Chapter 9, Problem 48RE

To determine

To evaluate : The improper integral or state that it diverges.

Answer to Problem 48RE

The answer is $\pi$ .

Explanation of Solution

Given information :

The given improper integral is ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}$ .

Calculation :

Since, if $f\left(x\right)$ is continuous on $\left(-\infty ,\infty \right)$ . Then,

  ${\displaystyle {\int }_{-\infty }^{\infty }f\left(x\right)}dx={\displaystyle {\int }_{-\infty }^{c}f\left(x\right)}dx+{\displaystyle {\int }_c^{\infty }f\left(x\right)}dx$ .

Therefore,

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\frac{4}{16}{\displaystyle {\int }_{-\infty }^{\infty }\frac{dx}{{\left(\frac{x}{4}\right)}^2+1}}$

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\frac{1}{4}\left[{\displaystyle {\int }_{-\infty }^0\frac{dx}{{\left(\frac{x}{4}\right)}^2+1}}+{\displaystyle {\int }_0^{\infty }\frac{dx}{{\left(\frac{x}{4}\right)}^2+1}}\right]$

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\frac{1}{4}\left[\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^0\frac{dx}{{\left(\frac{x}{4}\right)}^2+1}}+\underset{a\to +\infty }{\lim }{\displaystyle {\int }_0^a\frac{dx}{{\left(\frac{x}{4}\right)}^2+1}}\right]$

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\underset{a\to -\infty }{\lim }{\left[{\tan }^{-1}\left(\frac{x}{4}\right)\right]}_a^0+\underset{a\to +\infty }{\lim }{\left[{\tan }^{-1}\left(\frac{x}{4}\right)\right]}_0^a$

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\underset{a\to -\infty }{\lim }\left[{\tan }^{-1}\left(\frac{0}{4}\right)-{\tan }^{-1}\left(\frac{a}{4}\right)\right]+\underset{a\to +\infty }{\lim }\left[{\tan }^{-1}\left(\frac{a}{4}\right)-{\tan }^{-1}\left(\frac{0}{4}\right)\right]$

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\left[{\tan }^{-1}\left(0\right)-{\tan }^{-1}\left(-\infty \right)\right]+\left[{\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(0\right)\right]$

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\left[0+\frac{\pi }{2}\right]+\left[\frac{\pi }{2}-0\right]$

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\frac{\pi }{2}+\frac{\pi }{2}$

  ${\displaystyle {\int }_{-\infty }^{\infty }\frac{4dx}{x^2+16}}=\pi$

Hence,

The answer is $\pi$ .