Chapter 8 – Improper Integrals. Subject: Real Analysis (Mathematics) Level: M.Sc. Source: Syed Gul Shah (Chairman, Department of Mathematics, US Sargodha) Collected & Composed by: Atiq ur Rehman (
[email protected]), http://www.mathcity.org
We discussed Riemann-Stieltjes’s integrals of the form ò f d a b
a
under the
restrictions that both f and a are are def defin ined ed and and bound bounded ed on a fin finit itee inte interv rval al [ a, b] . To extend the concept, we shall relax these restrictions on f and a . Ø Definition b
ò f d a
The integral
is called an improper integral of first kind if a = - ¥ or
a
b = + ¥ or both i.e. one or both integration limits is infinite. Ø Definition b
ò f d a
The integral
is called called an improper improper integral integral of second kind if f ( x) is
a
unbounded at one or more points of a £ x £ b . Such points po ints are called singularities of f ( x ) . Ø Notations
We shall denote the set of all functions funct ions f such that f Î R(a ) on [a, b] by
R (a ; a, b) .
When a ( x ) = x , we shal shalll sim simply ply wri write R( a, b) for this set. The notation
onoton oniical cally incr increa easi sing ng on [ a, ¥) . a on [a, ¥) will mean that a is monot Ø Definition
Assume that f Î R(a ; a, b) for every b ³ a . Keep a,a and f fixed and define a function I on [a, ¥) as follows: b
I (b) =
ò f ( x) da ( x)
if b ³ a ………… (i)
a
The function I so defined is called an infinite ( or an improper ) integral of first kind and is denoted by the symbol The integral
¥
ò f d a
¥
ò
a
f ( x) da ( x) or by
¥
ò f d a . a
is said to converge if the limit
a
lim I (b) ………… (ii) b ®¥
exists (finite). Otherwise,
¥
ò f d a a
is said to diverge.
If the limit in ( ii) exists and equals A , the number A is called the value of the ¥
ò
integral and we write
a
f da = A
Ø Example
Consider
ò
1
x- p dx.
1- b - ) ( dx=
¥
1 p
b
ò 1
b
- p
x
if p ¹ 1 , the integral
p - 1
1
p > 1 , it converges and has the value
If p = 1 , we get
ò
x- p dx diverges if p < 1 . When
b
ò
1
1 p - 1
.
x-1 dx= log b® ¥ as b ® ¥ .
Þ
¥
ò 1
x-1 dx diverges.
ap.. ap
2
– mprop proper er n egra egra s
.
Ø Example b
ò sin2p xdx
Consider
0 b
Q
ò
sin2p xdx =
(1 - cos 2p b) 2p
0
® ¥ as b ® ¥ .
¥
\ the integral ò sin2p x dx diverges. 0
Ø Note ¥
a
If
ò f d a
and
-¥
ò f d a
are both convergent for some value of a , we say that
a
¥
the integral
ò f d a
is convergent and its value is is defined to be the the sum
-¥
¥
ò
¥
a
f da =
-¥
ò
f da +
-¥
ò f da a
The choice of the point a is clearly immaterial. ¥
If the integral
ò f d a
b
converges, its value is equal to the limit:
-¥
Ø
lim
b ®+ ¥
ò f d a .
-b
Theorem
Assume that a on [a, + ¥) and suppose that f Î R(a ; a, b) for every b ³ a . Assume that f ( x ) ³ 0 for each x ³ a . Then
¥
ò f d a a
converges if, and only if,
there exists a constant M > 0 such that b
ò f da
£ M for every b ³ a .
a
Proof b
We have
I ( b) =
ò f ( x) da ( x) ,
b³a
a
Þ I on [a, + ¥) 0 and the theorem follows Then lim I ( b) = sup { I ( b) | b ³ a} = M > b ®+¥
b
Þ
ò f da £ M for every
b ³ a whenever the integral converges.
a
]]]]]]]]]]]]]]]]
ap.
Ø
– mprop proper er n egra egra s
.
Theorem: (Comparison Test) Assume that a on [a, + ¥) . If f Î R(a ; a, b) for every b ³ a , if ¥
¥
ò g d a
0 £ f ( x) £ g ( x) for every x ³ a , and if
converges, then
a
ò f d a
converges
a
and we have ¥
¥
ò
f da £
ò g da
a
a
Proof b
I1 ( b) =
Let
b
ò f da
I2 ( b) =
and
a
ò g da
,
b³a
a
0 £ f ( x) £ g ( x)
Q
for every x ³ a
\ I1 ( b) £ I2 ( b) …………………. (i) ¥
Q
ò g d a
converges \ $ a constant M > 0 such that
a
¥
ò g da £ M
b ³ a …………………(ii)
,
a
I1 ( b) £ M
From (i) and (ii) we have
, b ³ a.
