Improper integral with logarithm

Evaluate the integral:

$$\int_0^\infty \frac{\log^2 (x+\frac{1}{x})}{x^2+1}\, {\rm d}x$$

Solution

 We present two useful lemmas that will guide us through the evaluations of this particular integral:

Lemma 1: $$ \int_0^{\pi/2} \ln^2 \tan \theta \, {\rm d}x = \frac{\pi^3}{8}$$

Proof:$$\begin{align*} I_2&=\displaystyle \int_{0}^{\frac{\pi}{2}}{\log^2(\tan{x})\,dx}\\ &\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,\tan{x}}\\ {dx\,=\,\frac{1}{1+t^2}\,dt}\\ \end{subarray}}\,\int_{0}^{+\infty}{\frac{\log^2{t}}{1+t^2}\,dt}\\ &=\int_{0}^{1}{\frac{\log^2{t}}{1+t^2}\,dt}+\int_{1}^{+\infty}{\frac{\log^2{t}}{1+t^2}\,dt}\\ &\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {s\,=\,\frac{1}{t}}\\ {-\frac{1}{s^2}\,ds\,=\,dt}\\ \end{subarray}}\,\int_{0}^{1}{\frac{\log^2{t}}{1+t^2}\,dt}-\int_{1}^{0}{\frac{\log^2{\tfrac{1}{s}}}{1+\frac{1}{s^2}}\,\frac{1}{s^2}\,ds}\\ &=\int_{0}^{1}{\frac{\log^2{t}}{1+t^2}\,dt}+\int_{0}^{1}{\frac{\log^2{s}}{1+s^2}\,ds}\\ &=2\,\int_{0}^{1}{\frac{\log^2{t}}{1+t^2}\,dt}\\ &\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,{\mathrm{e}}^{-x}}\\ {dt\,=\,-{\mathrm{e}}^{-x}dx}\\ \end{subarray}}\,-2\,\int_{+\infty}^{0}{\frac{(-x)^2\,{\mathrm{e}}^{-x}}{1+{\mathrm{e}}^{-2x}}\,dx}\\ &=2\,\int_{0}^{+\infty}{x^2\frac{{\mathrm{e}}^{-x}}{1+{\mathrm{e}}^{-2x}}\,dx}\\ &=2\,\int_{0}^{+\infty}{x^2\,\biggl({\mathop{\sum}\limits_{n=1}^{+\infty}{(-1)^{n+1}\,\mathrm{e}}^{-(2n-1)x}}\biggr)\,dx}\\ &=2\,\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({(-1)^{n+1}\int_{0}^{+\infty}{x^2\,\mathrm{e}}^{-(2n-1)x}\,dx}\biggr)\\ &\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,(2n-1)x}\\ {\frac{1}{2n-1}\,dt\,=\,dx}\\ \end{subarray}}\,2\,\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({(-1)^{n+1}\,\int_{0}^{+\infty}{\frac{t^2}{(2n-1)^3}\,{\mathrm{e}}^{-t}\,dt}}\biggr)\\ &=2\,\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({\frac{(-1)^{n+1}}{(2n-1)^3}\int_{0}^{+\infty}{t^{3-1}\,\mathrm{e}}^{-t}\,dt}\biggr)\\ &=2\,\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({\frac{(-1)^{n+1}}{(2n-1)^3}\,\Gamma(3)}\biggr)\\ &=4\,\mathop{\sum}\limits_{n=1}^{+\infty}{\frac{(-1)^{n+1}}{(2n-1)^3}}\\ &=4\,\frac{\pi^3}{32}\\ &=\frac{\pi^3}{8}\,. \end{align*}$$

We've taken this particular proof from the link.

