Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$

Question

Calculate $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$

I have tried to put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$

$$ \int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$

$$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)^2}dt$$

$$ \frac{1}{2}\bigg[-t\ln^2(t)\frac{1}{t^2-1}+\int^{\infty}_{0}\frac{\ln^2(t)}{t^2-1}+2\int^{\infty}_{0}\frac{\ln(t)}{t^2-1}dt\bigg]$$

How can I solve it?

Answer 1

You're definetly on the right track with that substitution of $x=\frac1t$

Basically we have: $$I=\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx=\int_0^\infty \frac{x^2\ln^2 x}{(1-x^2)^2}dx$$ Now what if we add them up? $$2I=\int_0^\infty \ln^2 x \frac{1+x^2}{(1-x^2)^2}dx$$ If you don't know how to deal easily with the integral $$\int \frac{1+x^2}{(1-x^2)^2}dx=\frac{x}{1-x^2}+C$$ I recommend you to take a look here.

Anyway we have, integrating by parts: $$2I= \underbrace{\frac{x}{1-x^2}\ln^2x \bigg|_0^\infty}_{=0} +2\underbrace{\int_0^\infty \frac{\ln x}{x^2-1}dx}_{\large =\frac{\pi^2}{4}}$$ $$\Rightarrow 2I= 2\cdot \frac{\pi^2}{4} \Rightarrow I=\frac{\pi^2}{4}$$ For the last integral see here for example.

Answer 2

This is a variant of TheSimpliFire's approach given in a comment.

By letting $x=e^t$ we get $$\begin{align*} \int_0^\infty\frac{\ln^2(x)}{(1-x^2)^2}\,dx &=\int_{-\infty}^\infty\frac{t^2e^{t}}{(1-e^{2t})^2}\,dt\\ &=\int_{0}^{+\infty}\frac{t^2e^{-t}}{(1-e^{-2t})^2}\,dt+\int_{0}^\infty\frac{t^2e^{-3t}}{(1-e^{-2t})^2}\,dt\\ &=\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+1)t}\,dt+\sum_{n=0}^\infty (1+n)\int_0^\infty t^2e^{-(2n+3)t}\,dt\\ &=\sum_{n=0}^\infty\frac{2(1+n)}{(2n+1)^3}+\sum_{n=0}^\infty\frac{2(1+n)}{(2n+3)^3}\\ &=\left(\sum_{n=0}^\infty\frac{1}{(2n+1)^2}+\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\right)+\left(\sum_{n=0}^\infty\frac{1}{(2n+1)^2}-\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\right)\\ &=2\sum_{n=0}^\infty\frac{1}{(2n+1)^2}=2\left(\sum_{n=0}^\infty\frac{1}{n^2}-\sum_{n=0}^\infty\frac{1}{(2n)^2}\right)\\&=2\left(1-\frac{1}{4}\right)\sum_{n=0}^\infty\frac{1}{n^2}=\frac{3}{2}\cdot \frac{\pi^2}{6}=\frac{\pi^2}{4}. \end{align*}$$

Answer 3

Here is yet another slight variation on a theme.

Let $$I = \int_0^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx$$ then \begin{align} I &= \int_0^1 \frac{\ln^2 x}{(1 - x^2)^2} \, dx + \int_1^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx = \int_0^1 \frac{(1 + x^2) \ln^2 x}{(1 - x^2)^2} \, dx \tag1, \end{align} after a substitution of $x \mapsto 1/x$ has been enforced in the second of the integrals.

