1.6 Improper Integrals

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We will compute definite integrals involving infinity.

Introduction

In this section we will consider definite integrals involving infinity. There are two fundamentally different types of improper integrals. The first involves infinity as an “endpoint” of integration and the second involves vertical asymptotes at or between the endpoints of integration. Examples of the first type include $\int _0^\infty e^{-2x} \; dx \; \; \text { and } \; \; \int _{-\infty }^\infty \frac {1}{1+x^2}\; dx.$ The second of the above examples is called “doubly improper” since it involves both $\infty$ and $-\infty$ . Examples of the second type include $\int _0^{\pi /2} \tan (x)\; dx \; \; \text { and } \; \; \int _0^1 \frac {1}{x} \; dx.$ If the value of an improper integral is a number, then we say that the integral converges. Otherwise, we say that the integral diverges. The method of computing improper integrals involves replacing an endpoint with a variable and then taking an appropriate limit.

Infinity as an endpoint

Type 1 Improper Integral Suppose $f(x)$ is continuous on the interval $[a, \infty )$ . Then we compute the improper integral at infinity as follows: $\int _a^\infty f(x) \; dx = \lim _{b \to \infty } \int _a^b f(x) \; dx.$ The improper integral at negative infinity is defined similarly.

example 1 Compute the improper integral $\int _0^\infty e^{-2x} \; dx.$

We will begin by replacing $\infty$ in the improper integral with the variable $b$ , a typical choice for an upper endpoint of integration. We compute

$\begin{align*} \int_0^b e^{-2x} \; dx &= \left(-\frac12 e^{-2x} \right) \bigg|_0^b\\ &= \left(-\frac12 e^{-2b} \right) - \left(-\frac12 e^{0} \right)\\ &=\frac12 -\frac12 e^{-2b}. \end{align*}$

To complete the problem, we now take a limit as $b \to \infty$ .

$\begin{align*} \int_0^\infty e^{-2x} \; dx &= \lim_{b \to \infty}\int_0^b e^{-2x} \; dx\\ &=\lim_{b \to \infty} \left(\frac12 -\frac12 e^{-2b}\right)\\ &= \frac12 -\frac12 e^{-\infty}\\ &= \frac12 - 0 \\ &= \frac12. \end{align*}$ Since the limit is a number, we say that the integral converges. More precisely, we can say that the integral converges to $1/2$ .

(problem 1a) Compute the improper integral $\int _0^\infty 2e^{-5x} \; dx = \answer {2/5}$

converge diverge

(problem 1b) Compute the improper integral $\int _{-\infty }^1 3e^{x/2} \; dx = \answer {6e^{1/2}}$

converge diverge

example 2 Compute the improper integral $\int _1^\infty \frac {3}{x^2} \; dx.$

We will begin by replacing $\infty$ in the improper integral with the variable $b$ , a typical choice for an upper endpoint of integration. We compute

$\begin{align*} \int_1^b \frac{3}{x^2} \; dx &= \int_1^b 3x^{-2} \; dx\\ &= \left(3\cdot \frac{x^{-1}}{-1} \right) \bigg|_1^b\\ &= \left(-\frac{3}{x} \right) \bigg|_1^b\\ &= \left(-\frac{3}{b} \right) - \left(-\frac{3}{1} \right)\\ &=3 - \frac{3}{b}. \end{align*}$

To complete the problem, we now take a limit as $b \to \infty$ .

$\begin{align*} \int_1^\infty \frac{3}{x^2} \; dx &= \lim_{b \to \infty}\int_1^b \frac{3}{x^2} \; dx\\ &=\lim_{b \to \infty} \left(3 - \frac{3}{b}\right)\\ &= 3 -\frac{3}{\infty}\\ &= 3 - 0 \\ &= 3. \end{align*}$ Since the limit is a number, we say that the integral converges. More precisely, we can say that the integral converges to $3$ .

(problem 2a) Compute the improper integral $\int _2^\infty \frac {5}{x^3} \; dx = \answer {5/8}$

converge diverge

(problem 2b) Compute the improper integral $\int _1^\infty \frac {5}{(3x+1)^2} \; dx = \answer {5/12}$

converge diverge

example 3 Compute the improper integral $\int _1^\infty \frac {4}{x} \; dx.$

We will begin by replacing $\infty$ in the improper integral with the variable $b$ , a typical choice for an upper endpoint of integration. We compute

$\begin{align*} \int_1^b \frac{4}{x} \; dx &= \int_1^b \frac{4}{x} \; dx\\ &= \left(4\ln|x| \right) \bigg|_1^b\\ &= \left(4\ln|b| \right) - \left(4\ln|1| \right)\\ &=4\ln|b|. \end{align*}$

To complete the problem, we now take a limit as $b \to \infty$ .

