has a point Indeed, sinfe sinO frcosO
sinusoidal curve y
of inflexion at the point 0. sin/i -h. This difference is negative for 0. positive for h
7&>0 and
is
<
Similarly, the cubic parabola y of inflexion at the point 0.
from theorem on one
It follows
curve lies the point 2.
then
1 that
a? possesses a point
if
f"(c)
^
0,
then the
side of the tangent. Hence not a point of inflexion. In other words:
(locally) c is
// a point c is a point of inflexion of a curve y /"((?)
f(x),
- 0.
This theorem cannot be reversed: the second deriva-
may vanish at a point c although the point c is riot a point of inflexion (just as the first derivative may vanish although there is no extremum at the given point); = a? has a minimum at the point e.g. the curve y tive
and so this point is not a point of inflexion of this function although the second derivative I2x2 vanishes for x 0. To get a more precise theorem, further derivatives
=
have to be considered. This leads to the following theorem:
=
=
-
^
3. If f"(c) but fW(c) 0, /'"(c) 0, ... p-V(c) then in the case of an odd n the curve y f(x) possesses a point of inflexion at the point c; in the case of an even ,
=
a point of inflexion. (We assume the continuity of the ^-th derivative at the point c).
n, c is not
Indeed, applying the Taylor formula (24)
we
obtain,
DERIVATIVES OF HIGHER ORDERS
8.
199
Let us assume n to be an odd number and let /(n>(c) > (n are analogous). (the arguments in the case / >(
<
>
Hence we have
0.
for h
<
(n)
Hence the curve
ip(h)
>
for
>
fe
and
Thus, the point c is a point of iny(h) flexion of the curve y = f(x). On the other hand, if n is an even number, then hn > independently of the sign of 7&, whence y(h) has the same sign as
/
(o).
0.
lies
tangent to the curve at the point point c is not a point of inflexion.
on one [c, /(
side of the
Thus, the
Be mark. As whether
c is
in the case of the extrema, the question a point of inflexion may always be answered
Theorem
3) for functions expansible in a neighbourhood of the point c. At the Taylor same time the points of inflexion of the curve y = f(x) are extrema of the curve y = f'(x).
(according to
series in
8
Exercises on 1.
Prove the (Halphen) formula:
2.
Find the w-th derivative of the functions:
1O& T*
x
, '
& cos x. r
3
.
,.
hm ro 4.
-
Evaluate:
~ 2x --
e~ x
e?
^
;
,.
lim
.
sin a?
x-o
Expand the
-&
#
sin# =
/I lim -r
,. .
o
,
9
\
cot 2 o?l
\^
.
/
functions sinh# and cosh# in power
series. j.
Sketch the graphs of the functions #*, & log sin a?, showing their extrema, points of inflexion, asymptotes, 5.
,
points of one-side discontinuity or points of both-sides discontinuity
(if
any
exist).
200
III.
6.
DIFFERENTIAL CALCULUS
Prove that
an n times differentiable function
if
vanishes at n+1 different points of a given interval, then there exists in this interval a point at which the n-th derivative vanishes. 7.
/'(O)
Let i(x)
=
fined
0.
=
Deduce
1
x
e
--*
x for
^
x
and
/(O)
=
0.
as a corollary that the function g de-
by the conditions _j_
g(x)
= e~*
x
for
possesses at the point to (cl 8.5, Eemark 8.
Prove that
Prove that
en
<
and
=
#(0)
=
derivatives of all orders equal (a)).
(2n + 1)
tyin .
Deduce that
IV
lim - \ nl '
=
-
.
CHAPTER
IV
INTEGRAL CALCULUS 9.
INDEFINITE INTEGRALS
9.1. Definition
F
A
is called a primitive junction of a funcfunction tion / defined in an open interval (finite or infinite) ,
=
f(x) for
every x. For example the function sin a? is a primitive function of the function cos#. Any function of the form sin a? + (7, where C is a constant, is also a primitive function of the
if
F'(x)
function cos a?.
a function / is defined in a closed interval a < x < fc, then the function F is called its primitive function, if - f(a) and JFL(6) = f(b). F'(x) = /(a?) for a < x< 6, F+(a) If
1. // too functions F and G are primitive functions a of function f in an interval ab (open or closed), then these two functions differ by a constant.
Indeed, if F'(x) = (?'(#), then, according to Theorem 4 of 7.5, there exists a constant C such that G(x) ^F(x) + C for every x.
Conversely, a function obtained by adding a constant to a primitive function of a function / is also a primitive function of the function /. Thus the expression F(x) + C
the general form of a primitive function of the function /. We indicate this expression by the symbol / f(x)dx ("the integral f(x)dx") and we call it the indefinite integral is
of the function
(1)
ff(x)dx
f.
So we have
- F(x) + 0,
where
F'(x)=
f(x)
,
202
IV.
INTEGRAL CALCULUS
(2)
,
The evaluation
of the indefinite integral of a function
/,
the calculation of the primitive function of a function / is called the integration of the function f. So ini.e.
tegration is an inverse operation to differentiation. It follows from the definition of the indefinite integral,
that any formula for the derivative of a function automatically gives a formula for the integral of another function
(namely, the derived function).
formula
=
-=
ClX
cos#,
For example from the
we obtain
J
cosxdx
=
sin#
+ C.
In general, however, the problem of the calculation of the integral of a continuous function, which we do not know to be a derivative of a certain function, is more difficult
than the problem
seen in
7,
of differentiation.
As we have
the differentiation of functions which are
compositions of elementary functions does not lead out of their domain; yet an analogous theorem for indefinite integrals would not be true. It is known only that every continuous function possesses an indefinite integral
However, this theorem gives us no practical of procedure evaluating the indefinite integral of a given continuous function. 9.2).
(cf.
EXAMPLES. Applying well-known formulae from the differential calculus
we obtain immediately the
formulae: (4)
(5) (6)
(7)
(8)
J
adx
= ax + C
f
,
l
+ C,
= mix+C / mixdx = cos#+(7,
/ cos x da;
,
following
INDEFINITE INTEGRALS
9.
dx -
f (9)
7
I
I/ 1
I
dx
C (13)
1
y
=
Eemark.
F(x) + C
+C
a
if ,
^
1
and
o?
>
.
the domain of x for which the equation satisfied is not an interval (finite or in-
cannot be stated that the expression primitive functions of the function /
it
gives
domain
in this
xa+l
If
is
then
finite),
,
= arctano?+C,
-
+ 07"
= f(x)
= arcshiff+C
nr&
-
aPdx
(14)
F'(x)
,
e*efa do/'
(12)
6
X/
/
(11)
-t-
=
f*? IT
(10)
= tana?
203
all
of arguments.
For example,
log|o?|
+ C gives all
primitive functions of
the function - in each of the two domains x
< and x > 0. x separately, but not in the whole domain of x real and different from 0. Namely, the function G(x) defined as and as log|o?|+l for x ^ is a primitive log o? for x < function of the function - for all x ^ although it is |
|
not given by the formula log|o?| + 6 Let us complete now Theorem 1 as follows: 2. Let a point XQ be given inside an interval ab and Y
.
an
number y Q be given. If a function f a possesses primitive function in the interval aft, then it and only one primitive function F such that one possesses let
arbitrary real
Let P(x) be an arbitrary primitive function of the function f(x) in an interval ab. Let us write F(x) = P(x)
-P(o?
F(xQ
)
)
=
+ ^o2/0-
Then we have Hence the function
of the theorem.
F'(x)
F
=
P'(x)
satisfies
=
f(x)
and
the conditions
204
INTEGRAL CALCULUS
IV.
it is the only function satisfying these consince any other primitive function of the funcditions, and this implies tion / is of form F(x) C, where
Moreover,
+
0^0,
Geometrically, this theorem means that given any point on the plane with the abscissa belonging to the interval ab, there exists an integral curve (i.e. the graph of a primitive function) passing through this point. The integral curves being parallel one to another, only one integral curve of a given function / may pass through a given point on
the plane. 9.2.
The
of the limit. Integrability of continuous
integral
functions
We
have proved
that given a sequence of such that the derivatives are continuous and uniformly convergent in the interval ab to a function g(x) and that the sequence Fn (c) is convergent for a certain point c belonging to the interval ab, functions
Fn (x),
the sequence
a
in
7.9
< x < b,
Fn (x)
is
convergent for every x belonging
to this interval; moreover, writing F(x)
have F'(x)
= ]imFn (x), we 71
= g(x).
= 00
Hence, the following lemma follows: 1. If the functions f n (%) are continuous and uniformly convergent in an interval ab to a function f(x) and if they possess primitive functions, then the function f(x) pos-
sesses also
a primitive function.
The primitive functions
Fn (x) may
be chosen in such
F
a
=
way (according to 9.1, 2) that the equation n (c) holds for a certain point c of the interval ab, for each value of n. Then
f f(x)dx=\imFn (x) + C, i.e.
the integral of the limit equals to the limit of the
integrals.
9.
INDEFINITE INTEGRALS
205
Indeed, to obtain our lemma it is sufficient to subabove formulation fn in place of F'n and /
stitute in the
in place of g. 2. THEOKEM. Every function continuous in an interval ab possesses a primitive function in this interval.
By the theorem proved in 6.4, every function continuous in the interval ab is a limit of a uniformly convergent sequence of polygonal functions. Hence, according to the lemma it remains only to prove that any polygonal function
possesses a primitive function. According to the definition of a polygonal function /
there exists a system of n+1 points aQ < al < ... < an an = fc, and two systems of numbers where a a, ^
6'i> 6'a?
/(#)
?
and d1? d2
n
= ck x+
for
die
>
?
dn such that
< x < ak
ak -i
(k
=1
,
2
,
...
,
??/)
.
Let us write -**(#)
where
=
et
2
i CA- a?
+ dk x + e k
Then
fc
<
fc
,
and
^ 1. Fk (ak = I fc+i(a
for
a -i
for
7
)
fc ),
whence the
collection of the
functions I\,]<\, ...,Fn defines one function equal to each of the functions of this collection in the suitable interval, respectively.
Fk F is
Differentiating the function f(x), i.e. the function
J?'(x)
=
to the function 9.3.
we
obtain immediately
primitive with respect
/.
General formulae for integration
Let us assume the functions
Then the following formulae
since
f(x)doc+
/
and g to be continuous.
hold:
j g(x)dx\
206
INTEGRAL CALCULUS
IV.
2.
=
af(x)dx
/
a j f(x) dx
^ (a f f(x)dx\ - a^ J
since
EXAMPLE. The
,
/(#)*&
-
integral of a polynomial:
n+ formula for the integration by parts:
T/ie
3.
(if
the functions
We write and z = g(x)
and
/'
g'
dx
y
side.
taking y
==
f(x)
= 7/s
--
l
J
c?.r
.
da?
this formula, let us differentiate its right
We obtain:
function of the is
briefly,
as follows:
J ^da?
To prove
are continuous).
formula more
this
f'(x)g(x)dx
yz'
+ y'z
left side of
=
yz'. This is the integrated the formula. Thus the formula
zy'
proved.
A
special case of formula 3
is the following formula the formula 3 obtained from by substitution z = x:
3'.
I
ydx
EXAMPLE. 4.
J
=
yx
logxdx
I
x~^dx.
= #log#
J
x-dx~x\ogxx+ C.
The formula
formula
for integration by substitution the for change of the variable):
x^ where on the right
side
it
(i.e.
g(y)dy,
has to be substituted y
after evaluation of the integral.
the
=
f(x)
9.
INDEFINITE INTEGRALS
We rewrite this formula z = g(y), as follows:
more
207
=
briefly, writing y
f(x)
and
Let us write G(y) = j g(y)dy. To prove the formula 4 it has to be shown that the derivative of the function #[/(#)] i g equal to the integrated function on the left side of this formula. Now, -
--
df(x) ldG(y)\ dx \ Ay Jy-Hx)
i
)
dx
<*/(*) = t/L/v g\](x)\ =
y. ^
.
dx
fj.
JJ.
As is seen, from the point of view of the calculation we may reduce by the differential. The formula 3 may be written also in the following form: I
EXAMPLES, Namely,
(a)
ydz
= yz
/ g(ax)dx
substituting
zdy
j
= ~j =
ax
the
in
=
ax.
formula
4,
where y
g(y)dy,
d
f(x)
.
we have tv
Similarly
(P)
I
=II
we
f(x + a)dx
I
g(ax)aAx
'
g(y)dy
J
.
j^rove that
= j
f(y)dy
where y
,
/y
*
(
)
integral
J JL
*.
C
J
x (
)
We
xdx l+o?2
To
example
~j~
^
we
substitute
obtain:
=
/*
___
J
1
1 d(x*)
dx
l+x*'2'~dx~
simplify the calculation, we omit the constant and in the further examples.
(y)
.
/7 /y
*
To evaluate the
x
(y)
= x+ a
C
in the
208
IV.
Since
as
we
INTEGRAL CALCULUS by the
said
integration the differential
df(x) may be substituted in place of f(x)dx, i.e. in the given case, d(x*) in place of 2xdx, the above calculation may be performed in a little shorter way:
xdx
We
(S)
dx2
f 1
have for w
>
1
/*
J f,
,
ta>nxdx
=
/'
J
J
The formulae
(y)
-
sin# ax cos#
and
(e)
-
/*cZcos# J
=
.
cos #. log tol ,
,
cos#
l
are easily obtained from the
following general formula:
y dx
J
Substituting y
y
= 1+
.T
2
in formula (S)
we deduce
this
formula easily from the following general formula:
J seen, the method of integration in the above examples consists in finding a function y f(x) such that the given integral will be transformed in an integral
As
is
=
with respect to y (which
is
easier
to
calculate).
This
makes it possible to replace the calculation of the integral on the left side of the formula 4 by the calculation of the integral on the right side of this formula. Sometimes, the
reverse
procedure,
side
by the
left
integral / g(y)dy
tion
y
=
f(x)
side is
i.e.
we look
such
the
more
replacing of the right suitable; to calculate the
for a strictly
that the integral
monotone func/ g[/(#)]
^
dx
a/x
be easily calculated. Denoting by x = Ji(y) the function inverse to y = f(x) and by F(x) and 0(y) the integrals
may
9.
on the
left
respectively,
side
INDEFINITE INTEGRALS and on the
209
right side of the formula 4,
we have then
In other words,
where in the integral on the right stituted x = h(y).
EXAMPLE. Let us x = sinJ, - 2 < t <
side has to
be sub-
substitute in the following integral
^:
dt = J yi but cos = |(l + cos2<), whence / co&tdt = Since sin2J = 2sincos = 2sinJ |/1 sin 2
J
2
^,
%t
we
obtain
hence:
E em ark. We
verify the correctness of the integration by differentiating the function obtained as the result of the integration. After the differentiation we should obtain
the integrated function. 5. Recurrence methods for calculation of integrals. The recurrence method for calculation of the integral / fn (x)dx
=
1 (or for consists in calculation of the integral for n 1-th n 0) and in reducing the n-th integral to the n
=
one
(or
an even
earlier one).
JC
EXAMPLES,
(a) e~ x dx
Evaluate the integral ln = / e~ x xn dx. e~ x Moreover, integrating by parts,
Now, I = / we have (for w>0): ard(e-
.
x
)
=
tfn e~ x
+J
dotf*
e~ x -=--dx
210
IV.
In particular, r
i
by
=
1^
i,
INTEGRAL CALCULUS xe- x
^
T ^
i
+I
w
Q
I2
^
... by and adding, we obtain
JT
plying sively,
n!,
,
,
=
x*e~ x
+ 2I
1
w
Irn _! T, by
-
...
,
Multi-
-
_
?
succes-
-
/
-
/fj/p (1 + x we havee
for
n
>
__
Now,
Ij
= arctan#. Moreover,
1
fl + ^
2
r
.
)
x* j
f
__ r
"~J d+^r
J
^ oT^' ^2
Since
_ "" 9 we
--
obtain integrating r
j
a?dx
(i+x r
_i
by parts 1
~
2
A (
r
2n-2j
,
xd
-1
(i+ay-
1
-1 Consequently,
x
1
2n-3 This (y)
is
the required recurrence formula.
Let y = /(a?), 2 = gf(a?). We calculate the integral applying n times the integration by parts: J
2i/<
n + 1 ) do?
J 2>(
J
w ) ^a?
z^y'dx
= J ^'i/ dx = z'y(n -v J z'Yn ~ (n)
(n)
5?j/
,
= #(n y >
J
z( n
l)
+v y dx
dx
.
,
9.
INDEFINITE INTEGRALS
211
Hence
+
...
+ (-If*
+ (-l) n+ f 1
(n)
y
This formula may be applied e.g. to the previously calculated integral / e~ x x^dx after taking into account that
dxk
'
9.4. Integration of rational functions
Let a rational function
be given, where P(x) and Q(x) are two polynomials. The integral / j(x)dx has to be evaluated. We may assume that the degree of the numerator is less than the degree of the denominator. For, if this were not so, we could divide the numerator by the denominator and then we have to integrate the sum of a polynomial (eventually reduced to a constant) and of a fraction, the numerator and the denominator of which are polynomials, and the degree of the numerator is less than the degree of the denominator. The first of these two terms is easy to integrate (as a polynomial); thus our task is to integrate the second component, i.e. a quoP(x) where the polynomial P(x) is tient of the form nf ,
of a degree less
in
than the degree of the polynomial Q(x). rational functions is proved
The following theorem on books on algebra: Let us
call
an expression
A (x-pf where
A
fraction.
and
PJ or C,
or
of the
^
D, q and
form
Cx + D
r are real numbers, a partial
212
IV.
Then a
INTEGRAL CALCULUS
rational
function n/
~{
is
a sum
of
partial
fractions the denominators of which are the factors of the
polynomial Q(x).
To decompose a
rational function into partial fracthe tions, polynomial Q(x) has to be factorized into prime factors. Then the coefficients appearing in numerators of the partial fractions! have to be found. first
The
factorization of the polynomial Q(x) in factors is obtained applying the theorem known from algebra stating that if p is a (real or complex) root of the equation Q(x) = 0, then a? p is a factor of the polynomial Q(x)\ k moreover, if p is a root of the multiplicity n, then (x p)
a factor of this polynomial for each Jc < n. Finally, is a complex root of the equation Q (x) = with
is if
u + iv
iv is a root of this equation, real coefficients, then u too (of the same multiplicity), and so the polynomial Q(x) has the product (x u iv)(xu + iv) = (a? u) 2 -\-v2 as
a factor.
