Chapter 8.4, Problem 46E

i.

To determine

To calculate: To solve the integral ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}}dx$ diverges.

Answer to Problem 46E

Yes, the integral diverges

Explanation of Solution

Given information: The given information as follows:

The integral ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}}dx$ diverges and show that $\underset{b\to \infty }{\lim }{\displaystyle {\int }_{-b}^b\frac{2x}{x^2+1}}dx=0$ .

Calculation:

Let us suppose that

  $\begin{array}{c}{x}^2+1=u\\ 2xdx=du\end{array}$

The values can be written as:

  $\begin{array}{c}Letx=0,\\ 0+1=u\\ u=1\\ Letx=\infty ,\\ \infty +1=u\\ u=\infty \end{array}$

  $\begin{array}{c}{\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}}dx={\displaystyle {\int }_1^{\infty }\frac{1}{u}}du\\ ={\left[\ln u\right]}_1^{\infty }\\ =\ln \infty -\ln 1\\ =\infty \end{array}$

Hence, the integral diverges.

ii.

To determine

To calculate: To solve the integral ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}}dx$ diverges.

Answer to Problem 46E

Diverges.

Explanation of Solution

Given information: The given information as follows:

The integral ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}}dx$ diverges and show that $\underset{b\to \infty }{\lim }{\displaystyle {\int }_{-b}^b\frac{2x}{x^2+1}}dx=0$ .

Calculation:

The given integral ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{x^2+1}}dx$

can be calculated as:

Let us suppose that

  $\begin{array}{c}{x}^2+1=u\\ 2xdx=du\end{array}$

The values can be written as:

  $\begin{array}{c}Letx=-\infty ,\\ -\infty +1=u\\ u=-\infty \\ Letx=\infty ,\\ \infty +1=u\\ u=\infty \end{array}$

  $\begin{array}{c}{\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{x^2+1}}dx={\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{u}}du\\ ={\left[\ln u\right]}_{-\infty }^{\infty }\\ =\ln \left(-\infty \right)-\ln \left(\infty \right)\\ =\infty \end{array}$

Hence, the integral diverges.

iii.

To determine

To calculate: To solve the integral ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}}dx$ diverges.

Answer to Problem 46E

Yes, the integral diverges

Explanation of Solution

Given information: The given information as follows:

The integral ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}}dx$ diverges and show that $\underset{b\to \infty }{\lim }{\displaystyle {\int }_{-b}^b\frac{2x}{x^2+1}}dx=0$ .

Calculation:

The given integral $\underset{b\to \infty }{\lim }{\displaystyle {\int }_{-b}^b\frac{2x}{x^2+1}}dx=0$ .

Let us suppose that

  $\begin{array}{c}{x}^2+1=u\\ 2xdx=du\end{array}$

The values can be written as:

  $\begin{array}{c}Letx=-\infty ,\\ -\infty +1=u\\ u=-\infty \\ Letx=\infty ,\\ \infty +1=u\\ u=\infty \end{array}$

  $\begin{array}{c}{\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{x^2+1}}dx={\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{u}}du\\ ={\left[\ln u\right]}_{-\infty }^{\infty }\\ =\ln \left(-\infty \right)-\ln \left(\infty \right)\\ =\infty \end{array}$

Hence, the integral diverges.