Question

$\underset{0}{\overset{\infty }{\int }}\log \left(x+\frac{1}{x}\right)\frac{1}{1+x^2}dx=$

(a) π ln 2
(b) −π ln 2
(c) 0
(d) $-\frac{\mathrm{\pi }}{2}\ln 2$

Answer 1

(a) π ln 2

${\int }_0^{\infty }\log \left(x+\frac{1}{x}\right)\frac{1}{1+x^2}dx$
Substitute x = tan θ
⇒ dx = sec² θ dθ.
when,
x = 0 ⇒ θ = 0
$$\begin{gathered} x=\infty \Rightarrow \theta =\frac{\mathrm{\pi }}{2} \\ {\int }_0^{\frac{\mathrm{\pi }}{2}}\left(\tan \theta +\frac{1}{\tan \theta }\right)\frac{1}{1+{\tan }^2\theta }\times {\sec }^2\theta d\theta \\ {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{{\tan }^2\theta +1}{\tan \theta }\right)\frac{1}{1+{\tan }^2\theta }\times {\sec }^2\theta d\theta \\ \Rightarrow {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{{\sec }^2\theta }{\tan \theta }\right)\frac{1}{{\sec }^2\theta }\times {\sec }^2\theta d\theta \left[\because 1+{\tan }^2\theta ={\sec }^2\theta \right] \\ \Rightarrow {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{{\sec }^2\theta }{\tan \theta }\right)d\theta \\ \Rightarrow {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{1}{\sin \theta .\cos \theta }\right)d\theta \\ \Rightarrow -{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin \theta .\cos \theta \right)d\theta \\ \Rightarrow -{\int }_0^{\frac{\mathrm{\pi }}{2}}\left[\log \sin \theta +\log \cos \theta \right]d\theta \\ \Rightarrow -{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \theta d\theta -{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos \theta d\theta \end{gathered}$$
Let us consider,
$$\begin{gathered} {\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \theta d\theta =I.....\left(\mathrm{i}\right) \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin \left(\frac{\mathrm{\pi }}{2}-\theta \right)\right)d\theta \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos \theta d\theta .....\left(\mathrm{ii}\right) \\ \mathrm{Adding}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right) \\ 2\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \theta d\theta +{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos \theta d\theta \\ ={\int }_0^{\frac{\pi }{2}}\log \left(\sin \theta .\cos \theta \right)d\theta \\ ={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin 2\theta \right)d\theta -{\int }_0^{\frac{\mathrm{\pi }}{2}}\log 2d\theta \\ \mathrm{Let}\mathrm{us}\mathrm{consider}2\theta =t \\ 2d\theta =dt \\ 2\mathrm{I}=\frac{1}{2}{\int }_0^{\mathrm{\pi }}\log \left(\sin t\right)dt-\frac{\mathrm{\pi }}{2}\log 2 \\ 2\mathrm{I}=\frac{2}{2}{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin t\right)dt-\frac{\mathrm{\pi }}{2}\log 2\left[\because \sin \theta \mathrm{is}\mathrm{positive}\mathrm{in}\mathrm{both}{1}^{\mathrm{st}}\mathrm{and}{2}^{\mathrm{nd}}\mathrm{quadrants}\right] \\ 2\mathrm{I}=\mathrm{I}-\frac{\mathrm{\pi }}{2}\log 2 \\ 2\mathrm{I}-\mathrm{I}=-\frac{\mathrm{\pi }}{2}\log 2 \\ \mathrm{I}=-\frac{\mathrm{\pi }}{2}\log 2,\mathrm{where}\mathrm{I}={\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \theta d\theta \\ \mathrm{Now}, \\ -{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \left(\sin \theta \right)d\theta -{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \cos \theta d\theta \\ -2{\int }_0^{\frac{\mathrm{\pi }}{2}}\log \sin \theta d\theta =-2\times I \\ =-2\times -\frac{\mathrm{\pi }}{2}\log 2\left[\because \mathrm{where}\mathrm{I}=-\frac{\mathrm{\pi }}{2}\log 2\right] \\ =\mathrm{\pi }\log 2 \end{gathered}$$