The
numerators of partial
coefficients appearing in the
fractions are usually found
by the so called method of the undetermined coefficients, which we shall learn by examples.
The denominator
of the rational function
:r
-.
(x 2
possesses the factors (x 2) (x 2) and our rational function is of the form ,
A
x-l (x-2)*(x-3)
To
B
3)
(#3). Thus,
C
I
(x
B
calculate the coefficients A, and 0, we reduce the on the right side of the above equation to
fractions
a
2
2) (x
common
denominator.
We
obtain
07-1
(x-2)*(x-3)
(x-2)*(x-3)
INDEFINITE INTEGRALS
9.
213
The numerators on the side of the
any value
of x.
Hence
it
follows that
-3A + 65 + 4(7 = -1. equations give the values A, B and 0: = 2. A=-l, B = -2,
B + C = 0, These
left side and on the right above identity are equal one to another for
A-5B-4:C = 1,
22
Consequently,
0-1
We
-1
note that the coefficients
found in a to
the
JL, jB,
and C might be
as follows. After reducing
slightly simpler way common denominator we
obtain the following
identity:
=
Substituting in this identity x 3 gives C = the substitution x x = 1, we conclude J5 = 2.
=
we obtain
2,
2.
A=
1;
Hence, substituting
of the decomposition of a rational function in partial fractions is given by:
Another example
x-1
Ax + B
+ ,
Cx + D
.E ,
Eeducing to the common denominator we obtain the identity
whence we
easily find:
The theorem on the decomposition
of a rational func-
tion into partial fractions reduces the problem of integration of rational functions fcto the integration of partial fractions. Now, we shall see that the partial fractions may be integrated by the methods which
we have
learnt already.
214
IV.
INTEGRAL CALCULUS p we obtain
Indeed, substituting y ==x
.
.
7 fc==1
for
or
1 = A 7 --TT-T k~ -
The evaluation
is
k
for
l
1-fc y
>1
>
respectively.
,
of the integral
reduced to the evaluation of two integrals:
dx
r J
xq,
Substituting y
r
ana
Ka-tf + i*?
reduced to the integral /
the 8
\C'
(x-g)dx
J
[(-8)
first of
a
.
fc
,
these integrals
which
ii
reduced by th<
is
/
I
2 /fj/y 1 ^
which w<
,
already
know
9.3, 5 (p)).
(cf.
The second integral we find by the formula and (YJ) substituting y = (# 2) 2 + ^2 namely,
9.3'
(
;
(xq)dx
=
or
according to whether fc = 1 or fc ^ 1. Hence, the integration of rational functions can alwayi be carried out (of course, if the roots of the equatioi
Q(x)
=0
are known).
EXAMPLES,
(a)
I
=
/y
/
J
X
y
/
~\~ X ~t~ JL
.
Since
9.
INDEFINITE INTEGRALS
so substituting first y
=
y ?
C J
2 /
2 = x + -1 and then z = -i=.y, we obtain
2
'
1/3
_
%dy ^ s
r
=
d!t/ y
215
2
2
dz
r i
+
arctanz
1/3 -/^: (a?
+ -| 2]
1/3 \
-dx
=
A
ff
J
C
3
.
D
dx
[(a?-
side of this equation to the
Eeducing the right
common
A = 3, B == 4, C = 2, D =
denominator, we easily find: Integrating, we obtain
1.
K em ark.
Analysing the method of integration of rational functions, we state that the integral of a rational function is of the form
W + A log U (x) + B arc tan.7 (x)
(x)
,
where W(x), U(x) and V (x) are rational functions and and -B are constant coefficients. 9.5. Integration
We
of irrational functions of the second degree
shall consider
I First
of
A
all,
=
let
an integral 2
J i/a#
of the
+ bx + cdx
form
.
us consider the case
Then we may assume that 6^0,
when a
=
0.
since otherwise the
function under the sign of the integral would be reduced to a constant. Hence, we have to evaluate the integral
216
We -=-
substitute
=T
Then the
.
o
ay
INTEGRAL CALCULUS
IV.
t
=
bx + c, is
integral
#
i.e.
_
=
,
whence
transformed into the fonew-
ing integral:
Now,
us assume that a
let
i=-
We
0.
consider two cases:
4ac > 0. As is known, this assumption implies (1) 6 bx + c = that the equation ax* possesses real roots of is of second the form ax2 + bx + o the trinomial and degree a(x p)(x g), where p and # are the roots of the 2
+
=
above equation. Thus, we have
=\
\/ax*
+ bx + cdx =
Let us substitute a
We
x x
o
(x
I
=
x
i.e.
*, '
p
=
ao
t -
a
i
obtain f
T f
I
J
*aqap
f
*
a
.
t
2
dx .
,
,
/+ ' fl
dt
This integral being an integral of a rational function /7 /7*
(since -=-
a rational function, as
is
is
easily seen),
it
ctt
be calculated by the methods given in 9.4. 2 6 In 4ac? 0. it this case < be assumed that (2) may
may a
>
0, since otherwise the trinomial constituting the function under the sign of the root, which is of the form 2
+
s
would be always negative. Euler substitution: following a [(x
a)
/3 ],
i.e.
that
is
F-c
/y
________________
b-2t}/ a'
bx + c
We
=
apply the
INDEFINITE INTEGRALS
9.
Since
j/W
2
+ bx + c
as well as
217
are rational functions
-=r cut
J, so our integral is reduced to an integral a rational function. The above Euler substitution may be applied also in Jie case (1), if only a > 0. If c> 0, the second Euler substitution may be applied:
)f
the variable
>f
2
]/a#
+ bx + o = xt + j/c
ax + b
i.e.
,
= xt + 2t j/0 2
,
ijhat is
x
=
2t j/c t
-b .
Similar to the previous case,
we
see that this substi-
tution reduces the integral / to an integral of a rational function with respect to the variable t. t>e
[t
i
we conclude
Consequently, evaluated for follows also
R(u,
v) is
all
that the integral I
values of the constants
from the above considerations
a, 6,
may
and
c.
easily that
a rational function of the variable u and
v,
jhen the integral
J is
E (yax + bx + c 2
,
x}
dx
reduced to the integration of a rational function (thus,
bhe integration is performable).
EXAMPLES,
(a)
Let I
=
(Iff*
,.
for
xz
>
c.
We
shall
Euler substitution in a slightly modified form. Namely, let us substitute }/x2 + c = t x, whence z t -c dx , ipply the
first
=
and
Hence
x=
t
2t
t
2
c
+c,
and cat
_
= log|t| + C = log \x+ ]/a? + c\ + C
.
218
IV.
(P)
I
=
2
i/x
INTEGRAL CALCULUS
+ cdx
evaluated by means it
may
gration
x2
>
c. This integral may be of the Euler substitution. However,
for
be calculated more simply by reduction (by inteby parts) to the previous integral. Namely,
Let us note that the formulae for the derivatives of the functions inverse to th& hyperbolic .functions ( 7, (22) and (23)) make it possible to write directly the formulae for the integrals considered in the
example
(a) for c
=
1.
Namely, dx
dx
+0,
= arcosho?+C =log#+ }x2
1
+ (7
for
x>l
Be mark. As we have seen, the integration of irrational functions of second degree is always performable. However, replacing the quadratic ax* bx c under the
+
sign of the root
degree, the problem (i.e.
+
of the third or higher of integrability becomes in general
by a polynomial
without special
on the coefficients) of elementary functions.
assumptions
impossible to solve in the
domain
In particular, the integral
dx
cannot be expressed by means of elementary functions. Thus, it leads outside the domain of elementary functions
9.
INDEFINITE INTEGRALS
219
/7/v
(similarly as the integral
J
of rational functions). It
is
domain
leads outside the CO
the so-called
elliptic integral.
Similarly, the integrals
x dx
r
r
J |/cosi2a?'
oc doc
2
J l/l
fc
sin2 #
cannot be expressed by means of elementary functions (and lead also to elliptic integrals). 9.6. Integration of trigonometric functions
denote a rational function of two variables of integration of the function on based the substitution J = is jR(sin#, cos#)
Let R(u,
u and
We
v.
v)
The method
conclude from the identity 2
that cos
= ^aj
and, since C08a,
COS
cosx =
2 = ___
2*
2cos*|#
1,
l-< __. 2
At the same time, sin 2 Jo?
- 1 - cos
sina?
2 sin
2
Jo?
=
t
2 ,
whence
^ cos \x
Finally, the equation
#
dte
_ "~
// "^
=
2t .
2
2 arc tan t implies
2
1+T
2
'
The above three formulae give
(in, >*)*
and the
last
-J
integral,
function of variable
J,
being
may
an
integral of a rational calculated be by methods
with which we have already become acquainted.
220
INTEGRAL CALCULUS
IV.
EXAMPLES, tution
t
=J
Let I
(a)
= tan Jo? we
Applying the substi-
.
COS t>C
obtain
2dt i- Jfl. *-*[ J i-* i+p i-< 2
2
,1-t
But ~^.*
=T
= tan(# + JTT).
T
T
tan % x
1
i
Consequently,
/ cos# Hence f
= cos(o?
since sina?
f
|TT)
The
last
may
_ "~
da?
J sino?
,
. '
be deduced easily from
also
9.3, (e)):
(cf.
log[coso?|
f
I
-
.,
^TT)
J
,
_
~
-
cotiP^
f
2 J
J 2sm|a7cos|a?
=
.
^rAr^iog * &l tanio?
cos(o?
since
f
so i:r)
S
J
integral;
J tan#d# =
and
f d(x
~dte =
J sma?
the integrals
=
tanjo?d(ja?)+
= log|sin#|
si (
,
?
sin^cos
I
/
/
If
we know the
--
the integral / COS 00 (P)
The
reduced
integral &
,
Jf
// 7* , '
we
substituting cos#
find immediately J
=
B
irrational functions of second degree may be easily to trigonometric integrals of the form
9.
221
Namely, the functions
12(sin#, cos a?).
may
INDEFINITE INTEGRALS
be reduced to the functions of this type by subx = sin/, x = tan/, x sec/; in any of these
stitutions
three cases the derivative
and
of the functions sin/
A
=-T7
sin2 /
yi
+1 =
/i 57T-q i/ tan2 /
-=-
is
cos/.
Namely, dsin/
= cos/, .
^
sec/
~
= cos/,
dtan/
!
=
^
,
=
1
dt
cos 2 /
sin/
d!sec/
sin/
cos/'
dt
cos 2 /
cos/
-1 = tan/ =
also a rational function
,
'
e.g.
J
a
i/ r
+lflte=
Sometimes instead of x
it is
dx .
J
VV+I Since -
Hence we
#
= tan/.
more convenient to substitute x = cos/ a? == cot/ instead of # = tan/; e.g. let
sin/ or
us substitute x r
where
r^,
J cos 3 /'
=
r
= cot/ .
sm/
in the integral
dcoit
J
,4
a/
r 5/
=
= tan/ = -
i
u
m
log tan*/
J sin/
dt
-
|JT
,
so
find the integral
= log calculated previously in another way. Now, we shall give some other types of integrals of trigonometric functions, being of importance in applications.
The integral In =
/
8inn xdx.
222
INTEGRAL CALCULUS
IV.
We
shall give the recurrence formula.
f sin
L*
x Cv c*ob x
I
~~~
sinn-l $/ oos
P
i~
j
;W f cosa? a
OAJ-L
*/
ci/v^
i
J
*/
.,
do?
5-
dx
l-l) f 1) In -2
i
since cos
2
=
a?
2
sin
1
#
(n
]
.
Hence the required formula: In
-In -z
si^^^cosaJH
/I Similarly,
we
n>2).
(for
find
/r
%l
~~
n
^i
f ""
J
The integrals
smma?sinno?aa?
J
,
sinmo?cosw#a#
r
J
,
cos mx cos nx dx
can be evaluated applying the formulae 1 ir*f~isi / wn QITI w QiTi w 10 ^ r*no I II I tv livtJU olll tv f v Oil A win \j\J\5 ^" C/vlo y fin r c/t*/
siumxeosnx
cosw#cos^#
known from the
We
^y* JL>
I
-
/
Yt
\
= ^[sin(m + n)^ + sin(m = |[cos(m + n)o? + cos(m
'W \ o^T j^ IV i \J(J J I
n)x]
,
'S
trigonometry.
find easily:
sin
mx sin nx dx =
m 1
/ [
x
2
sin2mo?\ --
v ,,.
,
,
m+n
1 cos2mo? 2
.
\
2m
w 2[
-
for
m+n
n
2m
y/w
w
v
^^
,y
m
n
f or
m==n
^
^
for
m=n
'
'
J
v
,
9.
I
INDEFINITE INTEGRALS
=
cosmxcosnxdx
+
mn^^
"
for
m=n
.
I
J
calculate this integral,
6'sin(#+<9).
m?
J
dx
The integral of the function
for
v
+ x\
To
223
we
write the denominator
under the sign of the integral in the form we find from the of c and
The values
identity:
i.e.
that
_
is
c
We
= ]/ a + 6 2
2
,
=
arctan -
.
a
obtain
f
_1
cte
J asin#
+ 6cos#
Eemark. In
f
~~
c J
r
(l
+x
__
sin(a?+ 0)
9.2, 5
for the integral /
d(x+0) we gave .
1 c
the recurrence formula
This integral
calculated
maybe
)
m applying the formula for the integral / cos xdx. = tan/, we obtain Namely, substituting x also
f
J
*% n = JrcOB^^^'= f dt J 2
(l+x (1+aO'
Exercises on
)
9
Find the indefinite integrals of the following functions :
5.
arctan#,
6.
o^(loga?)
w ,
224
INTEGRAL CALCULUS
IV.
1 7.
8
-rV sm # # -# + 4 >
2
2
9.
'
1A
11.
+c = x+
2
13.
substituting i/#
,
9.5,
(cf.
14.
15.
/"
(m an
integer),
/
(m an
integer),
1
Example
(a)),
16. 2 (a?
+c
2
71
)
|P^2 + 2^ + r
E em ark. two
of
Prove that then variables,
H
=====
,
where
J7
E R (x
if
and
a rational function 2 = |/ao? + bx + c) is
,
W
are rational functions.
Hence deduce (by
14-16) a general method for integration of irrational functions of second degree.
17
Oot^
o^
20.
'T*
oo "1ft
eostf- cos2a?
cos3#,
a a
1 "1
Q
o-i
SlTl^^ ^OOfi^
/
7*
arc sin a?
21.
.
DEFINITE INTEGRALS
10.
10.1. Definition
SOO^ 'T
and examples
Let a function y = /(#) continuous in a closed interval be given. By the definite integral f(x)dx from to b we understand
<x
b
(1)
10.
where tion
/
DEFINITE INTEGRALS
225
F
is an arbitrary primitive function of the funcin the interval aft, i.e. F'(x) == f(x) for a x ft,
<
<
F'+(a)=f(a) and JFL(6) =/(&).
The definite integral does not of the function JP; this means that function of the function /, then
depend on the choice if
G
also a primitive
is
G(b)-G(a)=F(b)-F(a). This follows immediately from the fact that the difference between the functions G and F is constant in the interval a ^oo ^b (cf 7.5, 4, Eemark). .
The formula (1) will be assumed when < a. Hence we have
also
in
the
case
ft
a
b
f f(x)dx = f f(x)dx a
(2)
b
and (3)
From Theorem 2, of 9, 2 it follows immediately that every function continuous in an interval a ^. x ^ b b
possesses the integral
/ f(x)dx, that
is,
as
we
say,
it
is
a
integrable in this interval.
To
indicate the difference
F(b)F(a), we
use also
the symbol
Be mark.
It
follows
from the definition that the
definite integral depends on the function / and on the limits of integration. However, it is not a function of
the variable x integration). if
we
(x appears here only as a variable of of the integral will not be changed,
The value
replace the variable x
by another
b
b
f
= f
f(x)dx
variable, e.g.
t:
f(t)dt.
15
226
INTEGRAL CALCULUS
IV.
b
EXAMPLES,
/ cdx
(a)
=
c(b
Indeed,
a).
we have
a b
J cdx
= cx + 0,
=
whence / cdx
= c(b
[cxfa
a
a).
I
=
J,
/ sinxdx
=
/ x*dx
(P)
2 / x dx
because
=
#3 +
(7,
whence
o
o 71/2
(y)
The
(S)
/2
cos#]*
[
cos(7r/2)
definite
following
11.7) are easily found for the indefinite integrals cf.
two
=
+ cosO =
1
.
integrals (Fourier's, of the formulae
on the basis
m
if
9.6):
(
and n indicate
different positive integers, then Jt
7T
=
wimx&mnxdx =
J n
for arbitrary positive integers
cos mx cos nx dx,
J n
m
and
n:
7T
J
smmxvosnxdx
~
;
7t
for arbitrary positive integer
m:
n
J
TT
sin2 mo?^
= = TU
cos 2 ma?da?
.
7t
re
(e)
J
For an arbitrary positive integer w the formulae 7T/2
(4)
I
m&n xdx =
71
*2
'
7t/2
/hold.
These formulae formula ( 9.6): n xdx
/&in
=
follow
easily
n sin'^aJCOSOJH
from the recurrence
I
n J
&mn ~ 2 xdx
(n v
> 2)
10.
DEFINITE INTEGRALS
227
and from the formulae: f
--1
=
n
L 71/2
7T/2
sm
J
,
Jo
l
ocdx
=1
J
,
o
sin
o
Let us note that if we replace sine by cosine, then the formulae (4) remain true. 10.2. Calculation formulae
The
first
three general formulae for the indefinite
integrals ( 9.3) lead immediately to the following three calculation formulae for the definite integrals (we assume
that the functions under the signs of the integral are continuous in the closed interval of integration): 6
1-
/[/(aO0(ff)]*= $ 1(x)dxj a
a
c-
J
f(x)dx
=
c
J
f(x)dx
,
a
a
b
b
3.
g(x)dx,
a
b
b
2.
b
b
/ 1(x)g'(x)dx = U(x)g(x)?
The formula
for integration
by
substitution leads to
the following formula: b
/(&) /
4.
