Roberto’s Notes on Integral Calculus Chapter 6: Improper and ......Integral Calculus Chapter 6: Improper and approximate integration Section 2: Improper integrals with one issue

Integral Calculus Chapter 6: Improper and approximate integration Section 2: Improper integrals with one issue at one limit Page 1

X

Y

1 2 3 4 5

3

6

9

0

Roberto’s Notes on Integral Calculus

Chapter 6: Improper and approximate integration Section 2

Improper integrals

with one issue at one limit

What you need to know already: What you can learn here:

The Fundamental Theorem of Calculus.

What an improper integral is.

How to give meaning to an improper integral

whose problem arises at one of the limits of

integration.

Here is the first case of how to handle an improper integral. Notice that the

secret is to move away from the problem value and then reach it by using limits.

Definition

If y f x is a function that is continuous for

a x b , but has a single discontinuity at x b , the

improper integral ( )

b

a

f x dx is defined as:

( ) lim ( )

b k

k b

a a

f x dx f x dx

If this limit exists, the integral is convergent,

otherwise it is divergent.

Example:

4

2

1

4dx

x

Technically speaking, this integral is not well

defined, since the function is not continuous at

the upper limit. However, that is the only

discontinuity and we can try to use this

definition of improper integral to see if a

meaning can still be given to it: 4

24 4

2 2

1 1lim lim 2 4

4 4

kk

k kdx dx x

x x

4lim 2 4 2 2 2 2k

k

Since this limit exists, we conclude that this integral is convergent and its

value can be taken to be 2 2 .


Of course we can use the same approach to deal with a discontinuity that occurs

at the lower limit.

Definition

If y f x is a function that is continuous for

a x b , but has a single discontinuity at x a , the

improper integral ( )

b

a

f x dx is defined as:

( ) lim ( )

b b

k a

a k

f x dx f x dx

If this limit exists, the integral is convergent,


Example:

4

3/2

2

1

2dx

x

This time the integral is improper because the

integrand is discontinuous at the lower limit.

But again, we use the limit approach to try to

give it meaning:

4 4 4

3/2 3/2 1/22 2

2

1 1 2lim lim

2 2 2k k

kk

dx dxx x x

1/2

2

2 2lim

2 2k k

Therefore, in this case the integral is divergent and we cannot assign a value

to it even with this approach.

Why do we even bother trying to assign a value to an integral that clearly should not have one? After all, in both previous examples the integral would represent the area of an infinitely long region. So, shouldn’t such area be considered infinite anyway?

Remember that an improper integral may come up in applications, such as arc

length, that involve constructing a complex formula with a discontinuity starting

from a continuous function. But even from the point of view of computing areas, the

convergence or divergence of the improper integral tells us something qualitative

about the region whose area we are trying to compute.

Notice that in the first example, the one that converges, the curve approaches

the asymptote very rapidly. In the second, the one that diverges, it also approaches

the asymptote, but not as fast. The convergence of the integral gives us a way to

compare the two rates of approach in a quantitative way. If you prefer to look at this

from a more practical perspective, we now know that we may be able to paint the

region corresponding to a convergent integral with a finite amount of paint, but no

amount of paint will suffice for a divergent one.

Wanting to paint an infinite region does not seem convincing!

Ok, if theoretical considerations do not appeal to you, here is another example.

Example: 1 , 0 1y x x

This curve is well defined and its length is

clearly finite and may have a practical

importance, so it is legitimate to want to

compute it. But such length is given by the

integral:

X

Y

1 2 3 4 5

3

6

9

0

X

Y

-1 1 2

-1

1

2

0


1 12

0 0

1 11 1

4 42 1dx dx

xx

This integral is improper, since the integrand is discontinuous at the upper

limit, but its geometric meaning tells us that it must be convergent. See if you

can confirm that by using the FTC.

Now that we have a method for dealing with a discontinuity at one limit, we

can extend its use to the case where one of the two integration limits is not, as the

definition requires, a number, but is infinite.

Definition

If the function y f x is continuous for x a , we

define the improper integral:

( ) lim ( )

k

k

a a

f x dx f x dx

In the same way, if the function y f x is

continuous for x b , we define the improper

integral:

( ) lim ( )

b b

k

k

f x dx f x dx

If either limit exists, the integral is convergent,


Example:

3/2

4

1

2dx

x

This time the integrand is continuous on the

whole domain considered, but the upper limit

is not a finite value. So, we try to approach it

by using limits.

3/2 3/2 1/2

44 4

1 1 2lim lim

2 2 2

k k

k kdx dx

x x x

1/2

2 2lim 2

22k k

Therefore in this case the integral is convergent and we can assign to it the

value 2 .

Example:

2

1

4dx

x

Here too, the integrand is continuous on the

whole domain considered, but the lower limit

is not a finite value. So, we try to approach it

by using limits.

