
## Section5.10Improper Integrals

###### Motivating Questions
• What are improper integrals and why are they important?

• What does it mean to say that an improper integral converges or diverges?

• What are some typical improper integrals that we can classify as convergent or divergent?

So far we have considered integrals of continuous functions over finite intervals of integration. However, in many applications, we may need to integrate over an unbounded interval. That is, at least one of the limits of integration is infinite. In other cases, we may have an integrand that becomes infinite at a point within the interval or at an endpoint of the interval. Such cases are called improper integrals.

In this section we learn how to evaluate these integrals. It may be surprising that despite having regions that appear infinite, in many cases the area is finite. We first look at some examples of how such integrals may arise.

###### Example5.95

Consider a company that manufactures incandescent light bulbs. Based on a large volume of test results, they have determined that the fraction of light bulbs that fail between times $t = a$ and $t = b$ of use (where $t$ is measured in months) is given by

\begin{equation*} \int_a^b 0.3 e^{-0.3t} \, dt\text{.} \end{equation*}

For example, the fraction of light bulbs that fail during their third month of use is given by

\begin{align*} \int_2^3 0.3e^{-0.3t} \, dt \amp =\left. -e^{-0.3t} \right|_2^3\\ \amp = -e^{-0.9} + e^{-0.6}\\ \amp \approx 0.1422\text{.} \end{align*}

Thus about 14.22% of all lightbulbs fail between $t = 2$ and $t = 3\text{.}$

We can adjust the limits of integration to measure the fraction of light bulbs that fail during any time period of interest. For example, to find the fraction of light bulbs that fail eventually, we wish to find

\begin{equation*} \lim_{b \to \infty} \int_0^b 0.3e^{-0.3t} \, dt\text{,} \end{equation*}

for which we will also use the notation

$$\int_0^\infty 0.3e^{-0.3t} \, dt\text{.}\label{E-ImpInt1}\tag{5.20}$$

The integral in Example5.95 can be interpreted as the area of an unbounded region, as pictured at right in Figure5.96.

###### Improper integrals

An integral $\int_a^b f(x) \ dx$ is improper if one of the following holds.

1. The interval of integration is infinite. That is, $a = \infty$ and/or $b = \infty \text{.}$

2. The integrand is unbounded on $[a,b] \text{.}$

### SubsectionImproper Integrals Involving Unbounded Intervals

In view of the above example, we see that we may want to integrate over an interval whose upper limit grows without bound.

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

\begin{equation*} \int_1^{\infty} \frac{1}{x^2} \, dx, \ \ \int_{-\infty}^0 \frac{1}{1+x^2} \, dx, \ \ \text{and} \int_{-\infty}^{\infty} e^{-x^2} \, dx \end{equation*}

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,

\begin{equation*} \int_0^\infty f(x) \, dx = \lim_{b \to \infty} \int_0^b f(x) \,dx\text{.} \end{equation*}

We first attempt to evaluate $\int_0^b f(x) \,dx$ using the First Fundamental Theorem of Calculus4.4, and then evaluate the limit. Is it even possible for the area of an unbounded region to be finite? The following examples explore this issue and others in more detail.

###### Example5.97

A company with a large customer base has a call center that receives thousands of calls a day. After studying the data that represents how long callers wait for assistance, they find that the function $p(t) = 0.25e^{-0.25t}$ models the time customers wait in the following way: the fraction of customers who wait between $t = a$ and $t = b$ minutes is given by

\begin{equation*} \int_a^b p(t) \, dt\text{.} \end{equation*}

Use this information to answer the following questions.

1. Determine the fraction of callers who wait between 5 and 10 minutes.

2. Next, let's study the fraction who wait up to a certain number of minutes:

1. What is the fraction of callers who wait between 0 and 5 minutes?

2. What is the fraction of callers who wait between 0 and 10 minutes?

3. Let $F(b)$ represent the fraction of callers who wait between $0$ and $b$ minutes. Find a formula for $F(b)$ that involves a definite integral, and then use the First Fundamental Theorem of Calculus4.4 to find a formula for $F(b)$ that does not involve a definite integral.

4. What is the value of the limit $\lim\limits_{b \to \infty} F(b)\text{?}$ What is its meaning in the context of the problem?

