Integrate \(\int_{0}^{\infty} \mathrm{ln}^{2}(x)\,(1+x^{2})^{-1} \mathrm{d} x \)

The following integral was a question on Mathematics Stack Exchange

\begin{equation}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{2}(x)}{1+x^{2}} \mathrm{d} x
\label{eq:1608131}
\tag{1}
\end{equation}

Let us evaluate it via the Mellin transform
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:1608132}
\tag{2}
\end{equation}

where
\begin{equation}
f(x) = \frac{1}{1+x^{2}}
\label{eq:1608133}
\tag{3}
\end{equation}

Applying the Mellin transform, yields
\begin{equation}
\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} \frac{x^{s-1}}{1+x^{2}} \mathrm{d} x = \frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right)
\label{eq:1608134}
\tag{4}
\end{equation}

And thus
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{2}(x)}{1+x^{2}} \mathrm{d} x & = \frac{\mathrm{d}^{2}}{\mathrm{d}s^{2}} \int\limits_{0}^{\infty} \frac{x^{s-1}}{1+x^{2}} \mathrm{d} x |_{s=1} \\
& = \frac{\mathrm{d}^{2}}{\mathrm{d}s^{2}} \frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right) |_{s=1} \\
& = \frac{\pi^{3}}{16}[\cos(\pi s) + 3]\csc^{3}\left(\frac{\pi}{2}s\right) |_{s=1} \\
& = \frac{\pi^{3}}{8}
\label{eq:1608135}
\tag{5}
\end{align}

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