As $$\frac{1}{1 - x^2} = \sum_{n = 0}^\infty x^{2n}, \qquad |x| < 1,$$ differentiating with respect to $x$ gives $$\frac{1}{(1 - x^2)^2} = \sum_{n = 1}^\infty n x^{2n - 2}.$$ On substituting the above series expansion in (1), after interchanging the order of the summation with the integration we have $$I = \sum_{n = 1}^\infty n \int_0^1 (x^{2n - 2} + x^{2n}) \ln^2 x \, dx.$$ Integrating by parts twice, we are left with $$I = \sum_{n = 1}^\infty \left (\frac{2n}{(2n - 1)^3} + \frac{2n}{(2n + 1)^3} \right ).$$ As the series is absolutely convergent, terms can be rearranged without changing its sum. Doing so we have \begin{align} I &= \sum_{n = 1}^\infty \left [\frac{1}{(2n - 1)^2} + \frac{1}{(2n - 1)^3} + \frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\ &= \sum_{n = 1}^\infty \left [\frac{1}{(2n - 1)^2} + \frac{1}{(2n - 1)^3} \right ] + \sum_{n = 1}^\infty \left [\frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\ &= \sum_{n = 0}^\infty \left [\frac{1}{(2n + 1)^2} + \frac{1}{(2n + 1)^3} \right ] + \sum_{n = 1}^\infty \left [\frac{1}{(2n + 1)^2} - \frac{1}{(2n + 1)^3} \right ]\\ &= 2 + 2 \sum_{n = 1}^\infty \frac{1}{(2n + 1)^2}\\ &= 2 \sum_{n = 0}^\infty \frac{1}{(2n + 1)^2}\\ &= 2 \left [\sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \frac{1}{(2n)^2} \right ]\\ &= 2 \left (1 - \frac{1}{4} \right ) \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= \frac{3}{2} \sum_{n = 1}^\infty \frac{1}{n^2}\\ &= \frac{3}{2} \cdot \frac{\pi^2}{6}\\ &= \frac{\pi^2}{4}, \end{align} as expected.

Answer 4

Integrate by parts

\begin{align} \int^{\infty}_{0}\frac{\ln^2x}{(1-x^2)^2} dx &=\int^{\infty}_{0}\frac{\ln^2x}{2x} \ d\left(\frac{x^2}{1-x^2}\right)\\ &\overset{ibp}= -\int^{\infty}_{0}\frac{\ln x}{1-x^2} dx +\frac12 \int^{\infty}_{0}\frac{\ln^2x}{1-x^2} dx \\ &= -(-\frac{\pi^2}4)+\frac12\cdot 0=\frac{\pi^2}4 \end{align}

where $\int^{\infty}_{0}\frac{\ln x}{1-x^2} dx =-\frac{\pi^2}4$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{1 - x^{2}}^{2}}\,\dd x} \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 8}\int_{0}^{\infty}{x^{-1/2}\,\ln^{2}\pars{x} \over \pars{1 - x}^{2}}\,\dd x \\[5mm] = &\ \left.{1 \over 8}\partiald[2]{}{\nu}\int_{0}^{\infty}{x^{-1/2}\pars{x^{\nu} - 1} \over \pars{1 - x}^{2}}\,\dd x \,\right\vert_{\ \nu\ =\ 0} \\[5mm] = &\ \left.{1 \over 8}\partiald[2]{}{\nu} \int_{0}^{\infty} {x^{\pars{\nu + 1/2} - 1}\,\,\, - x^{1/2 - 1} \over \pars{1 - x}^{2}}\,\dd x \,\right\vert_{\ \nu\ =\ 0} \end{align}

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\begin{align} &{1 \over \pars{1 - x}^{2}} = \sum_{k = 0}^{\infty}{-2 \choose k}\pars{-x}^{k} = \sum_{k = 0}^{\infty}{k + 1 \choose k}\pars{-1}^{k} \pars{-x}^{k} \\[5mm] = &\ \sum_{k = 0}^{\infty} \color{red}{\pars{1 + k}\Gamma\pars{1 + k}\expo{\ic\pi k}}{\pars{-x}^{k} \over k!} \end{align}

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\begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\ln^{2}\pars{x} \over \pars{1 - x^{2}}^{2}}\,\dd x} = {1 \over 8}\partiald[2]{}{\nu} \bracks{\Gamma\pars{\nu + {1 \over 2}} \pars{{1 \over 2} - \nu}\Gamma\pars{{1 \over 2} - \nu} \expo{-\ic\pi\pars{\nu + 1/2}}}_{\ \nu\ =\ 0} \\[5mm] = &\ -\,{\pi \over 8}\ic\,\partiald[2]{}{\nu} \bracks{\pars{{1 \over 2} - \nu}\bracks{1 - \ic\tan\pars{\pi\nu}}}_{\ \nu\ =\ 0} \\[5mm] = &\ {\pi \over 8}\,\partiald[2]{}{\nu} \bracks{\pars{\nu - {1 \over 2}}\tan\pars{\pi\nu}} _{\ \nu\ =\ 0} = {\pi \over 8}\,\partiald[2]{}{\nu} \pars{\pi\nu^{2}}_{\ \nu\ =\ 0} \\[5mm] = &\ \bbx{\pi^{2} \over 4}\\ & \end{align}