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.

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.

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

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

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

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 ten months since the changes were made and there have been no accidents. Were the changes effective or is the 10-month interval without an accident a result of chance?

Solution

Revise to: Let $x$ represent the amount of time it takes for the next accident to occur. We want to know how likely it is that $x>10$. Define the rate parameter $\lambda$ to be the average number of accidents per month. According to probability theory, for $a>0$,

$$P\left(x>a\right)={\displaystyle {\int }_a^{\infty }\lambda e^{-\lambda x}}dx,$$

In this example, since one accident happens every three months, on average, $\lambda =\frac{1}{3}$. The desired probability is:

$$\begin{array}{l}P(x>10)={\displaystyle {\int }_{10}^{\infty }}\frac{1}{3}{e}^{-\frac{1}{3}x}dx\\ =\underset{t\to \infty }{\lim }{\displaystyle {\int }_{10}^t}\frac{1}{3}{e}^{-\frac{1}{3}x}dx\\ =\underset{t\to \infty }{\lim }-e^{-\frac{1}{3}x}{\displaystyle {|}_{10}^t}\\ =\underset{t\to \infty }{\lim }(-e^{-\frac{t}{3}}+e^{-\frac{10}{3}})\\ \approx 0.0357\end{array}$$

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 $b.$

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

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.