Chapter 8.4, Problem 11E

To determine

To evaluate: The improper integral, ${\displaystyle {\int }_{-\infty }^{-2}\frac{2dx}{x^2-1}}$ or state that it diverges.

Answer to Problem 11E

The value of the integral is $\ln \left(3\right)$ .

Explanation of Solution

Given information: The given integral is ${\displaystyle {\int }_{-\infty }^{-2}\frac{2dx}{x^2-1}}$ .

Concept used: Consider that $f\left(x\right)$ is continuous on $\left(-\infty ,b\right]$ , then ${\displaystyle \underset{-\infty }{\overset{b}{\int }}f\left(x\right)dx=\underset{a\to -\infty }{\lim }}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}$ .

Calculation:

Assume $I={\displaystyle {\int }_{-\infty }^{-2}\frac{2dx}{x^2-1}}$ . In order to express the improper integral as a limit definite integral, consider a real number $a$ such that,

  $I=\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{-2}\frac{2dx}{x^2-1}}$

Rewrite the integral.

  $\begin{array}{l}{\displaystyle {\int }_a^{-2}\frac{2dx}{x^2-1}}={\displaystyle {\int }_a^{-2}\frac{2dx}{\left(x+1\right)\left(x-1\right)}}\\ \Rightarrow {\displaystyle {\int }_a^{-2}\frac{2dx}{x^2-1}}={\displaystyle {\int }_a^{-2}\frac{\left(\left(x+1\right)-\left(x-1\right)\right)dx}{\left(x+1\right)\left(x-1\right)}}\\ \Rightarrow {\displaystyle {\int }_a^2\frac{2dx}{x^2-1}}={\displaystyle {\int }_a^{-2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx}\end{array}$

Apply the rules of integration to determine the value of integral in terms of $a$ .

  $\begin{array}{l}{\displaystyle {\int }_a^{-2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx}={\left[\ln \left|x-1\right|-\ln \left|x+1\right|\right]}_a^{-2}\\ \Rightarrow {\displaystyle {\int }_a^{-2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx}={\left[\ln \left|\frac{x-1}{x+1}\right|\right]}_a^{-2}\\ \Rightarrow {\displaystyle {\int }_a^{-2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx}=\ln \left|\frac{-2-1}{-2+1}\right|-\ln \left|\frac{a-1}{a+1}\right|\\ \Rightarrow {\displaystyle {\int }_a^{-2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx}=\ln \left|3\right|-\ln \left|\frac{a-1}{a+1}\right|\end{array}$

Determine the value of $\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{-2}\frac{2dx}{x^2-1}}$ .

  $\begin{array}{l}\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{-2}\frac{2dx}{x^2-1}}=\underset{a\to -\infty }{\lim }\left[\ln \left|3\right|-\ln \left|\frac{1-\frac{1}{a}}{1+\frac{1}{a}}\right|\right]\\ \Rightarrow \underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{-2}\frac{2dx}{x^2-1}}=\ln \left(3\right)-\ln \left|\frac{1-\frac{1}{\infty }}{1-\frac{1}{\infty }}\right|\\ \Rightarrow \underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^{-2}\frac{2dx}{x^2-1}}=\ln \left(3\right)\end{array}$

Since the given improper integral has some value, it converges to $\ln \left(3\right)$ .