6.5: Improper Integrals

Improper Integrals Involving Unbounded Intervals

In view of the above examples, we see that we may want to integrate over an interval whose upper limit grows without bound. For example, to find the fraction of light bulbs that fail eventually, we wish to find

for which we will also use the notation

Such an integral can be interpreted as the area of an unbounded region, as pictured at right in Figure $6.5.1$.

We call an integral for which the interval of integration is unbounded improper. For instance, the integrals

are all improper because they have limits of integration that involve $\infty\text{.}$ To evaluate an improper integral we replace it with a limit of proper integrals. That is,

We first attempt to evaluate $\int_0^b f(x) \,dx$ using the First FTC, and then evaluate the limit. Is it even possible for the area of an unbounded region to be finite? The following activity explores this issue and others in more detail.

Activity $6.5.2$

In this activity we explore the improper integrals $\int_1^{\infty} \frac{1}{x} \, dx$ and $\int_1^{\infty} \frac{1}{x^{3/2}} \, dx\text{.}$

1. First we investigate $\int_1^{\infty} \frac{1}{x} \, dx\text{.}$
   1. Use the First FTC to determine the exact values of $\int_1^{10} \frac{1}{x} \, dx\text{,}$ $\int_1^{1000} \frac{1}{x} \, dx\text{,}$ and $\int_1^{100000} \frac{1}{x} \, dx\text{.}$ Then, use your computational device to compute a decimal approximation of each result.
   2. Use the First FTC to evaluate the definite integral $\int_1^{b} \frac{1}{x} \, dx$ (which results in an expression that depends on $b$).
   3. Now, use your work from (ii.) to evaluate the limit given by
      $$\lim_{b \to \infty} \int_1^{b} \frac{1}{x} \, dx\text{.}$$
2. Next, we investigate $\int_1^{\infty} \frac{1}{x^{3/2}} \, dx\text{.}$
   1. Use the First FTC to determine the exact values of $\int_1^{10} \frac{1}{x^{3/2}} \, dx\text{,}$ $\int_1^{1000} \frac{1}{x^{3/2}} \, dx\text{,}$ and $\int_1^{100000} \frac{1}{x^{3/2}} \, dx\text{.}$ Then, use your calculator to compute a decimal approximation of each result.
   2. Use the First FTC to evaluate the definite integral $\int_1^{b} \frac{1}{x^{3/2}} \, dx$ (which results in an expression that depends on $b$).
   3. Now, use your work from (ii.) to evaluate the limit given by
      $$\lim_{b \to \infty} \int_1^{b} \frac{1}{x^{3/2}} \, dx\text{.}$$
3. Plot the functions $y = \frac{1}{x}$ and $y = \frac{1}{x^{3/2}}$ on the same coordinate axes for the values $x = 0 \ldots 10\text{.}$ How would you compare their behavior as $x$ increases without bound? What is similar? What is different?
4. How would you characterize the value of $\int_1^{\infty} \frac{1}{x} \, dx\text{?}$ of $\int_1^{\infty} \frac{1}{x^{3/2}} \, dx\text{?}$ What does this tell us about the respective areas bounded by these two curves for $x \ge 1\text{?}$

Convergence and Divergence

Activity $6.5.2$ suggests that $\lim_{b \to \infty} \int_1^b f(x) \, dx$ is either finite or infinite (or it doesn't exist). With these possibilities in mind, we introduce the following terminology.

Note

If $f(x)$ is nonnegative for $x \ge a\text{,}$ then we say that the improper integral $\int_a^{\infty} f(x) \, dx$ converges provided that

$$\lim_{b \to \infty} \int_a^{b} f(x) \, dx$$

exists and is finite. Otherwise, we say that $\int_a^{\infty} f(x) \, dx$ diverges.

We will restrict our interest to improper integrals for which the integrand is nonnegative. Also, we require that $\lim_{x \to \infty} f(x) = 0\text{,}$ for if $f$ does not approach $0$ as $x \to \infty\text{,}$ then it is impossible for $\int_a^{\infty} f(x) \, dx$ to converge.

Activity $6.5.3$

Determine whether each of the following improper integrals converges or diverges. For each integral that converges, find its exact value.