Lemma 2: $$\int_0^{\pi/2} \ln^2 \sin \theta \, {\rm d}\theta = \frac{\pi^3}{24}+\frac{\pi \ln^2 2}{2}$$

Proof:Consider the integral:

$$\int_{0}^{1}\frac{u^{m-1}}{\sqrt{1-u^2}}du=\frac{1}{2}{\rm B}\left ( \frac{1}{2}, \frac{m}{2} \right )=\frac{1}{2}\frac{\Gamma \left ( \frac{1}{2} \right )\Gamma \left ( \frac{m}{2} \right )}{\Gamma \left ( \frac{m+1}{2} \right )}=\frac{1}{2}\frac{\sqrt{\pi}\,\Gamma \left ( \frac{m}{2} \right )}{\Gamma \left ( \frac{m+1}{2} \right )}$$

Differentiating with respect to $m$ twice and in continuity plugging $m=1$ we have:

$$\begin{align*} I(m)&=\int_{0}^{1}{\frac{u^{m-1}}{\sqrt{1-u^2}}\,du}=\frac{\sqrt{\pi}}{2}\frac{\Gamma\bigl({\tfrac{m}{2}}\bigr)}{\Gamma\bigl({\tfrac{m+1}{2}}\bigr)}\quad\Rightarrow\\ \frac{d^2}{dm^2}I(m)&=\int_{0}^{1}{\frac{u^{m-1}\log^2{u}}{\sqrt{1-u^2}}\,du}\\ &=\frac{\sqrt{\pi}}{8}\frac{\Gamma\bigl({\tfrac{m}{2}}\bigr)}{\Gamma\bigl({\tfrac{m+1}{2}}\bigr)}\Bigl({\psi'\bigl({\tfrac{m}{2}}\bigr)-\psi'\bigl({\tfrac{m+1}{2}}\bigr)+\Bigr({\psi\bigl({\tfrac{m}{2}}\bigr)-\psi\bigl({\tfrac{m+1}{2}}\bigr)}\Bigr)^2}\Bigr)\quad\Rightarrow\\ \int_{0}^{1}{\frac{\log^2{u}}{\sqrt{1-u^2}}\,du}&=\frac{\sqrt{\pi}}{8}\frac{\Gamma\bigl({\tfrac{1}{2}}\bigr)}{\Gamma(1)}\Bigl({\psi'\bigl({\tfrac{1}{2}}\bigr)-\psi'(1)+\Bigr({\psi\bigl({\tfrac{1}{2}}\bigr)-\psi(1)}\Bigr)^2}\Bigr)\\ &=\frac{\sqrt{\pi}}{8}\frac{\sqrt{\pi}}{1}\Bigl({\frac{\pi^2}{2}-\frac{\pi^2}{6}+({-2\log2-\gamma+\gamma})^2}\Bigr)\\ &=\frac{\pi}{2}\log^2{2}+\frac{\pi^3}{24}\,, \end{align*}$$

where $\psi$ is the digamma and $\gamma$ is the Euler - Masceroni constant.

We've taken this particular proof from the link.

Now back to our integral:

$$\begin{align*}
\int_{0}^{\infty}\frac{\log^2 \left ( x+ \frac{1}{x} \right )}{x^2+1}\, {\rm d}x &\overset{x=\tan \theta}{=\! =\! =\! =\! =\!}\int_{0}^{\pi/2}\frac{\ln^2 \left ( \tan \theta + \frac{1}{
\tan \theta} \right )}{\tan^2 \theta+1}\sec^2 \theta \, {\rm d}\theta \\
 &= \int_{0}^{\pi/2}\ln^2 \left ( \tan \theta + \frac{1}{\tan \theta} \right )\, {\rm d}\theta\\
 &= \int_{0}^{\pi}\ln^2 \left [ \tan \theta \left ( \cot^2 \theta +1 \right ) \right ]\, {\rm d}\theta\\
 &= \int_{0}^{\pi/2}\ln^2 \tan \theta \, {\rm d}\theta + \int_{0}^{\pi/2} \ln^2 \left ( \left ( \cot^2 \theta +1 \right ) \right )\, {\rm d}\theta \\
&=\frac{\pi^3}{8}+ \int_{0}^{\pi/2}\ln^2 \csc^2 \theta \, {\rm d}\theta \\
&= \frac{\pi^3}{8}+ 2 \int_{0}^{\pi/2} \ln^2 \sin \theta \, {\rm d} \theta \\
&= \frac{\pi^3}{8}+ \frac{\pi^3}{12}+ \pi \ln^2 2 \\
&= \frac{5\pi^3}{24}+ \pi \ln^2 2
\end{align*}$$

ending the calculations.