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.

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
$$\int_0^\infty\frac{(\ln z)^2}{1+z^2} dz = \frac{\pi^3}{8}$$
