Question

Solve ${\int }_0^{\infty }\log (x+\frac{1}{x})\frac{dx}{1+x^2}$

Answer 1

${\int }_0^{\infty }\log (x+\frac{1}{x})\frac{dx}{1+x^2}$

Put $x=\tan \theta$

$\therefore dx={\sec }^2\theta d\theta$

When $x=0\Rightarrow \theta =0$

When $x=\infty \Rightarrow \theta =\frac{\pi }{2}$

$\therefore I={\int }_0^{\pi /2}\log (\tan \theta +\cot \theta )d\theta$

$={\int }_0^{\pi /2}\log \frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }d\theta$

$=-{\int }_0^{\pi /2}\log \sin \theta \cos \theta d\theta$

We know that

${\int }_0^{\pi /2}\log \sin \theta =-\frac{\pi }{2}\log 2$

${\int }_0^{\pi /2}\log \cos \theta =-\frac{\pi }{2}\log 2$

$=-[{\int }_0^{\pi /2}\log \sin \theta d\theta +{\int }_0^{\pi /2}\log \cos \theta d\theta ]$

$=-[-\frac{\pi }{2}\log 2-\frac{\pi }{2}\log 2]$

$I=\pi \log 2.$