Chap 08 Real Analysis: Improper Integrals.

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

b a

f da under the

restrictions that both f and a are defined and bounded on a finite interval [a, b] . To extend the concept, we shall relax these restrictions on f and a . Ø Definition The integral

ò

b

f da 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

ò

The integral

b a

f da is called an improper integral of second kind if f ( x) is

unbounded at one or more points of a £ x £ b . Such points are called singularities of f ( x) . Ø Notations We shall denote the set of all functions f such that f ÎR(a ) on [a, b] by R(a ; a, b) . When a ( x ) = x , we shall simply write R(a, b) for this set. The notation

a - on [a, ¥) will mean that a is monotonically increasing on [a, ¥) .

Ø Definition Assume that f ÎR(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

¥ a

ò

¥ a

f ( x) da ( x ) or by

ò

¥ a

f da .

f da is said to converge if the limit lim I (b) ………… (ii) b®¥

exists (finite). Otherwise,

ò

¥ a

f da 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 b

1

x - p dx .

(1 - b ) dx = 1- p

òx 1

ò

b

-p

¥

if p ¹ 1 , the integral

p -1

-p

dx diverges if p < 1 . When

1

p > 1 , it converges and has the value If p = 1 , we get

òx

ò

b 1

1 . p -1

x -1 dx = log b ® ¥ as b ® ¥ .

Þ

ò

¥ 1

x -1 dx diverges.

2

Chap. 8 – Improper Integrals.

Ø Example b

Consider

ò sin 2p x dx 0

b

Q

ò sin 2p x dx = 0

(1 - cos 2p b) ® ¥ as b ® ¥ . 2p

¥

\ the integral ò sin 2p x dx diverges. 0

Ø Note ¥

a

If

ò

ò f da

f da and

-¥

are both convergent for some value of a , we say that

a

¥

the integral

ò

f da is convergent and its value is defined to be 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

b

f da converges, its value is equal to the limit:

-¥

lim

b®+ ¥

ò f da .

-b

Ø Theorem Assume that a - on [a, + ¥) and suppose that f ÎR(a ; a, b) for every b ³ a . Assume that f ( x ) ³ 0 for each x ³ a . Then

ò

¥ a

f da 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, + ¥) Then lim I (b) = sup {I (b) | b ³ a} = M > 0 and the theorem follows b®+¥

b

Þ

ò f da £ M

for every b ³ a whenever the integral converges.

a

]]]]]]]]]]]]]]]]

3

Chap. 8 – Improper Integrals.

Ø Theorem: (Comparison Test) Assume that a - on [a, + ¥) . If f ÎR(a ; a, b) for every b ³ a , if ¥

ò g da

0 £ f ( x ) £ g ( x) for every x ³ a , and if

¥

converges, then

a

ò f da

converges

a

and we have ¥

ò f da

¥

£

a

ò g da a

Proof b

b

Let I1 (b) = ò f da

and

I 2 (b) = ò g da

,

b³a

a

a

Q 0 £ f ( x) £ g ( x ) for every x ³ a \ I1 (b) £ I 2 (b) …………………. (i) Q

¥

ò g da

converges \ $ a constant M > 0 such that

a ¥

ò g da £ M

,

b ³ a …………………(ii)

a

From (i) and (ii) we have I1 (b) £ M Þ lim I1 (b) exists and is finite.

, b ³ a.

b®¥ ¥

Þ

ò f da

converges.

a

lim I1 (b) £ lim I 2 (b) £ M

Also Þ

b®¥

b®¥

¥

¥

a

a

ò f da £ ò g da .