Þ lim I1 (b) exists and is finite. b®¥
¥
Þ
ò f d a
converges.
a
lim I1 ( b) £ lim I2 ( b) £ M
Also
b®¥
¥
Þ
b ®¥
¥
ò
f da £
a
ò g da . a
Theorem (Limit Comparison Test) Assume that a on [a, + ¥) . Suppose that f Î R(a ; a, b) and that g Î R (a ; a, b) for every b ³ a , where f ( x ) ³ 0 and g ( x ) ³ 0 if x ³ a . If
Ø
lim
x ®¥
¥
g ( x)
=1
¥
ò f d a
then
f ( x )
and
a
ò g d a
both converge or both diverge.
a
Proof
For all b ³ a , we can find some so me N > 0 such that f ( x ) g ( x)
Þ 1-e < Let e =
1 2
" x ³ N for every e > 0 .
- 1 < e
f ( x ) g ( x)
< 1 + e
, then we have 1
<
f ( x )
<
3
g ( x) 2 2 Þ g ( x ) < 2 f ( x ) …..…..(i) and 2 f ( x) < 3 g( x) ……....(ii)
ap.. ap
4 ¥
< 2 ò f da
a
a
¥
Þ
ò g d a
.
¥
ò g da
From (i)
– mprop proper er n egra egra s
¥
¥
ò
ò f d a
a
a
converges if f d a converges and
a
¥
ò
diverges if f d a a
diverges. ¥
From (ii) 2
ò
¥
f da < 3
a
ò g da a
¥
Þ
¥
ò f d a
converges if
a
¥
ò g d a
converges and
a
¥
ò g d a
ò
diverges if f d a
a
a
diverges. ¥
Þ The integrals
¥
ò f d a
and
a
ò g d a
converge or diverge together.
a
Ø Note
The above theorem also holds if lim
x ®¥
f ( x ) g ( x)
= c , provided that c ¹ 0 . If c = 0 ,
¥
we can only conclude that convergence of
ò g d a
¥
ò
implies convergence of f d a .
a
a
Ø Example ¥
For every real p , the integral
ò e
- x
x p dx dx converges.
1
¥
This can be seen by comparison of this integral with
1
ò x
2
dx .
1
- x p 1 e x Since lim = lim where f ( x) = e- x xp and g ( x) = 2 . x ®¥ g ( x) x®¥ 1 x x 2 p + 2 x (f )x - x p + 2 Þ lim = lim e x = lim x = 0 ®x¥ g ( x ) ®x¥ ®x¥ e
f ( x)
¥
and
Q
1
ò x
2
dx is convergent
1
¥
\ the given integral
ò e
- x
x p dx dx is also convergent.
1
Ø
Theorem ¥
Assume a on [a, + ¥) . If f Î R(a ; a, b) for every b ³ a and if
ò f a
¥
converges, then
ò f d a
also converges.
a
Or: An absolutely convergent integral is is convergent.
Proof If x ³ a , ± f ( x) £ f ( x) Þ f ( x) - f ( x) ³ 0 Þ 0 £ f ( x) - f ( x) £ 2 f ( x)
d a
ap.
– mprop proper er n egra egra s
.
¥
ò (
Þ
f - f ) da converges.
a
¥
Subtracting from
¥
ò f
d a we find that
a
ò f d a
converges.
a
( Q Difference of two convergent integrals is convergent ) Ø Note ¥
¥
ò f d a
is said to converge absolutely if
a
ò f
d a converges. It is said to be
a
¥
¥
ò
ò f
a
a
convergent conditionally if f d a converges but
d a diverges.
Ø Remark
Every absolutely convergent integral is convergent. Ø
Theorem
Let f be a posi positi tiv ve decr decrea easi sing ng func functi tion on defi define ned d on [ a, + ¥) such that f ( x ) ® 0 as x ® + ¥ . Let a be bounded on [a, + ¥) and assume that f Î R (a ; a, b) for every b ³ a . Then the integral
¥
ò f d a a
is convergent.
Proof Integration by parts gives b
ò
b
f da = f ( x) × a ( x)
b a
a
- ò a ( x) df a b
= f ( b) × a ( b) - f ( a) ×a ( a) + ò a d (- f ) a
It is obvious that f ( b) a ( b) ® 0 as b ® + ¥ (Q a is bounded and f ( x ) ® 0 as x ® + ¥ ) and
f ( a)a ( a) is finite. b
b
a
a
depen nds upon pon the the conv conver erg gence ence of ò a d ( - f ) . \ the convergence of ò f d a depe Actually, this integral converges absolutely. To see this, suppose a ( x) £ M for all x ³ a ( Q a ( x ) is given to be bounded ) b
ò a ( x)
Þ
a
b
d (- f ) £
ò M d (- f ) a
b
But
ò
M d (- f ) = M - f
b a
= M f (a ) - M f (b ) ® M f (a ) as b ® ¥ .
a
¥
Þ
ò M d (- f ) is convergent. a
Q
- f is an increasing function. ¥
\
ò a
d ( - f ) is convergent.
(Comparison Test)
a
¥
Þ
ò f d a
is convergent.
a ]]]]]]]]]]]]]]]]
ap.. ap
6 Ø
– mprop proper er n egra egra s
.