1. $\displaystyle \int_1^{\infty} \frac{1}{x^2} \, dx$
2. $\displaystyle \int_0^{\infty} e^{-x/4} \, dx$
3. $\displaystyle \int_2^{\infty} \frac{9}{(x+5)^{2/3}} \, dx$
4. $\displaystyle \int_4^{\infty} \frac{3}{(x+2)^{5/4}} \, dx$
5. $\displaystyle \int_0^{\infty} x e^{-x/4} \, dx$
6. $\int_1^{\infty} \frac{1}{x^p} \, dx\text{,}$ where $p$ is a positive real number

Improper Integrals Involving Unbounded Integrands

An integral is also called improper if the integrand is unbounded on the interval of integration. For example, consider

Because $f(x) = \frac{1}{\sqrt{x}}$ has a vertical asymptote at $x = 0\text{,}$ $f$ is not continuous on $[0,1]\text{,}$ and the integral represents the area of the unbounded region shown at right in Figure $6.5.2$.

We address the problem of the integrand being unbounded by replacing the improper integral with a limit of proper integrals. For example, to evaluate $\int_0^1 \frac{1}{\sqrt{x}} \, dx\text{,}$ we replace $0$ with $a$ and let $a$ approach 0 from the right. Thus,

We evaluate the proper integral $\int_a^1 \frac{1}{\sqrt{x}} \, dx\text{,}$ and then take the limit. We will say that the improper integral converges if this limit exists, and diverges otherwise. In this example, we observe that

so the improper integral $\int_0^1 \frac{1}{\sqrt{x}} \, dx$ converges (to the value 2).

We have to be particularly careful with unbounded integrands, for they may arise in ways that may not initially be obvious. Consider, for instance, the integral

At first glance we might think that we can simply apply the Fundamental Theorem of Calculus by antidifferentiating $\frac{1}{(x-2)^2}$ to get $-\frac{1}{x-2}$ and then evaluating from $1$ to $3\text{.}$ Were we to do so, we would be erroneously applying the FTC because $f(x) = \frac{1}{(x-2)^2}$ fails to be continuous throughout the interval, as seen in Figure $6.5.3$.

Such an incorrect application of the FTC leads to an impossible result ($-2$), which would itself suggest that something we did must be wrong. Instead, we must address the vertical asymptote at $x = 2$ by writing

We then evaluate two separate limits of proper integrals. For instance, doing so for the integral with $a$ approaching $2$ from the left, we find

since $\frac{1}{a-2} \to -\infty$ as $a$ approaches 2 from the left. Thus, the improper integral $\int_1^2 \frac{1}{(x-2)^2} \, dx$ diverges; similar work shows that $\int_2^3 \frac{1}{(x-2)^2} \, dx$ also diverges. From either of these two results, we can conclude that that the original integral, $\int_1^3 \frac{1}{(x-2)^2} \, dx$ diverges, too.

Activity $6.5.4$

For each of the following definite integrals, decide whether the integral is improper or not. If the integral is proper, evaluate it using the First FTC. If the integral is improper, determine whether or not the integral converges or diverges; if the integral converges, find its exact value.

1. $\displaystyle \int_0^1 \frac{1}{x^{1/3}} \, dx$
2. $\displaystyle \int_0^2 e^{-x} \, dx$
3. $\displaystyle \int_1^4 \frac{1}{\sqrt{4-x}} \, dx$
4. $\displaystyle \int_{-2}^2 \frac{1}{x^2} \, dx$
5. $\displaystyle \int_0^{\pi/2} \tan(x) \, dx$
6. $\displaystyle \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx$

Summary

• An integral $\int_a^b f(x) \, dx$ can be improper if at least one of $a$ or $b$ is $\pm \infty\text{,}$ making the interval unbounded, or if $f$ has a vertical asymptote at $x = c$ for some value of $c$ that satisfies $a \le c \le b\text{.}$ One reason that improper integrals are important is that certain probabilities can be represented by integrals that involve infinite limits.
• When we encounter an improper integral, we work to understand it by replacing the improper integral with a limit of proper integrals. For instance, we write
  $$\int_a^\infty f(x) \, dx = \lim_{b \to \infty} \int_a^b f(x) \, dx\text{,}$$

  and then work to determine whether the limit exists and is finite. For any improper integral, if the resulting limit of proper integrals exists and is finite, we say the improper integral converges. Otherwise, the improper integral diverges.

• An important class of improper integrals is given by
  $$\int_1^{\infty} \frac{1}{x^p} \, dx$$

  where $p$ is a positive real number. We can show that this improper integral converges whenever $p \gt 1\text{,}$ and diverges whenever $0 \lt p \le 1\text{.}$ A related class of improper integrals is $\int_0^1 \frac{1}{x^p} \, dx\text{,}$ which converges for $0 \lt p \lt 1\text{,}$ and diverges for $p \ge 1\text{.}$