Jflf[/(n/ a
()*=
f
g(y)dy.
/(a)
G(y) be a primitive function of the function g(y). Then (by formula 4 of 9.3), #[/(#)] is the primitive function of the function
Namely,
let
=
g(y)dy /(o)
.
228
IV.
INTEGRAL CALCULUS
ebb
The theorem on the division
5.
of the interval
of integration:
f
i(x)dx +
f
f(x)dx
= f f(x)dx
.
Namely, denoting by JF'a primitive function of the /, we have
function
6.
// /(#)
>0
and a
<
then
ft,
J
Indeed, denoting by F a primitive function of the function /, we have F'(x) = f(x) > 0; hence it follows that the function F is increasing (in the wider sense). b
So F(a)^F(b),
Formula
i.e.
/ f(x)dx
= F(b)-F(a) > 0.
6 implies the following slightly
more general
formula: b
7.
// /(#)
By
^g(x) and a
assumption, g(x)
<
ft,
then
b
j
j(x)dx< / g(x)dx.
a
a
f(x)^0. Thus, by
6,
we have
b
1(x)~\dx
>
I
0,
whence by
g(x)dx
1,
j
As is seen from formula 7, the relation unchanged by the integration (when a < b). deduce from 7 the following two formulae: b
8.
|//(a?)te|< j\i(x)\dx, a
if
a
< remains We shall
10.
Namely,
DEFINITE INTEGRALS
follows (according to 7)
it
the double inequality
1/(#)|
b
by
integration of
(/(#)!
b
b
f(x)dx
f -\f(x)\dx^f
< / \i(x)\dx
,
a
a
a
<
229
i.e. b
b
-f a
\f(x)\dx
b
< / f(x)dx < / \1(x)\Ax a a
Hence we get the required formula
u^v^u 9.
implies the inequality
.
(since the inequality \v\
(cf.
1, (17)).
The mean-value formula. a+h
where
is
F
Let i.e.
a suitably chosen number satisfying the con-
<9<
dition
1.
denote a primitive function of the function
F'(x) =f(x).
/,
Then we have
a+h
according to the mean-value theorem of the differential calculus 10.
a
(
7.5,
(p)).
Let a function Let us write
/
be continuous in the interval
< x < b.
X
g (x)
(1)
=
//(*) dt
(i).
a
Then the following
basic formula holds: X
(2)
g'(x)
It
may
=/(),
i.e.
be also written g(x)
^/ =
f(t)dt
f f(x)dx.
= /(),
230
IV.
INTEGRAL CALCULUS
x i.e.
the integral
/
f(t)dt,
considered as a function of the
a
upper limit of integration, is a primitive function of the function under the sign of the integral. To prove this, let F denote an arbitrary primitive
= /(#). According 10.1, we have g(x) = F(x)F(a). (1) of we obtain g'(x) = F'(x) = f(x).
function of the function to the formula Differentiating,
Remarks, (a) At gration we have
F'(x)
the ends of the interval of inte-
and
g'+(a)=f(a) ((3)
let
/, i.e.
gL(b)=f(b).
Let us note that the function g
is
continuous as
a differentiate function. 11. If a sequence of continuous functions jn (x), a
<x<
a function f(x) in the interval
uniformly convergent to then the integral of the limit
is
6,
aft,
equal to the limit of the
is
integrals: b
(3)
b
/ i(x)dx
=
a
lim
fn (x)dx
*=aJ
,
b i.e.
J a
b
[lim/n (#)]$#
= lim
n=oo
J fn (x)dx
.
n=oo a
Since the limit of a uniformly convergent sequence is a continuous function ( 6.1,
of continuous functions
Theorem 1), the function / is integrable. Applying the formula 1 to the functions / and /n we have ,
b
(4)
b
b
/ f(x)dx-
/
fn (x)dx
=
At the same time, because vergence function
of the /,
such that
sequence
of
there exists for a given \f(x)
fn (x)\
<
e
the
of functions
for
n
>
Tc.
e
>
uniform
/!,/ a ,
...
con-
to the
a number
Hence, by
(4),
fc
8
10.
and
7,
DEFINITE INTEGRALS
231
we have b
6
\ff(a>)dx-ffn (x)dx
In this way we have proved that for n
>
k there
holds the inequality I
6
6
f
f
j fn (x)dx
J /(0)dte
e\b
a\
.
This means that equation (3) is satisfied. As a corollary of Theorem 11 we have:
a
oo
oo
b
b
w=0a
n=>0
assuming the functions
fn
and the
to be continuous
series
00
to
fn() a
be
uniformly
convergent
in
the
interval
< b.
EXAMPLES AND APPLICATIONS, / (1
integral
x
n
2 )
we
dx,
(a)
substitute x
To evaluate the
=
sintf.
Then we
o
have
=
sinO, 1
=
sin(7u/2).
1
Thus
(cf.
TC/2
IT/2
(
lim J cosna?do?
= = lim
J
si
10.1, 5),
232
IV.
INTEGRAL CALCULUS
Indeed, substituting y ? I
=
we obtain
nx,
**?
=
co$nxdx
-
cosydy
I
a
=
-
sinwa)
(sinnft
.
rca
b
Since |sinw&
<
sinwa|
2
,
we have lim / oo n
cosn^ete
=
.
a
b
it
Analogously
is
minxdx
proved that lim / n~ oo a
The following more general formula
0.
holds: b
6
lim j f(x)c.o&nxdx
(Y)
=
= lim
=
J
/ is an arbitrary continuous function. Let a number e > be given. According to the uniform continuity of the function / in the interval a < x < b there exists an m such that dividing the segment ab
where
m
equal segments, to any pair of points x and x' belonging to the very same from among these segments the inequality \f(x) f(x')\ < e holds. Let a 1? a 2 ..., am _ x in
m
,
be the points of the division; moreover, By formula 5, we have
m
b
j
f(x)cosnxdx
=
^
k=~l
J
Thus
i
(cf.
cosnajcte+
formula
m v^
f(x)o,osnxdx
because
f(x)cosnxdx
<
J
8),
b
a
J a1c _ l
ak
r I
ak
ak
m
>
ic=*l
let
/( afc)r
c
I
I
-
aQ
=
a,
am
6.
DEFINITE INTEGRALS
10.
233
we have proved,
Since, as
1, -
cosnxdx
I
n
n
M the upper bound
so denoting by in the interval a
< x < 6, we
of the function \f(x)\
obtain
b
S
Whence choosing n
2 -
n
Mm + s(b
a)
Mm <
-
.
e
we have
(y) follows.
The proof
so large that
IV
J
Hence the of the second
() not
the equations
first of
one
is
analogous.
If the indefinite
known
formly convergent to integrate
(e. g.
Theorem 12 makes integral by a series. (cf.
9.5,
which we are able then the application
series of functions
in a
of
In this
of a given function is may be expanded in a uni-
integral
but this function
way we
power
series),
possible to express the definite
it
calculate the elliptic integral
shall
Eemark): 7T/2
fc
I/I
As we know
8.4):
(cf.
1
2-4-6
2-4
2
Hence, replacing
2
^
by
_ = 1 + -2 &
2
A;
sin x
sin2
#+
we have: ft
4
sin
2-4 1 3 5 -\
2.4-6
234
IV.
uniformly convergent in the interval (even on the whole X-axis), since the series
This series
<
x
INTEGRAL CALCULUS
< 7T/2
representing
is
is
.
1/1J.
/2 I
\
uniformly convergent in the interval
|fc|
by term and applying
K/2
_____._, we obtain 7T/2
fc
2
sin2 x
sums
10.3. Definite integral as a limit of
Now, we
shall
prove theorems which give important
applications of the definite integral. 1.
Lei a continuous function y
/(#),
a
<x<
fe,
a number e> be given. Then there exists a number 6 > < possessing the following property: if the points a Q < < ... < am , where a = a and am = 6, satisfy the inequality
%
ajcajc-i
=
<
x,...,xm
m
and d for each fc 1, 2, ..., are chosen from the intervals
respectively
(i.e.
ak -i
<
afc
if
the
points
< a&)
Indeed, according to the uniform continuity of the function / in the interval a < x 6 (cf. 6.4, 1) there
<
exists
a
d
>
such that the condition \x
x'\
<
d implies
DEFINITE INTEGRALS
10.
\f(x)-f(x')\
<
j. a
o
By
235
the formulae 5 and
10.2,
9,
we have 6
*%
ai
=/ J f(x)dx
f(x)dx + ...+ ...
/
i(x)dx
+i(x'm )(am -am ~i)
^
a suitably chosen number in the interval a&-i Since the points xk and x'k belong to the interval ^k-i akj the distance between these points is not greater 5. than the length of this interval, i.e. \xk ~x'k a^-i
where xk
is
-
<%
Hence
it
follows that \i(xk )
f(xk )\
<
g
T
.
Consequently
//*
I /c=l
m
In this way Theorem 1
is
proved.
us consider instead of one partition of the Now, segment ab (defined by the points a a 15 ..., am ), a sequence of partitions of this segment in smaller segments. let
,
Let the n-th partition be defined by the points an n an>0 = a, an ln = ... where n,o,
W
,
,
,i
,
fe
.
,
Let rn denote the length of the greatest segment of We say that the above sequence of partitions is normal, if lim rn = 0. The following theorem holds:
the n-ih partition.
236
IV.
2.
INTEGRAL CALCULUS
=
Let a continuous function y
a normal sequence of partitions of
xn>k belongs
If the point
partition
n
= 1,
2,
(i.e. ...),
/
(6)
an>fc _!
<
the
to
<
xn>k
for
,
<x<
6,
segment be given. interval of the n-th
Jc-th
an k
a
/(#),
this
Jc
=
1, 2,
..., ln
;
tt^
/(0)
- Mm j
/(avi,fc)(n,*~n,*-i)
e > >N
be given. We choose JV" in such a way the inequality rn < d holds, i. e. that all intervals of the n-ih partition have the length < d (where d has the same meaning as in the previous theorem). Let an
that for
n
Then by Theorem
1
we have
In
an,*-l)
This means that equality
~
<
J
(6)
is
It follows in particular that
if
8
satisfied. all
segments of the
n-th partition are of the same length, namely, 2~ n (b
a),
then 2n
6
f(x)dx
Eemark. The
= (b-a)lim2~ n notation
/ ydx a
V f(a is
n>k ).
closely related to
Theorem 2, so from the point of view of notions as from the historical point of view. The symbol /, the modified letter S, had to indicate the summation of infinitely
many
<
'infinitesimal"
terms of the form ydx; this con-
cept, which was not well-defined from the logical point of view, becomes now expressed strictly in Theorem 2. It is easily seen that this theorem represents the definite
integral not as an infinite sum but as a limit of of an increasing number of terms.
sums
In 10.11 we shall turn to the definition of the integral based on formula (6).
10.
10.4.
The
DEFINITE INTEGRALS
integral as
Let i(x)
>
237
an area
for a
< x < b.
b
The
integral
is the
/ f(x)dx
area of the region
P
con-
a
sisting of the points x, y of the plane satisfying the conditions:
a^x^b, It is really so,
<#
when the function
/ is
linear.
Then
the region P is a trapezoid the bases of which (parallel to the Y-axis) have the length /(a) and /(ft), respectively, and the hight has the length b a. Thus the area of the trapezoid
is
MM(
fe
_a).
Zt
On is of
the other hand, the function cx + d. Hence the form f(x)
/
as a linear function
b
/ &>+*&>_.,, since f(a)
=ca + d and
Thus we
see that
f(b)
=
in the case
when the
function
/
is linear, the notion of the area known from the elementary geometry is equivalent to the above defined notion of the
area as the definite integral. This equivalence holds also if / is a polygonal function (cf. 6.4),
in a wider sense,
when the graph
i.e.
of the function
is
a polygonal
line.
Namely, the region between this polygonal line and the X-axis consists of a certain number of trapczoids and is the sum of the areas of these trapezoids. we as have stated above, the area of each of since, these trapczoids is a suitable integral, so the area of the
hence
its
area
Yet
b
whole region
is
the
sum
of these integrals,
i.e.
/ f(x)dx a
(according to
10.2, 5).
In the general case
if
function, the function /
is
/ is an arbitrary continuous the limit of a uniformly con-
238
INTEGRAL CALCULUS
IV.
vergent sequence of polygonal functions
= \imfn (x).
Hence, by
noo
6.4): f(x)
10.2, 11,
b
b
f(x)dx = lim J
J
fn (x)dx
.
n=oo a
a
Then we
(cf.
see that the area of the region
determined by the b
=
curve y
(defined as the integral
j(x)
/ j(x)dx)
is
the
a
limit (cf.
areas
of
of
Fig. 10 in
"inscribed"
polygons
in
this
curve
6.4).
Thus, there
here a complete analogy to the area
is
of a circle, defined as the limit of areas of inscribed (or circumscribed) polygons. Similarly, we define the area of
the region
P
given by the curve y
/(#) as the integral
Remark.
This area could also be defined on the basis measure theory, which we shall not consider here. Then this definition could be proved to be equivalent to
of the
the definition of the area by means of the definite integral (we have performed such a proof in the most elementary / is linear
if
namely,
case,
EXAMPLES,
Let y
(a)
=
or polygonal). 2
|/V
#2
r
,
<
x
< r. We
shall
r
calculate
/
2
x* dx.
|/V
r
For
this
purpose we shall find the indefinite integral. z = ~ we obtain /Y
Substituting
9.3, 4)
(cf.
=r
2
x 1 #_ -arc sin- + --!// 1
/I \2
.
r
2 r
i
\
\
Hence r 2
J]/r
x*d%
= r2(^arcsinl-~|arcsin(
1))
~f
since arcsinl
=
^TT
and arcsin(
1)
=
|TC.
10.
DEFINITE INTEGRALS
239
In this way we have obtained the formula for the r, known from ele-
area of the semicircle with a radius
mentary geometry. ((3) The area of the region contained between the arc of the parabola y = x2 the .X-axis and the straight line ,
a
x
a
is
2 given by the integral / x dx
=
Ja
3 .
o z
(y)
The area
of
the ellipse
x
2
y + jr =
1
culated as follows: the above equation gives y
will
be
cal-
= b 1 / 1 --a
,
I
1/1 --&^dx. a \ -p
whence the area
of the ellipse
is
equal to 2b /
Applying, similarly as in the example for the indefinite integral / j/1
z*dz
(a),
we
the formula
find
a
26
X ~
1
/
a2
() The area between the sinusoid and the segment on the X-axis is
J
sinxdx
=
[
COS^]Q
=2
Or:
.
Let two continuous functions u(x) and v(x) defined be given. Moreover, let u(x) segment a x b. The contained between for < < region
a<#<&
in the
a is
<x
,
N
u(x)
< y < v(x)
called a normal region. b
The area
of the
normal region
N
is
/ [v(x)-~ u(x)]dx.
a
N
is Indeed, u(x) > everywhere, then the region curve obtained from the region contained between the y = v(x) and the X-axis by subtraction of the region if
240
IV.
INTEGRAL CALCULUS
contained between the curve y
=
u(x) and the X-axis;
b
the area of the
first
/ v(x)dx and the area of
is
region
a b
the second one
is
/ u(x)dx.
N
Hence the area
is
the
a
difference of these integrals.
Now, all
x,
if
does not hold for the inequality u(x) > the lower bound of the
then we denote by
m
and by N the normal m and v(x) m. The region defined by the curves u(x) region N is obtained from the region N by translation and so NQ is congruent to N. Thus, function u in the interval a
<x<
b
b
area
N = area N = J
{[# (x)
m]
[u (x)
m] } dx
a b
= Eemarks.
J [y(x) a
u(x)]dx
.
Considering the area of the normal region it is meaningless whether we take its perimeter into account. Namely, we assume the perimeter to have
the area
0.
(a)
This follows from the fact that the curve
be included in a normal region with an arbitrarily small area, namely, in the region contained between the arcs y = /(#) + and y = i(x) e; the area y
f(x)
may
of this region (P)
is
2e(b
a).
The symbol ydx
is
called the element of the area
7.13).
(cf.
6
(y)
The
integral
/ f(x)dx a
may
be interpreted as an
area also without the assumption f(x) > 0, if we agree upon considering the area of regions lying below the X-axis to be negative (where a < b). Thus, if the interval ab
may
be divided in smaller intervals in such a way
that in everyone of these intervals the function
is
either
b
everywhere nonnegative or nonpositive, then J
/ (x)
dx
is
10.
DEFINITE INTEGRALS
241
the algebraic sum of the areas defined by the arcs of the curve y = /(#), in each of these intervals separately. For example for the curve y = sin#, < x < 2?r, we 2-K into two halves. The sine curve divide the interval TC of the X-axis gives a region together with the interval 2 area and with the with an interval TU, 2jc, together a region with an area 2. The algebraic sum of these ,
,
areas equals
0, i.e.
J
(8)
smxdx =
,
Interpreting the definite integral as an area, many proved theorems become of a clear
of the previously
geometrical content (especially, if the function under the sign of the integral is nonnegative); e.g. Theorem 9 of 10.2 means that the region defined by the arc?/ = /(a?), a < x < b, the X-axis and the perpendicular segments lying on the straight lines x = a and x = b has the same area as a rectangle having the segment a < x < b as the base and the ordinate of a suitable chosen point on the arc?/ = /(#), a < x < />, as the height. It follows
from Theorem
10, of
10.2
(cf.
remark
that the area of a region defined by the arc?/ a < x < x', varies continuously together with x'.
Theorem 11
of
10.2
means that
if
the functions
fn
are uniformly convergent in an interval ab to a function /, then the areas of the regions defined by fn are convergent to the area of the region defined 10.5.
by
/.
The length of an arc
Let a function y = /(a?), a < x < 6, having a continuous /'(a?) be given. The integral
derivative
16
242
IV.
INTEGRAL CALCULUS
gives the length of the arc formed satisfying the conditions
x,
y
a^x^b.
y=/(a),
(8)
by the points
a linear function, i.e. if the considered curve is a segment joining the points [a, /(a)] and [&,/(&)]. The length of the segment is equal to the It is indeed
distance of
its
if
so,
ends,
/ is
i.e.
Applying the integral we obtain the same result, cx + d. Hence f(c) = c. Thus, Namely, let us write f(x) b
a
since G
In the general case
b-a we do not assume the
linearity find the explanation of the above definition of the length of an arc in the following
of the function
assumed
= (if
we
/)
construction.
