- #1

Tony1

- 17

- 0

I am very surprise it has a very simple closed form. I was expecting something else, probably involving a few other constants, like $\ln a$, $\pi$ ...

I don't know how to go about to show that it is $\pi\over 4$. I did try a substitution of $u=\ln(2\cos x)$, which make the integral looked more harder than it is...

$$\int_{0}^{\infty}{\mathrm du\over \tan x}\cdot{u\over x^2+u^2}$$

Can anyone help. Thank you!