Evaluate the integral $\int_0^{\infty} \left(\frac{\log x \arctan x}{x}\right)^2 \ dx$

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?

\,{\rm d}x

W|A tells the integral evaluates $0$ but this is not true.
Then how do I exactly compute it?

Solutions Collecting From Web of "Evaluate the integral $\int_0^{\infty} \left(\frac{\log x \arctan x}{x}\right)^2 \ dx$"

Related problems: (I), (II), (III). Denoting our integral by $J$ and recalling the mellin transform
$$ F(s)=\int_{0}^{\infty}x^{s-1} f(x)\,dx \implies F”(s)=\int_{0}^{\infty}x^{s-1} \ln(x)^2\,f(x)\,dx.$$

Taking $f(x)=\arctan(x)^2$, then the mellin transform of $f(x)$ is

$$ \frac{1}{2}\,{\frac {\pi \, \left( \gamma+2\,\ln\left( 2 \right) +\psi
\left( \frac{1}{2}+\frac{s}{2} \right)\right) }{s\sin \left( \frac{\pi \,s}{2}
\right)}}-\frac{1}{2}\,{\frac {{\pi }^{2}}{s\cos\left( \frac{\pi \,s}{2}
\right) }},$$

where $\psi(x)=\frac{d}{dx}\ln \Gamma(x)$ is the digamma function. Thus $J$ can be calculated directly as

$$ J= \lim_{s\to -1} F”(s) = \frac{1}{12}\,\pi \, \left( 3\,{\pi }^{2}\ln \left( 2 \right) -{\pi }^{2}+24
\,\ln \left( 2 \right) -3\,\zeta \left( 3 \right) \right)\sim 6.200200824 .$$

The function is positive and continuous (except perhaps at the origin) with isolated zeros. The integral is therefore some nonzero, positive quantity.

That Wolfram | Alpha indicates otherwise is quite possibly an error. If you change the upper limit of the integral from $\infty$ to, say $5$, Wolfram | Alpha gives you a positive quantity, further indicating that there may be an error in the way the site is evaluating this integral.

\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{\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{\ket}[1]{\left\vert #1\right\rangle}
\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{\verts}[1]{\left\vert\, #1 \,\right\vert}
$\ds{\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x:\
{\large ?}}$

\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}

&\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd 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
{\ln^{2}\pars{x} \over \pars{x^{2}y^{2} + 1}\pars{x^{2}z^{2} + 1}}\,\dd x
\,\dd y\,\dd z
{y^{-2}z^{-2} \over z^{-2} – y^{-2}}
{\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}

However, with $\ds{a > 0}$:
&\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
\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}
$\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x = {\pi^{3} \over 8}}$
is a well known result.

Replacing $\pars{2}$ in $\pars{1}$:
&\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
Both integrals can be trivially evaluated:
\int_{0}^{1}\int_{0}^{1}{\dd y\,\dd z \over y + z}
{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}

\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