We know
6.4) that
(cf.
the conditions
may
(8)
an arc
by
be approximated by polygonal
lines inscribed in this arc (cf.
shall
of a curve given
6.4,
Fig. 10).
Now, we
L
the length of this arc and by L19 L2 ... the length
prove that, denoting by
expressed by the integral (7) of the successive polygonal lines,
L = lim Ln
,
we have .
n=-oo
Namely,
let
the n-th polygonal line be obtained by join/(n,o)L On,l, /(n,l)L -> l>Wn? f( an,ln )]
ing the pOlIltS [>n
,
,
by segments, successively, where an>Q = a, an j n = b and the length of the greatest interval of the n-ih partition as n tends to oo (i.e. the sequence of these tends to partitions is normal in the sense of 10.3). The length
10.
DEFINITE INTEGRALS
243
of the n-ih polygonal line is the sum of the length of the particular segments, which constitute this polygonal line, i.e. In
__
Applying the mean-value theorem we obtain i
Ln =
n
^
(Om*-0n,*-l) V'l
+ tf'teri,*)]
2
,
fc=i
where
Hence by Theorem 2 after substituting I/ 1
of
0^-1
10.3 (formula
+ [/'(#)?
< xn
,k
(6)),
we conclude
in place of the function /(#)
that b
f ^
lim 1/TWMP * = n-oo
The remarks noted
=i
.
in the case of the definition of
the area are valid in the case of the definition of the length of an arc, too. The notion of the length of a curve may be defined on the basis of the general measure theory
and
be proved that for arcs of the form (8) this consistent with the definition given here by means of the integral (7). However, since we do not use here the measure theory, we define the length of the arc (8) as equal to the integral (7). it
may
definition
is
The approximation sidered above
an arc by the which we have con-
of the length of
length of inscribed polygonal lines
the generalization of the approximation by the perimeters of the inscribed polygons, known from the elementary geois
of the length of a circumference
metry. circle
We
prove that the length of an arc of the in the sense of elementary geometry is equal to shall
the length in the sense of the definition assumed here.
<
EXAMPLE. Let y = j/V2 The considered /3 < n).
#2 rcos/? < x < rcosa (Q < a arc is a part of the semicircle ,
244
INTEGRAL CALCULUS
IV.
with the radius r and the centre integral
y
We
(7).
x = -7===, 2 j/r -^ 2
r
,
whence
'
Let us calculate the
= rcost rdx l/A2
,
+ V^W>
i/l
J
Substituting x
0.
have
we
=
(y')
z
dx
rdx
r
=
/
y r*-x*
J
find ^
rt
=
x
rarccos-. y
/y2
Hence
r COB ft
in accordance with the formula known from elementary geometry. Let us write
The function
s(x), i.e.
the length of the arc treated
as a function of the abscissa of the right (varying) end of the arc is an increasing function. Thus, there exists
a function x = x(s) inverse to s(x), which represents x as a function of the length of the arc. There hold the formulae:
Denoting by a the angle between the tangent to the curve and the X-axis, we have -- = tana and so -jdx ds 7
=
cos a. This
means that the derivative
cosine of the tangent to the curve y to the X-axis.
=
the direction with respect
is
f(x)
This follows also from the fact that
dy -j-
as
=
dy dx ax as
==
,
tan a cos a
= sina .
-
and
i |
w^
\
)
,
/
^/
\
+1-^1
=
10.
DEFINITE INTEGRALS
245
Bemark. The represented ds*
=
"length element" may be symbolically analogously to the Pythagorean formula:
dx2 + dy*, being of a clear geometrical content
notion of the differential,
the
(cf .
7.13).
The curvature. Let us denote as previously by a the angle between the tangent and the X-axis. This angle is a function of #; considering x as a function of the length s we express a as a function a(s) of the variable 8. The derivative &
=
is
called
the
curvature
the
of
as
FIG. 23
curve y
= f(x)
at a given point, the inversion
p=-
f
=
Q
is
\lc\
called the radius of curvature.
Geometrically, the curvature fc is interpreted as follows: Let us draw the tangents to the considered curve at
the points Po
= I> (*o)
9
V (o)]
and
ph
= [0 (* + h)
,
t/
(*
+
fc)]
.
The angles between these tangents and the positive direction of the X-axis are a = a(s Q and ah = a(SQ + h), )
246
IV.
INTEGRAL CALCULUS
The angle d^ between these two tangents is is the same angle as the angle between a/i OQ. the normals to the curve at the points p Q and p h Let
respectively.
This
.
us consider the quotient of this angle by the length of the arcpoP/ij ie. by h. The curvature of the curve y f(x) at the point p is the limit of this quotient as h tends to 0.
y
In particular, if our curve 2 x2 r < # < r, then h j/V
=
,
is
=
sidered quotient has the constant
same value has the curvature curvature
fc,
an arc of the circle rd^ and so the convalue
.
Thus the
and the radius
of the
equal to the radius of the circle. In the case of a straight line the normals are parallel. Thus dh always and the curvature is equal to 0; hence is
=
the radius of the curvature
^
is
infinite.
then the point lying on the normal in the distance Q from the point p Q on the same side of the tangent on which the curve lies locally, is called the centre of curvature. We may prove that this point is the limit position to which the point of the intersection of the normals at the points p Q and p h tends as h tends to 0. The curvature k may be expressed in the coordinates x If
it
and y as
0,
and basing on formula da ds
_
=
Taking into account that a
follows.
(9)
da dx dx ds
arctan(-p)
we obtain
dx-
1
+
i.e.
NDT The centre
of mass. Let a system of n material points with the coordinates (xl9 #1), ..., (xn yn ) and with ,
10.
the masses
mass
of
To
m
ly ...,
DEFINITE INTEGRALS
mn
247
be given on the plane. The centre is the point with the coordinates
of this system
define
< <
the
centre
mass of an
of
arcj/
=
/(#),
with a constant density g we proceed as follows. Let us represent a? as a function of the length s, i.e. x = x(s) = /[#($)] = y(s). Let $ denote the (as above); hence y = of the arcj/ /(#), a < a? < ft. Let us divide this length a
oo
arc into
fe,
n equal
of the interval
parts. This division constitutes a partition into n equal intervals; let the points
0$
of the partition be
=S
nt Q
,
8 Ut i
,
...
,
Sntn
The point with the coordinates
=$
.
#(* nv *)>
^(^n,A;)
is
the
end of the fc-th arc of the n-th partition; the mass of this arc corresponds to this point. This mass is the product of the density by the length of the arc, i.e. it is equal to g
m
(8n,k
system
is
sn,fc-i)'
Thus given
n,
the centre of mass of the
of points
the point with the coordinates
mass of the arc is equal to gS. limit position of the point (xnj y n ) as n tends to oo is the position of the centre of mass of the con-
since the
The
sidered arc.
(12)
x
Thus the coordinates
= limxn = n-oo
s 1 f -~
^^
xds
and
of the centre of
_ y
mass are s 1 f
= lim y n = -~ n=oo
AJ
-T
2/a^
.
248 10.6.
IV.
INTEGRAL CALCULUS
The volume and surface area of a
solid of revolution
Let a continuous function y = /(#), a < x < &, be given. 10.4 by P Moreover, let j(x) > 0. Let us denote as in the region constituted by points x, y satisfying the conditions a < x < 6,
its
volume by W.
We
have
b
(13)
a
W=
-
In order to explain -this formula, normal sequence of partitions (see
us
let
10.3,
consider 2).
Let
FIG. 24
f(n,k) be the upper bound of the function / in the interval an ,fc-i? &n,&- Replacing the arc of our curve lying
over this interval by the segment y = f(xn ,k) we obtain by the rotation around the X-axis a cylinder with the base 7c/(o?n,&) 2 and with the height (in> k ^n k-u an
Proceeding in this way for each Tc (n being constant) we replace the considered solid of revolution by another one, circumscribed on the given solid and made up of a certain number of cylinders. The volume of this solid
DEFINITE INTEGRALS
10.
is
the
sum
249
volumes of the particular cylinders: volume by TFn we have then
of the
denoting this
By Theorem
,
10.3
2 of
we
obtain in the limit
W = lim Wn
.
Analogously, replacing the upper bound by the lower is bound in the above construction we prove that the limit of the volumes of certain solids inscribed in the considered solid of revolution everyone of which is
W
a
sum
of a certain
number
of cylinders.
Let us denote by B the area of the surface obtained /(a?), a < x < ft, i.e. the by the rotation of the arcy area of the lateral surface of the considered solid of revolution.
We
have the formula
B=2
(14)1
a
(assuming the continuity of the derivative To establish this formula we approximate the considered surface
by
surfaces the area of which
from the elementary geometry. For
this
is
purpose
known let
us
consider a sequence of polygonal lines approximating the 10.5 (cf. Fig. 10). defined in arci/ /(a?), a < x < 6,
=
The
surface obtained
line
around the X-axis constitutes of a
by the rotation
of such a polygonal finite number of
truncated cones. Since the generator of a truncated cone is the segment joining the points l> n ,A;-i? /(,*-i)] [
and the /(fln,fc-i)j
is
radii of the bases
respectively,
equal to
are of the length /(an ,&) area of its lateral surface
so the
250
IV.
from
follows
as
INTEGRAL CALCULUS a
formula
known from elementary
geometry. Hence, denoting by Bn the area of the surface obtained by the rotation of the n-th polygonal line, we get
fc-1
where a Utk -i
We
shall
^ xUtk ^ an>k
.
prove that
B=
(15)
For
this
purpose
let
lim
En
.
us compare
Bn
with
In
Cn
=
27C
By Theorem
2 of
10.3 with the function /(#) replaced
Thus
it
remains
by
+ [/'(a?)] we conclude that B = lim Cw noo to prove that lim (B n Cn = 0. 2
the product j(x) ]/l -
.
)
n=-oo
Now,
According to the assumption that the considered sequence of partitions of the interval of integration is a normal one we may assume that if the points x and x' belong to the same interval of the n-th partition, then /(#) /(#')! |
<
e
(where
e is
a given positive number). Thus
<e
and
|
/(an,*)
-/tea)
I
<
DEFINITE INTEGRALS
10.
M
Denoting by
251
the upper bound of the function
+ [/'(#)? in the interval ab we obtain immediately \Bn -Cn < 2nsM(b-a). Hence lim (Bn -Cn = 0.
1/1
\
71
)
= 00
the formula (15) is proved. formula for the surface of a solid of resimpler volution is obtained treating a? as a function of the length 10.5. Namely, applying the of the arc, similarly as in first of equalities (9) we have
In this
way
A
S
6
B = 27r / yyi+W)*dx = 2n f
(16)
yds
,
where 8 denotes the length of the arcy = /(#), a < x Formula (16) may be transformed as follows:
<
b.
B = 8-2n
(17)
6
where y denotes the ordinate of the centre of mass of the arcy = f(x), a < x < b (cf. (12)). In other words, the area of the surface obtained by a rotation of the arc y f(x) around the X-axis is the product of the length of this arc and the path of the centre of mass (Guldin's theorem).
EXAMPLES, value
r for
(a)
<
x
A <
cylinder. Let h.
f(x)
be of a constant
The region P constituted by the < x < h and < y < r is a rec-
points #, y such that tangle. By a full rotation of this rectangle around the JC-axis, a cylinder with the height h and with the radius of the
base r
is
Hence the volume of this The same result will be obtained
obtained. 2
cylinder is equal to applying formula (13), since 7rr A.
252
IV.
INTEGRAL CALCULUS
(14) gives for the lateral surface of the cylinder (taking into account that y' 0) the formula
The formula
=
h
B=
known formula from geometry.
in accordance with the
A P
=2
rdx
27i
=
< x < "h. In this case the the rotation, a cone with the By region = ch is obtained. height h and with the radius of the base r By formula (13) the volume of this cone is equal to ((3)
cone. Let y is a triangle.
cx
9
h TT
J
(PaPdx
According to formula is
-
-
JTCCW
2
J7cr
fe
.
(14) the lateral surface of the
cone
equal to h 2-Tt
J
ex
o /
where Z
Z
denotes the length
of
the generator (because
= (y)
A
sphere. Let y
=
2
#2
]/r
r^x^r.
,
The
set
points defined by these conditions is a semicircle. Rotating the semicircle around the JT-axis we obtain a sphere with the radius r. The volume of this sphere is of
r 7i
J
r 2
x*)dx
(r
=
r
7i
j
r
r2 dx
7t
r
To evaluate the
may
be applied. For
2 f x dx
=
3
27rr3
|7ir
s,
i.e.
y
=
.
o
rsin-.
We
the following value of the surface of the sphere: rtr
j rsin^ds
3
frcr
surface of the sphere, formula (16) this purpose we express y as a func-
tion of the length of the arc
2n
=
r
TT
= 2^
J smtdt
=
obtain
DEFINITE INTEGRALS
10.
10.7.
Two
253
mean-value theorems
FIRST THEOREM. Let two continuous functions
<
<
x b. g be given in an interval a be a constant Then of sign. function g b
f let
Moreover,
and the
b
= J(t)fg(x)dx, ff(x)g(x)dx
(18)
a
a
where f a < f <
a suitable chosen value satisfying the condition
is &.
let
Indeed,
bound
m
respectively.
M indicate the lower and the upper
and
function
the
of
the
in
/
a
interval
<
x
<
fc,
Then we have
m
Let us assume that g(x) > everywhere. Multiplying the above double inequality by g(x) we obtain then
mg(x) ^f(x)g(x)
and hence, by integration
I
10.2, 7),
(cf.
6
?
?
m
<Mg(x)
g(x)dx
a
<
f(x)g(x)dx
J
<
a
M
J a
g(x)dx.
Hence we conclude (assuming the function not identically equal to
0,
g to be
which can be obviously sup-
posed) that b
b
m<
|
J f(x)g(x)dx: J a
g(x)dx}
<M
.
a
According to the Darboux property of the function / (which is a consequence of its continuity), this function assumes all values contained between m and M. Therefore there exists a f in the interval ab such that b
b
i.e.
that formula (18)
is
satisfied.
254
IV.
INTEGRAL CALCULUS
<
In the case when g(x) completely analogous.
Be mark. Assuming
everywhere, the proof
that g(x)
^
0,
is
always we obtain
a shorter proof of the above theorem applying the Cauchy theorem stated in 7.5, (12). Let us denote by H(x) the primitive function of the function 1(x)g(x) and the primitive function of the function g(x). Then
by G(x) we have
6
6
f(x)g(x)dx
= H(b)-H(a)
/
,
=
g(x)dx
G(b)-G(a)
.
a
By
the Cauchy theorem,
g(ft)-g(g) G(b)~G(a)
and formula
!T(f)
~
Q'(g)
(18) follows.
SECOND THEOREM.
// the function f is continuous and is monotone and possesses the second g function derivative continuous in the interval a < x < fe, then
the
b
b
c
(19)
J a
c
f(x)g(x)dx
g(a) j f(x)dx
+ g(b)
c
J
f(x)dx,
a
a suitable chosen value belonging to the interval ab. Let us denote by F(x) a primitive function of the function /(#), i.e. F'(x) = f(x). Since the derivative of a monotone function is of a constant sign (cf. Theorem 2 of 7.7), so we may apply the first mean- value theorem where
is
66
to the product of the functions F(x) and g'(x). Hence, applying the rule for integration by parts, we obtain 6
b
since
f g'(x)dx
= g(b)
g(a)
f
DEFINITE INTEGRALS
10.
255
Consequently, b
= g(a)tF(!)-F(a)] + g(b)[F(b)-F(t)-]
f(x)g(ai)dx
whence we obtain formula
,
account that
(19) taking into
b
*
F()-F(a) = J f(x)dx a
F(b)-F() = f f(x)dx
and
.
$
EXAMPLES AND APPLICATIONS. The mean-value theorems are often applied in order to estimate the value of integrals. shall consider some of their applications.
We
(a)
<
Let
<
a
b.
We
shall
prove that
6
sin wo? -
lim
(20)
00 n==oo
For
this
^/f
0?
a
purpose we shall prove that b
4 \j--dx < --,
/m\
f sin 00
I
(21)
We apply the f(x) = sin oo? and ?
-
f sin - co? J
#
,
?
?
-
jj
,
a J
s
1
ao
<
/
& J
1
= Since |cos/|
.
second mean- value theorem substituting g(x) = l/o?. We obtain
If-sinoo?ao? + -1
,
ax
c>
if
1
(cos ca
and a
cos c)
<
ft,
+ -r(cos cE bo
we conclude
cos co)
.
that
b
"
sin co?
n .
0?
The formula
be
ao
formula (20) immediately. Eemark. The formula (20) is also an immediate con10.2. sequence of the formula (y) in the first mean- value theorem, the Taylor (P) Applying formula may be deduced in the following way. (21) implies the
256
IV.
We
INTEGRAL CALCULUS
substitute in the formula
9.3, 5, (y), z
proved in
=
+ (-ip #) # = /(#)? n
(6
,
to possess a continuous x 6. the interval a
function
/
n+l-th
assuming the derivative in
< <
Integrating from a to
we obtain
ft
6
^ On
the other hand, the function (b a constant sign in the interval a < x < 6, the first mean-value theorem
n x)
we
being of obtain by
r
J (*Comparing the obtained formulae we get the Taylor formula with the remainder in the Lagrange-form. 10.8.
Methods of approximate
integrations.
Lagrange
inter-
polation
n+I points
Given values
2/oj
?/u
?
2/nj
#
,
xn on the X-axis and n+I may be defined a polynomial
x19
there
...
,
w(x) of the n-th degree assuming at these points the values 2/ y ly ..., ynj i.e. such that ,
(22)
w(xk )=yk
we
for
i
= 0,1,
...,.
a polynomial uk (x) of n-ih degree with the property that u k (xk ) = 1 and = for each ^lc. First
shall define for
m
each
Tc
10.
DEFINITE INTEGRALS
257
Namely, we write
uk{xv
= (x-x )-...-(x-xk
->
By means
'
...-(xk
of the polynomials
we
Uk(x),
xn )
'
define the
polynomial w(x): w(x)
(23)
=y
u
(x)
+y
1
u l (x)
+
.