Ø Theorem (Limit Comparison Test) Assume that a - on [a, + ¥) . Suppose that f ÎR(a ; a, b) and that g ÎR (a ; a, b) for every b ³ a , where f ( x ) ³ 0 and g ( x ) ³ 0 if x ³ a . If f ( x) lim =1 x ®¥ g ( x ) ¥

then

ò f da a

¥

and

ò g da

both converge or both diverge.

a

Proof For all b ³ a , we can find some N > 0 such that f ( x) -1 < e " x ³ N for every e > 0 . g ( x) f ( x) Þ 1-e < <1+ e g ( x) 1 Let e = , then we have 2 1 f ( x) 3 < < g ( x) 2 2 Þ g ( x ) < 2 f ( x ) …..…..(i) and 2 f ( x ) < 3 g ( x ) ……....(ii)

4

Chap. 8 – Improper Integrals.

¥

¥

ò g da

From (i)

< 2 ò f da

a

a

¥

Þ

¥

ò g da

converges if

¥

ò f da

converges and

a

a

¥

ò f da

ò f da

diverges if

a

a

diverges. ¥

¥

a

a

From (ii) 2 ò f da < 3ò g da ¥

Þ

¥

ò f da

converges if

ò g da

¥

converges and

a

a

¥

ò g da

diverges if

a

ò f da a

diverges. ¥

Þ The integrals

ò f da

¥

and

a

ò g da

converge or diverge together.

a

Ø Note f ( x) = c , provided that c ¹ 0 . If c = 0 , x ®¥ g ( x )

The above theorem also holds if lim

¥

we can only conclude that convergence of

ò g da

¥

implies convergence of

a

ò f da . a

Ø Example ¥

For every real p , the integral

òe

-x

x p dx converges.

1

¥

This can be seen by comparison of this integral with

1

òx

2

dx .

1

-x

p

1 f ( x) e x = lim where f ( x) = e - x x p and g ( x) = 2 . x ®¥ g ( x ) x®¥ 1 x x2 f ( x) x p+2 Þ lim = lim e - x x p+ 2 = lim x = 0 x®¥ g ( x ) x ®¥ x ®¥ e

Since lim

and Q

¥

1

òx

2

dx is convergent

1

¥

\ the given integral

òe

-x

x p dx is also convergent.

1

Ø Theorem Assume a - on [a, + ¥) . If f ÎR(a ; a, b) for every b ³ a and if

¥

ò a

¥

converges, then

ò f da

also converges.

a

Or: An absolutely convergent integral 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 )

f da

5

Chap. 8 – Improper Integrals.

¥

ò(

Þ

f - f ) da converges.

a

¥

Subtracting from

ò

¥

f da we find that

a

ò f da

converges.

a

( Q Difference of two convergent integrals is convergent ) Ø Note ¥

ò f da

¥

is said to converge absolutely if

a

ò

f da converges. It is said to be

a

¥

¥

ò f da

convergent conditionally if

converges but

a

ò

f da diverges.

a

Ø Remark Every absolutely convergent integral is convergent. Ø Theorem Let f be a positive decreasing function defined on [a, + ¥) such that f ( x ) ® 0 as x ® +¥ . Let a be bounded on [a, + ¥) and assume that f ÎR(a ; a, b) for every b ³ a . Then the integral

ò

¥

a

f da is convergent.

Proof Integration by parts gives b

ò f da = a

b

f ( x ) × a ( x) a - ò a ( x ) df b

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

\ the convergence of

ò f da a

b

depends upon the convergence of ò a d (- f ) . a

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

b

a b

a

ò a ( x) d ( - f ) £ ò M d ( - f )

Þ

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. ¥

\

òa

d (- f ) is convergent.

(Comparison Test)

a ¥

Þ

ò f da

is convergent.

a

]]]]]]]]]]]]]]]]

6

Chap. 8 – Improper Integrals.

