Find $\int_0^\infty \frac{\ln ^2z} {1+z^2}{d}z$ How to find the value of the integral $$\int_{0}^{\infty} \frac{\ln^2z}{1+z^2}{d}z$$ without using contour integration - using usual special functions, e.g. zeta/gamma/beta/etc.
Thank you.
 A: Here's another way to go: 
$$\begin{eqnarray*}
\int_0^\infty dz\, \dfrac{\ln ^2z} {1+z^2} 
&=& \frac{d^2}{ds^2} \left. \int_0^\infty dz\, \dfrac{z^s} {1+z^2} \right|_{s=0} \\
&=& \frac{d^2}{ds^2} \left. \frac{\pi}{2} \sec\frac{\pi s}{2} \right|_{s=0} \\
&=& \frac{\pi^3}{8}.
\end{eqnarray*}$$
The integral $\int_0^\infty dz\, z^s/(1+z^2)$ can be handled with the beta function.
See some of the answers here, for example.
A: With $ \int_0^\infty \frac{2y\ln y }{(y^2+z^2)(y^2+1)}{dy}= \frac{\ln^2 z}{z^2-1}$
\begin{align}
&\int_0^\infty\frac{\ln^2 z}{1+z^2}dz\\
=&\ \int_0^\infty \frac1{1+z^2} \left(\int_0^\infty\frac{2(z^2-1)y\ln y}{(y^2+z^2)(y^2+1)}\ dy \right) dz\\
 =& \int_0^\infty\int_0^\infty
\frac{2y\ln y}{(y^2+z^2)(y^2-1)}-\frac{4y\ln y}{(1+z^2)(y^4-1)}\ dz \ dy\\
=& \ \pi\int_0^\infty \frac{\ln y}{y^2-1} {dy}- 2\pi\int_0^\infty \frac{y\ln y}{y^4-1}\overset{y^2\to y} {dy}\\
=&\ \frac\pi2 \int_0^\infty \frac{\ln y}{y^2-1}\ dy
 = \frac\pi4 \int_0^\infty \int_0^1 \frac{x}{1+(y^2-1)x^2}dx\>dy\\
= &\ \frac{\pi^2}4\int_0^1 \frac1{\sqrt{1-x^2}}dx=\frac{\pi^3}{8}
\end{align}
A: You can integrate by substitution with $z = e^u$. This yields
$$
\int_0^\infty\frac{(\ln z)^2}{1+z^2} dz = \int_{-\infty}^{+\infty}\frac{u^2e^u}{1+e^{2u}}du = 2\int_0^\infty \frac{u^2e^{-u}}{1+e^{-2u}}du
$$
Now, expand the series:
$$
\frac{u^2e^{-u}}{1+e^{-2u}} = \sum_{n=0}^\infty (-1)^n u^2e^{-u(2n+1)}
$$
Interverting the $\int$ and $\sum$ (use Fubini's theorem), we have
$$
\int_0^\infty\frac{(\ln z)^2}{1+z^2} dz = 2\Gamma(3)\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3} = 4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}
$$
The last sum can be computed using Fourier series, yielding
$$\int_0^\infty\frac{(\ln z)^2}{1+z^2} dz = \frac{\pi^3}{8}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\color{#f00}{\int_{0}^{\infty}{\ln^{2}\pars{z} \over 1 + z^{2}}\,\dd z} =
\lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty}{z^{\mu} \over 1 + z^{2}}\,\dd z
\end{align}
With $\ds{z \equiv \pars{{1 \over t} - 1}^{1/2}}$:
\begin{align}
&\color{#f00}{\int_{0}^{\infty}{\ln^{2}\pars{z} \over 1 + z^{2}}\,\dd z} =
\half\,\lim_{\mu \to 0}\partiald[2]{}{\mu}
\int_{0}^{1}t^{-\mu/2 - 1/2}\pars{1 - t}^{\mu/2 - 1/2}\,\dd t
\\[3mm] = &\
\half\,\lim_{\mu \to 0}\partiald[2]{}{\mu}\bracks{%
\Gamma\pars{-\mu/2 + 1/2}\Gamma\pars{\mu/2 + 1/2} \over \Gamma\pars{1}} =
\half\,\ \overbrace{%
\lim_{\mu \to 0}\partiald[2]{\bracks{\pi\csc\pars{\pi\bracks{\mu + 1}/2}}}{\mu}}
^{\ds{=\ {\pi^{3} \over 4}}} =
\color{#f00}{{\pi^{3} \over 8}}
\end{align}
A: \begin{align}J&=\int_0^\infty \frac{\ln^2 x}{1+x^2}dx\\
K&=\int_0^\infty\int_0^\infty \frac{\ln^2(xy)}{(1+x^2)(1+y^2)}dxdy\\
&\overset{u(x)=xy}=\int_0^\infty\int_0^\infty \frac{y\ln^2 u}{(u^2+y^2)(1+y^2)}dudy\\
&=\frac{1}{2}\int_0^\infty \ln^2 u\left[\frac{\ln\left(\frac{1+y^2}{u^2+y^2}\right)}{u^2-1}\right]_{y=0}^{y=\infty}du\\
&=\int_0^\infty \frac{\ln^3 u}{u^2-1}du=\int_0^1 \frac{\ln^3 u}{u^2-1}du+\underbrace{\int_1^\infty \frac{\ln^3 u}{u^2-1}du}_{z=\frac{1}{u}}=2\int_0^1 \frac{\ln^3 u}{u^2-1}du\\
&=2\int_0^1 \frac{\ln^3 u}{u-1}du-2\underbrace{\int_0^1 \frac{u\ln^3 u}{u^2-1}du}_{z=u^2}=2\int_0^1 \frac{\ln^3 u}{u-1}du-\frac{1}{8}\int_0^1 \frac{\ln^3 z}{z-1}dz=\frac{15}{8}\int_0^1 \frac{\ln^3 z}{z-1}dz\\
&=\frac{15}{8}\times 6\zeta(4)=\boxed{\frac{\pi^4}{8}}
\end{align}
On the other hand,
\begin{align}K&=\int_0^\infty\int_0^\infty\frac{\ln^2 x}{(1+x^2)(1+y^2)}dxdy+\int_0^\infty\int_0^\infty\frac{\ln^2 y}{(1+x^2)(1+y^2)}dxdy+\\&
2\underbrace{\int_0^\infty\int_0^\infty\frac{\ln x\ln y}{(1+x^2)(1+y^2)}dxdy}_{=0}\\
&=2J\underbrace{\int_0^\infty \frac{1}{1+x^2}dx}_{=\frac{\pi}{2}}=\boxed{\pi J}
\end{align}
Therefore,
\begin{align}\boxed{J=\frac{\pi^3}{8}}\end{align}
NB:
I assume,
\begin{align}\int_0^1 \frac{\log^3 x}{x-1}dx=6\zeta(4)=\frac{\pi^4}{15}\end{align}
