TITLE: Double Integral Math Olympiad Problem
QUESTION [13 upvotes]: I was taking a Math Olympiad test and one of the questions was to calculate the following double integral:
$$\int_0^\infty\int_0^\infty\frac{\log|(x+y)(1-xy)|}{(1+x^2)(1+y^2)}\ \mathrm{d}x\ \mathrm{d}y$$
Here, as usual, $\log a$ and $|a|$ are the natural logarithm and absolute value of $a$ respectively.
I'm guessing that you're not supposed to solve it analytically, but rather find some symmetry argument or clever simplification that would make it straightforward. Since I don't even know where to start, any help is welcome.
In case you want to know, this was taken from the 2016 Rio de Janeiro State Math Olympiad, known in Portuguese as OMERJ.

REPLY [5 votes]: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\color{#f00}{\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\bracks{x + y}\verts{1 - xy}} \over \pars{1 + x^{2}}\pars{1 +y^{2}}}
\,\dd x\,\dd y}\ =\ \overbrace{\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{x} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y}^{\ds{=\ 0}}
\\[5mm] +&\
\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{1 + y/x} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y +
\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\verts{1 - xy}} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y
\end{align}
In the second integral, in the RHS, lets $\ds{x\ \mapsto\ 1/x}$:
\begin{align}
&\color{#f00}{\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\bracks{x + y}\verts{1 - xy}} \over \pars{1 + x^{2}}\pars{1 +y^{2}}}
\,\dd x\,\dd y}
\\[5mm] = &\
\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{1 + y/x} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y +
\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\verts{1 - y/x}} \over \pars{1 + x^{2}}\pars{1 + y^{2}}}\,\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{\infty}{1 \over x^{2} + 1}\int_{0}^{\infty}
{\ln\pars{\verts{1 - y^{2}/x^{2}}} \over 1 + y^{2}}\,\dd y\,\dd x
\,\,\,\stackrel{y/x\ \mapsto\ y}{=}\,\,\,
\int_{0}^{\infty}{x \over x^{2} + 1}\int_{0}^{\infty}
{\ln\pars{\verts{1 - y^{2}}} \over 1 + x^{2}y^{2}}\,\dd y\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}\ln\pars{\verts{1 - y^{2}}}\ \overbrace{\int_{0}^{\infty}
{x \over \pars{x^{2} + 1}\pars{y^{2}x^{2} + 1}}\,\dd x}
^{\ds{\ln\pars{y} \over y^{2} - 1}}\ \,\dd y\ =\
\int_{0}^{\infty}{\ln\pars{\verts{1 - y^{2}}}\ln\pars{y} \over y^{2} - 1}\,\dd y
\end{align}
Now, we split the integral along $\ds{\pars{0,1}}$ and $\ds{\pars{1,\infty}}$.
Later on, we make the substitution $\ds{y \mapsto 1/y}$ in the second integral
$\pars{~\mbox{along}\ \pars{1,\infty}~}$:
\begin{align}
&\color{#f00}{\int_{0}^{\infty}\int_{0}^{\infty}
{\ln\pars{\bracks{x + y}\verts{1 - xy}} \over \pars{1 + x^{2}}\pars{1 +y^{2}}}
\,\dd x\,\dd y}
\\[5mm] = &\
\int_{0}^{1}{\ln\pars{1 - y^{2}}\ln\pars{y} \over y^{2} - 1}\,\dd y +
\int_{0}^{1}{\bracks{\ln\pars{1 - y^{2}} - 2\ln\pars{y}}\ln\pars{y} \over
y^{2} - 1}\,\dd y
\\[5mm] = &\
2\int_{0}^{1}
{\ln\pars{1 - y^{2}}\ln\pars{y} - \ln^{2}\pars{y} \over y^{2} - 1}\,\dd y
\,\,\,\stackrel{y^{2}\ \mapsto\ y}{=}\,\,\,
2\int_{0}^{1}
{\ln^{2}\pars{y}/4 - \ln\pars{1 - y}\ln\pars{y}/2 \over 1 - y}\,{1 \over 2}\,y^{-1/2}\dd y
\\[5mm] = &\
{1 \over 4}\ \underbrace{%
\int_{0}^{1}{y^{-1/2}\ln^{2}\pars{y} \over 1 - y}\,\dd y}
_{\ds{=\ 14\zeta\pars{3}}}\ -\
{1 \over 2}\ \underbrace{\int_{0}^{1}{y^{-1/2}\ln\pars{y}\ln\pars{1 - y} \over 1 - y}\,\dd y}_{\ds{=\ -\pi^{2}\ln\pars{2} + 7\zeta\pars{3}}}\ =\
\color{#f00}{{1 \over 2}\,\pi^{2}\ln\pars{2}}
\end{align}

The last two integrals can be straightforward reduced to derivatives of the Beta Function and evaluated at suitable limits.