Ø Theorem (Cauchy condition for infinite integrals) ¥

Assume that f ÎR(a ; a, b) for every b ³ a . Then the integral

ò f da

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 da

be convergent. Then $ B > 0 such that B

a

¥

b

ò f da - ò f da a

<

e for every b ³ B ………..(i) 2

<

e …………….. (ii) 2

a

b

c

Also for c > b > B , ¥

c

ò f da - ò f da a

a

c

b

a

a

c

¥

¥

b

a

a

a

a

c

¥

Consider c

ò f da

=

b

= £

ò f da - ò f da ò f da - ò f da + ò f da - ò f da ¥

ò f da - ò f da a

b

ò f da - ò f da

+

a

a

a

<

e e + =e 2 2

c

Þ

ò f da

<e

when c > b > B .

b

Conversely, assume that the Cauchy condition holds. a +n

Define an =

ò

f da

if n = 1,2,......

a

The sequence {an } is a Cauchy sequence Þ it converges. Let lim an = A n®¥

Given e > 0 , choose B so that

c

ò

f da <

b

e 2

if c > b > B .

e 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

=

a

ò

b

f da - A +

ò

f da

a+ N

a

b

£ aN - A +

ò

a+ N ¥

Þ

ò f da = A a

This completes the proof.

f da <

e e + =e 2 2

a+N a

B

b

c

7

Chap. 8 – Improper Integrals.

Ø Remarks

ò

It follows from the above theorem that convergence of lim ò

b +e

b®¥ b

¥

a

f da implies

f da = 0 for every fixed e > 0 .

However, this does not imply that f ( x ) ® 0 as x ® ¥ . Ø Theorem Every convergent infinite integral

ò

¥

a

f ( x) da ( x) can be written as a convergent

infinite series. In fact, we have ¥

¥

a

k =1

ò f ( x) da ( x) = å a

k

a +k

where ak =

Proof Q

¥

ò f da

converges, the sequence

a

a +n

But

ò

{ò

f ( x) da ( x) ……….. (1)

}

a +n

f da also converges.

a

n

f da = å ak . Hence the series k =1

a

ò

a + k -1

¥

¥

åa

k

k =1

converges and equals

ò f da . a

Ø Remarks It is to be noted that the convergence of the series in (1) does not always imply k

convergence of the integral. For example, suppose ak = ak = 0 and

ò sin 2p x dx . Then each

k -1

åa

k

converges. ¥

However, the integral

1 - cos 2p b diverges. b®¥ 2p

b

ò sin 2p x dx = lim ò sin 2p x dx = lim b®¥

0

0

IMPROPER INTEGRAL OF THE SECOND KIND Ø Definition Let f be defined on the half open interval ( a, b] and assume that f ÎR(a ; x, b) for every x Î ( a, b] . Define a function I on ( a, b] as follows: b

I ( x ) = ò f da

if x Î ( 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 da .

a+

b

The integral

ò f da

is said to converge if the limit

a+

lim I ( x ) ……...(ii) exists (finite).

x ®a + b

Otherwise,

ò f da

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+

8

Chap. 8 – Improper Integrals.

Similarly, if f is defined on [a, b) and f ÎR(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-

denoted as

ò

f da and is convergent if lim I ( x ) exists (finite). x ® b-

a

Ø Example f ( x) = x - p is defined on (0, b] and f ÎR ( x, b) for every x Î (0, b] . b

I ( x ) = ò x - p dx x b

=

òx

-p

if x Î (0, b] b

ò

dx = lim e ®0

0+

x - p dx

0 +e

b

x1- p = lim e ®0 1 - p

b1- p - e 1- p 1- p

= lim e ®0

e

,

( p ¹ 1)

é finite , p < 1 =ê ë infinite , p > 1 b 1 When p = 1 , we get ò dx = log b - log e ® ¥ as e ® 0 . x e b

Þ

òx

-1

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 da

and

a+

ò f da

both converge, we write

c

b-

ò

c

b-

a+

c

f da =

a+

ò f da + ò f da

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 + ò f da

a+

¥

which can be written as

ò f da .

a+

b

Ø Example ¥

Consider

òe

-x

x p -1 dx

,

( p > 0)

0+

This integral must be interpreted as a sum as ¥

òe

0+

-x

x

p -1

1

dx =

òe

0+

-x

x

p -1

¥

dx + ò e - x x p -1 dx 1

= I1 + I 2 ………..….…… (i) I 2 , the second integral, converges for every real p as proved earlier. 1 1 To test I1 , put t = Þ dx = - 2 dt x t

9

Chap. 8 – Improper Integrals.