$\begin{align*} \int_1^\infty \frac{4}{x} \; dx &= \lim_{b \to \infty}\int_1^b \frac{4}{x} \; dx\\ &=\lim_{b \to \infty} 4\ln|b|\\ &= 4\ln(\infty)\\ &= \infty. \end{align*}$ Since the answer is a not a finite number, we say that the integral diverges.

(problem 3) Compute the improper integral $\int _3^\infty \frac {7}{\sqrt x} \; dx = \answer {\infty }.$ Does the improper integral converge or diverge?

converge diverge

Based on the previous examples and problems, we can expect the following theorem about improper integrals of the form $\int _1^\infty \frac {1}{x^p} \; dx$

$p$ -integrals The improper $p$ -integral $\int _1^\infty \frac {1}{x^p} \; dx$ converges if $p>1$ and diverges if $p \leq 1$ The result is the same if lower endpoint is replaced by any other positive number.

This theorem will be of value to us in future sections, but for now, we use it to help us with another improper integral.

example 4 Use the theory of $p$ -integrals to determine if the integral converges or diverges $\int _2^\infty \frac {2}{\sqrt x} \; dx$ This is a (multiple of a) $p$ -integral with $p = 1/2$ . Since $1/2 \leq 1$ , this integral diverges.

(problem 4a) Use the theory of $p$ -integrals to determine if the integral converges or diverges $\int _4^\infty \frac {3}{\sqrt [3] x} \; dx$ This is a (multiple of a) $p$ -integral with $p = \answer {1/3}$
Does the improper integral converge or diverge?

converge diverge

(problem 4b) Use the theory of $p$ -integrals to determine if the integral converges or diverges $\int _{0.3}^\infty 2x^{-5} \; dx$ This is a (multiple of a) $p$ -integral with $p = \answer {5}$
Does the improper integral converge or diverge?

converge diverge

example 5 Use the theory of $p$ -integrals to determine whether the improper integral convergs or diverges $\int _2^\infty \frac {1}{x\ln (x)} \; dx.$ Using a $u$ -substitution with $u = \ln (x)$ and $du = \frac {1}{x} \, dx$ , the integral becomes $\int _2^\infty \frac {1}{x\ln (x)} \; dx = \int _{\ln (2)}^\infty \frac {1}{u} \; du.$ This is a $p$ -integral in the variable $u$ with $p=1$ . Since $p \leq 1$ , the theorem says that this integral diverges, and hence $\int _2^\infty \frac {1}{x\ln (x)} \; dx \;\; \text { diverges}.$

(problem 5a) Use the theory of $p$ -integrals to determine if the improper integral converges or diverges: $\int _3^\infty \frac {1}{x\ln ^2(x)} \; dx$

Make a substitution
$u = \answer {\ln (x)} \;\; du = \answer {1/x} \, dx$
The endpoints become
$x = 3 \implies u = \answer {\ln (3)}$ and $x= \infty \implies u = \answer {\infty }$
In terms of $u$ , the integral is $\int _3^\infty \frac {1}{x\ln ^2(x)} \; dx = \int _{\ln (3)}^\infty \answer {1/u^2} \; du$ This is a $p$ -integral with $p = \answer {2}$
According to the theory of $p$ -integrals, this integral

converges diverges

(problem 5b) Use the theory of $p$ -integrals to determine the values of $p$ for which the improper integral converges $\int _2^\infty \frac {1}{x \ln ^p(x)} \; dx$ This integral will converge for $p > \answer {1}$ .

Vertical Asymptotes in the Interval of Integration

In this section we examine definite integrals in which the integrand has a vertical asymptote at an endpoint of integration.

Type 2 Improper Integral Suppose $f(x)$ is continuous on the interval $(a, b]$ and that $f(x)$ has a vertical asymptote at $x = a$ . Then $\int _a^b f(x) \; dx = \lim _{t \to a^+} \int _t^b f(x) \; dx.$ The improper integral is defined similarly if the vertical asymptote is at the upper endpoint of integration.

example 6 Compute the improper integral $\int _0^1 \frac {3}{\sqrt x} \; dx.$ The integral is improper because the integrand has a vertical asymptote at $x = 0$ . Hence $\begin{align*} \int_0^1 \frac{3}{\sqrt x} \; dx &= \lim_{a \to 0^+} \int_a^1 \frac{3}{\sqrt x} \; dx\\ &= \lim_{a \to 0^+} \int_a^1 3x^{-1/2} \; dx\\ &= \lim_{a \to 0^+} \left( 6x^{1/2} \right) \bigg|_a^1\\ &= \lim_{a \to 0^+} \left( 6\sqrt{1} - 6\sqrt{a} \right) \\ &= 6 - 0 \\ &= 6. \end{align*}$ Therefore, the improper integral converges to 6.