Theorem (Cauchy condition for infinite integrals) ¥
ò f d a
Assume that f Î R(a ; a, b) for every b ³ a . Then the integral
converges
a
if, and only if, for every e > 0 there exists a B > 0 such that c > b > B implies c
ò f ( x) da ( x)
< e
b
Proof ¥
Let
ò f d a
be convergent. Then $ B > 0 such that B
a
¥
b
ò
f da -
a
ò
f da <
e
f da <
e
for every b ³ B ………..(i)
2
a
c
b
Also for c > b > B , ¥
c
ò
f da -
ò
a
a
c
b
a
a
c
¥
a
a
c
¥
2
…………….. (ii)
Consider c
ò f da
=
b
= £
ò f da - ò f da ¥
b
ò f da - ò f da + ò f da - ò f d a ò
f da -
a
ò
a
a
¥
ò
f da +
a
b
f da -
a
ò
f da <
a
e 2
+
e 2
=e
c
Þ
ò f d a
< e when c > b > B .
b
Conversely, assume that the Cauchy condition holds. a+n
Define an =
ò f d a
if n = 1,2,......
a
The sequence {an } is a Cauchy sequence Þ it converges. Let
lim an = A n ®¥
c
Given e > 0 , choose B so that
ò
f d a <
b
e 2
if c > b > B .
whenever a + n ³ B . 2 Choose an integer N such that a + N > B i.e. N > B - a Then, if b > a + N , we have and also that
an - A <
a+ N
b
ò f da - A = ò a
b
f da - A +
ò f da a+ N
a
b
£ a N - A +
ò a + N
¥
Þ
a+N a+N
e
ò f da = A a
This completes the proof.
f da <
e 2
+
e 2
= e
a
B
b
c
ap.
Ø
– mprop proper er n egra egra s
7
.
Remarks ¥
ò f d a
It follows from the above theorem that convergence of lim
b +e
ò
b®¥ b
a
implies
f d a = 0 for every fixed e > 0 .
However, er, this does not imply ply that f ( x) ® 0 as x ® ¥ . Ø
Theorem
Every conv onvergen rgentt infinit nite integr egral
¥
ò a
f ( x) da ( x) can be written as a convergent
infinite series. In fact, we have ha ve ¥
¥
ò f ( x) da ( x) = å a
k
a + k
where ak =
k =1
a
ò
f ( x ) da ( x ) ……….. (1)
a + k -1
Proof ¥
Q
{ ò
a+n
ò
f d a converges, the sequence
a
a a +n
But
ò
f da =
Ø
¥
¥
n
å a . Hence the series å a k
k
k =1
a
}
f d a also converges. converges and equals
k =1
ò f d a . a
Remarks
It is to be noted that the convergence of the series in (1) does not always imply k
ò sin2p x dx . Then each
convergence of the integral. For example, suppose ak = ak = 0 and
åa
k
k -1
converges. ¥
However, the integral
ò
b
x ò sin 2p =xldim
sin 2p
b ®¥
0
0
dx =xlim
1 - cos 2p b
b ®¥
2p
diverges.
IMPROPER INTEGRAL OF THE SECOND KIND Ø
Definition
Let f be defined on the half open interval ( a, b] and assume that f Î R(a ; x, b) for every xÎ ( a, b] . Define a function I on ( a, b] as follows: b
I ( x) =
ò f da
if xÎ ( a, b] ……….. (i)
x
The function I so defined is called an improper integral of the second kind and b
is denoted by the symbol
ò f ( t) da (t)
a+
b
or
ò f d a .
a+
b
The integral
ò f d a
is said to converge if the limit
a+
lim I ( x ) ……...( ii) exists (finite).
x ®a + b
Otherwise,
ò f d a
is said to diverge. If the limit in ( ii) exists and equals A , the
a+
b
number A is called the value of the integral and we write
ò f da = A. a+
ap.. ap
8
– mprop proper er n egra egra s
.
Similarly, if f is defined on [a, b) and f Î R(a ; a, x) " x Î [ a, b) then x
I ( x) =
ò f da
if x Î[ a, b) is also an improper integral of the second kind and is
a b-
ò f d a
denoted as
and is conv onvergen rgentt if lim I ( x) exists (finite). x ® b -
a
Ø
Example define ned d on (0, (0, b] and f Î R ( x, b) for ever every y x Î (0, (0, b] . f ( x) = x- p is defi b
ò x
- p
I( x) =
if x Î (0, b]
dx dx
x b
b
ò x
- p
=
e ®0
0+
= lim e ®0
ò x
- p
dx = lim
0 +e
b
x1- p
1- p
b
= lim
1 - p
dx
e ®0
e
- e 1- p 1 - p
,
( p ¹ 1)
é finite , p< 1 =ê infini inite , p > 1 ë inf b
1
ò x dx = log b - log e ® ¥
When p = 1 , we get
as e ® 0 .
e
b
ò x
-1
Þ
dx also diverges.