1. $\int_5^{10} 0.25 e^{0.25t} \, dt = e^{-5/4} -e^{-10/4} \approx 0.2044$
1. $1-e^{-5/4}\approx 0.7135$
2. $1-e^{-10/4}\approx 0.9179$
2. \begin{equation*} F(b)=\int_0^b p(t) \, dt=1-e^{-b/4} \end{equation*}
3. \begin{equation*} \lim_{b \to \infty} F(b)=1 \end{equation*}

The limit has the following meaning: All customers, or 100% of customers, have to wait at least 0 minutes.

Solution

The general antiderivative of $p(t)$ can be found by using the substitution5.5 $u=-0.25t \ , du=-0.25 dt\text{.}$ Then $\int p(t) \, dt= -e^{-0.25t} + C\text{.}$ We'll use the general antiderivative to efficiently find the definite integrals in parts (a), (b), (c), and (d).

1. \begin{align*} \int_5^{10} 0.25 e^{0.25t} \, dt&=\left.\left(-e^{-t/4}\right) \right|_{5}^{10} \\ &=-e^{-10/4}+e^{-5/4} \\ &= e^{-5/4} -e^{-10/4} \approx 0.2044 \end{align*}
1. \begin{align*} \int_{0}^{5} 0.25 e^{0.25t} \, dt&=\left.\left(-e^{-t/4}\right) \right|_{0}^{5} \\ &=-e^{-5/4}+e^{0/4} \\ &=1-e^{-5/4}\approx 0.7135 \end{align*}
2. \begin{align*} \int_{0}^{10} 0.25 e^{0.25t} \, dt&=\left.\left(-e^{-t/4}\right) \right|_{0}^{10}\\ &=-e^{-10/4}+e^{0/4} \\ &=1-e^{-10/4}\approx 0.9179 \end{align*}
2. \begin{equation*} F(b)=\int_0^b p(t) \, dt = \left.\left(-e^{-t/4}\right) \right|_{0}^{b}=1-e^{-b/4} \end{equation*}
3. \begin{align*} \lim_{b \to \infty} F(b)&= \lim_{b \to \infty} \left(1-e^{-b/4} \right)\\ &=1-\lim_{b \to \infty} e^{-b/4} \\ &=1-\lim_{b \to \infty} \frac{1}{e^{b/4}} \\ &=1 \end{align*}

The limit has the following meaning: All customers, or 100% of customers, have to wait at least 0 minutes.

###### Example5.98

In this example 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 Fundamental Theorem of Calculus 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 tool to compute a decimal approximation of each result.

2. Use the First Fundamental Theorem of Calculus to evaluate the definite integral $\int_1^{b} \frac{1}{x} \, dx$ which results in an expression that depends on $b\text{.}$

3. Now, use your work from (ii.) to evaluate the limit given by

\begin{equation*} \lim_{b \to \infty} \int_1^{b} \frac{1}{x} \, dx\text{.} \end{equation*}
2. Next, we investigate $\int_1^{\infty} \frac{1}{x^{3/2}} \, dx\text{.}$

1. Use the First Fundamental Theorem of Calculus 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 Fundamental Theorem of Calculus 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

\begin{equation*} \lim_{b \to \infty} \int_1^{b} \frac{1}{x^{3/2}} \, dx\text{.} \end{equation*}
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 $x \ge 1\text{?}$

Hint
1. $\frac{1}{x} = x^{-1}\text{.}$

2. $\frac{1}{x^{3/2}} = x^{-3/2}$

3. Compare how quickly the curves approach the $x$-axis as $x \to \infty\text{.}$

4. Remember that $\lim_{b \to \infty} \frac{1}{b^k} = 0$ as long as $k \gt 0\text{.}$

1. $\int_1^{10} \frac{1}{x} dx = \ln(10)$ $\int_1^{1000} \frac{1}{x} dx = \ln(1000)$ $\int_1^{100000} \frac{1}{x} dx = \ln(100000)$

2. $\int_1^b \frac{1}{x} dx = \ln(b)\text{.}$

3. $\lim_{b \to \infty} \int_1^b \frac{1}{x} dx = \lim_{b \to \infty} \ln(b) = \infty$

1. $\int_1^{10} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{10}}$ $\int_1^{1000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{1000}}$ $\int_1^{100000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{100000}}$

2. $\int_1^b \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{b}}\text{.}$

3. $\lim_{b \to \infty} \int_1^b \frac{1}{x^{3/2}} dx = \lim_{b \to \infty} \left( 2 - \frac{2}{\sqrt{b}} \right) = 2$

