Chapter 5, Problem 11RQ

To determine

To Find : if ${\displaystyle \underset{0}{\overset{4}{\int }}\frac{x}{{}^{x^{{}^2}-1}}}dx=\frac{1}{2}\ln 15$ .

Answer to Problem 11RQ

The limit does not exist as a finite number and so the improper integral ${\displaystyle \underset{0}{\overset{4}{\int }}\frac{x}{{}^{x^{{}^2}-1}}}dx$ is divergent

Explanation of Solution

Given information: the integral is given as ${\displaystyle \underset{0}{\overset{4}{\int }}\frac{x}{{}^{x^{{}^2}-1}}}dx$

Formula used: $\int \frac{1}{x}dx=\text{ln}\left|x\right|+\text{C}$

${\displaystyle \underset{0}{\overset{4}{\int }}\frac{x}{{}^{x^{{}^2}-1}}}dx$

The integral can be written as:

  ${\displaystyle \underset{0}{\overset{4}{\int }}\frac{x}{(x-1)(x+1)}dx}$

Observe that the line $x=1$ is a vertical asymptote of the integrand. Since it occurs in

the middle of the interval [0, 4]. For this, we must use the definition of improper

integral.

If $f$ has a discontinuity at c, where $a <c <b$ , and both ${\displaystyle \underset{a}{\overset{c}{\int }}f(x)dx}$ and

  ${\displaystyle \underset{c}{\overset{b}{\int }}f(x)dx}$ are convergent, then we define

  $\begin{array}{l}{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}={\displaystyle \underset{a}{\overset{c}{\int }}f(x)dx}+{\displaystyle \underset{c}{\overset{b}{\int }}f(x)dx}\\ {\displaystyle \underset{0}{\overset{4}{\int }}\frac{x}{x^2-1}}dx={\displaystyle \underset{0}{\overset{1}{\int }}\frac{x}{x^2-1}}dx+{\displaystyle \underset{1}{\overset{4}{\int }}\frac{x}{x^2-1}}dx\end{array}$

Where = ${\displaystyle \underset{0}{\overset{1}{\int }}\frac{x}{{}^{x^{{}^2}-1}}}dx$

First, we will find $\int \frac{x}{x^2-1}dx$ integration

If $u=x^2-1$

Then $du=2xdx$

  $\int \frac{1}{2}\frac{du}{u}=\frac{1}{2}\text{ln}$ | $u|$

Put $u=x^2-1$

  $\begin{array}{l}\int \frac{x}{x^2-1}dx=\frac{1}{2}\text{ln}\left|x^2-1\right|\\ \underset{t\to 1^{-}}{\lim }=\underset{\text{t}\to -1}{\lim }\text{ln}|x^2-1{|}_0^t\\ \underset{t\to 1^{-}}{\lim }\text{ln}|t^2-1|-\text{ln}\left|-1\right|\\ \underset{t\to 1^{-}}{\lim }\ln \left(1-t^2\right)=-\infty \end{array}$

because $1-t^2\to 0^{+}$ as $t\to 1^{-}$

The limit does not exist as a finite number

thus $$ is divergent.

This implies that ${\displaystyle \underset{0}{\overset{4}{\int }}\frac{x}{{}^{x^{{}^2}-1}}}dx$ is divergent. [ we do not need to evaluate ${\displaystyle \underset{1}{\overset{4}{\int }}\frac{x}{{}^{x^{{}^2}-1}}}dx$ ].