$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$ Evaluate Integral Anyone remember the method for this? I think this should been done on the site
$$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$$
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$

$\ds{\large\tt\mbox{It's almost done without any integral evaluation !!!.}}$

\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + a^{2}}\,\dd x}
={1\over \verts{a}}\int_{0}^{\infty}{\ln\pars{\verts{a}x} \over x^{2} + 1}\,\dd x
\\[3mm]&={\ln\pars{\verts{a}}\over \verts{a}}\
\overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}}^{\ds{=\ {\pi \over 2}}}\
+\
{1\over \verts{a}}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x
\\[3mm]&={\pi\ln\pars{\verts{a}} \over 2\verts{a}}\quad
+\quad{1 \over \verts{a}}\
\underbrace{\quad\bracks{\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\,\dd x
+\
\overbrace{%
\int_{1}^{0}{\ln\pars{1/x} \over 1/x^{2} + 1}\,\pars{-\,{\dd x \over x^{2}}}}
^{\ds{=-\int_{0}^{1}{\ln\pars{x} \over x^{2} + 1}\,\dd x}}}\quad}
_{\ds{\color{#c00000}{\LARGE =\ 0}}}
\\[3mm]&=\color{#00f}{\large{\pi\ln\pars{\verts{a}} \over 2\verts{a}}}
\end{align}
A: Let the considered integral be $I$. Use the substitution $x=a\tan\theta \,d\theta$ to obtain:
$$I=\frac{1}{a}\int_0^{\pi/2} \ln(a\tan\theta)\,d\theta=\frac{1}{a}\int_0^{\pi/2} \ln a\,d\theta+\frac{1}{a}\int_0^{\pi/2}\ln(\tan\theta)\,d\theta$$
It is easy to see that
$$\int_0^{\pi/2}\ln(\tan\theta)\,d\theta=0$$
(The above can be shown by using the substitution $\theta=\pi/2-t$).
Hence,
$$I=\frac{\pi}{2a}\ln a$$
A: Related problems: (I), (II), (III), $(4)$. Let us consider the integeral

$$ \int_{0}^{\infty}\frac{x^{s-1}}{x^2+a^2}\text{d}x,  $$

which is nothing but the Mellin transform of the function $ \frac{1}{x^2+a^2}$ and it is given by 
$$ F(s)=\int_{0}^{\infty}\frac{x^{s-1}}{x^2+a^2}\text{d}x = \frac{1}{2}\frac{\pi a^{s-2}}{\sin(\pi s/2)} $$

$$ \implies F'(s)=\int_{0}^{\infty}\frac{x^{s-1}\ln(x)}{x^2+a^2}\text{d}x = \frac{d}{ds}\frac{1}{2}\frac{\pi a^{s-2}}{\sin(\pi s/2)}. $$

Taking the limit as $s \to 1$ the desired result follows

$$ \int_{0}^{\infty}\frac{\ln(x)}{x^2+a^2}\text{d}x = \frac{\pi \ln(a)}{2a}. $$

A: I think @kiwi ment
$$
\fbox {$I$} = \int_0^\infty \frac {\ln x}{x^2+a^2} dx = \left | u = \frac {a^2}x \Longrightarrow\left\{\begin{array}{c}
\ln x = 2 \ln a - \ln u \\
dx = -\frac {a^2du}{u^2}
\end{array}\right\} \right | = -\int_\infty^0 \frac{2\ln a - \ln u}{\frac {a^4}{u^2}+a^2}\frac {a^2 du}{u^2} = \\
2\int_0^\infty \frac{\ln a}{u^2+a^2}du-\int_0^\infty \frac{\ln u}{u^2+a^2}du = \fbox{$2\int_0^\infty \frac{\ln a}{u^2+a^2}du - I$}
$$
From last part it's clear that
$$
I = \int_0^\infty \frac{\ln a}{u^2+a^2}du
$$
This integral can be easily found
$$
I = \ln a\int_0^\infty \frac {du}{u^2+a^2} = \frac {\ln a}a \ \left.\mbox{atan}\ \frac ua \right|_0^\infty = \frac {\pi \ln a}{2a}
$$
A: Or simply let $x=ay$
$$\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{dx}=\frac{\ln a}{a}\int_0^{\infty}\frac{1}{y^2+1}\mathrm{dy}+\frac{1}{a}\int_0^{\infty}\frac{\ln y}{y^2+1}\mathrm{dy}=\frac{\pi\ln a}{2a}$$
because letting $y=1/z$ in $\int_0^{1}\frac{\ln y}{y^2+1}\mathrm{dy}=-C$ (Catalan's constant) , we get $\int_1^{\infty}\frac{\ln z}{z^2+1}\mathrm{dz}=C$. Now, add up the $2$ integrals and get that $\int_0^{\infty}\frac{\ln y}{y^2+1}\mathrm{dy}=0$.
Chris.