1. Both graphs have a vertical asymptote at $x = 0$ and for both graphs, the $x$-axis is a horizontal asymptote. However, the graph of $y = \frac{1}{x^{3/2}}$ will ''approach the $x$-axis faster'' than the graph of $y = \frac{1}{x}\text{.}$

2. The area bounded by the graph of $y = \frac{1}{x}\text{,}$ the $x$-axis, and the vertical line $x = 1$ is infinite or unbounded. However, The area bounded by the graph of $y = \frac{1}{x^{3/2}}\text{,}$ the $x$-axis, and the vertical line $x = 1$ is equal to 2.

Solution
1. $\int_1^{10} \frac{1}{x} dx = \ln(10)$ $\int_1^{1000} \frac{1}{x} dx = \ln(1000)$ $\int_1^{100000} \frac{1}{x} dx = \ln(100000)$

2. $\int_1^b \frac{1}{x} dx = \ln(b)\text{.}$

3. $\lim_{b \to \infty} \int_1^b \frac{1}{x} dx = \lim_{b \to \infty} \ln(b) = \infty$

1. $\int_1^{10} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{10}}$ $\int_1^{1000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{1000}}$ $\int_1^{100000} \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{100000}}$

2. $\int_1^b \frac{1}{x^{3/2}} dx = 2 - \frac{2}{\sqrt{b}}\text{.}$

3. $\lim_{b \to \infty} \int_1^b \frac{1}{x^{3/2}} dx = \lim_{b \to \infty} \left( 2 - \frac{2}{\sqrt{b}} \right) = 2$

1. Both graphs have a vertical asymptote at $x = 0$ and for both graphs, the $x$-axis is a horizontal asymptote. However, the graph of $y = \frac{1}{x^{3/2}}$ will ''approach the $x$-axis faster'' than the graph of $y = \frac{1}{x}\text{.}$

2. The area bounded by the graph of $y = \frac{1}{x}\text{,}$ the $x$-axis, and the vertical line $x = 1$ is infinite or unbounded. However, The area bounded by the graph of $y = \frac{1}{x^{3/2}}\text{,}$ the $x$-axis, and the vertical line $x = 1$ is equal to 2.

### SubsectionConvergence and Divergence

Example5.98 suggests that $\lim\limits_{b \to \infty} \int_1^b f(x) \, dx$ is either finite or infinite (or it doesn't exist). With these possibilities in mind, we introduce the following terminology.

###### Convergence of Improper Integrals

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

\begin{equation*} \lim_{b \to \infty} \int_a^{b} f(x) \, dx \end{equation*}

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

Similarly, if $f(x)$ is nonnegative for $x \le b\text{,}$ then we say that the improper integral $\int_{-\infty}^{b} f(x) \, dx$ converges provided that

\begin{equation*} \lim_{a \to -\infty} \int_a^{b} f(x) \, dx \end{equation*}

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

###### Example5.99

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

1. $\int_1^{\infty} \frac{1}{x^2} \, dx$

2. $\int_0^{\infty} e^{-x/4} \, dx$

3. $\int_2^{\infty} \frac{9}{(x+5)^{2/3}} \, dx$

4. $\int_4^{\infty} \frac{3}{(x+2)^{5/4}} \, dx$

5. $\int_0^{\infty} x e^{-x/4} \, dx$

6. $\int_1^{\infty} \frac{1}{x^p} \, dx\text{,}$ where $p$ is a positive real number

Hint
1. $\frac{1}{x^2} = x^{-2}\text{.}$

2. Recall that for $k \gt 0\text{,}$ $e^{-kt} \to 0$ as $t \to \infty\text{.}$

3. $\frac{9}{(x+5)^{2/3}} = 9(x+5)^{-2/3}\text{.}$

4. Compare (c).

5. Compare (b), after integrating by parts.

6. Try two cases: $0 \lt p \lt 1$ and $1 \lt p\text{.}$

1. $\int_1^\infty \frac{1}{x^2} dx = 1$

2. $\int_0^\infty e^{-x/4} dx = 4$

3. $\int_2^\infty \frac{9}{(x+5)^{2/3}} dx = \infty$

4. $\int_4^\infty \frac{3}{(x+2)^{5/4}} dx = \frac{12}{6^{1/4}}$

5. $\int_0^\infty x e^{-x/4} dx = 16$

6. If $0 \lt p \lt 1\text{,}$ $\int_1^\infty \frac{1}{x^p} dx$ diverges, while if $p \gt 1\text{,}$ the integral converges.