1

Þ I1 = lim ò e- x x p -1 dx = lim

1

e ®0

e ®0

òe

e

Take f (t ) = e

-1

t

1

- 1 1- p t

t

e

æ 1 ö ç - 2 dt ÷ = lim e ®0 è t ø

-1

\

òe

- 1 - p -1 t

t

e

òe

- 1 - p -1 t

t

dt

1

t - p -1 and g (t ) = t - p -1 ¥

f (t ) e t × t - p -1 Then lim = lim - p -1 = 1 and since t ®¥ g (t ) t ®¥ t ¥

1

òt

- p -1

dt converges when p > 0

1

dt converges when p > 0

1

¥

Thus

òe

-x

x p -1 dx converges when p > 0 .

0+

When p > 0 , the value of the sum in (i) is denoted 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

1 = (1 - n)( x - a )n-1 =

Which tends to

1 (1 - n)

b

a +e

æ 1 1 ö ç (b - a ) n-1 - e n-1 ÷ è ø

1 or + ¥ according as n < 1 or n > 1 , as e ® 0 . (1 - n)(b - a) n-1

Again, if n = 1 , b

dx

òa+e x - a = log(b - a) - log e ® + ¥ b

Hence the improper integral

dx

ò ( x - a )n

as e ® 0 .

converges iff n < 1 .

a

]]]]]]]]]]]]]]]]

10

Chap. 8 – Improper Integrals.

Ø Question Examine the convergence of 1 dx (i) ò 1 (ii) 2 3 1+ x 0 x

(

1

1

dx ò0 x 2 (1 + x)2

)

(iii)

òx 0

dx 1

2

(1 - x )

1

3

Solution 1

ò0 x 13

(i)

dx

(1 + x ) 2

Here ‘0’ is the only point of infinite discontinuity of the integrand. We have 1 f ( x) = 1 x 3 1 + x2

(

Take g ( x) =

)

1 1

x 3 1 f ( x) Then lim = lim =1 x ®0 g ( x ) x ®0 1 + x 2

ò

Þ

0

1

Q

1

f ( x) dx and

dx

ò0 x 13

ò

1

0

g ( x ) dx have identical behaviours. 1

converges \

ò0 x 13

dx

(1 + x )

also converges.

2

1

dx ò0 x 2 (1 + x)2 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 ( )

(ii)

Þ

ò

1

0

f ( x) dx and

ò

1

0

g ( x) dx behave alike.

But n = 2 being greater than 1, the integral

ò

1

0

g ( x) dx does not converge. Hence

the given integral also does not converge. 1

(iii)

òx 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

11

Chap. 8 – Improper Integrals.

1

the improper integrals

1

2

ò0

ò 1

f ( x ) dx and

f ( x ) dx .

2

1

2

1

òx

To examine the convergence of

1

0

2

(1 - x)

1

1

dx , we take g ( x) = 3

x

1

2

Then f ( x) 1 = lim =1 g ( x) x®0 (1 - x) 13

lim x ®0

1

Q

1

2

1

ò0 x 12 dx

2

ò

converges \

0

1 1

x (1 - x) 2

1

To examine the convergence of

ò 1

1 3

dx is convergent.

1 1

2

x 2 (1 - x)

1

dx , we take g ( x) = 3

1 (1 - x)

1

3

Then f ( x) 1 = lim 1 = 1 g ( x) x®1 x 2

lim x ®1

Q

1

1

dx 1 ò 1 (1 - x ) 3

Q

converges

2

Hence

ò

1

0

1

1

dx 1 ò 1 1 x 2 (1 - x ) 3

is convergent.

2

f ( x ) dx converges.

Ø Question 1

Show that

m -1 ò x (1 - x ) dx exists iff m , n are both positive. n-1

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

ò0 x (1 - x ) m-1

1

n -1

dx and

n -1 m-1 ò1 x (1 - x ) dx

at ‘0’ and ‘1’

2

respectively. Convergence at 0: We write f ( x) = x

As

(1 - x )

n -1

(1 - x) n-1 = x1-m

and take g ( x) =

1 1- m

x

f ( x) ® 1 as x ® 0 g ( x)

Then 1

m-1

2

1

ò0 x1-m dx

is convergent at 0 iff 1 - m < 1 i.e. m > 0 1

We deduce that the integral

2

n -1 m-1 ò0 x (1 - x ) dx

is convergent at 0, iff m is +ive.

12

Chap. 8 – Improper Integrals.

Convergence at 1: We write f ( x ) = x

(1 - x )

n -1

x m-1 = (1 - x)1-n

and take g ( x) =

1 (1 - x )1- n

f ( x) ® 1 as x ® 1 g ( x)

Then 1

1

dx 1-n ò 1 (1 - x )

As

m-1

is convergent, iff 1 - n < 1 i.e. n > 0 .

2

1

We deduce that the integral

n -1 m-1 ò1 x (1 - x ) dx converges iff

n > 0.

2

1

Thus

n-1 m -1 ò0 x (1 - x ) dx exists for positive values of m , n only.

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 sin 2 x x log x 2 1 (i) ò sin dx (ii) ò 2 dx (iii) ò dx (iv) 2 x x (1 + x ) 1 0 1 Solution 1 1 (i) Let f ( x ) = sin 2 and g ( x) = 2 x x

1

ò0 log x × log(1 + x) dx

2

æ sin y ö sin 2 1x f ( x) lim = lim 1 = lim ç ÷ =1 x ®¥ g ( x ) x ®¥ y ®0 2 è y ø

then

x

¥

ò1 f ( x) dx

Þ

¥

¥

and

1

ò1 x 2 dx

behave alike. ¥

1 1 Q ò 2 dx is convergent \ ò sin 2 dx is also convergent. x x 1 1 ¥

sin 2 x (ii) ò 2 dx x 1 sin 2 x 1 and g ( x) = 2 2 x x 2 sin x 1 £ 2 " x Î (1, ¥ ) sin 2 x £ 1 Þ x2 x ¥ ¥ 1 sin 2 x and ò 2 dx converges \ ò 2 dx converges. x x 1 1 Take f ( x ) =

Ø Note 1 sin 2 x sin 2 x dx lim = 1 so that ‘0’ is not a point is a proper integral because ò0 x2 x ®0 x2 ¥

sin 2 x of infinite discontinuity. Therefore ò 2 dx is convergent. x 0

13

Chap. 8 – Improper Integrals.

1

x log x

ò0 (1 + x)2 dx

(iii)

Q log x < x ,

x Î (0,1)

\ x log x < x 2 x log x x2 Þ < 2 2 (1 + x ) (1 + x ) 1

x2

ò (1 + x )

Now

2

dx is a proper integral.

0

1

x log x

ò (1 + x )

\

2

dx is convergent.

0

1

(iv)

ò0 log x × log(1 + x) dx

Q log x < x \ log( x + 1) < x + 1 Þ log x × log(1 + x) < x ( x + 1)

Q

1

1

ò0 x ( x + 1) dx

is a proper integral \

ò0 log x × log(1 + x) dx

is convergent.

Ø Note a

(i)

¥

(ii)

1

ò0 x p dx 1

òa x p

diverges when p ³ 1 and converges when p < 1 .

dx converges iff p > 1 .

UNIFORM CONVERGENCE OF IMPROPER INTEGRALS Ø Definition Let f be a real valued function of two variables x & y , x Î [a, + ¥) , y ÎS 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 ÎS

a

the integral is said to converge 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 ÎS .

a

( Pointwise convergence means convergence when y is fixed but uniform convergence is for every y ÎS ).

