Chapter 9.4, Problem 46E

a.

To determine

Toshow: that the integral ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}dx}$ is diverges.

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}dx}$ .

Proof: to show that the integral is diverges, first find the value of integral,

  $\begin{array}{l}{\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}dx}={\left[\ln \left(x^2+1\right)\right]}_0^{\infty }\text{ [}{x}^2+1=t\Rightarrow 2xdx=dt\text{ and }{\displaystyle \int \frac{1}{x}dx=\ln x}]\\ {\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}dx}=\ln \left({\left(\infty \right)}^2+1\right)-\ln \left({\left(0\right)}^2+1\right)\\ \ln \left(x^2+1\right)=\text{ not defined}\end{array}$

Thus, the integral is diverges.

b.

To determine

To state: that the result of the part(a) conclude that ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{x^2+1}dx}$ diverges.

Answer to Problem 46E

The integral of ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{x^2+1}dx}$ is diverges.

Explanation of Solution

Given information:

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

from the part (a) it can be observed that ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}dx}$ is diverges, the integral of ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{x^2+1}dx}$ can be written in the form of ${\displaystyle {\int }_0^{\infty }\frac{2x}{x^2+1}dx}$ .

Thus, the ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{x^2+1}dx}$ is diverges.

c.

To determine

To show: that $\underset{b\to \infty }{\lim }{\displaystyle {\int }_{-b}^b\frac{2x}{x^2+1}dx}=0$

Explanation of Solution

Given information:

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

Proof: since , first substitute $x=-x$ , in the function of the integral,

  $\begin{array}{l}\frac{2x}{x^2+1}=\frac{2\left(-x\right)}{{\left(-x\right)}^2+1}\text{ [substitute }x=-x]\\ \frac{2x}{x^2+1}=\frac{-2x}{x^2+1}\text{ [simplify]}\end{array}$

Thus, the function is odd function.

So, $\underset{b\to \infty }{\lim }{\displaystyle {\int }_{-b}^b\frac{2x}{x^2+1}dx}=0$

d.

To determine

To state: that why the result in part (C) does not contradict part(b).

Answer to Problem 46E

The result in part (C) does not contradict part(b).

Explanation of Solution

Given information:

The result of part (c) and part (b).

the result of part (c) and part (b) does not contradict because in part (b) there is no definite integral and in part (c) the improper integral first converted into definite integral and then solve it.