Solution
1. \begin{align*} \int_1^\infty \frac{1}{x^2} dx \amp = \lim_{b \to \infty} \int_1^b \frac{1}{x^2} dx\\ \amp = \lim_{b \to \infty} \left.-\frac{1}{x} \right|_1^b\\ \amp = \lim_{b \to \infty} \left( -\frac{1}{b} + 1 \right)\\ \amp = 1 \end{align*}
2. \begin{align*} \int_0^\infty e^{-x/4} dx \amp = \lim_{b \to \infty} \int_0^b e^{-x/4} dx\\ \amp = \lim_{b \to \infty} \left.\left( - 4e^{-x/4} \right) \right|_0^b\\ \amp = \lim_{b \to \infty} \left( -4e^{-b/4} + 4\right)\\ \amp = 4 \end{align*}
3. \begin{align*} \int_2^\infty \frac{9}{(x+5)^{2/3}} dx \amp = \lim_{b \to \infty} \int_2^b \frac{9}{(x+5)^{2/3}} dx\\ \amp = \lim_{b \to \infty} \left. 27(x+5)^{1/3} \right|_2^b\\ \amp = \lim_{b \to \infty} \left( 27(b+5)^{1/3} - 27(7)^{1/3} \right)\\ \amp = \infty \end{align*}
4. \begin{align*} \int_4^\infty \frac{3}{(x+2)^{5/4}} dx \amp = \lim_{b \to \infty} \int_4^b \frac{3}{(x+2)^{5/4}} dx\\ \amp = \lim_{b \to \infty} \left( -\frac{12}{(b+2)^{1/4}} + \frac{12}{6^{1/4}} \right)\\ \amp = \frac{12}{6^{1/4}} \end{align*}
5. \begin{align*} \int_0^\infty x e^{-x/4} dx \amp = \lim_{b \to \infty} \int_0^b x e^{-x/4} dx\\ \amp = \lim_{b \to \infty} \left. \left( -4x e^{-x/4} -16e^{-x/4} \right) \right|_0^b\\ \amp = 16 \end{align*}
6. We assume that $p$ is a positive real number.

\begin{align*} \int_1^\infty \frac{1}{x^p} dx \amp = \lim_{b \to \infty} \int_1^b x^{-p} dx\\ \amp = \lim_{b \to \infty} \left( -\frac{b^{1-p}}{p-1} + \frac{1}{p-1} \right) \end{align*}

We now use two cases:

• (Assume $0 \lt p \lt 1\text{.}$) In this case, $1 - p \gt 0$ and so $\lim_{b \to \infty} b^{1-p} = \infty$ and

\begin{equation*} \lim_{b \to \infty} \left( -\frac{b^{1-p}}{p-1} + \frac{1}{p-1} \right) = \infty \end{equation*}

Using this and the equation above , we see that $\int_1^\infty \frac{1}{x^p} dx$ diverges.

• (Assume $p \gt 1\text{.}$) In this case, $1 - p \lt 0$ and so $\lim_{b \to \infty} b^{1-p} = 0$ and

\begin{equation*} \lim_{b \to \infty} \left( -\frac{b^{1-p}}{p-1} + \frac{1}{p-1} \right) = \frac{1}{p-1} \end{equation*}

Using this and the equation above, we see that $\int_1^\infty \frac{1}{x^p} dx$ converges to $\dfrac{1}{p-1}\text{.}$

###### Strategies: When the interval of integration is unbounded
1. To evaluate the improper integral $\int_a^{\infty} f(x) \, dx\text{,}$ use a new variable $b$ for the upper bound of integration and let $b \to \infty \text{.}$ In other words, find the definite integral $\int_a^b f(x) \, dx$ then take the limit as $b \to \infty \text{:}$

\begin{equation*} \int_a^{\infty} f(x) \, dx= \lim_{b \to \infty} \int_a^b f(x) \, dx\text{.} \end{equation*}
2. To evaluate the improper integral $\int_{-\infty}^{b} f(x) \, dx\text{,}$ use a new variable $a$ for the lower bound of integration and let $a \to -\infty \text{.}$ In other words, find the definite integral $\int_a^b f(x) \, dx$ then take the limit as $a \to -\infty \text{:}$

