Is there a simple proof for $\int_1^{\infty}\frac{2x^2\log^2 x}{(x^2-1)^2}dx=\frac{1}{4}(7\zeta(3)+\pi^2)$?

Question

This morning I've computed easy computations with simple integral representations for Apéry constant and I find a (conjecture) formula using an online integrator (Wolfram Alpha), I woluld like if it is possible find a proof of such fact, since I believe that it could be easy, since this online tool know how compute.

Question. Can you give a proof of $$\int_1^{\infty}\frac{2x^2\log^2 x}{(x^2-1)^2}dx=\frac{1}{4}(7\zeta(3)+\pi^2)?$$ Please, if you known yet the result reference the work, and of course if you believe that it is false say me. Thanks in advance.

Early, I will choose the more simple proof of this fact.

My context: I don't know if it was in the literature or if there are mistakes. Feel free to add useful comments if you find mistakes. This morning I was reading the section about simple integral representations for Apéry constant, here. We can compute easily that $$\zeta(3)=\frac{4}{7}\int_{1}^{\infty}\frac{\log^2 x}{x^2-1}dx,$$ after, if there are no mistakes this gives, using integration by parts $u=\frac{\log x}{x^2-1}$, and $dv=\log x dx$, $$\frac{4}{7}\left(\frac{1}{2}-\int_{1}^{\infty}\frac{-x^2+1+3x^2\log x-\log x-2x^2\log^2 x}{(x^2-1)^2}dx\right)$$

Answer 1

I don't know if it is the most simple way, but it works. Let $x=1/t $. We get $$I=-2\int_{0}^{1}\frac{\log^{2}\left(t\right)}{\left(1-t^{2}\right)^{2}}dt=2\sum_{k\geq0}\left(k+1\right)\int_{0}^{1}t^{2k}\log^{2}\left(t\right)dt $$ and now integrating by parts twice we have $$\int_{0}^{1}t^{2k}\log^{2}\left(t\right)dt=-\frac{2}{2k+1}\int_{0}^{1}t^{2k}\log\left(t\right)dt=\frac{2}{\left(2k+1\right)^{2}}\int_{0}^{1}t^{2k}dt=\frac{2}{\left(2k+1\right)^{3}} $$ so $$I=4\sum_{k\geq0}\frac{k+1}{\left(2k+1\right)^{3}}=4\sum_{k\geq1}\frac{k}{\left(2k-1\right)^{3}} $$ now observe that $$4\sum_{k\geq1}\frac{k}{\left(2k-1\right)^{3}}=2\sum_{k\geq1}\frac{1}{\left(2k-1\right)^{2}}+2\sum_{k\geq1}\frac{1}{\left(2k-1\right)^{3}} $$ and, due to the absolute convergence, we have $$\sum_{k\geq1}\frac{1}{\left(2k-1\right)^{2}}=\zeta\left(2\right)-\sum_{k\geq1}\frac{1}{4k^{2}}=\frac{3}{4}\zeta\left(2\right)=\frac{\pi^{2}}{8} $$ and $$\sum_{k\geq1}\frac{1}{\left(2k-1\right)^{3}}=\zeta\left(3\right)-\sum_{k\geq1}\frac{1}{8k^{3}}=\frac{7}{8}\zeta\left(3\right) $$ hence $$I=\frac{1}{4}\left(7\zeta\left(3\right)+\pi^{2}\right). $$

Answer 2

I would start with an integration by parts with $v=x\log^2(x)$ and $u'=\frac{2x}{(x^2-1)^2}$ . We get (the boundary term vansihs)

$$ I=\int_1^{\infty}dx\frac{\log(x)^2}{x^2-1}+2\int_1^{\infty}dx\frac{\log(x)}{x^2-1} $$

using your previous result we get

$$ I=\frac{7}{4}\zeta(3)+2\int_1^{\infty}dx\frac{\log(x)}{x^2-1} $$

The last integral can be caluclated by for example expanding the denominator as a geometric series (after performing an transforamtion $x\rightarrow 1/x$) resulting in a sum $\sum_{n=1}^{\infty} \frac{1}{(2n+1)^2}$ which is easily shown to be $\frac{\pi^2}{8}$. Therefore

