Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$
Solution 1:
Here is a solution based on real methods. Note that $\frac{1}{(x+1)^2+1}=\frac{x^2-2x+2}{x^4+4}$. Then
\begin{align} &\int_0^\infty \frac{\ln^2x}{(x+1)^2+1}dx \\ = & \int_0^\infty {\frac{x^2\ln^2x }{x^4+4}}\overset{x^2=1/t^2}{dx} -2\int_0^\infty {\frac{x\ln^2x }{x^4+4}} \overset{x^2=2t}{dx} + 2\int_0^\infty { \frac{\ln^2x }{x^4+4} } \overset{x^2=2t^2}{dx} \\ = & -\frac18\ln^22\int_0^\infty \frac{1}{t^2+1}dt -\frac18\int_0^\infty \frac{\ln^2t}{t^2+1}dt \\ & +\frac{1}{2\sqrt2} \ln^22 \int_0^\infty \frac{1}{t^4+1}dt + \sqrt2 \int_0^\infty \frac{\ln^2 t}{t^4+1}dt \\ = & -\frac{\ln^22}8\left(\frac\pi2\right) -\frac18\left(\frac{\pi^3}8\right) +\frac{\ln^22 }{2\sqrt2}\left(\frac\pi{2\sqrt2}\right) + \sqrt2 \left(\frac{3\pi^3\sqrt2}{64}\right) \\ =& \frac{5\pi^3}{64}+\frac\pi{16}\ln^22 \end{align}
Solution 2:
$$I(a)=\int_0^\infty\frac{x^{-a}}{x^2+2x+2}dx\overset{x\to 1/x}{=}\int_0^\infty\frac{x^a}{2x^2+2x+1}dx$$
$$=\Im\int_0^\infty\frac{(1+i)x^a}{1+(1+i)x}dx\overset{(1+i)x=u}{=}\Im \frac{1}{(1+i)^a}\int_0^\infty\frac{u^a}{1+u}du$$
$$=-\Im\frac{\pi\csc(a\pi)}{(1+i)^a}=2^{-a/2}\csc(a\pi)\sin(\frac{\pi}{4}a),$$
and your integral is $I''(0).$
Solution 3:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{}}$
With $\ds{r \equiv -1 + \ic = \root{2}\expo{-\ic\pi/4}}$: \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x + 1}^{2} + 1}\,\dd x} = \left.\partiald[2]{}{\nu}\int_{0}^{\infty}{x^{\nu - 1} \over \pars{x - r}\pars{x - \overline{r}}}\,\dd x \,\right\vert_{\ \nu\ = 1^{\large -}} \\[5mm] = & \left.\partiald[2]{}{\nu}\int_{0}^{\infty}x^{\nu - 1}\pars{% {1 \over r - x} - {1 \over \overline{r} - x}} {1 \over \overline{r} - r}\,\dd x \,\right\vert_{\ \nu\ = 1^{\large -}} \\[5mm] = &\ -\left.\partiald[2]{}{\nu}\Im\int_{0}^{\infty} {x^{\nu - 1} \over r - x}\,\dd x \,\right\vert_{\ \nu\ = 1^{\large -}} \\[5mm] = &\ -\partiald[2]{}{\nu}\Im\pars{{1 \over r}\int_{0}^{\infty} {x^{\pars{\color{red}{\nu} - 1}} \over 1 - x/r}\,\dd x} _{\ \nu\ = 1^{\large -}} \end{align}
Note that $\ds{{1 \over 1 - x/r} = \sum_{k = 0}^{\infty}\pars{x \over r}^{k} = \sum_{k = 0}^{\infty}\color{red}{\expo{\ic\pi k}\Gamma\pars{1 + k} \over r^{k}}{\pars{-x}^{k} \over k!}}$
With Ramanujan's Master Theorem; \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{x + 1}^{2} + 1}\,\dd x} \\[5mm] = & -\partiald[2]{}{\nu}\Im\bracks{{1 \over r}\, \Gamma\pars{\nu}\,{\expo{-\ic \pi\nu}\,\,\Gamma\pars{1 - \nu} \over r^{-\nu}}}_{\ \nu\ = 1^{\large -}} \\[5mm] = &\ \partiald[2]{}{\nu}\Im\braces{\pars{-r}^{\nu - 1}\, \bracks{\Gamma\pars{\nu}\Gamma\pars{1 - \nu}}} _{\ \nu\ = 1} \\[5mm] = &\ \partiald[2]{}{\nu}\Im\bracks{% \pars{1 - \ic}^{\nu - 1}\, {\pi \over \sin\pars{\pi\nu}}}_{\ \nu\ = 1} \\[5mm] = &\ \pi\partiald[2]{}{\nu}\bracks{2^{\nu/2 - 1/2}\,\,\sin\pars{\bracks{1 - \nu}\,{\pi \over 4}} \csc\pars{\pi\nu}}_{\ \nu\ = 1} \\[5mm] = &\ \bbx{{5\pi^{3} \over 64} + {\pi \over 16}\ln^{2}\pars{2}} \approx 2.5167 \\ & \end{align}