We will compute definite integrals involving infinity.

### 1 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 The second of the
above examples is called “doubly improper” since it involves both and .
Examples of the second type include 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.

### 2 Infinity as an endpoint

We will begin by replacing in the improper integral with the variable , a typical choice for an upper endpoint of integration. We compute \begin{align*} \int _0^b e^{-2x} \; dx &= \left (-\frac 12 e^{-2x} \right ) \bigg |_0^b\\ &= \left (-\frac 12 e^{-2b} \right ) - \left (-\frac 12 e^{0} \right )\\ &=\frac 12 -\frac 12 e^{-2b}. \end{align*}

To complete the problem, we now take a limit as . \begin{align*} \int _0^\infty e^{-2x} \; dx &= \lim _{b \to \infty }\int _0^b e^{-2x} \; dx\\ &=\lim _{b \to \infty } \left (\frac 12 -\frac 12 e^{-2b}\right )\\ &= \frac 12 -\frac 12 e^{-\infty }\\ &= \frac 12 - 0 \\ &= \frac 12. \end{align*}

Since the limit is a number, we say that the integral converges. More precisely, we can say that the integral converges to .

We will begin by replacing in the improper integral with the variable , 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 . \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 .

We will begin by replacing in the improper integral with the variable , 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 . \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.

Based on the previous examples and problems, we can expect the following theorem about improper integrals of the form

Does the improper integral converge or diverge?

Does the improper integral converge or diverge?

Make a substitution

The endpoints become

and

In terms of , the integral is This is a -integral with

According to the theory of -integrals, this integral

### 3 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.

Therefore, the improper integral converges to 6.

Therefore, the improper integral diverges.

Therefore, the improper integral converges.

### 4 Doubly Improper Integrals

Doubly improper integrals have the form To compute a doubly improper integral, we let be any number (typically 0) and split it into two improper integrals: If either of these improper integrals diverges, then we say that the doubly improper integral diverges. Otherwise, we say that it converges.