2 2

21 1lim lim 2 4

4 4k k k

k

dx dx xx x

lim 2 2 2 4k

k

X

Y

1 2 3 4 5 6 7 8 9100

X

Y

-10 -8 -6 -4 -2 20


Since this limit is infinite, we conclude that this integral is divergent. Notice

again how much thicker the tail of this region is with respect to the previous

example. HOWEVER, do not rely on such visual considerations to reach

your conclusion: your eyes may deceive you!

Let us revisit the motivational application I mentioned in the previous section.

Example:

The amount of work needed to lift an object to a height h above the Earth’s

surface is given by an integral of the form:

2

R h

R

cW dy

y

What does this integral become if we literally want to go to the end of the

universe? It becomes:

2 2lim lim lim

k k

k k kRR R

k c c c c cW dy dy

y y y k R R

Since this limit is finite, we conclude that this integral is convergent. But this

means that we need a finite amount of work to go to the end of the universe!

And that does not even account for the gravitational assist we may get from

other planets and stars. It looks like not even the sky is the limit!

Of course the work may be finite, but not the time!

True, but still we have established an important conceptual fact.

It is time for you to absorb this concept through some practice before moving

on to more complex situations.

Summary

An improper integral in which the difficulty lies at one of the end points can be given meaning by changing that limit of integration with a possible value and using the limit

process to arrive at the needed value, be it finite or infinite.

The value of an improper integral may have a practical meaning or it may provide a qualitative assessment of some concrete property of the integrand.

Common errors to avoid

Do not rely on your intuition or on visual assessment to determine the convergence or divergence of an improper integral, as these are likely to be wrong.


Learning questions for Section I 6-2

Review questions:

1. Describe how to give meaning to an improper integral that is such only because of an issue at one of its limits of integration.

Memory questions:

1. If f x is discontinuous only at x a , what does ( )

b

a

f x dx mean?

2. If f x is discontinuous only at x b , what does ( )

b

a

f x dx mean?

3. What is the definition of the improper integral ( )a

f x dx

?

4. What is the definition of the improper integral ( )

b

f x dx

?

Computation questions:

For each of the improper integrals presented in questions 1-18, use proper notation and proper steps to:

determine if it is convergent or divergent

in case of convergence, compute its value,

in case of divergence, determine the cause.

1. 3/1

20

xedx

x

2.

1

3 2

01

xdx

x 3.

4

01

x

x

edx

e


4. 2

0

4 5

xdx

x x

.

5.

0

cos xdx

6.

3

4 2

dx

x

7. 2

04

dx

x

8.

0

35

dx

x

9.

04x x

dx

e e

10. 0

2x x

dxe e

11. 1/

2

1

x

dxx

12.

1

x dx

13. 2ln

e

dx

x x

14.

2ln

e

dx

x x

15. 0

cosxe xdx

16.

12)3(

lndx

x

x

17.

/2 2

20

cos

1 sind

18. x x dx2

1

ln

Theory questions:

1. If ( ) ( ) 0f x g x are two functions that are continuous on 0, and

0

( )f x dx

is convergent, does it follow that 0

( )g x dx

is convergent?

2. If f(x) is a continuous function and

0

( )f x dx

is convergent, does it follow

that 2

0

( )f x dx

is also convergent?


3. If y f x is a continuous function and 1

f x dx

is divergent, is it

possible for 2

1

f x dx

to be convergent?

4. If f x and g x are positive, continuous functions with f x g x

and 1

g x dx

is divergent, is it possible for 1

f x dx

to be convergent?

5. If a function f(x) is continuous and goes to 0 as x goes to infinity, does it

follow that f x dx( )

10

is convergent?

Proof questions:

1. Determine the values of p for which the integral

1

px dx

is convergent.

2. Determine the values of p for which the integral

1

0

px dx is convergent.

3. Determine the values of n for which the integral 1

1n

dxx

is convergent.

4. Determine the values of n for which the integral

1

0

1n

dxx is convergent.

5. Show that if 0f x g x for x a and the improper integral

a

f x dx

converges, then a

g x dx

converges as well.

6. Show that if 0f x g x for x a and the improper integral

a

g x dx

diverges, then a

f x dx

diverges as well.

7. Show that the improper integral

2

2

1

sin xdx

x

converges by using the

conclusions of the previous two questions and relating this integral to another

one whose convergence is easily verifiable.

8. The Gamma function is defined by the formula 1

0

x zz e x dx

, where z

can be any complex number. Prove that if n is a real positive integer, then

1 !n n .


Application questions:

1. Is it possible to paint the region in the first quadrant bounded by

1,1y x

x with a finite amount of paint?

2. Consider the solid obtained by rotating the region in the first quadrant bounded

by 1

,1y xx

around the x axis. Determine whether it is possible to

fill the solid with a finite amount of paint. This is called Gabriel’s horn: how do

you explain what you found in questions 1 and 2?

Templated questions:

1. Construct an improper integral and determine whether it is convergent or divergent.

What questions do you have for your instructor?

Documents

Roberto’s Notes on Integral Calculus Chapter 6: Improper and ......Integral Calculus Chapter 6: Improper and approximate integration Section 2: Improper integrals with one issue