Question

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

Answer 1

Substitute $x=\tan \theta \therefore dx={\sec }^2\theta d\theta .$
The limits are adjusted as 0 to $\pi /2$
$\therefore I={\int }_0^{\pi /2}\frac{\log {\sec }^2\theta }{{\sec }^2\theta }{\sec }^2\theta d\theta ={\int }_0^{\pi /2}\log {\sec }^2\theta d\theta$
or $I=2{\int }_0^{\pi /2}\log \sec \theta d\theta =-2{\int }_0^{\pi /2}\log \cos \theta d\theta$
$=-2.\frac{\pi }{2}\log \frac{1}{2},$
$=-\pi \log \frac{1}{2}=\pi \log 2.$