\begin{equation*} \int_{-\infty}^{b} f(x) \, dx= \lim_{a \to -\infty} \int_a^b f(x) \, dx\text{.} \end{equation*}
3. To evaluate the improper integral $\int_{-\infty}^{\infty} f(x) \, dx\text{,}$ use new variables $a \text{,}$ $b \text{,}$ and $c \text{.}$ The variable $a$ will represent the lower bound tending towards $-\infty \text{.}$ The variable $b$ will represent the upper bound tending towards $\infty \text{.}$ The variable $c$ will represent any real number. Find the definite integral $\int_a^c f(x) \, dx$ then take the limit as $a \to -\infty \text{,}$ and find the definite integral $\int_c^b f(x) \, dx$ then take the limit as $b \to \infty \text{:}$

\begin{equation*} \int_{-\infty}^{\infty} f(x) \, dx= \lim_{a \to -\infty} \int_a^c f(x) \, dx + \lim_{b \to \infty} \int_c^b f(x) \, dx \text{.} \end{equation*}

### SubsectionImproper Integrals Involving Unbounded Integrands

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

\begin{equation*} \int_0^1 \frac{1}{\sqrt{x}} \, dx\text{.} \end{equation*}

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 Figure5.100.

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,

\begin{equation*} \int_0^1 \frac{1}{\sqrt{x}} \, dx = \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx\text{.} \end{equation*}

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

\begin{align*} \int_0^1 \frac{1}{\sqrt{x}} \, dx &= \lim_{a \to 0^+} \int_a^1 \frac{1}{\sqrt{x}} \, dx\\ &= \lim_{a \to 0^+} \left. 2\sqrt{x} \right|_a^1\\ &= \lim_{a \to 0^+} 2\sqrt{1} - 2\sqrt{a}\\ &= 2\text{,} \end{align*}

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

\begin{equation*} \int_1^3 \frac{1}{(x-2)^2} \, dx\text{.} \end{equation*}

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 Fundamental Theorem of Calculus because $f(x) = \frac{1}{(x-2)^2}$ fails to be continuous throughout the interval, as seen in Figure5.101.

Such an incorrect application of the Fundamental Theorem of Calculus 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

\begin{equation*} \int_1^3 \frac{1}{(x-2)^2} \, dx = \lim_{a \to 2^-} \int_1^a \frac{1}{(x-2)^2} \, dx + \lim_{b \to 2^+} \int_b^3 \frac{1}{(x-2)^2} \, dx\text{.} \end{equation*}

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

\begin{align*} \int_1^2 \frac{1}{(x-2)^2} \, dx&= \lim_{a \to 2^-} \int_1^a \frac{1}{(x-2)^2} \, dx\\ &= \lim_{a \to 2^-} \left. -\frac{1}{(x-2)} \right|_1^a\\ &= \lim_{a \to 2^-} -\frac{1}{(a-2)} + \frac{1}{1-2}\\ &= \infty\text{,} \end{align*}

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.

###### Strategies: When the integrand is unbounded on the interval of integration
1. If $f(x)$ is unbounded at $x=a \text{,}$ then $\int_a^b f(x) \, dx$ is an improper integral. To evaluate the improper integral $\int_a^b f(x) \, dx \text{,}$ use a new variable $d$ for the lower bound of integration and let $d \to a^{+}$ ($d$ approaches $a$ from the right). Find the definite integral $\int_d^b f(x) \, dx$ then take the limit $d \to a^+\text{:}$

\begin{equation*} \int_a^b f(x) \, dx= \lim_{d \to a^+} \int_d^b f(x) \, dx \end{equation*}
2. If $f(x)$ is unbounded at $x=b \text{,}$ then $\int_a^b f(x) \, dx$ is an improper integral. To evaluate the improper integral $\int_a^b f(x) \, dx \text{,}$ use a new variable $c$ for the upper bound of integration and let $c \to b^{-}$ ($c$ approaches $b$ from the left). Find the definite integral $\int_a^c f(x) \, dx$ then take the limit $c \to b^-\text{:}$

