TITLE: Evaluating $ \int^{\infty}_0\frac{\ln x}{x^2+\pi^2} ~dx$
QUESTION [5 upvotes]: The question is

Evaluate 
  $$ \int \limits^{\infty}_0\dfrac{\ln x}{x^2+\pi^2} dx $$

I have no idea what to do.Tried integration by parts, didn't work.
Help would be appreciated.
Thanks

REPLY [3 votes]: As I've hinted, setting $x=\pi\tan y $
We have $$ \int \limits^{\infty}_0\dfrac{\ln x}{x^2+\pi^2} dx =\frac1\pi\int_0^{\frac\pi2}(\ln\pi\tan y)dy $$
Using $$\int_a^b f(y)dy=\int_a^b f(a+b-y)dy$$
$$\pi I=\int_0^{\frac\pi2}(\ln\pi\tan y)dy=\int_0^{\frac\pi2}\ln\pi\tan\left(\frac\pi2+0- y\right)dy=\int_0^{\frac\pi2}(\ln\pi\cot y)dy $$
$$\implies \pi(I+I)=\int_0^{\frac\pi2}(\ln\pi\tan y)dy+\int_0^{\frac\pi2}(\ln\pi\cot y)dy$$
$$=\int_0^{\frac\pi2}(\ln\pi\tan y+\ln\pi\cot y)dy$$
$$=\int_0^{\frac\pi2}[\ln(\pi\tan y\cdot\pi\cot y)]dy$$
$$2\pi\cdot I=\ln \pi^2\int_0^{\frac\pi2}dy $$
Can you take it home from here?