14

Chap. 8 – Improper Integrals.

Ø Theorem (Cauchy condition for uniform convergence.) The integral

ò

¥

a

f ( x, y ) da ( x ) converges uniformly on S , iff, for every e > 0

there exists a B > 0 (depending on e ) such that c > b > B implies c

ò f ( x, y ) da ( x)

<e

" y ÎS .

b

Proof Proceed as in the proof for Cauchy condition for infinite integral

ò

¥

a

f da .

Ø Theorem (Weierstrass M-test) Assume that a - on [a, + ¥) and suppose that the integral

ò

b

a

f ( x , y ) da ( x )

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

¥

a

M ( x) da ( x) converges and

f ( x, y ) £ M ( x) for each x ³ a and every y in S , then the integral

ò

¥

a

f ( x, y ) da ( x ) converges uniformly on S .

Proof Q f ( x, y ) £ M ( x) for each x ³ a and every y in S . \ For every c ³ b , we have c

c

b

b

c

ò f ( x, y) da ( x) £ ò

f ( x, y ) da ( x) £ ò M da ………… (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

a

2

…………… (ii)

Also if c > b > B , then c

ò M da - I

< e

a

c

Then

ò M da

=

b

c

b

a

a

2

…………… (iii)

ò M da - ò M da c

=

b

ò M da - I + I - ò M da a

a

c

£

ò M da - I a

b

+

ò M da - I a

< e +e =e 2 2

(By ii & iii)

c

Þ

ò f ( x, y) da ( x)

< e ,

c > b > B & for each y ÎS

b

Cauchy condition for convergence (uniform) being satisfied. Therefore the integral

ò

¥

a

f ( x, y ) da ( x ) converges uniformly on S . ]]]]]]]]]]]]]]]]

15

Chap. 8 – Improper Integrals.

Ø Example ¥

ò0 e

Consider

- xy

sin x dx

(Q

e - xy sin x £ e - xy = e - xy

e- xy £ e - xc M ( x ) = e - cx

and Now take

¥

if ¥

ò0 M ( x) dx = ò0 e

The integral

- cx

sin x £ 1 )

c£ y

dx is convergent & converging to ¥

\ The conditions of M-test are satisfied and uniformly on [c, + ¥) for every c > 0 .

ò0 e

- xy

1 . c

sin x dx converges

Ø Theorem (Dirichlet’s test for uniform convergence) b

Assume that a is bounded on [a, + ¥) and suppose the integral

òa f ( x, y) da ( x)

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) < 4M ( Q g ( x ) is +ive and ® 0 as x ® ¥ \ g ( x ) - 0 < If c > b , integration by parts yields c

òb f da =

e 4M

for x ³ B )

c

f ( x, y ) × a ( x) b - ò a df c

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) +

òa 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 )

16

Chap. 8 – Improper Integrals.

£ M g (c) + M g (b) + M g (b) + M g (c) = 2M [ g (b) + g (c)]

e ù é e < 2M ê + =e ú 4 M 4 M ë û c

Þ

ò f da

<e

for every y in S .

b

¥

Therefore the Cauchy condition is satisfied and uniformly on S .

òa f ( x, y) da ( x) converges

Ø Example ¥

Consider

ò0

e - xy sin x dx x

e - xy if x > 0 , y ³ 0 . Take a ( x) = cos x and f ( x, y ) = x 1 If S = [0, +¥ ) and g ( x) = on [e , +¥ ) for every e > 0 then x i) f ( x, y ) £ f ( x¢, y ) if x¢ £ x and a ( x) is bounded on [e , +¥ ) . ii) g ( x ) ® 0 as x ® +¥ e - xy 1 iii) f ( x, y ) = £ = g ( x) " y ÎS . x x So that the conditions of Dirichlet’s theorem are satisfied. Hence ¥ - xy ¥ - xy e e òe x sin x dx = + òe x d (- cos x) converges uniformly on [e , +¥ ) if e > 0 . sin x Q lim =1 x ®0 x ¥

Þ

ò0 e

- xy

e

\

- xy òe 0

sin x dx converges being a proper integral. x

sin x dx also converges uniformly on [0, +¥ ) . x

Ø Remarks Dirichlet’s test can be applied to test the convergence 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

òa f ( x) dx be bounded when X ³ a .