We
see at once that this is a polynomial of the n-th degree, satisfying the condition (22). This is the so called
Lagrange interpolation polynomial. It is worth while noting that the polynomial w(x) is determined uniquely, i.e. if a polynomial w of n-th degree satisfies the condition (22), too, then it is identical with w] for their difference
is
a polynomial of degree
n+1 points (namely, at the points # When a function expressing a relation
at
xl9 ..., xn ). between certain ,
physical quantities is known only in an empirical way, i.e. if only a finite number of values of this function
(by means of measurements) is known, then an approximate evaluation of the integral of this function is applied, e.g. the Lagrange interpolation, and then the integral of the obtained polynomial is evaluated. We obtain an especially simple formula if n = 2 and
x
=
a,
=
x
-
2
,
#2
=
b,
i.e.
when we draw a parabola
through the given three points. As can be easily calculated,
we
obtain b
(24)
This
-~(
J is
the Simpson foimula giving an approximate b
value of the integral f f(x) dx. a
The approximation
will
become more accurate if we divide the interval ab into smaller intervals and if we apply the Simpson formula to everyone of them. Namely, dividing the interval ab 17
258
INTEGRAL CALCULUS
IV.
2n equal intervals by means of points # #1? ... where XQ = a, x2n = b, we obtain the general Simpson formula for the approximate value of an integral: into
,
EXAMPLE. Let us apply the Simpson formula to evaluate cdx
log 2
/
,
We
dividing the interval 1
<
x
<
2 into halves.
a/
i
have
=
#o
l,
3/0
=
1,
#!
=
1.5,
2/!
#2
Thus formula
(24) gives the
-0.667
=
=
2
0.667... ,
y2
,
= 0.5
.
approximate value of log 2:
+ 0.5) =
0.694...
we have Iog2 = 0.6931... Now we shall mention the so called
Actually,
"trapezoid method" of the evaluation integral of a funcapproximate in an interval ab. We divide the interval ab tion f(x) > of the
by the points XQ < x < ... < ocn and we draw a polygonal line through the points (#&,2//e), where & = 0, 1, ..., n (cf. Fig. 10). The area s n between this polygonal line and the X-axis is the sum of areas of n trapezoids T1? ..., Tnj where the into
n equal
where x
=
a,
intervals
xn
=
,
b
area of the trapezoid T&
is
equal to
Thus,
a
_b
n ~~
n
This is the approximate value of the area between the curve y = /(#), a < x < ft, and the X-axis; speaking otherwise, this is the approximate value of the integral 6
Sf(x)dx.
10.
,orx (25)
259
formula Q)
10.9. Wallis
We
DEFINITE INTEGRALS
prove the following formula: 2 r */ 2-4-...-2n \ 7u - lim ^ n =oo^\l-3.....(2n-l)/ shall base our proof on the formulae shall
--
.
s
We
TC/2 /'
,
.
sin2na?dte
= l-3-....(2n-l) o-i ^
2-4-...-2w
J
TT
o 2' ?
o 7T/2
('
si
J
proved in
From
10.1, 5.
these formulae
it
follows:
(26) 7T/2
We
shall
TT/2
prove that n/2 C
7T/2
lim
C
i
i
J
n=oo
in
Indeed,
sin^xdx: J
Q
the
interval
<x<
n/2
we have the
inequalities
whence
and
7T/2
7T/2
Tt/2
so
r
1<J
mi^xdx:
r J
TT/2
< (*)
(
si
7T/2
sin ^- 1 ^^: 2
J
John Wallis
i
J
sin^+^dte
<
^!
+ = 1+ 2n
2n
(1616-1703), an English mathematician.
.
260
INTEGRAL CALCULUS
IV.
Thus the considered quotient limit to
of integrals tends in the
1.
Hence we conclude by '/ 2TT
(26) that
< 27)
This formula gives formula (25) immediately, since n 1
Bemarks.
(a)
We
apply the Wallis formula also in
the following form:] (28)
J/TC
=
Namely, we have 1-2- ...-2n
__ (2n)l
n\-2
2-4-...-2n
n
'
Substituting the above formulae in formula (25) and taking a square root we obtain equality (28).
The Wallis formula may be written form of an infinite product: ((3)
n
-
2
n
1
4 ^2
1 4n 2 -l
~
n=l
also in
the
??*!_-
l'3'3'5'5V"'
Namely, denoting by p n the partial product 4>12
Pn ~"4-l 2
4n2 l'""'4^2 -l
-3-...-(2n-l)/
we
_ ~
/2 2\
\I'3/
2n
2n
\
\2n-I '2n+I/
2n+l'
have, according to (27), lim p
whence the formula
/
*"
(29) follows.
=5
At the same time
DEFINITE INTEGRALS
10.
261
10.10. Stirling formula C)
We shall deduce from the Wallis formula the following formula due to Stirling: ?' lim n n=ooj/27m^6T-
(30)
1
,
.
Let us write n\e n
nn ]/n So
is
it
to be proved that lim an
=
|/27c.
n=oo
we
First
prove the sequence al7 a a
shall
,
...
to
be
convergent. Since it is a sequence of positive terms, it suffices to prove that this sequence is decreasing, i.e. that
> 1.
Now, an
1
(n+l) n+1 + 1/2
~
1
/
e \
+ iyn-i/2
'
n)
whence
Since
for
7,
(cf.
<
a?
<
whence
1,
(54))
so substituting
>
1
a?
=
- we find
.
Thus, the sequence a1? a2 ... is convergent. Let us = We shall prove that g ^ 0. ,
write lim an
James
flf.
Stirling
(1692-1770)
noted Scottish mathematician.
262
IV.
For
this
INTEGRAL CALCULUS let
purpose
us note that ri+l
r
i\
since the arc of the hyperbola y
i /i
i
\
= - n^x^n +1,
lies
,
be-
30
low the segment joining the points n - and n +1 I
1
,
I
n]
\
\
,
,
I .
n+j
Hence
and
so
log
an
i.e.
,
arl
>e8
/*.
4:
remains for us to prove that g = 2^. To this end we now apply formula (28) taking into account the equations It
^ = ^.n ^w _ j
1
and
2
We
obtain
= lim
-
2
2-7=
=
02
^
i.e.
Remark. The practical importance of the Stirling formula is based on the fact that for large values of n the expression \/2nnnn e~~ n gives an approximate value of n\. We say that it is an asymptotic value of nl (which means that the quotient of these two expressions tends to
1).
10.11*.
Riemann
integral.
Upper
and
lower
Darboux
integrals
The theorems of 10.3 are the starting point for a generalization of the notion of a definite integral to certain classes of discontinuous functions. For this purpose we shall introduce the Darboux integrals.
DEFINITE INTEGRALS
10.
263
Let a bounded (but not necessarily continuous) function y = /(a?), a < x < 6, be given. Moreover, let a sequence n of partitions of the interval ab into 2 equal parts (n = 0, 1, ...) be given. Let us denote the points of the n-ih partition n,o
Let
,
#n,i
M
,
Htk
by ,
^n,2 w
where
,
a Wj0
=
a
,
^,2**
=6
.
denote the upper bound of the function
in the interval an ,/c-i <*
/
Finally, let 2*
$n
(31)
=
y^
^*n,fc
fc-1
The sequence {$n } is a decreasing sequence (in the wider sense). Indeed, for each ~k = 1, 2, ..., n the intervals #+i,2fc-2*Wi,2fe-i a ^d ^n+i,2A:-i tt Ti-fi,2A; are contained in the interval Oo^-i a n ,& Thus, .
i.e.
Hence
~
~~
-
2
2
Thus the sequence {Sn } is non-increasing. It is simultaneously a bounded sequence, since, denoting by the lower bound of the function /, we have obviously 8n ^m(b a). Hence it is a convergent sequence. Its limit
m
is
called
the upper Darboux integral
and denoted by
the symbol 6
(32)
m
Similarly, denoting by nt jc the lower bound of in a interval the n fc ~i^n,fc and writing /
function
,
,
(33)
sn
=
b
2W
a
v~i
-^r 2j /c-l
the
264
IV.
INTEGRAL CALCULUS
we prove the sequence
{s n }
to be increasing (in the wider
sense) and bounded, and so convergent. Its limit is called the lower Darboux integral and is denoted by the symbol
the upper and lower integrals are equal, then the
If
function
said to be integrable in the Riemann sense; value of these two integrals is called the
/ is
common
the
lim
/ i(x)dx
(34)
Riemann
integral
of
the
function
f
and
is
denoted by
6
/ f(x)dx as in the case of the definite integral of a cona
tinuous function. The same symbology may be applied, 10.3 imbecause (as follows from the theorem 2 of mediately) the Riemann integral of a continuous function is equal to the definite integral of this function (in the sense of the definition giA^eii in
10.1).
The functions integrable
the
in
sense
of
Eiemann
consist of a larger class of functions than the continuous functions. In particular, bounded functions having a finite
number
of points of discontinuity (we shall turn to such functions in 11) and monotone functions are integrable.
However, there in the sense of is
functions nonintegrable of such a function
exist (bounded)
An example
Eiemann.
the Dirichlet function which
we have already met
(cf. 4, 5), equal to 1 in rational points and to rational points; for this function we have always
and
sn
=
and
ir-
Sn =
1
so
b
/
in
b
i(x)dx
=1
,
a
J
i(x)dx
=
.
a
Let us add that there exists a definition
of
the
(Lebesgue) integral which ascribes the integral to a larger class of functions than the Eiemann integral ( x ). (*)
Cf.
S.
Saks, Theory
tyczne, vol. 7 (1937).
of
the Integral,
Monografie Matema-
DEFINITE INTEGRALS
10.
We
based
tlie
Darboux
definition of the
265 integrals
on
the consideration of a sequence of the "diadic" partitions of the interval of integration. We shall now show that
obtained considering an arbitrary normal More strictly, we shall prove the theorem: following Let a normal sequence of partitions of a segment ab
the same result
is
sequence of partitions.
defined by the points
<
(35)
<
M^ k
be given. Let
<
...
denote the upper bound of the function f
in the interval a,nk-iUn,k an d
8'n
(36)
where
,
let
M'ntk (a'ntk -
= k~l
Then b
= J f(x)dx. a
limS'n
(37)
n~oo
The proof ing lemma:
of this
theorem
be based on the follow-
will
Let two partitions of the interval ab,
a
=a
...
and
r
a
= aj < a[ <
...
< a^ = b
be given. Let d and A' indicate the length of the greatest interval of the first or of the second partition, respectively. Let k and k denote the upper bounds of the function f
M
M
in the intervals a k -iak or
(38)
S
Finally, for every
Then (39) holds.
M (a
=
k
k
a'k -ia'k
^ak ^)
,
and
respectively,
S'
and
let
=
M satisfy the inequality M>
the number x belonging to
let
the relation
S'
the interval ab.
\f(oo)
266
IV.
The in
r+1
INTEGRAL CALCULUS
intervals of the second partition may be classified classes; a given interval of this partition will
belong to the fc-th class (for fc = 1, 2, ..., r), if it is contained in the interval ak -idk and to the r+l-th class, if it is contained in none of the intervals of the first partition, i.e. if it contains inside one of the points
a 1?
2
,
...,
ar -i. It
clear that
is
some
of these classes
may
be empty. If the fc-th class (for k ^r) is non-empty, then let the first and j k the last index i satisfying the denote pk condition ak -\ < a'i < ak Since for the intervals a^i aj of .
M
so the &-th class there holds the inequality Mi ^ k k the sum over all intervals of the &-th denoting by ,
^
we have
class
ftfaJ-nf--!)
<
M
k
^
k
M
-a . = f
(a'i
i
k (a'ik
l ]
- a'Pk
)
M^ak-a^^M^a'^
M (a k
k
ak -i
the A-th class is empty, then let ]? = 0. Finally us note that there are at most r 1 intervals of the k
If let
y+l-th
class.
Thus,
Consequently,
,- a whence the formula Thus our lemma
)
+
...
(39) follows. is
proved.
In order to prove the theorem
let
e
>
/ f(x) dx
+
us consider an I
and
let
us choose an index q such that
$a
<
e.
a
Since the considered sequence of partitions
is
normal,
DEB'INITE INTEGRALS
10.
we have
<
d'n
-
-
for
3 2 Ju. according to the lemma,
267 n.
large
sufficiently
Hence,
we have b
f(x) dx
let
Moreover,
+ 2e
.
us choose to each n an index
holds < ol that the inequality -^r~ ^Tlvf JXL A' n
such
jn
-
n
whence b
6
J
<
f(x)dx
S'n
+
e
because
,
J
a
f(x)dx
< 8jH
.
CL
Consequently, S
5
f(x)dx
f
<S'n + e < f
The number formula
f(x)dx + 3e.
a
a e
being arbitrary,
this
inequality implies
(37).
Thus, our theorem is completely proved. There is a similar theorem for the lower integral. The following formula is an important consequence of
our theorem: b
c
6
J f(x)dx
= / 1(x)dx+ / f(x)dx, a c
f(x)dx
= f f(x)dx+ f f(x)dx,
a if
a
< c < b. n
us divide the segments ac and cb into 2 us denote the points of the partition by Let equal parts. Indeed,
(41)
a
let
= aw < ,
n ,l
<
.-
< #n,2w =
==
268
INTEGRAL CALCULUS
IV.
Let
M
denote the upper bound of the function / and PU)k the upper bound in
n>k
in the interval an k -iantk ,
the interval b n k -ib njk
Then we have
.
,
2n
C
lim
J f(x)dx
"-<.
a
i
and 2n
b
f(x)dx
/ c
=
lim
y
-*-i
Moreover,
2n
since the points (41) define a
normal sequence
Hence the
of partitions
the formulae (40) segment follows. The proof of the second one is analogous. It follows from the formulae (40) that if the function / is integrable in the intervals ac and cb (where a < c < b), of the
then
mula
ab.
it is
5 of
first of
also integrable in the interval ab and the for10.2 (on the division of the interval of in-
tegration) holds.
Conversely, if a function f(x) is integrable in the interval ab, then it is also integrable in both intervals ac and cb (and so in each interval contained in aft, too).
Indeed,
we have by the assumption
f f(x)dx = Comparing
//(0)
this pair of equations
c
(/ f(x)dx- /
with the formulae (40)
66
we obtain c
= f f(x
f(x)dx)
+ / f(x)*x- f /()*) = (
.
DEFINITE INTEGRALS
10.
Hence we conclude by the obvious
269
inequalities
~
b
c
/
i(x)dx
< / i(x)dx
/
,
a
a
I
i(x)dx
c
c
66
that c
c
i(x)dx
/
a
= / i(x)dx a
and
< / i(x)dx
/
i(x)dx
=
ff(x)dx. c
c
Just as for continuous functions
we assume the
for-
and
mulae
(2) (3) for arbitrary integrable functions and 10.2. In place prove theorems 1, 2, 6, 7 and 8 of
we
of the mean-value theorem
we have
for arbitrary integrable
functions the double inequality 6
m(b-a)
(42)
< / f(x)dx < M (b-a)
,
a
M
where a < b and where and m denote the upper bound and the lower lound of the function / in the interval ab respectively.
Theorem 10
of
10.2 on the differentiability of the
X
function g(x)
=
j a
f(x)dx does not hold for arbitrary in-
tegrable functions. However, function g(x) is continuous.
it
may
be proved that the
M
the upper bound of the funcIndeed, denoting by /(a?) in the interval ab we have (by the formulae (40)
tion
|
and
|
(42)):
X+h
-g(x)\
=
\f f(t)dta
x+h
=
IJ /(*)* X
whence limg(x + h)
= g(x).
f
f(t)dt
270
INTEGRAL CALCULUS
IV.
Exercises on
10 7T/2
rr/2
1.
r
2.
Prove that
n
xdx
1 T-I x -cu Evaluate the sum -
apply the
Deduce the
=/ 1 -
-
sin
n
#&.
/tl\
/M\
.
1
1
+ g ln\ 2 (J substitution x = costf to
1
mtf:
3.
cos
J
(
M\
,
...
(Q )
^+1 (J
)
the integral
formula
(Cavalieri) 1
lim
-
" +1 1
applying the integral of
m / x dx
also Exercises 2
(cf.
and 3
1).
4.
"
Evaluate lim n
(
n=oo
function
5.
\W
+
+ -^n
...
~rJ-
-\-
the n^j (integrating J
).
= limf-^ +-^ + - + ^1 \n+l n + 2 2w/
Prove that
Iog2. &
ns=00
6.
stant
Give a geometrical interpretation of the Euler con(cf.
log ^
),
/
rr
7.
-r*
7,
Example
18)
C
-
lim
=
applying the equation / x i J.T_
r
^.
Prove that
-
^sno? -
l
+ cos
I
o
_
r- ao? 2
=
7i .
4
a?
second integral x
Prove that
=
n
logsin^dte
7t2
=
logn.
^ Decompose ^
,.
the in-
t.
f J o
- + ...H--i+ n 2 o
and substitute in the
terval of integration into halves
8.
(l+
w=oo \
/ logcos#do?
=
= -^
7T
-.Iog2. (First
*
7i2
airi2ir
/ log
-
dx).
prove that
10.
DEFINITE INTEGRALS
271
Find the length of the arc of the parabola given #2 lying between the points with the equation y y 9.
,
tie
abscissae
and
Given an
a.
with the excentricity e and with tie large axis of length 2, express its perimeter by means f a power series with respect to e. 10.
ellipse
The circle given by the equation (xa) 2 + y* r* on the JTY-plane rotates (in the space) around the
11.
ring
Evaluate the area of the surface of the solid btained in this way, called the anchor ring (we assume ^-axis.
>r). 12. Lrcle
rid b
Evaluate the area of the common part of the with the radius r and an ellipse with the axes a having the same centre as the circle.
13. *
Applying the Simpson
x
formula to the integral
fl J. ~i
oo
2
,
evaluate the 5
ansion of the number 14.
Show
of the decimal ex-
first figures JTT.
that the Simpson formula (24) gives the
rea with an error less than
M
where
-
j^.
,
denotes
bound of |/"'(#)| in the interval a < x < To prove this, introduce an auxiliary function:
he upper
b.
c+t g(t)
=
J/^)^-|[/(o-) + 4/(c) + /(o + t)]
?
c-t
,
where
nd deduce the inequality lie
required result, 15.
