Chapter 9.4, Problem 39E

To determine

Tofind: the value of integral and state that it is diverges or not.

Answer to Problem 39E

The integral value is $2{\pi }^2$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle {\int }_0^{\infty }\frac{16{\tan }^{-1}vdv}{1+v^2}}$ .

Calculation: to find the value of integral, first integrate then simplify,

  $\begin{array}{l}{\displaystyle {\int }_0^{\infty }\frac{16{\tan }^{-1}vdv}{1+v^2}}={\left[16\frac{{\left({\tan }^{-1}v\right)}^2}{2}\right]}_0^{\infty }{\text{ [first tan}}^{-1}v=u\Rightarrow \frac{1}{1+v^2}dv=du\text{ and simplify and }{\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}}]\\ {\displaystyle {\int }_0^{\infty }\frac{16{\tan }^{-1}vdv}{1+v^2}}=8{\left({\tan }^{-1}\infty \right)}^2-8{\left({\tan }^{-1}0\right)}^2\\ {\displaystyle {\int }_0^{\infty }\frac{16{\tan }^{-1}vdv}{1+v^2}}=8\left(\frac{{\pi }^2}{4}\right)+16\left(0\right)\text{ [simplify]}\\ {\displaystyle {\int }_0^{\infty }\frac{16{\tan }^{-1}vdv}{1+v^2}}=2{\pi }^2\end{array}$

Thus, the integral value is $2{\pi }^2$ .