# 3.7 Improper Integrals

### Learning Objectives

• Evaluate an integral over an infinite interval.
• Evaluate an integral over a closed interval with an infinite discontinuity within the interval.
• Use the comparison theorem to determine whether a definite integral is convergent.

Is the area between the graph of and the x-axis over the interval finite or infinite? If this same region is revolved about the x-axis, is the volume finite or infinite? Surprisingly, the area of the region described is infinite, but the volume of the solid obtained by revolving this region about the x-axis is finite.

In this section, we define integrals over an infinite interval as well as integrals of functions containing a discontinuity on the interval. Integrals of these types are called improper integrals. We examine several techniques for evaluating improper integrals, all of which involve taking limits.

### Integrating over an Infinite Interval

How should we go about defining an integral We know how to evaluate for any value of so it is reasonable to look at the behavior of this integral as we substitute larger values of In Figure 1 below is interpreted as area below the graph of over an interval for various values of In other words, we may define an improper integral as a limit, taken as one of the limits of integration increases or decreases without bound.

### Definition

1. Let be continuous over an interval Then

provided this limit exists.
2. Let be continuous over an interval of the form Then

provided this limit exists.

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.

3. Let be continuous over We define as

provided that both and converge.

If either of these two integrals is divergent, then diverges. (It can be shown that, in fact, for any value of )

In our first example, we return to the question we posed at the start of this section: Is the area between the graph of and the -axis over the interval finite or infinite?

### Finding an Area

Determine whether the area between the graph of and the x-axis over the interval is finite or infinite.

#### Solution

We first do a quick sketch of the region in question, as shown in the following graph.

We can see that the area of this region is given by Then we have

Since the improper integral diverges to the area of the region is infinite.

### Finding a Volume

Find the volume of the solid obtained by revolving the region bounded by the graph of and the x-axis over the interval about the -axis.

#### Solution

The solid is shown in Figure 3 below. Using the disk method, we see that the volume V is

Then we have

The improper integral converges to Therefore, the volume of the solid of revolution is

In conclusion, although the area of the region between the x-axis and the graph of over the interval is infinite, the volume of the solid generated by revolving this region about the x-axis is finite. The solid generated is known as Gabriel’s Horn.

### Chapter Opener: Traffic Accidents in a City

In the chapter opener, we stated the following problem: Suppose that at a busy intersection, traffic accidents occur at an average rate of one every three months. After residents complained, changes were made to the traffic lights at the intersection. It has now been eight months since the changes were made and there have been no accidents. Were the changes effective or is the 8-month interval without an accident a result of chance?

Probability theory tells us that if the average time between events is the probability that the time between events, is between and is given by

Thus, if accidents are occurring at a rate of one every 3 months, then the probability that the time between accidents, is between and is given by

To answer the question, we must compute and decide whether it is likely that 8 months could have passed without an accident if there had been no improvement in the traffic situation.

#### Solution

We need to calculate the probability as an improper integral:

The value represents the probability of no accidents in 8 months under the initial conditions. Since this value is very, very small, it is reasonable to conclude that the changes were effective.

### Evaluating an Improper Integral over an Infinite Interval

Evaluate State whether the improper integral converges or diverges.

#### Solution

Begin by rewriting as a limit using the definition. We have:

The improper integral converges to

### Evaluating an Improper Integral over

Evaluate State whether the improper integral is convergent or divergent.

#### Solution

Start by splitting up the integral:

If either or diverges, then diverges. Compute each integral separately. For the first integral,

The first improper integral converges. For the second integral,

Thus, diverges. Since this integral diverges, diverges as well.

Evaluate State whether the improper integral converges or diverges.

converges

### Integrating a Discontinuous Function

We now examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form where is continuous over and discontinuous at Since the function is continuous over for all satisfying the integral is defined for all such values of Hence, it makes sense to consider the values of as approaches for That is, we define provided this limit exists. The figure below illustrates as areas of regions for values of approaching

We use a similar approach to define where is continuous over and discontinuous at We now proceed with a formal definition.

### Definition

1. Let be continuous over Then,

2. Let be continuous over Then,

In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.
3. If is continuous over except at a point in then we define as

provided both and converge. If either of these two integrals diverges, then diverges.

The following examples demonstrate the application of this definition.

### Integrating a Discontinuous Integrand

Evaluate if possible. State whether the integral converges or diverges.

#### Solution

The function is continuous over and discontinuous at 4. Using the above definition, we rewrite as a limit:

Because the limit exists, the improper integral converges.

### Integrating a Discontinuous Integrand

Evaluate State whether the integral converges or diverges.

#### Solution

Since is continuous over and is discontinuous at zero, we can rewrite the integral as a limit using the definition of improper integral of the corresponding type:

The improper integral converges.

### Integrating a Discontinuous Integrand

Evaluate State whether the improper integral converges or diverges.

#### Solution

Since is continuous at every point of except zero, we use the corresponding definition to write

Our integral converges if both integrals on the right converge. If either of the two integrals on the right diverges, then the original integral diverges as well. Begin with

Therefore, diverges, and hence diverges regardless of the behavior of .

Evaluate State whether the integral converges or diverges.

diverges

### Comparison Theorem

It is not always easy or even possible to evaluate an improper integral directly; however, by comparing it with another carefully chosen integral, it may be possible to determine its convergence or divergence. To see this, consider two continuous functions and satisfying for . In this case, we may view integrals of these functions over intervals of the form as areas, so by the comparison property for definite integrals, we have the relationship

Thus, if then

as well.