=
-~2M
integrating
this inequality.
Prove the Schwarz inequality for 6
c
6
b *
f
integrals:
Get
272
IV.
INTEGRAL CALCULUS b
(Apply the inequality /
[/(a?)
+c- g(x)fdx
>
which
a
holds for every
c).
Give the formula for the radius of curvature at the point #, y for the following curves: 16.
hyperbola xy
3)
parabola y = 4p# the curve y = x*.
5)
17.
.2
= 1,
1)
2
4)
,
is
a
|*2
hypocycloid x%
Find on the exponential curve y
which the curvature
+ |i = ^
2) ellipse ~i
=
0*
+ y% = a*,
the point at
maximum.
18. Given the hypocycloid from the exercise 16, 4). Evaluate its length, the area of the region contained
the volume and the area of the solid of by the rotation of this hypocycloid around the X-axis. inside of
it,
revolution obtained
A
system of functions /l? ...,/n, is called linearly if no system of n constants clj ...,cn exists of the system of n zeios) such ttat the equality except 19.
independent,
holds for every value of x. Prove that a necessary and sufficient condition for the linear independence of a system of n continuous functions in an interval ab is that the inequality
au
. . .
a
b
holds, where akm
=
/ ik(^)fm (x)dx.
a
(Apply the conditions for the existence of non-trivial solutions of a system of
with n unknowns).
n equations
DEFINITE INTEGRALS
10.
20. is
A
273
system of functions
finite or infinite
called orthogonal in the interval ab,
=
^
for
fc
for
k
/i
?
/2J
...
the conditions
if
^m =m
,
are satisfied.
Give examples of orthogonal systems among the trigonometric functions. 21. Prove that the polynomials (so called Legendre ( l ) polynomials) defined by the formula
n(
n
'
2 nl
constitute
and orthogonal system
in the interval +1
(First
prove that
if
m<
then /
n,
1,
+1.
w n (x}*x m dx =
0.)
IMPROPER INTEGRALS AND THEIR CONNECTION WITH INFINITE SERIES
11.
11.1. Integrals
an unbounded
with
Let a continuous function y
of
interval
= f(x)
integration
be given in an X
infinite interval
x
> a.
Then the
integral
/ f(z}dz exists a
for every
x
>
a.
If there exists X
lim F (x)
where
,
*-
F(x)
=
f f (z) dz
,
a oo
then this limit called
is
denoted by the symbol
the improper integral
function f from a to
oo.
(of
the
first
J a
f(x)dx and
kind) of the
Assuming the existence
of this
Adrien Legendre (1752-1833), a great French mathematician.
274
IV.
INTEGRAL CALCULUS
we have then
limit,
(1)
*-o
a
We
say in this case also that the improper integral
is
convergent. X
00
lim ff(z)dz^= oo (or -co), then
If
we
write / f(x)dx
scoo a
=
a
oo); in this case
oo (or
Similarly,
/(a?)flte==
lim X--00 X
and
J
j(x)dx
have by the
is
J f(x)dx.
oo
definition:
)-F(a)'}
F
f(x)dx+
J
oo
where
is
a
a
We
integral
oo. +00 or we define:
divergent to
(2)
we say that the
- ]\mF(x)-F(a)
a primitive function of the function
,
/.
E.g. dx
&
~^-
J
Hence
>
s
is
this
=
integral
t \
is
r hm x-co
convergent.
More
generally,
for
convergent. for s
=
l
we obtain an 00
I
X\l
[
1
=
1 the integral
But
(4)
11
I"
integral divergent to oo:
IMPROPER INTEGRALS
11.
An example
of
an
275
integral having neither a finite nor
oo
an
infinite limit is
sin x dx, since lim
/
cos# does not
exist.
= 00
In the last example the essential fact is that the function under the sign of the integral assumes positive as well as negative values. In fact, the following theorem holds: 00
3.
either
>
// f(x)
for
x
>
a, then the integral
f(x)dx has
J a
a finite or an infinite value (according
to
whether
X
the
F(x)
function
=
bounded
is
/ f(z)dz
unbounded
or
a
in the region
x^a).
the function
Indeed, F(x) being non-decreasing, it lim limit the F(x) finite or infinite according possesses to whether this function
bounded
is
or
unbounded
(cf.
4.7, 1).
Theorem immediately 2.
A
of
4'
4.7
the
implies
following theorem,
:
necessary and sufficient condition for the convergence oo
of the integral
/ f(x)dx
(to
a finite limit)
is,
that to
any
x>
and
a s
> >r
x'
a number r
imply
exists
such that the conditions
r
the inequality
_')dz
<e
.
x'
Indeed, the existence of the limit lim F(x)
is
equivalent
x=oo
to the condition that to every
such that \F(x)~ F(x')\
<
e,
e
>
number r and x' > r.
there exists a
whenever
x>
r
Since
z, Xf
so
<
276
IV.
INTEGRAL CALCULUS oo
Geometrically we
interprete the integral / f(x)dx simia
>
larly as the integral in finite limits: if /(a?) 0, then this means the area of the unbounded integral region made of such x a that and up by points #, y y f(x).
< <
>
11.2. Integrals of functions not defined at one point
Let a continuous function / be given in an interval < x < b (the interval without the point b). Thus for any x satisfying the above condition the definite integral a
X exists. If there exists
f f(z)dz
the limit
lim F(x), where re=6
J?(x)
=
J f(z)dz,
then this limit
is
denoted by the symbol
a b
/ f(x)dx, as in the case considered in the previous section, a
namely, as in the case when the function / is defined and continuous in the whole interval a < x < b (together with the ends). In the last case the definition given in 10 (formula (1)) is consistent with the definition of the b
symbol
given at present. In other words,
ff(x)dx
if
a
a function f(x)
is
continuous in the interval
a
<x<
6,
b
then the integral in
10,
formula
/ f(x)dx
(in
the sense of the definition
a
(1)) satisfies the
condition x
b
(5)
Jf(x)dx=z
Mm
F(x),
where
This equation means that F(b)
F(x) .= j f(z)dz.
=
lim F(x),
i.e.
that
a?=6-0
is (left-side) continuous at the point &; but this follows from Theorem 10 of 10 (Eemark (p)).
the function F(x)
The a
<x<
integral of a function / defined in an interval b is called an improper integral (of the second
IMPROPER INTEGRALS
11.
kind). If this integral exists,
i.e.
277
lim F(x) exists, then
if
x=*b-0
we say that the If the
is
integral
function
/
convergent.
and continuous
defined
is
an
in
b
<
interval a
x
<
6,
then the integral ff(x)dx
defined
is
a
analogously: 6
?
=
ff(x)dx a
lim ff(z)dz. x-a+0 x
More
generally, if a function / is defined and conin tinuous an open interval a x b and if the improper
<
integrals
J
<
b
c
f(x)dx and
a
J
f(x)dx are convergent,
c
being
c
an arbitrary point
a
of the interval
<
x
<
bf
then by
6
the integral
J
f(x)dx
we understand
a b
c
/ /()*= //(
(6)
It is easily seen that this
the choice of the point
c.
Yet more generally,
if
sum
does not depend on
the function
< < x
a
continuous in an interval
b
defined and
/ is
except
of
a finite
not defined
of points (in which the function discontinuous), then by the (improper) integral of the function / in this interval we understand the sum
number or
is
is
b
fj(x)dx= f f(x)dx + f a a
(7)
On
az
cii
f(x)dx + ...+
//(a?) Ac, an-\
i
a < a x < a 2 < ... < an = 6, and all points such or discontinuous at these points are undefined / a nin contained the system of the points a 2 1?
where a that
=
is
?
,
>
b
So the integral
J
f(x)dx
is
convergent,
if
all integrals
a
on the right side of foimula
(7) are
convergent.
278
IV.
1.
INTEGRAL CALCULUS and
If a junction f(x) is bounded
if it is
defined and
continuous except of a finite number of points of the interval b
is
ab, then the integral ff(x)dx
convergent.
a
According to the formulae (6) and (7) it is sufficient to prove this theorem in the special case when the function / is defined and continuous for a < x < b. For this purpose let us note that there exists a necessary
and
the convergence of improper
sufficient condition of
integrals of the second kind analogous to the condition given for improper integrals of the first kind ( 11.1, 2): 2.
A
and
necessary
condition
sufficient
con-
the
for
b
mergence of the integral / f(x}dx of a function defined
and
continuous in a < x < b is, that to any exists such that the conditions 6 >
and
a
x
<
r
d
> b
a number
x
<
6
imply the inequality
(8)
ff(z)ti
Indeed,
s
<
writing
F(x)
=
<
8
.
/ f(z)dz
we conclude from
a
Theorem 4
of
4.7 that the existence of the limit lim
F(x)
x=*b-0
equivalent to the inequality \F(x) F(x')\ < s (when x x' satisfy the above mentioned conditions). But
is
and
F(a?)-W)= x'
whence the formula Let us
an
e
>
a number 6
0,
<
conditions
equality
b
<x<
x
<
to the proof of Theorem 1. Given has to be chosen so that the
> d
and
<
x
f
<
imply the
in-
by assumption bounded
in
b
d
(8).
The function a
(8) follows.
now proceed
b.
f(x)
is
Thus, a number
M exists such that M >
\f(x)\
11.
for every
IMPROPER INTEGRALS
x in the considered
Now,
279
interval. Let us write 6 == e/M.
X
\ji(z)dz\<\x-x'\M x'
<
<
<
<
b b x x' d imply d and inequalities d e/M, whence the required formula (8) follows. |##'| In this way Theorem 1 is proved. Thus, the notion
and the
~
<
of an improper integral makes possible a generalization of the notion of the definite integral to bounded functions
having a
finite
EXAMPLES,
number (a)
-=
}dx 6
=-
/ O
of points of discontinuity.
S
JL
0(j
[X
I
~S ]Q
= ---L
$
for
s
<
1.
oo.
ft i
(p) J o
^ sin-dte exists ft
by Theorem
under the sign of the integral for < x < 1. (y) Similarly,
is
the integral / o
the
1,
since the function
continuous and bounded
-dx
& be
exists.
In this case
is considered to seemingly improper; the function under the sign of the integral is although riot defined at the point 0, it possesses a limit at this point. Assuming this limit to be the value of the integrated
integral
we obtain a definite integral function at the point (of a continuous function) in the usual sense. Analysing the definition of the (improper) integral and continuous in an open / defined < b we see that this integral may be defined as the difference F(b)F(a), where F(x) is an arbitrary function continuous in the closed interval a < x ^ b and satisfying the condition F'(x) = f(x) in the open interval a < x < b (of course, if such a function F exists). T fj w E. g. the integral f which expresses the length (S)
a
function interval a < x
of
tv*
,
2
2
is an improper integral (the denominator of the integrated function vanishes at
of a semicircle with a radius r
280
INTEGRAL CALCULUS
IV.
the ends of the interval of integration). However, since the primitive function of the integrated function, i.e. F(x) = rare cos (x/r) is continuous in the whole interval
the
r<#-
value
the
of
considered
expressed by the difference F(r) F(r) the example considered in 10.5).
=
integral
is
nr (just as in
Bemark. An
improper integral of a function bounded of a finite number of points is equal to the Eiemann integral of this function. According to the theorem on the division of the interval of integration, the proof is reduced to the case when / is continuous in the interval a < x < b. Moreover, according to the continuity of the Riemann integral treated as a function of the upper limit of integration (cf. 10.11) and to the of function in the the sense of Kiemann, / integrability the Eiemann integral of the function / in the interval ab
and continuous except
x is
equal to
x
b
lim ff(x)dx,
i.e.
=
x**bQ a
ff(x)dx (the symbols / a
a
b
and of
J
are applied here in the sense of the definition
10.1
and
11.2).
11.3. Calculation formulae
Let the functions f(x) and g(x) be continuous and < x < 6, where b may also denote oo. Let us assume that the integrals (improper of the first or of defined for a
b
b
the second kind)
J a
Then the formulae
f(x)dx and / g(x)dx are convergent. a
1
and 2
of
10.2 remain valid, for
x
b
f[f(x)g(x)]dx=\imf[f(x)g(x)-]dx a
x
= lim J f(x)dxlim J x~b x=b a
b
a
b
= j f\x)dxJ g(x)dx.
g(x)dx
IMPROPER INTEGRALS
281
proved similarly that the formula 2 of
10.2
11.
It
is
holds.
The formula formula
for the
integration by parts (i.e. the remains also true assuming that f(b)g(b) de-
3)
notes limf(x)g(x). x=b
Similarly, denoting lim/(#) by j(b) (where b as well x~b as f(b) may be also oo) the formula 4 for integration by substitution holds. Namely, assuming the integral /(/>)
J
g(y}dy
b e convergent, we have
^
/(a)
m
v
/(*)
-lim /
g(y)dy
=
lim
/
g(y)dy
=
j g(y)dy
,
/(a)
since the integral treated as a function of the upper limit of integration is continuous ( 10.2, 10). Assuming the strict monotony of the function /, the 6
convergence of the integral
g[f(W()dx
f a
implies the
/(&)
convergence of the integral
J
g(y)dy.
Hence the
pre-
1(0)
viously proved relation may be reversed; namely, according to the equation lim h(y) 6, where h(y) denotes
=
y=J(b)
the function inverse to
/(#),
the previous argument
may
be performed also in the reverse direction. Other calculation formulae: the formula 5 on the division of the interval of integration and the formulae 6-8 may be also generalized easily to the case of improper integrals.
Eemark.
Let a function
/
be piecewise continuous in
This means (cf. 5.1) that there exists =a a system of points aQ < a^ < a 2 < ... < an where = that the function /* defined by the conand an fe, such
an interval
ab.
,
282
INTEGRAL CALCULUS
IV.
= j(x) for a*_i < x < ak = 0) is continuous in /(a* fk (ak = A:-l<^^% (& 1,2, ...,tt).
ditions fk (x)
and
,
)
the whole interval
Then we have ak
a
n
k**l ak _i
ak
fc=l o fc _ 1
Using this equation and taking into account the fact that
J fk (x)dx is
a proper integral
we
state easily that
the formula for the integral of a sum remains valid if the functions / and g are piecewise continuous and that the formula for the integration by substitution remains also true assuming the functions g [/(#)] and f(x) to be piecewise continuous. 11.4.
Examples
(a)
x
To evaluate the
= tank
Since tanO
integral
=
and
00 00
1
I
"\
\
J. ~|
rf> JU
\
/Jf it/v
F
/
I
The same
dx
and taking
___
=
oo,
we have
(
1
n
may
>
1
be evaluated applying the
9.3, 5,
x
1
into account that
=
substitute
/
for
integral
recurrence formula
we
,
)
Tt/2
-J -
r
x (I lim tantf
7T/2
f
/
+^
f b
and
2n
3 r
dx
11.
IMPROPER INTEGRALS
283
mn
/*
(P)
we
To prove the convergence
of the integral
apply the second mean- value theorem
shall
We
creasing.
*^ (
taking into account the fact that the function oc
'T*
- dx
*
/
10.7) is
de-
obtain
6 /'
J
sin# -dx=~1
x
>
Let an
f
the conditions b
1 f
,
.
smxdx + T
bJ
aJ
.
7 smxdx s
.
be given. An r has to be defined so that > a > r imply (cf. 11.1, 2): b
sin a?
<
dx
e.
Now,
Thus,
We
_/l 1\ 4 4 ^2 - + r)<-<-. r a
t'smxj dx
smxdx
x
J
it
assume
suffices to
r
\a
=-
o/
.
note that a
C
,.
lim
(9)
OT,=aOO
co
m\nx X
J
sino? f ---
^
,
dx
,
~.
.
dx
X
J
(a>0).
o
o
Namely, let us substitute y
-
f sin no? (10)
J
x
=
We
n^.
obtain
na ,,
do?
=
11
/
J
r
if
<%
y
and
-
sin?/
because lim na
oo.
noo
Eemark. 00
J
sin*
&
dx
Let us write In
=
In n-0
.
This
is
=
sin a?
/
x
dx.
an alternating
Then we have
series
(which
is
284
IV.
also
easily
y
=
mn IT
terms have
decreasing
geometrically on the graph of
to illustrate
the function
y
INTEGRAL CALCULUS
= xnn we
cf
,
x tending
to
with
25)
Fig.
.
0.
(absolutely)
Indeed,
substituting
J
J
y
o
and C
sin?/
J y
+ rnc
dv J
r
sin y
<J
7
~^r~ dy
T In -i
=--
(nI)n
-
3jt
PIG. 25
Hence
it
follows easily that for every a
>
0,
we have
the formula _
(11)
/sin# J[/
^
dx \
C sin x
X
*J
,
dx
.
This formula together with formula (10) gives
(12)
(y)
x
dx
(
smx x
J
dx -j
.
Similarly as in the previous example,
we prove
00
the convergence of the integral Jsin(#2 )d# (the Fresnel integral applied in optics) although
we do not know the
IMPROPER INTEGRALS
11.
indefinite integral. Substituting x*
t
285
we obtain
CO
CO
r sin
,.
Applying the second mean-value theorem to the
we
integral
last
find that b
6
fsinJ, 4 dK
=
i'
i
1 C .,. -^i sin<& .
+
ai
\
1
C
tr \
b
.-.
.
sin t
,
i cvnf
co
and we prove the convergence
of the integral
"=-dt yt
f i
by the same arguments
as in the previous example. Thus,
OO
the integral
sin(x*)dx
J
convergent, too, and so does
is
i
CO
the integral
1
sin(#
j
2
)
2
since the integral
f
sm(x )dx
is
b
6
a proper one. (S)
We
have proved previously the convergence of the
CO
-
integral /
dx.
Now, we
shall j)rove that
CO '
sin o?
,
(13)
TU
t=-
i*
o
According to
(9), TC/2
/'
--
sin a?
X
.'
= Tlim n=0o
, f sin^wc dx
X
.'
,
whence TT/2
sin,T f -T
/
We j o
shall
= hm
X
n=co
reduce the last
^^^ --
sin ( sn ^__
dte
f sin(2H -
integral
to
the
integral
,., IVT J7 we '~dx, which we evaluate now. Namely, !
?
,
i
286
INTEGRAL CALCULUS
IV.
shall
show that 71/2
C sin(2n+l)#-,
^14^
(
)
tfz
J it
Indeed,
dx
=
TU
o A
Dill JU
follows from the formula
1,
(cf.