0+
Hence the integral converges when p < 1 and diverges when p ³ 1 . Ø
Note b-
c
If the two integrals
ò f d a
and
a+
ò f d a both converge, we write c
b-
ò
b-
c
f da =
a+
ò
f da +
a+
ò f da c
The definition can be extended to cover the case of any finite number of sums. We can also consider mixed combinations such as ¥
b
ò
f da +
a+
Ø
¥
ò f da
which can be written as
ò f d a .
a+
b
Example ¥
Consider
ò e
- x
x p -1 dx
( p > 0)
,
0+
This integral must be interpreted as a sum as ¥
ò
¥
1
e
-x
x
p-1
ò
dx =
0+
e
-x
0+
x
p-1
ò
dx + e - x x
p-1
dx
1
= I1 + I 2 ………..….…… ( i) I 2 , the second integral, converges for every real p as proved earlier.
To test I 1 , put t =
1 x
Þ dx = -
1 2
t
dt
ap.
– mprop proper er n egra egra s
1
1
Þ I1 = lim ò e x -x
e ®0
p -1
e ®0
-1
t t - p -1
Take f ( t) = e Then lim t ®¥
¥
t
æ 1 ö ç - 2 dt ÷ = lim e ®0 è t ø
e
1 e
ò e
- 1 - p -1 t
t
dt
1
¥
- p -1
e t × t
t ®¥
g (t )
- 1 - p -1 t
\ ò e t
= lim
1
- 1 1- p t
g (t ) = t - p -1
and -1
f ( t)
ò e
dx = lim
e
9
.
ò t
= 1 and since
- p -1
t
- p -1
dt converges when p > 0
1
dt converges when p > 0
1
¥
Thus
ò
e- x x p -1 dx converges when p > 0 .
0+
When p > 0 , the value of the sum in ( i) is den denoted oted by G( p ) . The function so defined is called the Gamma function. Ø
Note
The tests developed to check the behaviour of the improper integrals of Ist kind are applicable to improper integrals of IInd kind after making necessary modifications. Ø
A Useful Comparison Integral b
dx
ò ( x - a )
n
a
We have, if n ¹ 1 , b
dx
ò + ( x - a )
n
=
a e
= Which tends to
b
1 (1 - n) ( x - a )n-1
a +e
æ 1 1 ö ç ÷ (1 - n) è (b - a ) n-1 e n -1 ø 1
1 (1 - n) (b - a ) n-1
or + ¥ according as n < 1 or n > 1 , as e ® 0 .
Again, if n = 1 , b
dx
ò + x - a = log(b - a) - log e ® + ¥
as e ® 0 .
a e
b
Hence the improper integral
dx
ò ( x - a )
n
converges iff n < 1 .
a
]]]]]]]]]]]]]]]]
ap.. ap
10
– mprop proper er n egra egra s
.
Question
Ø
Examine the convergence of 1
ò x
(i)
1
dx 1
0
3
(
1 + x2
ò x (1 + x)
(ii)
)
1
dx
2
(iii)
2
0
ò x 0
dx 1
2
(1 - x )
1
3
Solution 1
ò x
(i)
dx 1
0
3
(1 + x ) 2
Here ‘0’ is the only point of infinite discontinuity of the integrand. We have 1 f ( x ) = 1 x 3 1 + x 2
(
Then lim 1
ò
Þ
dx
ò x
Q
1
0
1
g ( x)
= lim x ®0
1 1 + x2
f ( x) dx dx and
0
1
1
x 3 f ( x )
x ®0
(ii) ii)
1
g ( x) =
Take
)
=1
1
ò g (x ) dx 0
1
converges
3
\
ò x 0
have identical behaviours. dx
1
3
(1 + x ) 2
also converges.
dx
ò x (1 + x) 2
2
0
Here ‘0’ is the only point of infinite discontinuity of the given integrand. We have 1 f ( x ) = 2 x (1 + x ) 2 1 Take g ( x) = 2 x f ( x) 1 Then lim = lim =1 2 x ®0 g ( x ) x ®0 (1 + x )
Þ
1
dx ò f ( x) dx 0
and
1
ò g (x )dx 0
behave alike.
But n = 2 being greater than 1, the integral
1
ò g (x )dx does not converge. Hence 0
the given integral also does not converge. 1
iii) (iii)
ò x 0
dx 1
2
(1 - x )
1
3
Here ‘0’ and ‘1’ are the two points of infinite discontinuity of the integrand. We have 1 f ( x ) = 1 1 x 2 (1 - x ) 3 We take any number between 0 and 1, say 1 , and examine the convergence of 2
ap.
1
– mprop proper er n egra egra s
1
2
dx ò f ( x) dx
the improper integrals
11
.
dx . ò f ( x) dx
and
1
0 1
To examine the convergence of
2
2
ò x
1 1
0
2
(1 - x )
1
dx , we take g ( x) =
1 3
x
1
2
Then f ( x )
lim
x ®0
1
Q
g ( x)
= lim x ®0
1 (1 - x )
=1
1 3
1
2
1
ò x 0
1
ò x
\
dx converges
2
2
1 1
0
2
(1 - x )
1
To examine the convergence of
ò x
1
1 3
dx is convergent.