(problem 6) Compute the improper integral $\int _2^6 \frac {4}{\sqrt {x-2}} \; dx = \answer {16}$ Does the improper integral converge or diverge?

converge diverge

example 7 Compute the improper integral $\int _0^{\pi /2} \tan (x) \; dx.$ The integral is improper because the integrand has a vertical asymptote at $x = \pi /2$ . Hence $\begin{align*} \int_0^{\pi/2} \tan(x) \; dx &= \lim_{b \to \frac{\pi}{2}^-} \int_0^b \tan(x) \; dx\\ &= \lim_{b \to \frac{\pi}{2}^-} \int_0^b \frac{\sin(x)}{\cos(x)} \; dx\\ &= \lim_{b \to \frac{\pi}{2}^-} \ln|\sec(x)| \bigg|_0^b\\ &= \lim_{b \to \frac{\pi}{2}^-} \big( \ln|\sec(b)| - \ln|\sec(0)| \big) \\ &= \infty - 0 \\ &= \infty. \end{align*}$ Therefore, the improper integral diverges.

(problem 7) Compute the improper integral $\int _0^{\pi /2} \cot (x) \; dx$ The integrand has a vertical asymptote at

0 $\pi /2$

The anti derivative is $\int \cot (x) \; dx = \answer {\ln |\sin (x)|} + C$ The improper integral

converges diverges

example 8 Compute the improper integral $\int _0^\infty xe^{-x} \; dx$ We use integration by parts with $u = x$ and $dv = e^{-x} dx$ to obtain $\begin{align*} \int_0^\infty xe^{-x} \; dx &= \lim_{b \to \infty} \int_0^b xe^{-x} \; dx\\ &= \lim_{b \to \infty} \left[ -xe^{-x}\bigg|_0^b + \int_0^b e^{-x} \; dx \right] \\ &= \lim_{b \to \infty} -xe^{-x}- e^{-x}\bigg|_0^b \\ &= \lim_{b \to \infty} xe^{-x} + e^{-x}\bigg|_b^0 \\ &= \lim_{b \to \infty} (0+1) - (be^{-b} + e^{-b}) \\ &= 1 - 0 \\ &= 1 \end{align*}$ Therefore, the improper integral converges.

L’Hopital’s rule is used to compute the limit of $be^{-b}$ : $\lim _{b \to \infty } be^{-b} = \lim _{b \to \infty } \frac {b}{e^b} = \frac {\infty }{\infty }$ and now, by L’Hopital’s rule, we get $\lim _{b \to \infty } \frac {b}{e^b} = \lim _{b \to \infty } \frac {1}{e^b} = 0.$

Doubly Improper Integrals

Doubly improper integrals have the form $\int _{-\infty }^\infty f(x) \; dx$ To compute a doubly improper integral, we let $a$ be any number (typically 0) and split it into two improper integrals: $\int _{-\infty }^\infty f(x) \; dx = \int _{-\infty }^a f(x) \; dx + \int _a^\infty f(x) \; dx$ If either of these improper integrals diverges, then we say that the doubly improper integral diverges. Otherwise, we say that it converges.

example 9 Compute the doubly improper integral $\int _{-\infty }^\infty \frac {1}{1+x^2} \; dx$ We will rewrite this integral as $\int _{-\infty }^\infty \frac {1}{1+x^2} \; dx = \int _{-\infty }^0 \frac {1}{1+x^2} \; dx + \int _0^\infty \frac {1}{1+x^2} \; dx$ The first integral is $\int _{-\infty }^0 \frac {1}{1+x^2} \; dx = \lim _{a \to -\infty } \int _{a}^0 \frac {1}{1+x^2} \; dx$ $= \lim _{a \to -\infty } \tan ^{-1}(x) \bigg |_a^0 = \lim _{a \to -\infty } \left [\tan ^{-1}(0) - \tan ^{-1}(a)\right ]$ $= 0 - \left (-\frac {\pi }{2}\right ) = \frac {\pi }{2}$ Thus the first integral converges. You can verify that the second integral also converges to $\frac {\pi }{2}$ , and so the double improper integral converges to $\pi$ , i.e. $\int _{-\infty }^\infty \frac {1}{1+x^2} \; dx = \frac {\pi }{2} + \frac {\pi }{2} = \pi$

Here is a detailed, lecture style video on improper integrals: