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.

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}
