Improper Integrals

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 7.8 – Improper Integrals

Expected Educational Results

• Objective 09–01: I can identify an improper integral.

• Objective 09–02: I can use limits to evaluate an improper integral.

• Objective 09–03: I can use the Comparison Theorem to determine if an improper integral converges or diverges.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Comparison Theorem

Theorem: Comparison Theorem

Suppose $f$$f$ and $g$$g$ are continuous functions with $f(x)\ge g(x)\ge 0$$f(x)\geq g(x)\geq 0$ for $x\ge a$$x\geq a$.

1. If ${\int }_a^{b}f(x)dx$$\int_a^b{f(x)\,dx}$ is convergent, then ${\int }_a^{b}g(x)dx$$\int_a^b{g(x)\,dx}$ is convergent.

2. If ${\int }_a^{b}g(x)dx$$\int_a^b{g(x)\,dx}$ is divergent, then ${\int }_a^{b}f(x)dx$$\int_a^b{f(x)\,dx}$ is divergent.

Procedure for Comparison Theorem

1. Use DESMOS to sketch a graph of the integrand function.

2. Use only some of the terms of the given function $f(x)$$f(x)$ to find another function, $g(x)$$g(x)$, so that

   • $f(x)\le g(x)$$f(x)\leq g(x)$ for all $x$$x$ on the given interval; or

   • $f(x)\ge g(x)$$f(x)\geq g(x)$ for all $x$$x$ on the given interval

3. Use the Comparison Theorem to determine if the integral is convergent or divergent. Explain.

Example 04: Consider ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{{\sin }^2(x)+1}{x}dx}$$\displaystyle\int_1^{\infty}{\frac{\sin^2{(x)}+1}{x}\,dx}$

NOTE: There is no analytic (i.e., functional) antiderivative for ${\displaystyle \int \frac{{\sin }^2(x)+1}{x}dx}$$\displaystyle\int{\frac{\sin^2{(x)}+1}{x}\,dx}$.

Here is the DESMOS graph:

Solution 04: We will compare ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{{\sin }^2(x)+1}{x}dx}$$\displaystyle\int_1^{\infty}{\frac{\sin^2{(x)}+1}{x}\,dx}$ to ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{1}{x}dx}$$\displaystyle\int_1^{\infty}{\frac{1}{x}\,dx}$.

The DESMOS graphs show ${\displaystyle f(x)=\frac{{\sin }^2(x)+1}{x}}$$\displaystyle {\color{gsured}f(x)=\frac{\sin^2{(x)}+1}{x}}$ [red function] and ${\displaystyle g(x)=\frac{1}{x}}$$\displaystyle {\color{gsublue}g(x)=\frac{1}{x}}$ [blue function] are continuous for $x\ge 1$$x\geq 1$.

Also, the DESMOS graphs show ${\displaystyle \frac{{\sin }^2(x)+1}{x}\ge \frac{1}{x}\ge 0}$$\displaystyle \frac{\sin^2{(x)}+1}{x}\geq \frac{1}{x} \geq 0$ for $x\ge 1$$x\geq 1$.

By Theorem, ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{1}{x^p}dx}$$\displaystyle\int_1^{\infty}{\frac{1}{x^p}\,dx}$ diverges, when $p=1$$p=1$.

Thus, since the conditions of the Comparison Theorem are true, and the “smaller” function diverges, then the “larger” function also diverges, i.e., ${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{{\sin }^2(x)+1}{x}dx}$$\displaystyle\int_1^{\infty}{\frac{\sin^2{(x)}+1}{x}\,dx}$ diverges.

Example 05: Consider ${\displaystyle {\int }_1^{\mathrm{\infty }}{e}^{-x^2}dx}$$\displaystyle\int_1^{\infty}{e^{-x^2}\,dx}$.

NOTE: There is no analytic (i.e., functional) antiderivative for ${\displaystyle \int e^{-x^2}dx}$$\displaystyle\int{e^{-x^2}\,dx}$.

Here is the DESMOS graph:

Solution 05: We will compare ${\displaystyle {\int }_1^{\mathrm{\infty }}{e}^{-x^2}dx}$$\displaystyle\int_1^{\infty}{e^{-x^2}\,dx}$ to ${\displaystyle {\int }_1^{\mathrm{\infty }}{e}^{-x}dx}$$\displaystyle\int_1^{\infty}{e^{-x}\,dx}$ [which we can evaluate].

The embedded DESMOS graph shows ${\displaystyle f(x)=e^{-x^2}}$$\displaystyle {\color{gsured}f(x)=e^{-x^2}}$ [red function] and ${\displaystyle g(x)=e^{-x}}$$\displaystyle {g(x)=e^{-x}}$ [blue function] are continuous for $x\ge 1$$x\geq 1$.

Also, the DESMOS graph shows ${\displaystyle e^{-x}\ge e^{-x^2}\ge 0}$$\displaystyle e^{-x}\geq e^{-x^2} \geq 0$ for $x\ge 1$$x\geq 1$.

We know, ${\displaystyle {\int }_1^{\mathrm{\infty }}{e}^{-x}dx=\underset{a\to \mathrm{\infty }}{lim}{(-e^{-x})|}_1^a=\frac{1}{e}}$$\displaystyle\int_1^{\infty}{e^{-x}\,dx} = \lim_{a\rightarrow\infty}\left.\left(-e^{-x}\right)\right|_1^a=\frac{1}{e}$, i.e., the integral converges.

Thus, since the conditions of the the Comparison Theorem are true, and the “larger” function converges, then the “smaller” function also converges, i.e., ${\displaystyle {\int }_1^{\mathrm{\infty }}{e}^{-x^2}dx}$$\displaystyle\int_1^{\infty}{e^{-x^2}\,dx}$ converges.

Investigation 08

For each of the following functions:

Use the Comparison Theorem to determine if the following integrals are convergent or divergent. Explain.

1. ${\displaystyle {\int }_2^{\mathrm{\infty }}\frac{{\cos }^2(x)}{x^2}dx}$$\displaystyle\int_2^{\infty}{\frac{\cos^2{(x)}}{x^2}\,dx}$ $\Rightarrow$$\text{\,}\Rightarrow\text{\,}$ Use ${\displaystyle {\int }_2^{\mathrm{\infty }}\frac{1}{x^2}dx}$$\displaystyle\int_2^{\infty}{\frac{1}{x^2}\,dx}$ for the comparison

2. ${\displaystyle {\int }_3^{\mathrm{\infty }}\frac{1}{x+e^x}dx}$$\displaystyle\int_{3}^{\infty}{\frac{1}{x+e^x}\,dx}$ $\Rightarrow$$\text{\,}\Rightarrow\text{\,}$ Use ${\displaystyle {\int }_2^{\mathrm{\infty }}\frac{1}{x^2}dx}$$\displaystyle\int_2^{\infty}{\frac{1}{x^2}\,dx}$ for the comparison

3. ${\displaystyle {\int }_3^{\mathrm{\infty }}\frac{1}{x-e^{-x}}dx}$$\displaystyle\int_{3}^{\infty}{\frac{1}{x-e^{-x}}\,dx}$ $\Rightarrow$$\text{\,}\Rightarrow\text{\,}$ Use ${\displaystyle {\int }_2^{\mathrm{\infty }}\frac{1}{x}dx}$$\displaystyle\int_2^{\infty}{\frac{1}{x}\,dx}$ for the comparison

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Created: Thursday, 17 September 2020 2:48 EDT Last Modified: Sunday, 12 June 2022 - 17:40 (EDT)