¥

Then

òa f ( x)f ( x) dx

is convergent.

Ø Example ¥

sin x dx x sin x Q ® 1 as x ® 0 . x

Consider

ò0

Chap. 8 – Improper Integrals.

17

\ 0 is not a point of infinite discontinuity. ¥

Now consider the improper integral

ò1

sin x dx . x

1 of the integrand is monotonic and ® 0 as x ® ¥ . x

The factor X

ò sin x dx

Also

= - cos X + cos(1) £ cos X + cos(1) < 2

1

X

ò1 sin x dx

So that

is bounded above for every X ³ 1 .

¥

sin x Þ ò dx is convergent. Now since x 1 ¥

that

1

sin x ò0 x dx is a proper integral, we see

sin x dx is convergent. x

ò0

Ø Example ¥

Consider

ò0 sin x

2

dx .

1 × 2 x × sin x 2 2x ¥ ¥ 1 Now ò sin x 2 dx = ò × 2 x × sin x 2 dx 2x 1 1 1 is monotonic and ® 0 as x ® ¥ . 2x

We write sin x 2 =

X

ò 2 x sin x

Also

2

dx = - cos X 2 + cos(1) < 2

1 X

ò1 2 x sin x

So that

2

dx is bounded for X ³ 1 .

¥

1 Hence ò × 2 x × sin x 2 dx i.e. 2x 1

¥

ò1 sin x

2

dx is convergent.

1

Since

ò0 sin x

2

dx is only a proper integral, we see that the given integral is

convergent. Ø Example ¥

Consider ò e - a x 0

Here e

-a x

¥

Hence

ò0 e

sin x dx , a > 0 x ¥

is monotonic and bounded and -a x

sin x ò0 x dx is convergent.

sin x dx is convergent. x ]]]]]]]]]]]]]]]]

18

Chap. 8 – Improper Integrals.

Ø Example ¥

Show that

ò0

sin x dx is not absolutely convergent. x

Solution

We need not take x because x ³ 0 .

np

sin x Consider the proper integral ò dx x 0 where n is a positive integer. We have np rp n sin x sin x dx = å ò x ò x dx r = 1 0 ( r -1)p

Put x = (r - 1)p + y so that y varies in [0,p ] .

We have sin[(r - 1)p + y ] = (-1)r -1 sin y = sin y rp

ò

\

( r -1)p

p sin x sin y dx = ò dy x r p + y ( 1) 0

Q rp is the max. value of [(r - 1)p + y ] in [0,p ] p

p

sin y 1 2 dy ³ \ ò sin y dy = ò (r - 1)p + y rp 0 rp 0 np

Þ

ò0

Q

n

Division by max. value é Q êë will lessen the value

n sin x 2 2 n 1 = å dx ³ å x p 1 r 1 rp

1

år ®¥

as n ® ¥ , we see that

1

np

sin x dx ® ¥ as n ® ¥ . x Let, now, X be any real number. There exists a +tive integer n such that np £ X < (n + 1)p .

ò0

X

We have

ò0

sin x dx ³ x

np

ò0

sin x dx x X

Let X ® ¥ so that n also ® ¥ . Then we see that ¥

So that

ò0

ò0

sin x dx ® ¥ x

sin x dx does not converge. x

Ø Questions Examine the convergence of ¥ x dx (i) ò (ii) (1 + x)3 1

¥

1 ò1 (1 + x) x dx

Solution x 1 x and take g ( x) = 3 = 2 3 x x (1 + x) 3 f ( x) x As lim = lim =1 x ®¥ g ( x ) x®¥ (1 + x ) 3

(i) Let f ( x) =

¥

(iii)

òx 1

dx 1

3

(1 + x )

1

2

19

Chap. 8 – Improper Integrals.