That is, if the area of the region between the graph of and the x-axis over is infinite, then the area of the region between the graph of and the x-axis over is infinite too.

On the other hand, if

for some real number then

must converge to some value less than or equal to since increases as increases and for all

That is, if the area of the region between the graph of and the x-axis over is finite, then the area of the region between the graph of and the x-axis over is also finite.
These conclusions are summarized in the following theorem.

### Comparison Theorem

Let and be continuous over Assume that for

1. If then
2. If where is a real number, then for some real number

### Applying the Comparison Theorem

Use the comparison theorem to show that converges.

#### Solution

The integrand is continuous over and we can also see that  for ,

so if converges, then so does To evaluate first rewrite it as a limit:

Since the limit is finite, converges, and hence, by the comparison theorem, so does

### Applying the Comparison Theorem

Use the comparison theorem to show that diverges for all

#### Solution

First we note that is continuous over . If then for all We already showed that Therefore, by the comparison theorem, diverges for all

Use the comparison theorem to show that diverges.

on

### Student Project: Laplace Transforms

In the last few chapters, we have looked at several ways to use integration for solving real-world problems. For this next project, we are going to explore a more advanced application of integration: integral transforms. Specifically, we describe the Laplace transform and some of its properties. The Laplace transform is used in engineering and physics to simplify the computations needed to solve some problems. It takes functions expressed in terms of time and transforms them to functions expressed in terms of frequency. It turns out that, in many cases, the computations needed to solve problems in the frequency domain are much simpler than those required in the time domain.

The Laplace transform is defined in terms of an integral as

Note that the input to a Laplace transform is a function of time, and the output is a function of frequency, Although many real-world examples require the use of complex numbers (involving the imaginary number in this project we limit ourselves to functions of real numbers.

Let’s start with a simple example. Here we calculate the Laplace transform of . We have

This is an improper integral, so we express it in terms of a limit, which gives

Now we use integration by parts to evaluate the integral. Note that we are integrating with respect to t, so we treat the variable s as a constant. We have

Then we obtain

Therefore, .
1. Calculate the Laplace transform of
2. Calculate the Laplace transform of
3. Calculate the Laplace transform of (Note, you will have to integrate by parts twice.)
Laplace transforms are often used to solve differential equations. Differential equations are not covered in detail until later in this book; but, for now, let’s look at the relationship between the Laplace transform of a function and the Laplace transform of its derivative.

4. Use integration by parts to evaluate (Let and
After integrating by parts and evaluating the limit, you should see that

Then,

It follows that differentiation in the time domain simplifies to multiplication by s in the frequency domain.
The final thing we look at in this project is how the Laplace transforms of and its antiderivative are related. Let Then,

5. Use integration by parts to evaluate (Let and Note that from the way we have defined )
As you might expect, you should see that

That is, integration in the time domain simplifies to division by s in the frequency domain.

### Key Concepts

• Integrals of functions over infinite intervals are defined in terms of limits.
• Integrals of functions over an interval for which the function has a discontinuity at an endpoint may be defined in terms of limits.
• The convergence or divergence of an improper integral may be determined by comparing it with the value of an improper integral for which the convergence or divergence is known.

### Key Equations

• Improper integrals

### Exercises

Evaluate the following integrals, if possible. If the integral diverges, answer “divergent.”

1.

divergent

2.

3.

4.

5.

6.

7. Without integrating, determine whether the integral converges or diverges by comparing the integrand with

converges

8. Without integrating, determine whether the integral converges or diverges.

Determine whether the improper integrals converge or diverge. If possible, determine the value of the integrals that converge.

9.

Converges to 1/2.

10.

11.

−4

12.

13.

14.

15.

diverges

16.

17.

diverges

18.

19.

20.

21.

diverges

22.

23.

diverges

24.

25.

diverges

Determine the convergence of each of the following integrals by comparison with the given integral. If the integral converges, find the number to which it converges.

26. compare with

27. compare with

Both integrals diverge.

Use the comparison theorem to determine whether the given improper integral converges or diverges.

28.

29.

diverges

30.

31.

(Hint: compare to .)

converges

32.

33.

converges

Evaluate the integrals. If the integral is divergent, answer “diverges.”

34.

35.

diverges

36.

37.

diverges

38.

39.

40.

41.

diverges

42.

43.

44.

45.

6

46.

47.

48.

49.

50.

51. Evaluate

52. Find the area of the region in the first quadrant between the curve and the x-axis.

53. Find the area of the region bounded by the curve the x-axis, and on the left by

7

54. Find the area under the curve bounded on the left by

55. Find the area under in the first quadrant.

56. Find the volume of the solid generated by revolving the region under the curve from to about the x-axis.

57. Find the volume of the solid generated by revolving the region under the curve in the first quadrant about the y-axis.

58. Find the volume of the solid generated by revolving the region under the curve in the first quadrant about the x-axis.

The Laplace transform of a continuous function over the interval is defined by (see the Student Project). The domain of F is the set of all real numbers s such that the improper integral converges. Find the Laplace transform F of each of the following functions and give the domain of F.

59.

60.

61.

62.

A function is a probability density function if it satisfies the following definition: The probability that a random variable x lies between a and b is given by

63. Show that

is a probability density function.