$$ I=\frac{7}{4}\zeta(3)+\frac{\pi^2}{4} $$

Answer 3

Simplify to known integrals

\begin{align} \int_1^{\infty}\frac{2x^2\log^2 x}{(x^2-1)^2}dx & \overset{x\to\frac1x} = \int_0^{1}\frac{2\log^2 x}{(1-x^2)^2}dx = \int_0^{1}\frac{\log^2 x}x d\left( \frac{x^2}{1-x^2}\right)\\ & \overset{IBP}=\int_0^{1}\frac{\log^2 x}{1-x^2}dx-2\int_0^{1}\frac{\log x}{1-x^2}dx \\ &=\frac{7}{4}\zeta(3)-2\cdot(-\frac{\pi^2}8)= \frac14[7\zeta(3)+\pi^2] \end{align}

where $\int_0^{1}\frac{\log^2 x}{1-x^2}dx\overset{x\to\frac1x}=\int_1^{\infty}\frac{\log^2 x}{x^2-1}dx =\frac74\zeta(3)$ and $\int_0^{1}\frac{\log x}{1-x^2}dx =-\frac{\pi^2}8$

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{1}^{\infty}{2x^{2}\ln^{2}\pars{x} \over \pars{x^{2} - 1}^{2}}\,\dd x} \,\,\,\stackrel{x\ \mapsto\ 1/x}{=}\,\,\, 2\int_{0}^{1}{\ln^{2}\pars{x} \over \pars{1 - x^{2}}^{2}}\,\dd x \\[5mm] = &\ \left. -\,{1 \over a}\,\partiald{}{a}\int_{0}^{1}{\ln^{2}\pars{x} \over a^{2} - x^{2}}\,\dd x\,\right\vert_{\ a\ = 1} = \left. -\,\partiald{}{a}\pars{{1 \over a}\int_{0}^{1/a}{\ln^{2}\pars{ax} \over 1 - x^{2}}\,\dd x}\,\right\vert_{\ a\ = 1} \\[5mm] = &\ \color{red}{\mrm{f}\pars{1} - \mrm{f}\, '\pars{1}} \\[2mm] &\ \mbox{where}\quad \mrm{f}\pars{a} \equiv \int_{0}^{1/a}{\ln^{2}\pars{ax} \over 1 - x^{2}}\,\dd x \label{1}\tag{1} \end{align}

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\begin{align} \mrm{f}\pars{a} & \equiv \int_{0}^{1/a}{\ln^{2}\pars{ax} \over 1 - x^{2}}\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{1/a}{\ln^{2}\pars{ax} \over 1 - x}\,\dd x - {1 \over 2}\int_{0}^{-1/a}{\ln^{2}\pars{-ax} \over 1 - x}\,\dd x \\[5mm] & = -\int_{0}^{1/a}\mrm{Li}_{2}'\pars{x}\ln\pars{ax}\,\dd x \\[2mm] &\ + \int_{0}^{-1/a}\mrm{Li}_{2}'\pars{x}\ln\pars{-ax}\,\dd x \\[5mm] & = \int_{0}^{1/a}\mrm{Li}_{3}'\pars{x}\,\dd x - \int_{0}^{-1/a}\mrm{Li}_{3}'\pars{x}\,\dd x \\[5mm] & = \mrm{Li}_{3}\pars{1 \over a} - \mrm{Li}_{3}\pars{-\,{1 \over a}} \end{align}

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\begin{equation} \left\{\begin{array}{rcl} \ds{\mrm{f}\pars{1}} & \ds{=} & \ds{\mrm{Li}_{3}\pars{1} - \mrm{Li}_{3}\pars{-1}} \\ && \ds{= \sum_{n = 1}^{\infty}{1 \over n^{3}} - \sum_{n = 1}^{\infty}{\pars{-1}^{n} \over n^{3}}} \\ & \ds{=} & \ds{{7 \over 4}\sum_{n = 1}^{\infty}{1 \over n^{3}} = \color{red}{{7 \over 4}\,\zeta\pars{3}}} \\[5mm] \ds{\mrm{f}\, '\pars{1}} & \ds{=} & \ds{-\mrm{Li}_{2}\pars{1} + \mrm{Li}_{2}\pars{-1} \\ \ds{=} -\,{\pi^{2} \over 6} + \pars{-\,{\pi^{2} \over 12}} = \color{red}{-\,{\pi^{2} \over 4}}} \end{array}\right. \label{2}\tag{2} \end{equation} With (\ref{1}) and (\ref{2}): $$ \bbx{\large\bbox[5px,#ffd]{\int_{1}^{\infty}{2x^{2}\ln^{2}\pars{x} \over \pars{x^{2} - 1}^{2}}\,\dd x} = {1 \over 4}\bracks{7\zeta\pars{3} + \pi^{2}}} \\ $$