(2))
-
that
sm# whence Tt/2
ain TQ-M. -J_1
^
^
-dx
TU
~
=
Tt/2
7T/2
r
r
TC
~~
2
J
J
2
'
since all integrals appearing in expression in the middle part of this formula vanish.
Hence the formula
is
(14)
proved. It remains to prove
that 7T/2
_ sin(2n+l)*K = Q
lim
^
n=*oo
J
^ i.
e.
that 7T/2
lim n
Now, the
oo
last
I
^
-. o?sina?
sin(2n
+I)xdx
=
.
o
equation
is
a direct consequence of the
formula 71/2
lim J
/ (x) sin nx dx
=
IMPROPER INTEGRALS
11.
which we have proved already
10.2,
(
287 after the
(y)),
substitution
=
^ ' /(a>) 1
The function
_Q
/y
defined in this
/
.
.
(cf.
and
way
is
/(O)
=
.
continuous, since
i
4.
J o
prove that if the function f is monotone and continuous in the interval < x < a (a > 0),
shall
is
..
^.
hm
(lo)
n=^-<x>
Let a number a
>
" 8.4,? (18)).
*
^
4.
We f
x
for
T^ ui Dirichlet integral:
x
(s)
if
/v
*-
xmix
i
T
___^L*
,,
/
e
x
/(a?)
J
sri^^ _ --rte
=
re
-
,
/(O)
.
Jv
>
be given.
We
choose
such
c in
that
way
0
and
I/(0)-/(0)|<
(16)
SI Tl
/*
In order to estimate the difference
J /(a?)
'W
'I*
dx
*^
o
~/(0), let us write
C J
,,
,sin^o?,
/(a?)
TC,,^
=
f J
c
TU\
,
-2
a?)
C
and
let
sin nx
,
~^^
us estimate each of the three terms on the right
side separately.
To the
first
value theorem,
term we may apply the second meansince
the
function g(x)
monotone. Taking into account that #(0) c
f J
= /(#) /(O) is = we obtain
c
sinno?
,
288
INTEGRAL CALCULUS
IV.
By
(12), I
tjiiLiviJU
x
j
sin o? , SI
-
l
-.
<2
dx
x
J
- ax
.
>
This inequality together with the inequality (16) gives U
f (17) ./
To estimate the second term and
us note that
let
by
(9)
10.2, (y)
by
(13), c
V lim f
noo
* Saiti
/M/,
=/t
dx
x
^
.
2
Hence we have f (18)
dx-
siiii
J
<e
a?
for sufficiently large w.
in the formula
Finally, replacing f(x) ^
and taking
into account
the inequality
< < c
a,
o?
we obtain v lim f n=c
^A
sn wo? sin
x
/(a?) ^/
/
whence we conclude that smtM? (19)
'
x
<
dx
holds for sufficiently large n.
The estimations
(17)-(19) give the required
tion for i*
f J
sn no?
whence we obtain formula
,
~~
(15).
TT
,
i( 0)
estima-
IMPROPER INTEGRALS
11.
Remark. The formula
(15)
remains valid,
place the assumption of continuity of
we
if
and that
/'
re-
of
by a weaker assumption, namely, by the
of /
monotony
289
assumption that the function / is piecewise continuous together with its derivative and piecewise monotone here /(O) has to be replaced by /(+0) (in the case when the ;
function
discontinuous at the point
/ is
0), i.e.
(20) 5.1) that there Indeed, our assumption means (cf. = a < al < a2 < ... < am = a a system of points such that the function /& defined by the conditions
exists
fk (x)
= f(x)
and
<x< = /(fc-i + 0)
for
/fc(a*-i)
continuous together with
is
ak
ajc-i
0)
and monotone
derivative
its
- f(ak
fk (a k )
,
for each k ^ m. whole interval a^-i <S x ^ of the interval on theorem the division the Applying
in the
<*>k ?
we obtain
to the improper integrals a
am
i
*
sinnx
C
_
.
.
'
minx
.
.
m-l
ao
Hence a ..
Inn
f
n=oo J
But
..
.
j(x)
-- dx
smnx
,
\^ T > Inn
--
*
^ *=-!
$>
this gives the
a)k
m n
oo
f
/
/
v
h(x)
--- dx
siiin.7;
..
.
^
/
a^.^
formula
(20),
sinn#
/A = ^// A = 7c^ o/i() 9/(+ ^ ^
since
by
(15)
we have
ai
f^/
v Inn n=oo
v
/i(0)
,
&0
*/
*/
v
,
\
)
a
and, on the other hand,
v lim n = oo
f
,
7
/A; *^
we have by ,
(x)
sinnic
ax
10.2, (y),
=
^
19
290
INTEGRAL CALCULUS
IV.
for each
>
k
the function -^
1,
being continuous in
-
iJL'
<#<#&.
the interval %-i
() The following formula which will be used in the theory of Fourier series follows easily from the formula (15):
here
/(aO
lim
(21)
we assume
monotone. For this purpose ^ n < v < 7t, then
I'.//
'
'
xSinwff
its
/
to be piecewise
derivative)
and piecewise
enough to note that
it is
/'
7
f(ac)rdx sm ;
;
'
as previously the function
continuous (together with
J
0
/(+0), ' V
a?
1*
we have
if
-
.sinwo? \
./
{
w v
--
sin# T ^/ v# /(#)a? a? sin J n-oo
= lim i-
:
=
.
s
8.4,
(cf.
(18))
and the inequality x
tute to the function ,.
.
,, function f(x)
a?
/(a?)
in the
<
TT,
sin
'y'
:
we may
formula
.
-/v
For, taking into account the equation lim
A
substi-
10.2, (y),
the
sina?
oo
Poisson
(*)
integral
x2 f e~ dx
=
?.
o
First of all let us note that applying the substitution
x
=
foo\ \LiJ)
^|/n,
we obtain I
/>
I
(/
z 2 /7/r ttol/
~ 1//W Iv I/
I I
/> t/
ft*2
/?* tw
l Simeon Poisson, an author of numerous works in the cal( ) culus of probability, in mechanics, physics and astronomy.
11
I.M
.
P
11
1
KU
I
NT
O HA LS
Jfi
In order to be able to compare the
201
last integral
with
the integrals
(23) a
and
*
2-4.....(2n-2)2
which we already know (cf. 10.2, we apply the double inequality
which follows from the inequality x2 and t = by substitution t
(5)
^ 1+
et a?
2 .
(the first inequality being valid for
/l
I
'
(l-.^)
^v
'
e~*fa<
I
t
8.4, (16))
(cf.
<
\x\
whence
1), CX)
~
u-'^Au-r: "
| '
'
(a)),
Then we have
OO
1
1
and Example
Taking into account the formulae
--.
^f (14^ #2 ') ^
and
(22)
(23)
we
conclude that r
2-4-...-2-U
M a'-
^r 2^TT)
-
J
Denoting by an the
^^
2 ^ '
3- 4...
left-side
term
.
2) 2
(a
of this
relation
00
and by for
b n the right-side one,
each n
= 1, 2,
we have an <
j
e~ x*dx^b n
o ...
Now, we have by the Wallis formula .
where
^n
=
1
(
10.9, (25)):
2-4-...-2M
----------2 1 n 1-3-.. .-(2^-1)
292
INTEGRAL CALCULUS
IV.
Hence an
cn
decreasing as
W
easily
""^
I
2n
and
-<^n?
i-
^
O\
**
is
~<2(?n5 hence
.
4>
C\nn
{cn }
O/M
1
we conclude that
Consequently,
Cn
-
seen,
A
(\
r
and, since the sequence
'^--
I
'2w+l
^,(U
T
,
x%
6 />
/7/-/
J
2'
^. Cn 2n-l *"""
/*
.
o
since
r lim
2n 7r~
L
2n
= 1 = lim
-~
we have
= limcn - ~-
e- a!a * I
^
11.5.
Gamma
The
function
F(x)
This integral possesses for
.
^
n=oo
=
/F^e-'di,
a?
>
<
1 both singularities: infinite interval of in-
a?
is an integral in an and moreover, the integrated function tends tegration
namely,
it
tends to
to oo as
t
we
decompose
shall
0. it
In order to prove into
two
00
x~l
J
t
and
e- t dt
shall
t
x~l
e~-
j
i
dt
i
o
and we
convergence
integrals:
1
(24)
its
prove the convergence of each of these
integrals separately.
To prove the convergence let
us note that
< x ^e" < i
1
of the first of these integrals since t 0. But the inte,
x~l
t
>
i
gral
J
t
x ~ l dt is
convergent, since 1
#<
1
(cf.
11.2, (a));
11.
IMPROPER INTEGRALS
293
hence we deduce the convergence of the
first
of
the
integrals (24).
Now, 6
whence we have
Hm~- -
-
(cf.
7 (27)),
J=--3
t*- 1 e~ l
< - for t
sufficiently large
tf,
and the
convergence of the second of the integrals (24) follows OO
from the convergence
of the integral
-t
/ -^
(cf.
(3)).
In this way the convergence of the integral F(x) is proved for every x > 0. Now, we shall prove for positive integers n the formula
r(n)-(n-l)I
(15)
For
by
this
purpose we apply the formula for integration
parts to the integral V(x) for
(26)
r<) =
x>
1:
-|H-'^-*
Hence we deduce tho formula
(25)
by induction, ap-
plying the equality no
/'(I)
jV'd* -
=-
1
-0!
n
Let us note that the Poisson integral considered in is one half of the value of the F function for x = J,
11.4
since substituting
(27)
t
= z*
we
find
294
IV.
11.6.
The
/(#),
x
INTEGRAL CALCULUS
relation between the convergence of an integral and the convergence of an infinite series THE CAUCHY-MACLAURIN THEOREM. Let the function
>
decreasing and positive. Then condition for the convergence of sufficient
be continuous,
a,
a necessary and oo
the integral
I
is the
f(x)dx
convergence of the infinite series
a
I /(+).
a
a+1
By the assumption, a f
n1 < x < a + n. afn I
FIG.
2(>
f(a
+ n)
a+n f(a
+ n) dx ^
C
/(.r)
f(a
+n
for
1)
a +11 / (x)
f a-f n
a f ;j-l
.
Hence dx
\
/
(a
-\-
n - 1 dx )
,
1
i.e.
Denoting by 8n the partial sum of the i.e.
$n
/(+!) +
...
series J^ f(a
+/(a + n), we have
o-J-n
:_!+ /(a).
m-l
+ m),
IMPROPER INTEGRALS
11.
295
we conclude
integral to be convergent,
Assuming the
00
that the considered series
is
bounded: 8n
< //(#) dx,
and
a
convergent (as a series with positive terms). Conversely, assuming the convergence of the series, we have
so
it is
a+n
oo
J^r-^ f(x)dx '
'
<
> /(a + m)
for
Jm^J
each n.
Hence oo
.r
J f(x)dx
-v
^
for
m)
f(a 4
every
.r,
m-O
a
denoting for a given x by n a positive integer such that x < a + n, we have since,
/r
a+n
x
f(x)dx
-:'
n
80 the function
J
J n
oo
\^ f(x)dx
<^
2^ ffa
+ m)
.
7M=0
f(x)dx
is
bounded and, consequently
a oo
the integral f f(x)dx
is
convergent
11.1,
(cf.
1).
a
Remarks,
(a) The assumption that the function / be omitted in the above theorem. Namely, if a decreasing function is negative, then the integral as well as the considered series are divergent to oo. is
positive
may
oo
(p) If
the integral //(a?) dx of a decreasing function a
is
convergent, then lim
/(a?)
=
0,
as
is
easily seen.
However,
x-oo
not decreasing, then the integral may be convergent although this inequality does not hold. if
the function
is
00
The evidence
gives here the integral
f
2 m\(x )dx which
6
we have considered already
in
11.4,
(y);
this integral
2 convergent but the limit of sin (a? ) as x tends to oo does not exist.
is
296
INTEGRAL CALCULUS
IV.
Assuming
>
f(x)
of isosceles triangles
vals
(0, 1), (,1 1J),
bases,
let
U,
...,
Let y
successively.
infinite
an even stronger singularity may us construct an infinite sequence with height 1 and having the inter-
0,
be obtained. Namely,
+ on> = /(a?)
as
be the graph of an
by the
line constituted
polygonal
the ^' axis
of
sides of these
2
1
FIG. 27
from the bases) and the segments of X-axis joining the successive triangles. Thus, the region contained between this polygonal line and the X-axis is constituted by the above triangles; hence its
triangles (different
the
area
is
the
sum
of their areas,
However, lim/(#) does not
i.e.
exist.
X = OO
00
EXAMPLES, for every *
>
The
(a)
x
1
(
series
(s)
=
J~j
since the integral /
),
*
is
is
convergent
convergent
11.1, (3)).
(cf.
00
(S)
The
series
X
1 7;
~ 8 is
convergent for
s>
1
and
nlz n(logn)
divergent for (*)
,9
= 1.
The (Riemann) function {()
theory of numbers.
is
of great
importance in the
11.
IMPROl'KH IN TWO HAL 8
297 00
To prove this, let us consider the integral ^
doc r^-
}
'
2
a?
(log a?)
Now,
JdX JU
1 1 f\ ff /v> \ \ \\JcL JU i
3 K/
\A> A*-'
/
/vt
/lfiO I
>
i
^"o
/
'Ori
**/
-A
-I.
1 -l~
i
I
loo* V^-^
o
o
i
jf
/?! */y
7
Since
lim
8
"1
=
(logo?)
oo
,
we have
Hence we conclude by the Cauchy-Maclaurin theorem that our series
is
convergent for
s
>
1.
Yet for
s
=
1 this
series is divergent, for
f J
dx
l^x
=
C J
According to the equation lim log(log^)
oo, this
implies
a;=c >
the divergence of the integral to to
/
2
(y) Similarly,
it
may
be proved that the integral dx
f or s
>
is
convergent
is
sufficiently large).
1 and divergent for Hence the series
s
=
1 (where a
00
V
*J
'
n~k
s > 1 and divergent for * = 1. More general results may be obtained considering the product of n successive iterations of the logarithm the is
convergent for
298
I
which
last of
example n
V.
I
NTEG
II
AL CALO
LI
L
L>
S
raised to the power s (in the previous the zero -iteration = x).
is
= 2,
This leads to further types of convergent series and to
an
sequence of infinite
infinite
oo
oo
v
YII ^ j^j n'
series
oo
v
i
\
i
>
7
.
PTC
.
^-J nlogr^log(logn)
divergent more and more slowly. (S)
criteria.
Logarithmic
The comparison
of the
oo
terms of a given oo
~[
~~~
leads to the following convergence criterion:
8
*/
a n with the terms of the series
series
an
>
(28)
then the aeries
a n iv convergent]
n=2
^Aaw
is
if
thin
limit
fv
>l,
divergent.
assume the inequality (28) to be satisdenote by ~s a number greater than the considered limit and less than 1. Then we have 8 > 1 and Indeed,
let us
and
us
fied
let
log('/m n )
<
for sufficiently large
na n
<
-#log(log-M) w-,
(logw)~%
-
log (log -w-)""
an
<
8
whence i.e.
-
--
.
s
n(]ogn)
Thus, by the theorem on the comparison of with positive terms, the convergence of the of the series the convergence implies l &
nt^ n(logn)
Yet
if
series
series
the limit considered in the formula (28)
greater than
1,
is
then for sufficiently large n we have
11.
log(nan )
>
r
>
~~
IMPROPER INTEGRALS
log(logw),
and
the
whence nan
1
(logw)""
the
of
divergence
niogn
>
299 ,
series
an
i.e.
5] n 2
>
=
nlogn
oo
implies the divergence of the series
The case when the considered
JT an
limit
.
is
equal to
1
for our logarithmic criterion a "doubtful" case, i.e. we are not able to state by this criterion whether the series
is
convergent or divergent. However, in this case stronger logarithmic criteria of the considered in Example (y), may be applied. Let us add that there exist series (with positive terms) which do not react to any logarithmic criterion. is
11.7. Fourier
(*)
series
Suppose that we are given a convergent trigonometric the form
series of
/(0)
(29)
=
K+
(<
We note that if the given series is uniformly convergent n <x < then the coefficients an and bn
in the interval
TT,
be expressed easily by means of the sum of the series, i.e. of the function /. For, according to the uniform con10.2, 12) vergence we have (cf.
may
/
J
J
7T
>
^-J
2
n=l
7t
7T
and
00
71
7T /
.
n
since
cosnxdx
I
=
0=1
sinnxdx
7t
1 (
)
and
I
~dx =
na
K
Jean Fourier (1768-1830). Fourier introduced the expansions known under his name in connection with his work in
of functions
the theory of the heat conduction.
300
INTEGRAL CALCULUS
IV.
we have TT
(30) TC
.
To evaluate anj we multiply both (29) by cosnx and we find TC
TC
f(x)eo&nxdx
I
sides of the identity
J
n
~cosnxdx-i
re
OO oo
TC
v
a-
m
r ( am cos mx cos nx
1
+ bm sin mx cos nx) dx
,
TC
just as in the previous calculation.
Since
10.1, (S))
(cf.
TC
cos2 no?d#
J
=
TC
TC 7T
TC
and
I
QQ&mxGiQ&nxdx
=
=
TC
m 7^ n,
for
TC
we have an
(31)
Similarly
we
(
=
71
The
f(x)eosnxdx.
find
bn
(32)
=
following
j
f(x)&i&nxdx.
question
may and
b n satisfying the
which
functions / an the coefficients (29),
arises:
be represented in the form
formulae (30)-(32) (the so-called Euler-Fourier formulae). If such an expansion exists, then it is called the Fourier series of the function f. Since the functions cos and sin are periodic, it is evident that the periodicity of the function / has to be
IMPROPER INTEGRALS
11.
301
assumed, too. Moreover, we shall assume that the function /
is
We
piecewise
shall
monotone
THEOREM. Every f(x
(i.e.
its
4,
(cf.
2).
prove the following
+ 2n)
=
periodic junction f with the period
2n
f(x)), piecewise continuous (together with
derivative )j piecewise
monotone and satisfying
(at the
points of discontinuity) the condition
(33)
expanded in a Fourier series. Let us denote by Sn (x) the ti-th partial sum of the
may
be
series (29), i.e.