1 1
2
2
(1 - x )
1
dx , we take g ( x) = 3
1 (1 - x )
1
3
Then lim
f ( x )
x ®1
1
Q
x®1
x
1
=1 2 1
1
ò (1 - x )
1
1
dx converges
ò x
Q
3
1
2
Hence Ø
g ( x)
1
= lim
1 1
2
2
(1 - x )
1
dx is convergent. 3
1
dx converges. ò f ( x) dx 0
Question 1
ò
Show that
n-1
xm-1 (1 - x)
dxexists iff m , n are both positive.
0
Solution
The integral is proper if m ³ 1 and n ³ 1 . The number ‘0’ is a point of infinite discontinuity if m < 1 and the number ‘1’ is a point of infinite discontinuity if n < 1 . Let m < 1 and n < 1 . We take any number, say 1 , between 0 & 1 and examine the convergence of 2 1
the improper integrals
2
ò
m-1
x
n -1
(1 - x)
1
dx and
ò 1
0
n -1
xm-1 (1 - x)
dx at ‘0’ and ‘1’
2
respectively. Convergence at 0: We write m-1
f ( x) = x f ( x)
Then 1
As
g (x) 2
1
ò x -
1 m
(1 - x)
n -1
=
(1 - x ) n-1
and take g ( x) =
x1-m
1 x1-m
® 1 as x ® 0
dx is convergent at 0 iff 1 - m < 1 i.e. m > 0
0
1
We deduce that the integral
2
ò 0
n -1
xm-1 (1 - x)
dx is convergent at 0, iff m is +ive.
ap.. ap
12
– mprop proper er n egra egra s
.
Convergence at 1: m-1
f ( x) = x
We write
f ( x )
Then
1
ò (1 - x) -
As
1 n
1
=
x
m -1
and take g ( x) =
1-n
(1 - x)
1 (1 - x )1-n
® 1 as x ® 1
g ( x)
1
(1 - x)
n -1
dx is convergent, iff 1 - n < 1 i.e. n > 0 .
2 1
1 1
n-1
ò
Thus
xm -1 (1 - x)
n -1
ò
We deduce that the integral
xm-1 (1 - x)
dxconverges iff n > 0 .
2
dxexists for positive values of m , n only.
0
It is a function which depends upon m & n and is defined for all positive values of m & n . It is called Beta function. Ø
Question
Show that the following improper integrals are convergent. ¥
¥
1
ò sin x dx 2
(i)
sin 2 x
ò x
(ii)
1
2
1
dx
x log x
ò (1 + x)
(iii)
1
2
1
dx
(iv)
0
ò log
x× log(1 + x) dx
0
Solution (i) Let f ( x ) = sin 2 then
lim
®x¥
1 x
(f )x g ( x)
= lim
¥
ò
Þ
f ( x) d x and
1
¥
sin 2 x1
æ sin öy = lim ç ÷ =1 ®y0 y è ø
1 x2
1
ò x
ii) (ii)
2
dx behave alike.
2
1 2
¥
dx is convergent \
1
¥
x 2
1
ò x
Q
g ( x) =
®x¥
¥
1
and
1
sin 2 x
ò x
ò
sin 2
2
1
dx is also convergent.
x
dx
1
Take f ( x ) = sin x £ 1 2
¥
Þ
ò x 2 dx
and
1
Ø
1
sin 2 x
and g ( x) =
x 2 sin 2 x x 2
£
1
x 2
" x Î (1, ¥ )
x2
¥
converges
1
\
sin 2 x
ò x
2
dx converges.
1
Note 1
sin 2 x
ò x
2
dx is a proper integral because lim x ®0
0
¥
of infinite discontinuity. Therefore
sin 2 x
ò x 0
2
sin 2 x x
2
= 1 so that ‘0’ is not a po int
dx is convergent.
ap.
1
x log x
ò (1 + x )
iii) (iii)
2
– mprop proper er n egra egra s
13
.
dx
0
Q
log x < x ,
x Î (0,1)
\ xlog x < x2 Þ 1
xlog x
(1 + x ) x 2
ò (1 + x )
Now
2
2
<
x2
(1 + x )
2
dx is a proper integral.
0
1
x log x
ò (1 + x )
\
2
dx is convergent.
0
1
iv) (iv)
ò log
x× log(1 + x) dx
0
log x < x \ log( x + 1) < x + 1 Þ log x× log(1 + x) < x( x + 1)
Q
1
Q
ò x( x+ 1) dx is a proper integral
\ ò log x× log(1 + x) dx is convergent.
0
Ø
1
0
Note a
(i)
1
ò x
p
dx diverges when p ³ 1 and converges when p < 1 .
0
¥
(ii)
1
ò x
p
dx converges iff p > 1 .
a
UNIFORM CONVERGENCE OF IMPROPER INTEGRALS Ø
Definition
Let f be a real valued function of o f two variables x & y , x Î [a, + ¥) , y Î S where S Ì ¡ . Suppose further that, for each y in S , the integral
¥
ò a
f ( x, y) da ( x)
is convergent. If F denotes the function defined by the equation ¥
F ( y) =
ò f (x, y ) da (x )
if y Î S
a
the integral is said to converge con verge pointwise to F on S Ø
Definiton
Assume that the integral
¥
ò a
f ( x, y) da ( x) converges pointwise to F on S . The
integral is said to converge Uniformly on S if, for every e > 0 there exists a B > 0 (depending only on e ) such that b > B implies b
F ( y) -
ò f (x , y ) da (x )
< e
" y Î S .
a
( Pointwise convergence means convergence when y is fixed but uniform convergence is for every y Î S ).
ap.. ap
14
– mprop proper er n egra egra s
.
Theorem (Cauchy condition for uniform convergence.)
Ø
The integral
¥
ò
f ( x, y) da ( x) converges uniformly on S , iff, for every e > 0
a
there exists a B > 0 (depending on e ) such that c > b > B implies c
ò f ( x, y) da ( x)
< e
" y Î S .
b
Proof Proceed as in the proof for Cauchy condition for infinite integral
¥
ò f d a . a
Theorem (Weierstrass M-test)
Ø
Assume that a on [a, + ¥) and suppose that the integral
b
ò f ( x, y) da ( x) a
exists for every b ³ a and for every y in S . If there is a positive function M defined on [ a, + ¥) such that the integral
¥
ò M ( x) da ( x) converges and a
f ( x, y) £ M ( x) for each x ³ a and every y in ¥
ò a
S ,
then the integral
f ( x, y) da ( x) converges uniformly on S .
Proof f ( x, y) £ M ( x) for each x ³ a and every y in
Q
S .
\ For every c ³ b , we have c
c
c
ò f ( x, y ) da ( x) £ ò f (x , y ) da (x ) £ ò M da b
b
………… (i)
b
¥
Q
I=
ò M da
is convergent
a
\ given e > 0 , $ B > 0 such that b > B implies b
ò M da - I < e 2 …………… (ii) a
Also if c > b > B , then c
ò M da - I < e 2 …………… (iii) a
c
Then
ò M da
=
b
=
c
b
a
a
ò M da - ò M da c
b
a
a
ò M da - I + I - ò M da c
£
ò M da - I a
b
+
ò M da - I < e 2 + e 2 = e
(By ii & iii)
a
c
Þ
ò f ( x, y) da ( x)
< e ,
c > b > B & for each y Î S
b
Cauchy condition for convergence (uniform) being satisfied. Therefore the integral
¥
ò a
f ( x, y) da ( x) converges uniformly on S . ]]]]]]]]]]]]]]]]
ap.
Ø
– mprop proper er n egra egra s
15
.
Example ¥
ò e
Consider
- xy
sin x dx
0
e
-
sin x £ e xy
e-
and Now take
= e-
xy
£ eM ( x) = e- cx
xc
¥
¥
ò
The integral
-
xy
(Q
xy xy
if
sin x £ 1 )
c£ y
ò
M ( x) dx = e-cx dx is convergent & converging to
0
0
1 c
.
¥
\ The conditions of M-test are satisfied and
ò e
- xy
sin x dx converges
0
uniformly on [c, + ¥) for every c > 0 . Ø
Theorem (Dirichlet’s test for uniform convergence) b
Assume that a is bounded on [a, + ¥) and suppose the integral
ò f ( x, y) da ( x) a
exists for every b ³ a and for every y in S . For each fixed y in S , assume that f ( x, y) £ f ( x¢, y ) if a £ x¢ < x < + ¥ . Furthermore, suppose there exists a positive function g , defined on [ a, + ¥) , such that g ( x) ® 0 as x ® + ¥ and such that x ³ a implies f ( x, y) £ g( x) for every y in S . Then the integral
¥
ò a
f ( x, y) da ( x) converges uniformly on S .
Proof Let M > 0 be an upper bound for a
on
[a, +¥ ) .
Given e > 0 , choose B > a such that x ³ B implies e g ( x) < 4 M ( Q g ( x ) is +ive and ® 0 as x ® ¥ \ g ( x ) - 0 < If c > b , integration by parts yields c
ò
f da = f ( x, y) × a ( x)
c b
e 4 M
for x ³ B )
c
- ò a df
b
b c
= f ( c, y)a ( c) - f (b, y )a (b) + ò a d (- f ) ………… (i) b
But, since - f is increasing (for each fixed y ), we have c
òa d (- f ) b
c
£ M ò d (- f )
( Q upper bound of a is M )
b
= M f (b, y ) - M f (c , y ) …………… (ii) Now if c > b > B , we have from ( i) and (ii) c
ò f da b
c
£ f (c, y )a (c) - f (b, y )a (b ) + ò a d (- f ) b
£ a (c) f (c, y) + f (b, y ) a (b) + M f (b, y) - f ( c, y) £ a (c) f (c, y) + a (b) f (b, y) + M f (b, y) + M f ( c, y)
ap.. ap
16
– mprop proper er n egra egra s
.
£ M g (c) + M g (b ) + M g (b ) + M g (c ) = 2 M [ g(b) + g ( c)] e ù é e < 2 M ê + = e ë 4 M 4M úû c
ò f d a < e
Þ
for every y in S .
b
¥
Therefor eforee the Cauchy condi ondittion is satisfi sfied and
ò f ( x, y) da ( x)
converges
a
uniformly on S . Ø
Example ¥
e - xy
ò x
Consider
sin xdx
0
Take a ( x) = cos x
and f ( x , y) =
If S = [0 , + ¥ ) and g ( x) =
ii)
g ( x ) ® 0 as x ® + ¥ f ( x, y) =
£
x
if x > 0 , y ³ 0 .
on [e , + ¥ ) for every e > 0 then x if x¢ £ x and a ( x ) is bounded on [e , + ¥ ) .
f ( x, y) £ f ( x¢, y )
iii)
x
1
i)
e - xy
e - xy
1 x
= g( x)
" y Î S .
So that the conditions of Dirichlet’s D irichlet’s theorem are satisfied. Hence ¥
ò
e-
e
Q
sin x dx= +
x
lim
=1
x
xy
d( - cos x) converges uniformly on
ò
e - xy
\
ò
e - xy
sin x
0
sin x x
0
Ø
ò x e
sin x
x ®0
e-
[e , + ¥ )
if e > 0 .
e
¥
Þ
¥
xy
x
dx converges being a proper integral.
dx also converges uniformly on [0 , + ¥ ) .
Remarks
Dirichlet’s test can be applied to test the convergence conver gence of the integral of a product. For this purpose the test can be modified and restated as follows: Let f ( x) be bounded and monotonic in [ a, + ¥ ) and let f ( x) ® 0 , when X
x ® ¥ . Also let
dx be bounded when X ³ a . ò f ( x) dx a
¥
Then
ò f ( x)f ( x) dx
is convergent.
a
Ø
Example ¥
Consider
sin x
ò x
dx
0
Q
sin x x
® 1 as x ® 0 .
ap.
– mprop proper er n egra egra s
.
17
\ 0 is not a point of infinite discontinuity. ¥
Now consider the improper integral
sin x
ò x
dx .
1
1
The factor
of the integrand is monotonic and ® 0 as x ® ¥ .
x
X
ò sin
Also
x d=x - cos
+Xcos(1) £ cos
+X co s(1) < 2
1 X
ò sin x dx
So that
is bounded above for every X ³ 1 .
1
¥
Þ
1
sin x
ò x
dx is convergent. Now since
1
¥
that
ò
Ø
ò x
dx is a proper integral, we see
0
sin x
dx is convergent.
x
0
sin x
Example ¥
ò
sin x 2 dx .
Consider
0
We write sin x2 = ¥
Now
ò
2 x ¥
sin x2 dx=
1
1
1
× 2 x× sin x2
1
ò 2 x
× 2 x× si sin x2 dx
1
is monotonic and ® 0 as x ® ¥ .
2 x
X
ò 2 xsin
Also
x dx = - cos X + cos(1) < 2 2
2
1 X
ò
2 x sin x2 dx is bounded for X ³ 1 .
So that
1
¥
1
ò 2 x
Hence
¥
× 2 x× sin x dx i.e. 2
1
ò
sin x2 dx is convergent.
1
1
Since
ò sin x dx is only a proper integral, we see that the given integral is 2
0
convergent. Ø
Example ¥
Consider
ò
e-a x
0
sin x x
dx , a > 0 ¥
Here e
-a x
is monotonic and bounded and
sin x
ò x
dx is convergent.
0
¥
Hence
ò
e- a x
0
sin x x
dx is convergent.
]]]]]]]]]]]]]]]]
ap.. ap
18 Ø
– mprop proper er n egra egra s
.
Example ¥
sin x
ò x
Show that
dx is not absolutely convergent.
0
Solution np
ò
Consider the proper integral
sin x
dx
x
0
We need not take x because x ³ 0 .
where n is a positive integer. We have np r p n sin x sin x dx = dx x x r =1 ( r -1)p 0
å ò
ò
x= ( r- 1)p + y so that y varies in
Put
sin[( r - 1)p + y ] = (-1) 1)
We have r p
ò -
\
p
sin x x
( r 1)p
dy r- 1)p + y
0
r p p is the max. value of [( r - 1)p + y] in
Q
p
sin y
ò (r - 1)p + y
\
np
ò
Þ
0 n
sin x
dy ³
dx ³
x
1
ò
sin y dy =
rp
n
2
å
0
2
=
pr
1
å r ® ¥
p
1
0
Q
sin y = sin y
sin y
ò (
dx =
r -1
0,p ] . [0,p
p
n
å
0,p ] [0,p
é êë Q
2 rp
Division by max. value will lessen the value
1
1
r
as n ® ¥ , we see that
1
np
ò 0
sin x
dx ® ¥
x
as n ® ¥ .
Let, now, X be any real number. There exists a +tive integer n such that np £ X < ( n + 1)p . X
ò
We have
np
sin x
0
dx ³
x
ò 0
sin x x
dx X
Let X ® ¥ so that n also ® ¥ . Then we see that
ò
sin x
0
¥
So that
ò 0
Ø
sin x
x
dx ® ¥
dx does not converge.
x
Questions Examine the convergence of ¥
(i)
x
¥
ò (1 + x)
dx 3
(ii)
1
¥
1
ò (1 + x) 1
x
Solution (i) Let f ( x ) = As lim
x ®¥
f( x) g (x)
x
(1 + x )3
= lim
x®¥
and take g ( x) = 3
x
(1 + x )
3
=1
x x 3
=
1 x2
dx
(iii)
ò x 1
dx 1
3
(1 + x )
1
2
ap.
– mprop proper er n egra egra s
¥
¥
1
ò x 1
We have lim
x ®¥
¥
x
ò x
3
1
3
dx is convergent.
(1 + x ) x f( x)
= lim
x ®¥
g ( x)
g ( x) =
and take x
1+ x
dx is convergent. Thus 2
1
=
x
3
2
ò (1 + x)
x
dx is convergent.
1
(iii) iii) Let f ( x ) = x
1
3
(1 + x ) 1
we take g ( x) = x x ®¥
x x
1
1
We have lim
1
=1 ¥
1
1
Ø
dx have identical behaviour
1
(ii) ii) Let f ( x ) =
and
2
1
ò (1 + x)
\
dx is convergent 2
1
ò x
dx and 3
1
for convergence at ¥ . Q
¥
x
ò (1 + x)
Therefore the two integrals
¥
19
.
1 3
×x
1
1
= 2
2
1 x
5
6
¥
f ( x )
= 1 and
g ( x)
ò x 1
¥
1 5
dx is convergent \ 6
dx is convergent. ò f ( x) dx 1
Question ¥
1
ò 1 + x -¥
Show that
dx is convergent.
2
Solution We have a é0 1 ù 1 d x d x d x = lim + ò- ¥ 1 + 2 x a®¥ êê -òa 1 + 2 x ò 0 1 + 2x úú ë û a éa 1 ù éa 1 ù 1 dx + ò dx ú = 2lim ê ò dx ú = lim ê ò 2 2 2 a ®¥ a®¥ x x x 1 1 1 + + + ë0 û ë0 û 0
¥
1
-1
= 2 lim tan x a ®¥
a
0
æ p ö = 2 ç ÷ = p è2ø
therefore the integral is convergent. Ø
Question ¥
tan -1 x
ò 1 + x
Show that
dx is convergent.
2
0
Solution Q ¥
(1 + x ) × 2
tan -1 x
ò 1 + x
2
Q
dx
(1 + x ) 2
= tan -1 x ®
&
1
ò 1 + x
2
p 2
as x ® ¥
dx behave alike.
-1
Here f ( x) = and
0
1
ò 1 + x 0
1
¥
0
¥
tan x
2
dx is convergent \ A given integral is convergent.
tan x 1 + x
g (x) = 1 + x
2
2
ap.. ap
20 Ø
– mprop proper er n egra egra s
.
Question ¥
Show that
sin x
ò (1 + x)
dx converges for a > 0 .
a
0
Solution ¥
X
ò sin x dx
ò sin xdx £ 2
is bounded because
0
" x > 0 .
0
1
Furthermore the function ¥
sin x
ò (1 + x )
Þ the integral
, a > 0 is monotonic on
(1 + x )a
[0, + ¥ ) .
dx is convergent.
a
0
Ø
Question ¥
Show that
ò e
- x
cos x dx is absolutely convergent.
0
Solution ¥
Q
e
- x
cos x < e
-x
ò
e - x dx = 1
and
0
\ the given integral is absolutely convergent. (comparison test) Ø
Question e - x
1
ò
Show that
0
1 - x
4
dx is convergent.
Solution - x 2 Q e < 1 and 1 + x > 1 e- x
\
1 - x4 1
Also
ò 0
1
<
(1 - x2 ) (1 + x2 ) 1-e
1 1 - x
2
dx = lim e ®0
ò 0
<
1 1 - x2
1 1- x
2
dx
= lim si sin -1 (1 - e ) = e ®0
1
Þ
e - x
ò 1 - x 0
4
p 2
dx is convergent. (by comparison test)
References:
(1) Lectures (Year 2003-04) Prof. Syyed Gul Shah Chairman, Chairman, Department of Mathematics. University of Sargodha, Sargodha. Sargodha.
(2) Book Mathematical Analysis Tom M. Apostol (John Wiley & Sons, Inc.)
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