TITLE: Evaluate the integral $\int_0^{\infty} \left(\frac{\log x \arctan x}{x}\right)^2 \ dx$
QUESTION [21 upvotes]: Some rumours point out that the integral you see might be evaluated in a 
straightforward way.
But rumours are sometimes just rumours. Could you confirm/refute it?
$$
\int_0^{\infty}\left[\frac{\log\left(x\right)\arctan\left(x\right)}{x}\right]^{2}
\,{\rm d}x
$$
EDIT
W|A tells the integral evaluates $0$ but this is not true.
Then how do I exactly compute it?

REPLY [10 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x:\
     {\large ?}}$

$$
\mbox{With the identity}\quad\int_{0}^{1}{\dd y \over x^{2}y^{2} + 1}
={\arctan\pars{x} \over x}\quad\mbox{we'll have}
$$

\begin{align}
&\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x
=\int_{0}^{\infty}\ln^{2}\pars{x}
\int_{0}^{1}{\dd y \over x^{2}y^{2} + 1}\int_{0}^{1}{\dd z \over x^{2}z^{2} + 1}\,\dd x
\\[3mm]&=\int_{0}^{1}\int_{0}^{1}
\int_{0}^{\infty}
{\ln^{2}\pars{x} \over  \pars{x^{2}y^{2} + 1}\pars{x^{2}z^{2} + 1}}\,\dd x
\,\dd y\,\dd z
\\[3mm]&=\int_{0}^{1}\int_{0}^{1}\braces{
{y^{-2}z^{-2} \over z^{-2} - y^{-2}}
\bracks{\int_{0}^{\infty}
{\ln^{2}\pars{x} \over  x^{2} + y^{-2}}\,\dd x
-\int_{0}^{\infty}{\ln^{2}\pars{x} \over  x^{2} + z^{-2}}\,\dd x}}\,\dd y\,\dd z\tag{1}
\end{align}

However, with $\ds{a > 0}$:
  \begin{align}
&\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + a^{-2}}\,\dd x
=a\int_{0}^{\infty}{\ln^{2}\pars{x/a} \over x^{2} + 1}\,\dd x
=a\int_{0}^{\infty}{\bracks{\ln\pars{x} - \ln\pars{a}}^{2} \over x^{2} + 1}\,\dd x
\\[3mm]&=a\bracks{\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x
-2\ln\pars{a}\
\overbrace{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x}^{\ds{=\ 0}}\
+\ \ln^{2}\pars{a}\int_{0}^{\infty}{\dd x \over x^{2} + 1}}
\\[3mm]&={\pi^{3} \over 8}\,a + {\pi \over 2}\,a\ln^{2}\pars{a}\tag{2}
\end{align}
  since
  $\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x = {\pi^{3} \over 8}}$
  is a well known result.

Replacing $\pars{2}$ in $\pars{1}$:
\begin{align}
&\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x
\\[3mm]&=\int_{0}^{1}\int_{0}^{1}\braces{{1 \over y^{2} - z^{2}}
\bracks{{\pi^{3} \over 8}\,\pars{y - z} + {\pi \over 2}\,y\ln^{2}\pars{y}
-{\pi \over 2}\,z\ln^{2}\pars{z}}}\,\dd y\,\dd z
\\[3mm]&={\pi^{3} \over 8}\int_{0}^{1}\int_{0}^{1}{\dd y\,\dd z \over y + z}
+{\pi \over 2}\int_{0}^{1}\int_{0}^{1}
{y\ln^{2}\pars{y} - z\ln^{2}\pars{z} \over y^{2} - z^{2}}\,\dd y\,\dd z
\end{align}
Both integrals can be trivially evaluated:
\begin{align}
\int_{0}^{1}\int_{0}^{1}{\dd y\,\dd z \over y + z}
&=2\ln\pars{2}
\\[3mm]\int_{0}^{1}\int_{0}^{1}
{z\ln^{2}\pars{z} - y\ln^{2}\pars{y} \over y^{2} - z^{2}}\,\dd y\,\dd z&=
4\ln\pars{2} - {1 \over 6}\,\pi^{2} - \half\,\zeta\pars{3}
\end{align}

Then,
  \begin{align}&\color{#66f}{\large
\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x}
={\pi^{3} \over 8}\,\bracks{2\ln\pars{2}}
+{\pi \over 2}\bracks{4\ln\pars{2} - {1 \over 6}\,\pi^{2} - \half\,\zeta\pars{3}}
\\[3mm]&=\color{#66f}{\large{1 \over 4}\bracks{\ln\pars{2} - {1 \over 3}}\pi^{3}
+ \bracks{2\ln\pars{2} - {1 \over 4}\,\zeta\pars{3}}\pi}
\approx 6.200200822
\end{align}