3.7 Improper Integrals

Integrating over an Infinite Interval

How should we go about defining an integral of the type ${\displaystyle {\int }_a^{+\infty }f\left(x\right)dx}?$ We can integrate ${\displaystyle {\int }_a^{t}f\left(x\right)dx}$ for any value of $t,$ so it is reasonable to look at the behavior of this integral as we substitute larger values of $t.$ Figure 3.17 shows that ${\displaystyle {\int }_a^{t}f\left(x\right)dx}$ may be interpreted as area for various values of $t.$ 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.

Figure 3.17 To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.

In our first example, we return to the question we posed at the start of this section: Is the area between the graph of $f\left(x\right)=\frac{1}{x}$ and the $x$-axis over the interval $[1,\text{+}\infty )$ finite or infinite?

Example 3.47

Finding an Area

Determine whether the area between the graph of $f\left(x\right)=\frac{1}{x}$ and the x-axis over the interval $[1,\text{+}\infty )$ is finite or infinite.

Solution

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

Figure 3.18 We can find the area between the curve $f(x)=1\text{/}x$ and the x-axis on an infinite interval.

We can see that the area of this region is given by $A={\displaystyle {\int }_1^{\infty }\frac{1}{x}dx}.$ Then we have

$$\begin{array}{ccccc}\hfill A& ={\displaystyle {\int }_1^{\infty }\frac{1}{x}dx}\hfill & & & \\ & =\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle {\int }_1^t\frac{1}{x}dx}\hfill & & & \text{Rewrite the improper integral as a limit.}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}\text{ln}\left|x\right||{}_{\begin{array}{c}\\ 1\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}(\text{ln}\left|t\right|-\text{ln}1)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =\text{+}\infty .\hfill & & & \text{Evaluate the limit.}\hfill \end{array}$$

Since the improper integral diverges to $+\infty ,$ the area of the region is infinite.

Example 3.48

Finding a Volume

Find the volume of the solid obtained by revolving the region bounded by the graph of $f\left(x\right)=\frac{1}{x}$ and the x-axis over the interval $[1,\text{+}\infty )$ about the $x$-axis.

Solution

The solid is shown in Figure 3.19. Using the disk method, we see that the volume V is

$$V=\pi {\displaystyle {\int }_1^{+\infty }\frac{1}{x^2}dx}.$$

Figure 3.19 The solid of revolution can be generated by rotating an infinite area about the x-axis.

Then we have

$$\begin{array}{ccccc}\hfill V& =\pi {\displaystyle {\int }_1^{+\infty }\frac{1}{x^2}dx}\hfill & & & \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle {\int }_1^t\frac{1}{x^2}dx}\hfill & & & \text{Rewrite as a limit.}\hfill \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}-\frac{1}{x}|{}_{\begin{array}{c}\\ 1\end{array}}^{\begin{array}{c}t\\ \end{array}}\hfill & & & \text{Find the antiderivative.}\hfill \\ & =\pi \underset{t\to \text{+}\infty }{\text{lim}}\left(\text{−}\frac{1}{t}+1\right)\hfill & & & \text{Evaluate the antiderivative.}\hfill \\ & =\pi .\hfill & & & \end{array}$$

The improper integral converges to $\pi .$ Therefore, the volume of the solid of revolution is $\pi .$

In conclusion, although the area of the region between the x-axis and the graph of $f\left(x\right)=1\text{/}x$ over the interval $[1,\text{+}\infty )$ 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.

Example 3.49

Chapter Opener: Traffic Accidents in a City

Figure 3.20 (credit: modification of work by David McKelvey, Flickr)

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 $k,$ the probability that $X,$ the time between events, is between $a$ and $b$ is given by

$$P\left(a\le x\le b\right)={\displaystyle {\int }_a^{b}f\left(x\right)dx}\text{where}f\left(x\right)=\{\begin{array}{c}0\text{if}x<0\\ k{e}^{\text{−}kx}\text{if}x\ge 0\end{array}.$$

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

$$P\left(a\le x\le b\right)={\displaystyle {\int }_a^{b}f\left(x\right)dx}\text{where}f\left(x\right)=\{\begin{array}{c}0\text{if}x<0\\ 3e^{\mathrm{-3}x}\text{if}x\ge 0\end{array}.$$

