TITLE: Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$.
QUESTION [10 upvotes]: I have managed to solve it in one way, but I became very interested in this failed attempt.
$$
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x 
 = \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x
- \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x
$$
We only have to show that those two on the right are equal. And numerical evaluations seem to suggest that they both are in fact $-\frac{\pi}{4}$ but I don't know how to break these down.

I am currently really interested in proving this
  $$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$

Anyway, here's my trivial solution using $u = \frac1x$:
$$
\begin{align}
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x & = \int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = \int_\infty^1 \frac{\frac{1}{u^2} \ln(\frac1u)}{(1+\frac{1}{u^2})^3} \frac{-1}{u^2} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = -\int_1^\infty \frac{\ln(u)}{u(u+\frac{1}{u})^3} {\rm d}u 
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = - \int_1^\infty \frac{u^2 \ln(u)}{(1+u^2)^3} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = 0
\end{align}
$$
I'm sure there are many more interesting methods for cracking this integral, since it's so closely related to the popular $\int_0^\infty \frac{\ln(x)}{1+x^2} {\rm d}x = 0$. Please share them if you do come up with any.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\Re\pars{\mu} > - 1}$ and
  $\ds{\Re\pars{\nu} > 0}$:

\begin{align}
I_{\mu\nu} & \equiv
\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{\mu}\ln\pars{x} \over \pars{1 + x^{2}}^{\nu}}\,\dd x}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 4}\int_{0}^{\infty}
{x^{\mu/2 - 1/2}\ln\pars{x} \over \pars{1 + x}^{\nu}}\,\dd x
\\[5mm] & =
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{0}^{\infty}
{x^{\alpha + \mu/2 - 1/2} \over \pars{1 + x}^{\nu}}\,\dd x
\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & \stackrel{x + 1\ \mapsto\ x}{=}\,\,\,
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{1}^{\infty}
{\pars{x - 1}^{\alpha + \mu/2 - 1/2} \over x^{\nu}}\,\dd x
\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{1}^{0}
{\pars{1/x - 1}^{\alpha + \mu/2 - 1/2} \over \pars{1/x}^{\nu}}\,
\pars{-\,{\dd x \over x^{2}}}
\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & =
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{0}^{1}
x^{\nu - \alpha - \mu/2 - 3/2}\pars{1 - x}^{\alpha + \mu/2 - 1/2}\,
\dd x\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & =
{1 \over 4}\,\partiald{}{\alpha}\bracks{%
\Gamma\pars{\nu - \alpha - \mu/2 - 1/2}\Gamma\pars{\alpha + \mu/2 + 1/2} \over \Gamma\pars{\nu}}_{\ \alpha\ =\ 0}
\\[5mm] & =
\bbx{{\Gamma\pars{\mu/2 + 1/2}\Gamma\pars{\nu - \mu/2 - 1/2} \over
4\Gamma\pars{\nu}}
\bracks{H_{\mu/2 - 1/2} - H_{\nu - \mu/2 - 3/2}}}
\end{align}

$\ds{H_{z}}$ is a Harmonic Number.


$$
\begin{array}{|c|c|}\hline
\ds{\mu \setminus \nu} & \ds{I_{\mu\nu}}
\\ \hline
\ds{0 \setminus 2} & \ds{-\,{\pi \over 4}}
\\ \hline
\ds{0 \setminus 3} & \ds{-\,{\pi \over 4}}
\\ \hline
\end{array}
$$


Note that $\ds{\bbx{I_{\mu,\mu + 1} = 0}}$ because, in such a case,
  $\ds{H_{\mu/2 - 1/2} = H_{\nu - \mu/2 - 3/2}}$.