Chapter 8.4, Problem 3E

a.

To determine

To express: The improper integral, ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{{\left(x^2+1\right)}^2}dx}$ as a limit of definite integrals.

Answer to Problem 3E

The limit of definite integral will be, $\underset{b\to -\infty }{\lim }{\displaystyle {\int }_b^0\frac{2x}{{\left(x^2+1\right)}^2}dx}+\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^b\frac{2x}{{\left(x^2+1\right)}^2}dx}$ .

Explanation of Solution

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

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

Calculation:

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

  $I=\underset{b\to -\infty }{\lim }{\displaystyle {\int }_b^0\frac{2x}{{\left(x^2+1\right)}^2}dx}+\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^b\frac{2x}{{\left(x^2+1\right)}^2}dx}$

So, the limit of definite integral will be, $\underset{b\to -\infty }{\lim }{\displaystyle {\int }_b^0\frac{2x}{{\left(x^2+1\right)}^2}dx}+\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^b\frac{2x}{{\left(x^2+1\right)}^2}dx}$ .

b.

To determine

To evaluate: The value of the improper integral, ${\displaystyle {\int }_{-\infty }^{\infty }\frac{2x}{{\left(x^2+1\right)}^2}dx}$ .

Answer to Problem 3E

The value of the given integral is 0.

Explanation of Solution

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

Concept used: The formula for determining the value of the integral is ${\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}+c$ .

The formula for determining the value of the integral is ${\displaystyle \int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx}=\ln \left|f\left(x\right)\right|+c$ .

Calculation:

From part a, the limit of integration is $\underset{b\to -\infty }{\lim }{\displaystyle {\int }_b^0\frac{2x}{{\left(x^2+1\right)}^2}dx}+\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^b\frac{2x}{{\left(x^2+1\right)}^2}dx}$ .

It is known that the derivative of $x^2+1$ is $2x$ .

Apply the rules of integration to determine the value of integral, ${\displaystyle {\int }_b^0\frac{2x}{{\left(x^2+1\right)}^2}dx}$ in terms of $b$ .

  $\begin{array}{c}{\displaystyle {\int }_b^0\frac{2x}{{\left(x^2+1\right)}^2}}={\left[\frac{-1}{x^2+1}\right]}_b^0\\ =\frac{-1}{0^2+1}-\left(-\frac{1}{b^2+1}\right)\\ =-1+\frac{1}{b^2+1}\end{array}$

Apply the rules of integration to determine the value of integral, ${\displaystyle {\int }_0^b\frac{2x}{{\left(x^2+1\right)}^2}dx}$ in terms of $b$ .

  $\begin{array}{c}{\displaystyle {\int }_0^b\frac{2x}{{\left(x^2+1\right)}^2}}={\left[\frac{-1}{x^2+1}\right]}_0^b\\ =\left(-\frac{1}{b^2+1}\right)-\left(\frac{-1}{0^2+1}\right)\\ =-\frac{1}{b^2+1}+1\end{array}$

Determine the value of $\underset{b\to -\infty }{\lim }{\displaystyle {\int }_b^0\frac{2x}{{\left(x^2+1\right)}^2}dx}+\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^b\frac{2x}{{\left(x^2+1\right)}^2}dx}$ .

  $\begin{array}{c}\underset{b\to -\infty }{\lim }{\displaystyle {\int }_b^0\frac{2x}{{\left(x^2+1\right)}^2}dx}+\underset{b\to \infty }{\lim }{\displaystyle {\int }_0^b\frac{2x}{{\left(x^2+1\right)}^2}dx}=\underset{b\to -\infty }{\lim }\left[-1+\frac{1}{b^2+1}\right]+\underset{b\to \infty }{\lim }\left[-\frac{1}{b^2+1}+1\right]\\ =\left[-1+\frac{1}{{\infty }^2+1}\right]+\left[-\frac{1}{{\infty }^2+1}+1\right]\\ =\left(-1+0\right)+\left(-0+1\right)\\ =0\end{array}$

Thus, the value of the given integral is 0.