\begin{equation*} \int_a^b f(x) \, dx= \lim_{c \to b^-} \int_a^c f(x) \, dx \end{equation*}
3. If $f(x)$ is unbounded at $x=p$ and $p$ lies inside the interval $[a,b] \text{,}$ then $\int_a^b f(x) \, dx$ is an improper integral. To evaluate the improper integral $\int_a^b f(x) \, dx \text{,}$ use new variables $c$ and $d$ to represent location where $f(x)$ is unbounded in the interval $[a,b]$ and split the original integral into two improper integrals. Find the definite integral $\int_a^c f(x) \, dx$ then take the limit $c \to p^-\text{,}$ and find the definite integral $\int_d^b f(x) \, dx$ then take the limit $d \to p^+\text{:}$

\begin{equation*} \int_a^b f(x) \, dx=\lim_{c \to p^-} \int_a^c f(x) \, dx + \lim_{d \to p^+} \int_d^b f(x) \, dx \end{equation*}
###### Example5.102

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 Fundamental Theorem of Calculus. If the integral is improper, determine whether or not the integral converges or diverges; if the integral converges, find its exact value.

1. $\int_0^1 \frac{1}{x^{1/3}} \, dx$

2. $\int_0^2 e^{-x} \, dx$

3. $\int_1^4 \frac{1}{\sqrt{4-x}} \, dx$

4. $\int_{-2}^2 \frac{1}{x^2} \, dx$

5. $\int_0^{\pi/2} \tan(x) \, dx$

6. $\int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx$

Hint
1. $\frac{1}{x^{1/3}}$ is undefined at $x = 0\text{.}$

2. This integral is a proper one.

3. $\frac{1}{\sqrt{4-x}}$ is undefined at $x = 4\text{.}$

4. Be careful about what happens at $x = 0\text{;}$ split the original integral into two integrals that involve $x = 0\text{.}$

5. $\tan(x)$ is undefined at $x = \pi/2\text{.}$

6. Recall that you know an antiderivative for $\frac{1}{\sqrt{1-x^2}}\text{.}$

1. $\int_0^1 \frac{1}{x^{1/3}}dx = \frac{3}{2}$

2. $\int_0^2 e^{-x} dx = 1 - e^{-2}$

3. $\int_0^4 \frac{1}{\sqrt{4-x}} dx = 4$

4. $\int_{-2}^2 \frac{1}{x^2} \, dx$ diverges.

5. $\int_0^{\pi/2} \tan(x) dx = \infty$

6. $\int_0^1 \frac{1}{\sqrt{1-x^2}} dx = \frac{\pi}{2}$

Solution
1. \begin{align*} \int_0^1 \frac{1}{x^{1/3}}dx \amp = \lim_{a \to 0^+} \int_0^1 \frac{1}{x^{1/3}}dx\\ \amp = \lim_{a \to 0^+} \left( \frac{3}{2} - \frac{3}{2} a^{2/3} \right)\\ \amp = \frac{3}{2} \end{align*}
2. \begin{align*} \int_0^2 e^{-x} dx \amp = \left. -e^{-x} \right|_0^2\\ \amp = 1 - e^{-2} \end{align*}
3. \begin{align*} \int_0^4 \frac{1}{\sqrt{4-x}} dx \amp = \lim_{b \to 4^-}\int_0^b \frac{1}{\sqrt{4-x}} dx\\ \amp = \lim_{b \to 4^-} \left. -2\sqrt{4-x} \right|_0^b\\ \amp = 4 \end{align*}
4. \begin{equation*} \int_{-2}^2 \frac{1}{x^2} dx = \int_{-2}^0 \frac{1}{x^2} + \int_{0}^2 \frac{1}{x^2} \end{equation*}

However, each of the improper integrals on the right side of the equation diverges. For example,

\begin{align*} \int_{0}^2 \frac{1}{x^2} dx \amp = \lim_{a \to 0^+} \int_{a}^2 \frac{1}{x^2} dx\\ \amp = \lim_{a \to +} \left( -\frac{1}{2} + \frac{1}{a} \right)\\ \amp = \infty \end{align*}

So $\int_{-2}^2 \frac{1}{x^2}$ dx diverges.

