Integral $\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16\sqrt 2}$

Following Mhenni's suggestion, I will calculate $$I(\mu) = \int_0^{\infty} x^{\mu}\frac{1+x^2}{1+x^4} \,dx $$ and then take $I''(0)$. By the ubiquitous formula $$\int_0^{\infty}\frac{x^a}{1+x^b} \,dx =\frac{\pi}{b \sin(\pi(a+1)/b)}$$ we obtain $$I(\mu)=\frac{\pi}{4} \left[ \frac{1}{\sin(\pi(\mu+1)/4)} + \frac{1}{\sin(\pi(\mu+3)/4)} \right]$$ This gives (after some simple algebra)

$$I''(0) = \int_0^{\infty} \frac{1+x^2}{1+x^4} \log^2 x \,dx = \frac{\pi}{4}\left[\frac{3 \pi ^2}{8 \sqrt{2}}+\frac{3 \pi ^2}{8 \sqrt{2}} \right] = \frac{3 \pi ^3}{16 \sqrt{2}}$$ as was to be shown.

A related problem. Recalling the Mellin transform of a function $f$

$$ F(s)=\int_{0}^{\infty} x^{s-1}f(x)dx \implies F''(s)=\int_{0}^{\infty} \ln(x)^2x^{s-1}f(x)dx .$$

Now the whole problem boils down to finding the Mellin transform of $\frac{1+x^2}{1+x^4}$, differentiating twice, and then taking the limit as $s \to 1$.

Can you finish it?

Here is another way to calculate. It is much simpler. Clearly \begin{eqnarray*} I&=&2\int_0^1 \log^2 x\frac{1+x^2}{1+x^4}dx\\ &=&2\int_0^1\sum_{n=0}^\infty(1+x^2)(-1)^nx^{4n}\log^2xdx\\ &=&2\int_0^1\sum_{n=0}^\infty(-1)^n(x^{4n}+x^{4n+2})\log^2xdx\\ &=&2\sum_{n=0}^\infty\int_0^1(-1)^n(x^{4n}+x^{4n+2})\log^2xdx\\ &=&4\sum_{n=0}^\infty(-1)^n(\frac{1}{(4n+1)^3}+\frac{1}{(4n+3)^3})\\ &=&4\sum_{n=-\infty}^\infty(-1)^n\frac{1}{(4n+1)^3}\\ &=&\frac{3\pi^3}{16\sqrt2} \end{eqnarray*} Here we use the following theorem $$ \sum_{n=-\infty}^\infty (-1)^nf(n)=-\pi \sum_{k=1}^m\text{Re}(\frac{f(z)}{\sin\pi z},a_k) $$ where $a_1,a_2,\cdots,a_m$ are poles of $f(z)$. For $f(z)=\frac{1}{(4z+1)^3}$, $z=-\frac{1}{4}$ is the only pole and $$ \text{Re}(\frac{f(z)}{\sin\pi z},-\frac{1}{4})=-\frac{3\pi^2}{64\sqrt2}. $$ Thus we have the result.

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I\equiv\int_{0}^{\infty}\ln^{2}\pars{x}\,{1 + x^{2} \over 1 + x^{4}}\,\dd x ={3 \pi^{3} \over 16 \root{2}}}$

\begin{align} I&=-\Im\bracks{\pars{1 - \ic} \int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + \ic}\,\dd x} \\[3mm]&=-\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty} x^{\mu}\int_{0}^{\infty}\expo{-\pars{x^{2} + \ic}\xi}\,\dd\xi\,\dd x} \\[3mm]&=-\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty}\expo{-\ic\xi}\ \overbrace{\int_{0}^{\infty}x^{\mu}\expo{-\xi x^{2}}\,\dd x}^{\ds{t \equiv \xi x^{2}\ \imp\ x = \xi^{-1/2}t^{1/2}}}\ \dd\xi} \\[3mm]&=-\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty}\expo{-\ic\xi}\ \int_{0}^{\infty}\xi^{-\mu/2}\ t^{\mu/2}\expo{-t}\xi^{-1/2}\,\half\,t^{-1/2}\,\dd t\, \dd\xi} \\[3mm]&=-\,\half\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\int_{0}^{\infty} \xi^{-\pars{\mu + 1}/2}\expo{-\ic\xi} \int_{0}^{\infty}t^{\pars{\mu - 1}/2}\expo{-t}\,\dd t\,\dd\xi} \\[3mm]&=-\,\half\,\Im\bracks{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\Gamma\pars{{\mu \over 2} + \half} \color{#c00000}{\int_{0}^{\infty}\xi^{-\pars{\mu + 1}/2}\expo{-\ic\xi}\,\dd\xi}} \tag{1} \end{align} where $\ds{\Gamma\pars{z}}$ is the Gamma Function.

Also, \begin{align} &\overbrace{% \color{#c00000}{\int_{0}^{\infty}\xi^{-\pars{\mu + 1}/2}\expo{-\ic\xi}\,\dd\xi}} ^{\ds{t \equiv \ic\xi\quad\imp\quad\xi = -\ic t = \expo{-\ic\pi/2}t}} =\int_{0}^{\infty\ic}\pars{\expo{-\ic\pi/2}t}^{-\pars{\mu + 1}/2} \expo{-t}\,\pars{-\ic\,\dd t} \\[3mm]&=-\ic\expo{\ic\pi\pars{\mu + 1}/4}\int_{0}^{\infty}t^{-\pars{\mu + 1}/2} \expo{-t}\,\dd t=-\ic\expo{\ic\pi\pars{\mu + 1}/4}\Gamma\pars{\half - {\mu \over 2}} \end{align}

Expression $\pars{1}$ is reduce to: \begin{align} I&=-\,\half\,\Im\braces{\pars{1 - \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\Gamma\pars{{\mu \over 2} + \half} \bracks{-\ic\expo{\ic\pi\pars{\mu + 1}/4}\Gamma\pars{\half - {\mu \over 2}}}} \\[3mm]&=\half\,\Im\bracks{\pars{1 + \ic} \lim_{\mu \to 0}\partiald[2]{}{\mu}\expo{\ic\pi\pars{\mu + 1}/4}\, {\pi \over \sin\pars{\pi\bracks{\mu/2 + 1/2}}}} \end{align} where we used Euler Reflection Formula ${\bf\mbox{6.1.17}}$.

\begin{align} I&={\root{2} \over 2}\,\pi\, \lim_{\mu \to 0}\partiald[2]{}{\mu}\bracks{\cos\pars{\pi\mu \over 4} \sec\pars{\pi\mu \over 2}} ={\pi \over \root{2}}\pars{-\,{\pi^{2} \over 16} + {\pi^{2} \over 4}} \end{align}

$$\color{#00f}{\large% I\equiv\int_{0}^{\infty}\ln^{2}\pars{x}\,{1 + x^{2} \over 1 + x^{4}}\,\dd x ={3 \pi^{3} \over 16 \root{2}}} $$