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.

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.