5. \begin{align*} \int_0^{\pi/2} \tan(x) dx \amp = \lim_{b \to \dfrac{\pi}{2}^+} \int_0^b \tan(x) dx\\ \amp = \lim_{b \to \dfrac{\pi}{2}^+} \left( -\ln(\cos(b)) + \ln(\cos(0)) \right)\\ \amp = \infty \end{align*}
6. \begin{align*} \int_0^1 \frac{1}{\sqrt{1-x^2}} dx \amp = \lim_{b \to 1+} \int_0^b \frac{1}{\sqrt{1-x^2}} dx\\ \amp = \lim_{b \to 1+} \left( \arcsin(b) - \arcsin(0) \right)\\ \amp = \frac{\pi}{2} \end{align*}

### SubsectionImportant Classes of Improper Integrals

There are three important classes of improper integrals that we will use in Section5.11. The three classes are $\int_1^{\infty} \frac{1}{x^p} \, dx$ where $p$ is a real number, $\int_0^1 \frac{1}{x^p} \, dx$ where $p$ is a real number, and $\int_0^{\infty} \frac{1}{e^{ax}}\, dx$ where $a$ is a real number. We will use the convergence behavior of these improper integrals which have simple integrands to gain understanding of the convergence behavior of improper integrals whose integrands are more complex. For each of these classes, the value of $p$ or $a$ will determine if the improper integral converges or diverges.

###### Example5.103

Determine the values of $p$ so that $\int_1^{\infty} \frac{1}{x^p} \, dx$ converges. In addition, determine the values of $p$ so that $\int_1^{\infty} \frac{1}{x^p} \, dx$ diverges.

$\int_1^{\infty} \frac{1}{x^p} \ dx$ converges if $p>1$ and diverges if $p\leq 1$

Solution

First, set up a limit to properly evaluate the improper integral.

\begin{equation*} \int_1^{\infty} \frac{1}{x^p} \, dx = \lim_{b \to \infty} \int_1^b x^{-p} \, dx \end{equation*}

Notice that the antiderivative is different if $p=1$ or $p \neq 1 \text{,}$ so we'll calculate these cases separately. Assume $p\neq 1 \text{.}$

\begin{align*} \int_1^{\infty} \frac{1}{x^p} \, dx &= \lim_{b \to \infty} \int_1^b x^{-p} \, dx \\ &= \lim_{b \to \infty} \left. \frac{x^{-p+1}}{-p+1} \right|_1^b \\ &= \lim_{b \to \infty} \frac{b^{-p+1}}{-p+1} - \frac{1}{-p+1} \\ &= -\frac{1}{-p+1} + \frac{1}{-p+1} \lim_{b \to \infty} b^{-p+1} \end{align*}

If $-p+1 >0$ (so $p <1$) then this limit diverges which implies that the improper integral diverges.

If $-p+1 < 0$ (so $p > 1$) then this limit converges which implies that the improper integral converges.

Now, assume $p=1 \text{.}$

\begin{align*} \int_1^{\infty} \frac{1}{x} \, dx &= \lim_{b \to \infty} \int_1^b \frac{1}{x} \, dx \\ & =\lim_{b \to \infty} \left. \ln |x| \right|_1^b \\ & =\lim_{b \to \infty} \ln|b| - \ln|1|\\ &= \lim_{b \to \infty} \ln |b| \end{align*}

This limit diverges, so the improper integral also diverges. Therefore $\int_1^{\infty} \frac{1}{x^p} \ dx$ converges if $p>1$ and diverges if $p\leq 1 \text{.}$

###### Example5.104

Determine the values of $p$ so that $\int_0^1 \frac{1}{x^p} \, dx$ converges. In addition, determine the values of $p$ so that $\int_0^1 \frac{1}{x^p} \, dx$ diverges.

$\int_0^{1} \frac{1}{x^p} \ dx$ converges if $p < 1$ and diverges if $p \geq 1$

Solution

The integrand is undefined at $x=0 \text{,}$ so we need to set up a limit that approaches 0 from the right for the lower bound of the integral.

\begin{equation*} \int_0^1 \frac{1}{x^p} \, dx = \lim_{a \to 0^{+}} \int_a^1 x^{-p} \, dx \end{equation*}

The gereral antiderivative is different if $p=1$ or $p \neq 1 \text{,}$ so we will handle each case separately. Assume $p \neq 1 \text{.}$

\begin{align*} \int_0^1 \frac{1}{x^p} \, dx &= \lim_{a \to 0^{+}} \int_a^1 x^{-p} \, dx \\ &= \lim_{a \to 0^{+}} \left. \frac{x^{-p+1}}{-p+1} \right|_{a}^{1} \\ &= \lim_{a \to 0^{+}} \frac{1}{-p+1} - \frac{a^{-p+1}}{-p+1} \\ &= \frac{1}{-p+1} -\frac{1}{-p+1} \lim_{a \to 0^{+}} a^{-p+1} \end{align*}

If $-p+1>0$ (so $p< 1$) then the limit converges so the improper integral converges.