(34)
S n (x)
=
|a
+ (a
1
coso?
+ ftiSina?) + ...+ + (an cosnx + b n smnx)
We
have to prove that conditions (30)-(32), then (35)
f(x)
if
= lim
the coefficients satisfy the
flf
noo
n (o?)
.
The above mentioned conditions make us to transform formula (34) as follows: 7t
7t
(36)
itS n (x)
TC
...+
I
n
=
\
t
o 28
.
-
m-
x
it
possible for
302
IV.
INTEGRAL CALCULUS
by the known formula
(ct.
1,
(2)): ,.
+ Go&nt =
...
Let us substitute
-~~
2n +1
=
2;.
We
7
obtain
(37) 2
Decomposing intervals
this integral into
from
to
-
two
(38)
-
and from
=
substituting in the second integral z
integrals in the
y,
and
to
we obtain
nSn (x) o
n+x .
/
J
Now, we
shall
According to (33)
(39)
^/
o
+ l)y sm^n ^-:--Ay. -
\
1(x2y) J/
;v
sm
TT prove the equation (35) for it is sufficient to prove that
<
x
<
TT.
lim
noo
and 2
(40)
lim
1(xSin?/
us write /(#+ 20) = gf(^). Then the function piecewise continuous (together with its derivative)
Given g(z) is
J
o?
?
let
11.
and
piece-wise
IMPROPKU INTEGRALS
monotone, Hence
it
303
satisfies (cf. (21))
the
formula lim
tf()fe=
if
TL
In this formula a the inequality
<
TT
<
may
#
<
TU
be replaced by
'T
1
because
,
A
<
implies the inequality
TT
-
/y
J
+ 0) =
/(#+0), formula (41) gives (39) immediately (we restrict ourselves from the sequence n = 1, 2, ... to the subsequence of the odd numbers). Formula (40) is obtained in an exactly similar way, writing f(x2y) = h(y) and taking into account that TU.
Since
flf(
Thus, om^ theorem is proved for mains to consider the case when x -=
Let
#~
TT.
According to (37)
=
i
J
TT
<
x
<
It re-
-K.
when x
n' (or
=
TT).
we have
4i
/( ;
Decomposing this integral into two integrals in the to ivr and from |TC to n and substituting intervals from y
^n
z in
we obtain
the second integral
'/ x(Tt
-
./
The
first
of these integrals tends to
|TT/(
previously and the second one tends to ^/(TT TT 0) (since the function / is periodic). |7r/(
Hence the formula
(35) holds also for
according to the periodicity, In this way our theorem
x
=
7r
+ 0)
0), i.e.
TT;
as to
hence,
it
holds for every
is
proved completely.
x.
304
IV.
11.8. Applications
INTEGRAL CALCULUS
and examples
Let f(x) denote a periodic function with the period given by the following conditions:
(a) 2ru
j(x)
- - ITT
for
=
for
}?:
The graph the length t/
=
TT
-7c
<x<
0, /(O)
<x<
TT,
/(
:r)
,
=0.
of this function consists of segments of
lying on the
lines
straight
alternately, and of points on the
JTT,
-
integer multiples of the number do not belong to the graph).
-2Ji
TC
y
=
|TI
and
^Y-axis being
(the ends of the segments
2JC
Jt
-Ji
FIG. 28
immediately seen that the function defined
It is
way
is
(namely,
it
this
<
in
piecewise continuous and piecewise monotone is monotone in intervals of the form (m l)rc <
m
<
x is an integer); moreover, the conWTC, where dition (33) is satisfied. The theorem on the expansion in Fourier series may therefore be applied to this function.
We
calculate the coefficients a n
and
bn
according to
formulae (30)-(32). We obtain 7T
ao
=
C
1
n J
TC
,
4
C
1
~dx + ~ ,
7i
7U
7 4
J
_
,
^-0,
7C 71
a n =^j{ 4;
I
I
cos nx dx
+
/
I
cosnxdxl
/
J
=0
(n>0),
-r- (2
2 cos me)
n TC
bn
= T! 4
I
J
sin/M?dfe+
I
J
sin ^a? da?} J
=
4:U
o
and
so 6 n
=
for
n even and
bn
=
1/w for
ti
odd.
.
IMPROPER INTEGRALS
11.
305
Consequently, f(a
(42)
^+*g2+... <
x satisfying the condition
So, for
be replaced by
In
)^ +
JTT.
x
substituting
particular,
=
|TT
x
<
TT,
/(#)
may
we obtain the
Leibniz expansion
*
=
4
i_l+i_i+ 3 7^'" ,5
which we already derived (cf. 7, (51)). with the period 2n defined the function be Let /(#) (P) as follows:
=
f(x)
=
We
see easily that the function / defined in this
\x
,
when
TT
<
#
<
TI,
/(
TT)
piecewise continuous and piece wise monotone
=/(TT)
(cf.
.
way
is
Fig. 29).
Moreover, the condition (33) is satisfied. Thus, the func/ may be expanded in a Fourier series. Calculating the coefficients an and b n according to the formulae (30)-(32) we find
tion
,. ox
(43)
x
=
Let us note that substituting x = |TT, we obtain the Leibniz expansion for JTC (just as in the previous example). Let us expand the function f(x) = \x\ in the interval For other x the function is defined by the condition f(oo + 2mn) f(x). Hence it is a continuous function on the whole X-axis. The graph of this function is a polygonal line. (y)
~n ^ x ^n.
20
306
IV.
We now
INTEGRAL CALCULUS
calculate the coefficients of the expansion
in the Fourier series: TC
an
=
1 /' n J \
TC
2
,,
= -I
\x \dx '
/'
xdx
TcJ
'
~ TU,
'
-TC TC
TC
an
=
-
|a?|cosna?
=
'
7U
J
TC
.'
xw&nxdx =
-
or TCfr
-TC
according as n
is
even (>0) or odd. For,
cccosnxdx
= -xsmnx n
J
= we
Similarly,
,
1 -
T:
-
4 /cos#
J
,1
.
-
ftsiiiwrH
find that b n
,
minxdx
n
J
\x\ainnxdx
cosSa?
,
cosSo?
Let us note that by substituting x
1
(45)
Hence we obtain
For, denoting
111
i.e.
fs
=
JTT, that is
,
we obtain
Euler formula
the
sum
of this series,
=
we have
111
.
6*
Thus,
^ + ^.1+...
tlie
s
0.
2
J?r
.
cos to, < ^ < TC and let us assume that (S) Let 7/ not an integer. The formulae (30)-(32) easily give: ==
is
by
=
=
O
TU
r>
a
"
t
IMPROPER INTEGRALS
11.
307
Thus, ,._.
n
.
.
.
cos 2^
. ,
(47)
=
In particular, substituting x sides of identity (47)
by
sinrctf
we
TU
and dividing both
obtain:
t
The formula
has the following interesting ap-
(48)
plication.
Let us write
Given a number x satisfying the condition the series S(t)
^t
^x
9
is
since
uniformly
~
-
-
<~
<
<
x
1,
convergent in the interval
1
;
and the
1
i
series
~
z
convergent, as is easily seen. Therefore the series S(t) may be integrated term term in the interval < t < x. Since
,
is
r J o
Hence
(cf.
2<
,
^-p dt
10.2, 12
=-
and
/,
,
l
i-
s
*
v
5,
(12)):
-l
On
the other hand,
TC
J
(
\
^
nt
log^
= log
in
by
308
INTEGKAL CALCULUS
IV.
Thus,
= log*
,.
-
sin nt
,
lim log B
nt
= log6 ,
nx
since ..
= log lim
K*
= logl = .
,.
.
.
lim log J- +
J= +
.
.
7T*
Taking into account the fact that the relation (48) < t ^ x and holds for every t satisfying the inequality has a right-side limit at the
that the function cotTrf point
=
namely lim (cot<- - }
0,
f- + 0\
(cf.
X
X 7t
8, (19)),
*/
=J (eotiet-^dt
I
'
8(t)dt
,
o
i.e.
therefore
III --2 / 11
iC \
sinToz?
(49)
TCO? /
'
^
x
, '
=
Let us note that by substituting in the formula (49)> ^, we obtain the Wallis formula already derived
=
10, (29)):
(cf.
(50)
-
=
4n*
/ f 2 ""i i 4n2 -i
Substituting ^
=
-??4466
""
J
,
we obtain TT
n=l
__41636
I'S'S'S'S'T'""" s'iB'ss'
.
...
3.5 7.9 8 12-12
11.
IMPROPER INTEGRALS
from which we conclude by
309
(50) that
10 10 _> 36 100 V/ 2 ~l-3'5-7'9-ll '"'"" 3 '35 *99"'" '
Exercises on -
1.
2.
11
-n
i Evaluate
arc tan a?,
r
-L
-r
J
:
J
#2 + 4'
,
j
Investigate the convergence of the integrals 00
1
dx
I
J
Let lim f(x) X
=
dx
I
.
J I
]/x r
o
3.
%
o
and
number
a
let
.
+ x* c>
be given.
00
Let us write a+cn
an
=
f(x)dx.
J a-fc(n-l)
If
the series a 1
+a
2
+...
is
convergent, then the integral 00
00
/ f(x)dx
is
and
convergent, too,
is
n
a
Apply the above theorem
an
equal to J
.
l
to the proof of the con-
vergence of the integral r
sm#
oy
x
,
dx
00
4
An
integral J f(x)dx
is
called absolutely convergent,
a
oo
if
the integral / \f(x)\dx
convergent. Prove that an
is
a
absolutely convergent integral usual sense.
is
also convergent in the
-
00
^
5.
Prove that the
series
~i
-is convergent (ifc> a).
310
INTEGRAL CALCULUS
IV.
6.
Prove that
(
1<*<1*> x
-i
in
fe
=
.
4
(Expand the function under the sign of the integral an infinite series and apply the formula (45)) 7.
mula 8.
-
Prove that
?>2+
42
+-^
j|
(
cf
the for "
-
(45)).
Expand the 1)
#2
functions: ,
2)
#cos#,
3) |sin#|,
5) cosh^o?
4) sinhte,
in Fourier series. 9.
Assuming that the system
of
functions
/ 1? / 2 ,
...
00
is
orthogonal and that
the
series
/(#)
=
a/ n (^)
is
?i-i
uniformly
convergent
(in
the
calculate the coefficients a n (by /,
A,
/,-)
considered
means
interval
aft),
of the functions
INDEX transformation 61
remainder 186 theorems 44,
theorem 60
155
Abel
57,
71,
75,
Absolute convergence of an integral 309 of series 68 Absolute value 19 d'Alembert criterion of convergence 63 Alternating series 59
Cavalieri formula 270
Alternative 130
Composed function 108
Anchor ring 271 Anharmonic series 59 Arc cos, arc sin, arc tan 88
Conditional convergence of a series 68
Ar
cosh, ar sinh, ar tanh 159
Area of a region 237
Argument Arithmetic
26,
of curvature 246 of
mass 246
Closed interval 82 of a series 68 Composition of functions 94
Commutativity
Conjunction 130 Continuity
Cauchy
of a function 81
mean
Centre
definition of 104
Heine definition
of 104
left-side 103
49
Ascoli theorem 53
of a function 102, 103
Asymptote 174
one-side 103
Asyrntotic value of n\ 262
piecewise 103, 281 principle 18 right-side 103
Bernoulli inequality 13
Bolzano-Weierstrass theorem 39
Bound (upper and
lower)
of a function 110
uniform 109 Continuous function 102, 103
Convergence absolute of a series 68
of a set 20
absolute of an integral 309 absolute of a product 78
Bounded function 97
sequence 32 series 58
d'Alembert criterion of 63
set 19
conditional of series 68
variation
1
Cauchy
02
criterion of 63
logarithmic
criterion
of
298
of integral 274, 277 of a product 74
Cauchy criterion of convergence 63 definition of continuity 104
of a sequence 29 of a series 55
-Hadamard formula
radius of 123
137
-Maclaurin theorem 294
uniform, of a sequence of func-
INTRODUCTION TO CALCULUS
312 tions 118
Elliptic integral 274, 277
uniform, of a series 121
Entier function 83 Euler
Criterion of convergence
d'Alembert 63
constant 180, 190
Cauchy 63
formula 300
logarithmic 298 Kaabe 67
gamma
function 292
Even function 102
Curvature 245 Cut 18
Existential quantifier 132 Exponential function 87
Cyclometric functions 88
Extremum
Darboux
Factors of an infinite product 74 Finite increments (theorem) 153
integrals
(upper
and lower)
263, 264 property 111 Decreasing function 83 sequence 28
Formula
Cauchy-Hadamard 137 Cavalieri 270
Dedekind continuity principle 18 Definite integral 224
Derivative 138 generalized 179 left-side 140
logarithmic 158 one- side 140 right-side 140
second 180 Differences quotient 138 Differentiate function 141 Differential 175
Direction cosine 244
function 92 integral 287 Divergence to infinity 44
above and below bounded 97 argument of a 81
continuous 102, 103 cyclometric 88 decreasing 83
Divergent sequence 29 series 55
differentiate 141 Dirichlet 92
elementary 86, 161 entier 83 even 102
47
Element of the area 240
exponential 87 extremum of a 150
of the length 243
Elementary function
Euler 300 de 1'Hospital 164 Leibniz 173, 184 Maclaurin 187 Simpson 258 Stirling 261 Taylor 185 Fourier integral 226 series 300 Fresnel integral 284 Function 81
bound of a 110 bounded 97 composed 108
Dirichlet
e
of a function 150
86,
161
7" of
Euler 292
INDEX hyperbolic 158 increasing 83 inverse 85
313
Indefinite integral 201
Indeterminate expression 163 Infinite
Lejeume-Dirichlet 92 limit of a 89 linear 86 maximum of a 149 minimum of a 150 monotone 83 odd 102 one-to-one 85 periodic 84 piecewise continuous 103, 281 piecewise monotone 83, 85 polygonal 127 primitive 201 prepositional 131 rational 87
second degree of a 86 trigonometric 87
uniform continuity of a 109 value of a 82
product 74 sequence 26 series 55
Inflexion point 197, 198 Integral
convergent 274, 277 Darboux (upper and lower) 263, 264 definite 224 Dirichlet 287 elliptic 274, 277 Fourier 226 Fresnel 284 indefinite 201 improper 273, 276 Poisson 290 Riemann 264 seemingly improper 279 Integration
by parts 206 Generalized derivative 179
Geometric interpretation
of
a
function
82
substitution 206
Interpolation 256 Interval closed 82
mean Graph
by
26, 51
of a function 82
Guldin theorem 251
Harmonic
mean 26 series 65
Heine definition of continuity 104 de 1' Hospital formula 164 Hyperbolic function 158 Hypocycloid 272 Implication 131
Improper
limit 45, 91
Increasing function 83
sequence 28 strictly 83
of
convergence
of
a
power
series 125
open 82 Inverse function 85
Improper integral 273, 276
Kummer
theorem 66
Lagrange interpolation 256 remainder 187
theorem 153 Lateral surface 249 Left-side
continuity 103 derivative 140 limit 91
INTRODUCTION TO CALCULUS
314
Newton binomal formula Normal
Legendre polynomials 273 Leibniz formula 173, 184 Length, of an arc 241 Limes 29
15,
192
region 239 to a curve 141
Numerical
inferior 53
line 11
superior 53
One
Limi^i 28
improper
45, 91
left-side 91
limit 91
one-side 91
interval 82 Orthogonal system 273
Open
right-side 91
Linear function 86
Partial
independence 272
fraction 211 product 74 sum 56 Pascal triangle 54
Lipschitz condition 117 Local property 105
Logarithm 87 Logarithmic criterion series
of
convergence
of
Piecewise
continuous function 103, 281 monotone function 83, 85 Point of inflexion 197, 198 Poisson integral 290 Polygonal function 127
Maclaurin formula 187 Mathematical induction 12 Maximum of a function 149
Mean
Polynomials 87
arithmetic 26, 49 geometric 26, 51
Positive integers 11 Power series 123
harmonic 26
Mean -value Lagrange theorem 153 theorem in the integral culus 229, 253 Method of indetermined
Quantifier 132
of a function 150
function 83
sequence 28 de Morgan laws 130, 132
Natural logarithm 49, 87 Necessary condition 42 Negation 130
cal-
Primitive function 201 Product of sentences 130 Propositional function 131
coeffi-
cients 212
Monotone
Periodic function 84
Permutation 68
298
derivative 158
Minimum
side
continuity 103 derivative 140
Raabe
criterion 67
Iladius
convergence of a power 123 of curvature 245 Rapidness of the increase of sequence 65 Rational of
series
function 87
number
11
INDEX bound of a 20 bounded 19
Keal number 11 Recurrence definition 27
Simpson formula 258
Remainder
Solid of revolution 248
Cauchy 186 Lagrange 187 the
in
315
formula 261 Subsequence 38
Stirling
Taylor formula
186,
Sufficient condition 42
187
Sum Sum
of a series 56
Riemann
of sentences 130 of the infinite series 55
integral 264
theorem 71
Tangent to a curve 139 Taylor formula 185
Right-side
continuity 103 derivative 140
Term
of sequence 27
Theorem
.limit 91
Abel
125
60,
Ascoli 53
Schwarz inequality
17,
271
Bolzano-Weierstrass 39
Secant 139
Cauchy
Sequence
Cauchy-Maclaurin 294
44, 57, 71, 75,
155
increments 153 Guldin 251 Kummer 66 Lagrange 153 finite
bounded 32 convergent 29 decreasing 28 divergent 29 increasing 28 infinite 26 monotone 28
mean-value (Lagrange) 153 mean-value (in the integral
rapidness of the increase of a 65
term of a 27
calculus) 229, 253 71
Riemann
Rolle 151 Weierstrass 39, 110
uniformly convergent 118
Uniform continuity
Series
of a function
109
alternating 59 an harmonic 59
Uniform convergence
bounded 58 of a 69
commutativity conditional convergence
of a sequence of functions 118 of series 121
Universal quatifier 132
of a 68
convergent 55 divergent 55 Fourier 300 infinite 55
Value of a function 82 Variable 82 Variation bounded 102
Volume 248
power 123 uniform convergence of a 121 Set
Wallis formula 259
Weierstrass theorem 110