To answer the question, we must compute $P\left(X\ge 8\right)={\displaystyle {\int }_8^{+\infty }3e^{\mathrm{-3}x}dx}$ 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:

$$\begin{array}{cc}\hfill P\left(X\ge 8\right)& ={\displaystyle {\int }_8^{+\infty }3e^{\mathrm{-3}x}dx}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle {\int }_8^{t}3e^{\mathrm{-3}x}dx}\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}{\text{−}{e}^{\mathrm{-3}x}|}_8^t\hfill \\ & =\underset{t\to \text{+}\infty }{\text{lim}}(\text{−}{e}^{\mathrm{-3}t}+e^{\mathrm{-24}})\hfill \\ & \approx 3.8\times {10}^{\mathrm{-11}}.\hfill \end{array}$$

The value $3.8\times {10}^{\mathrm{-11}}$ 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 the changes were effective.

Integrating a Discontinuous Integrand

Now let’s examine integrals of functions containing an infinite discontinuity in the interval over which the integration occurs. Consider an integral of the form ${\displaystyle {\int }_a^{b}f(x)dx,}$ where $f(x)$ is continuous over $[a,b)$ and discontinuous at $b.$ Since the function $f(x)$ is continuous over $[a,t]$ for all values of $t$ satisfying $a<t<b,$ the integral ${\displaystyle {\int }_a^{t}f(x)dx}$ is defined for all such values of $t.$ Thus, it makes sense to consider the values of ${\displaystyle {\int }_a^{t}f(x)dx}$ as $t$ approaches $b$ for $a<t<b.$ That is, we define ${\displaystyle {\int }_a^{b}f(x)dx=\underset{t\to b^{-}}{\text{lim}}{\displaystyle {\int }_a^{t}f(x)dx}},$ provided this limit exists. Figure 3.21 illustrates ${\displaystyle {\int }_a^{t}f(x)dx}$ as areas of regions for values of $t$ approaching $b.$

Figure 3.21 As t approaches b from the left, the value of the area from a to t approaches the area from a to b.

We use a similar approach to define ${\displaystyle {\int }_a^{b}f(x)dx,}$ where $f(x)$ is continuous over $(a,b]$ and discontinuous at $a.$ We now proceed with a formal definition.

The following examples demonstrate the application of this definition.

A 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 $f(x)$ and $g(x)$ satisfying $0\le f(x)\le g(x)$ for $x\ge a$ (Figure 3.22). In this case, we may view integrals of these functions over intervals of the form $[a,t]$ as areas, so we have the relationship

Figure 3.22 If $0\le f(x)\le g(x)$ for $x\ge a,$ then for $t\ge a,$ ${\displaystyle {\int }_a^{t}f(x)dx\le {\displaystyle {\int }_a^{t}g(x)dx}}.$

Thus, if

then

${\displaystyle {\int }_a^{+\infty }g\left(x\right)dx=\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle {\int }_a^{t}g\left(x\right)dx=\text{+}\infty }}$ as well. That is, if the area of the region between the graph of $f(x)$ and the x-axis over $[a,\text{+}\infty )$ is infinite, then the area of the region between the graph of $g(x)$ and the x-axis over $[a,\text{+}\infty )$ is infinite too.

On the other hand, if

${\displaystyle {\int }_a^{+\infty }g\left(x\right)dx=\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle {\int }_a^{t}g\left(x\right)dx=L}}$ for some real number $L,$ then

${\displaystyle {\int }_a^{+\infty }f\left(x\right)dx=\underset{t\to \text{+}\infty }{\text{lim}}{\displaystyle {\int }_a^{t}f\left(x\right)dx}}$ must converge to some value less than or equal to $L,$ since ${\displaystyle {\int }_a^{t}f\left(x\right)dx}$ increases as $t$ increases and ${\displaystyle {\int }_a^{t}f\left(x\right)dx\le L}$ for all $t\ge a.$

If the area of the region between the graph of $g(x)$ and the x-axis over $[a,\text{+}\infty )$ is finite, then the area of the region between the graph of $f(x)$ and the x-axis over $[a,\text{+}\infty )$ is also finite.

These conclusions are summarized in the following theorem.