If $-p+1<0$ (so $p> 1$) then the limit diverges so that improper integral diverges.

Assume $p=1\text{.}$

\begin{align*} \int_0^1 \frac{1}{x} \, dx &= \lim_{a \to 0^{+}} \int_a^1 \frac{1}{x} \, dx \\ & =\lim_{a \to 0^{+}}\left. \ln |x| \right|_a^1\\ & =\lim_{a \to 0^{+}} \ln (1) - \ln(a) \\ & =\lim_{a \to 0^{+}} 0 - \ln(a) \end{align*}

This limit diverges so the improper integral diverges. Therefore, $\int_0^{1} \frac{1}{x^p} \ dx$ converges if $p < 1$ and diverges if $p \geq 1 \text{.}$

###### Example5.105

Determine the values of $a$ so that $\int_0^{\infty} \frac{1}{e^{ax}} \, dx$ converges. In addition, determine the values of $a$ so that $\int_0^{\infty} \frac{1}{e^{ax}} \, dx$ diverges.

$\int_0^{\infty} \frac{1}{e^{ax}} \ dx$ converges if $a>0$ and diverges if $a\leq0$

Solution

Set up a limit to carefully calculate the improper integral.

\begin{equation*} \int_0^{\infty} e^{-ax} \, dx = \lim_{b \to \infty} \int_0^b e^{-ax} \, dx \end{equation*}

The antiderivative is different if $a=0$ or if $a \neq 0\text{,}$ so we'll calculate these cases separately. Assume $a \neq 0 \text{.}$

\begin{align*} \int_0^{\infty} e^{-ax} \, dx &= \lim_{b \to \infty} \int_0^b e^{-ax} \, dx \\ &= \lim_{b \to \infty} \left. -\frac{1}{a} e^{-ax} \right|_0^b \\ &= -\frac{1}{a} \lim_{b \to \infty} \left( e^{-ab} -1 \right) \\ &=\frac{1}{a} -\frac{1}{a} \lim_{b \to \infty} e^{-ab} \end{align*}

If $-ab \lt 0$ (so $a\gt0$), then the limit converges which implies the improper integral converges.

If $-ab \gt 0$ (so $a\lt0$), then the limit diverges which implies the improper integral diverges.

Assume $a=0\text{.}$

\begin{align*} \int_0^{\infty} e^{0} \, dx &= \lim_{b \to \infty} \int_0^b 1 \, dx \\ &= \lim_{b \to \infty} \left. x \right|_0^b\\ &= \lim_{b \to \infty} b \end{align*}

This limit diverges, so the improper integral diverges. Therefore, $\int_0^{\infty} \frac{1}{e^{ax}} \ dx$ converges if $a>0$ and diverges if $a\leq0 \text{.}$

###### Convergence Behavior of Important Classes of Improper Integrals
• $\int_1^{\infty} \frac{1}{x^p} \ dx$ converges if $p>1$ and diverges if $p\leq1$

• $\int_0^{1} \frac{1}{x^p} \ dx$ converges if $p < 1$ and diverges if $p \geq 1$

• $\int_0^{\infty} \frac{1}{e^{ax}} \ dx$ converges if $a>0$ and diverges if $a\leq0$

### SubsectionSummary

• 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

\begin{equation*} \int_a^\infty f(x) \, dx = \lim_{b \to \infty} \int_a^b f(x) \, dx\text{,} \end{equation*}

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.

• There are three important classes of improper integrals discussed in this section.

• One important class of improper integrals is given by

\begin{equation*} \int_1^{\infty} \frac{1}{x^p} \, dx \end{equation*}

where $p$ is a positive real number. We can show that this improper integral converges whenever $p \gt 1$ and diverges whenever $p \le 1\text{.}$

• A related class of improper integrals is

\begin{equation*} \int_0^1 \frac{1}{x^p} \, dx, \end{equation*}

which converges for $p \lt 1$ and diverges for $p \ge 1\text{.}$

• Another class of improper integrals is

\begin{equation*} \int_0^{\infty} e^{-ax} \, dx, \end{equation*}

which converges when $a \gt 0$ and diverges when $a \leq 0 \text{.}$