¥

¥

x 1 Therefore the two integrals ò dx and ò 2 dx have identical behaviour 3 (1 + x) x 1 1 for convergence at ¥ . ¥ ¥ 1 x Q ò 2 dx is convergent \ ò dx is convergent. 3 x (1 + x ) 1 1 1 1 1 and take g ( x) = = 3 (1 + x) x x x x 2 f ( x) x We have lim = lim =1 x ®¥ g ( x ) x ®¥ 1 + x

(ii) Let f ( x ) =

¥

and

1

ò1 x 3 2 dx

¥

is convergent. Thus

1

ò1 (1 + x)

x

dx is convergent.

1

(iii) Let f ( x ) = x

1

3

(1 + x ) 1

we take g ( x) =

1

1

1

=

2

1 5

x 3 ×x 2 x 6 ¥ f ( x) 1 We have lim = 1 and ò 5 dx is convergent \ x ®¥ g ( x ) 6 1 x

¥

ò1 f ( x) dx is convergent.

Ø Question ¥

1

ò 1+ x

Show that

2

dx is convergent.

-¥

Solution We have a é0 1 ù 1 1 dx + ò dx ú ê ò 1 + x2 dx = alim ®¥ ò 1 + x 2 1 + x 2 úû êë - a -¥ 0 ¥

a éa 1 ù éa 1 ù 1 = lim ê ò dx + ò dx ú = 2 lim ê ò dx ú 2 2 2 a ®¥ a®¥ 1+ x ë0 1+ x 0 û ë 01+ x û a æp ö = 2 lim tan -1 x = 2 ç ÷ = p 0 a ®¥ è2ø therefore the integral is convergent.

Ø Question ¥

Show that

tan -1 x ò0 1 + x 2 dx is convergent.

Solution Q (1 + x 2 ) ×

tan -1 x p -1 = tan x ® (1 + x 2 ) 2

¥

tan -1 x ò0 1 + x 2 dx

Q

¥

¥

&

as x ® ¥

1 ò0 1 + x 2 dx behave alike.

-1

Here f ( x ) = and

1 ò0 1 + x 2 dx is convergent \ A given integral is convergent.

tan x

1+ x 2 g (x) = 1 + x

2

20

Chap. 8 – Improper Integrals.

Ø Question ¥

Show that

sin x

ò0 (1 + x)a dx converges for a > 0 .

Solution ¥

X

ò0 sin x dx

ò0 sin x dx £ 2

is bounded because

1 , a > 0 is monotonic on [0, +¥ ) . (1 + x)a

Furthermore the function ¥

" x >0.

sin x

ò0 (1 + x)a dx is convergent.

Þ the integral Ø Question ¥

Show that

ò0 e

-x

cos x dx is absolutely convergent.

Solution Q e cos x < e -x

¥

-x

ò0 e

and

-x

dx = 1

\ the given integral is absolutely convergent. (comparison test) Ø Question e- x

1

ò0

Show that

1- x

4

dx is convergent.

Solution Q e - x < 1 and 1 + x 2 > 1 e- x

\

1 - x4 1

Also

ò0

<

1 (1 - x 2 )(1 + x 2 ) 1-e

1 1 - x2

e ®0 ò 0

dx = lim

<

1 1 - x2

1 1 - x2

dx

= lim sin -1 (1 - e ) = e ®0

1

Þ

ò0

e- x 1- x

4

p 2

dx is convergent. (by comparison test)

References:

(1) Lectures (Year 2003-04) Prof. Syyed Gul Shah Chairman, Department of Mathematics. University of Sargodha, Sargodha.

(2) Book Mathematical Analysis

Tom M. Apostol (John Wiley & Sons, Inc.)

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