Evaluating $ \int^{\infty}_0\frac{\ln x}{x^2+\pi^2} ~dx$ 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
 A: An alternative, $a > 0$:
$$\displaystyle\begin{align*} \int_0^{\infty} \frac{\ln x}{x^2 + a^2}dx & \overset{x\mapsto a \tan x}= \frac{1}{a}\int_0^{\frac{\pi}{2}} \ln a \ dx + \frac{1}{a}\underbrace{\int_0^{\frac{\pi}{2}}\ln \tan x \ dx}_{=0} \\ & = \frac{\ln a}{a}\cdot \frac{\pi}{2} \end{align*}$$
The latter integral can be shown to be equal to 0 by splitting it up like so:
$$\int_0^{\frac{\pi}{2}} \ln \tan x\ dx = \int_0^{\frac{\pi}{2}} \ln \sin x \ dx - \int_0^{\frac{\pi}{2}} \ln \cos x \ dx$$
and $$\int_0^{\frac{\pi}{2}} \ln \sin x \ dx \overset{x\mapsto \frac{\pi}{2} - x}= \int_0^{\frac{\pi}{2}} \ln \cos x \ dx$$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + \pi^{2}}\,\dd x:\ {\large ?}}$

\begin{align}
\color{#00f}{\large\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + \pi^{2}}\,\dd x}&=
{1 \over \pi}\int_{0}^{\infty}{\ln\pars{\pi x} \over x^{2} + 1}\,\dd x
={1 \over \pi}\,\ln\pars{\pi}\
\overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{=\ {\pi \over 2}}}\
+\ {1 \over \pi}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x
\\[3mm]&=\half\,\ln\pars{\pi}
+ {1 \over \pi}
\overbrace{\quad\bracks{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\,\dd x\
+\
\underbrace{\quad\int_{1}^{0}{\ln\pars{1/x} \over \pars{1/x}^{2} + 1}\,\pars{-\,{\dd x \over x^{2}}}\quad}_{\ds{=-\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\,\dd x}}}\quad}^{\ds{=\ 0}}
\\[3mm]&=\color{#00f}{\large\half\,\ln\pars{\pi}}
\end{align}

A: 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?
A: Method I
Split the integration region as follows:
$$\int_0^{\pi} dx \frac{\log{x}}{x^2+\pi^2} + \int_{\pi}^{\infty} dx \frac{\log{x}}{x^2+\pi^2}$$
In the second integral, sub $x=\pi^2/y$ and get that the integral is
$$\int_0^{\pi} dx \frac{\log{x}}{x^2+\pi^2} + \pi^2 \int_0^{\pi} \frac{dy}{y^2} \frac{\log{\pi^2/y}}{\pi^2+\pi^4/y^2}$$
or after some cancellation
$$\int_0^{\pi} dx \frac{\log{\pi^2}}{x^2+\pi^2}  = \frac12 \log{\pi}$$
Method II
As intimated in the comments, we may use the residue theorem.  Use a keyhole contour of radius $R$ $C$ about the positive real axis and consider the integral
$$\oint_C dz \frac{\log^2{z}}{z^2+\pi^2}$$
The integral over the circular arc vanishes as $R \to\infty$ and by the residue theorem, we are left with
$$\int_0^{\infty} dx \frac{\log^2{x}}{x^2+\pi^2} - \int_0^{\infty} dx \frac{(\log{x}+i 2 \pi)^2}{x^2+\pi^2} = i 2 \pi \left [\frac{(\log{\pi}+i \pi/2)^2}{i 2 \pi}- \frac{(\log{\pi}+i 3\pi/2)^2}{i 2 \pi}\right ]$$
or
$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2+\pi^2}+ 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2+\pi^2} = -i 2 \pi \log{\pi} + 2 \pi^2$$
The respective terms on the right on each side cancel, and we are left with
$$\int_0^{\infty} dx \frac{\log{x}}{x^2+\pi^2} = \frac12 \log{\pi}$$
