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$$I=\int_0^\infty{\frac{(\ln(x))^2}{\sqrt{x}(1-x)^2}}dx$$

I first substituted the $x=t^2$ to get
$$I= 8\int_0^\infty{\frac{(\ln(t))^2}{(1-t^2)^2}}dt$$

After this I don't have much of an idea how to proceed forward.

Any help would be appreciated!
The answer given is $2\pi^2$

Momo
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WHOOP
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    Using complex integration it is not difficult to show that $$J(\alpha)=\int_0^\infty{\frac{x^\alpha\ln x}{(1-x)^2}}dx=-\pi\cot\pi\alpha+\frac{\pi^2\alpha}{\sin^2\pi\alpha},,,,,\alpha\in(-1;1)$$ Then $I_n(\alpha)=\int_0^\infty{\frac{x^\alpha\ln^n x}{(1-x)^2}}dx=\frac{\partial^{n-1}}{\partial \alpha^{n-1}}J(\alpha)$. For example, $$I_2(\alpha)=\frac{2\pi^2}{\sin^2\pi\alpha}-\frac{2\pi^3\alpha,\cos\pi\alpha}{\sin^3\pi\alpha},;,, I=I_2\big(-\frac{1}{2}\big)=2\pi^2$$ I can present the solution if it brings some value. – Svyatoslav Apr 14 '22 at 15:02
  • @Svyatoslav Please add your solution, many may benefit. – Dhanvin Apr 17 '22 at 10:56
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    @4444 I posted the solution on Quora. – Svyatoslav Apr 17 '22 at 13:40
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    Michael Penn has done a video evaluating this integral today. See here. – KStarGamer Jul 20 '22 at 00:36
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    qncubed3 has done a video evaluating this integral via contour integration today. See here. – KStarGamer Jul 27 '22 at 23:25

3 Answers3

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The substitution $x = \exp(2u)$ is more natural to get rid of the $\ln$. Doing this gives $$ I = 8\int_{-\infty}^\infty\frac{u^2e^{u}}{(1-e^{2u})^2}du $$ Now we split the integral into positive and negative parts and take $u\rightarrow -u$ in the negative part to get $$ I = 8\int_0^\infty \frac{e^{2u} + 1}{(e^{2u} - 1)^2}u^2e^{u}du = 4\int_0^\infty u^2\coth(u)\mathrm{csch}(u)du $$ Conveniently, $\coth(u)\mathrm{csch}(u)du = -d(\mathrm{csch}(u))$, so we can integrate this by parts to get $$ I = 8\int_0^\infty u\,\mathrm{csch}(u)du $$ This is a noticeably simpler integral which can probably be evaluated by a number of methods. One way to proceed is that the difference of exponentials in the denominator suggests transforming to known zeta function integrals: $$ \mathrm{csch}(u)/2 = \frac{1}{e^{u} - e^{-u}} = \frac{e^{u}}{e^{2u}-1} = \frac{1}{e^{2u} - 1}+\frac{e^{u}-1}{e^{2u} -1} = \frac{1}{e^{2u} - 1} + \frac{1}{e^{u} + 1} $$ which gives $$ I = 16\left(\int_0^\infty \frac{u}{e^{2u} - 1}du + \int_0^\infty \frac{u}{e^u + 1}du\right) = 16\left[\frac{\zeta(2)\Gamma(2)}{4} + \frac{\zeta(2)\Gamma(2)}{2}\right] = 2\pi^2 $$ since $\Gamma(2) = 1$ and $\zeta(2)= \pi^2/6$.

eyeballfrog
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  • Nice simplification ! Now $$\int u,\mathrm{csch}(u),du=\text{Li}_2\left(-e^{-u}\right)-\text{Li}_2\left(e^{-u}\right)+u \left(\log \left(1-e^{-u}\right)-\log \left(e^{-u}+1\right)\right)$$ – Claude Leibovici Apr 14 '22 at 14:21
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$\newcommand{\d}{\mathrm{d}}$The integral can be reduced, as Eyeballfrog says, to showing: $$\int_0^\infty\frac{u}{e^u-e^{-u}}\d u=\frac{\pi^2}{8}$$

Begin geometrically: $$\int_0^\infty\frac{u}{1-e^{-2u}}\cdot e^{-u}\d u=\sum_{n\ge0}\int_0^\infty u\cdot e^{-(2n+1)u}\d u\overset{u\mapsto(2n+1)u}{=}\sum_{n\ge0}\frac{1}{(2n+1)^2}\int_0^\infty u\cdot e^{-u}\d u$$

It is easy to show by integration by parts, or by knowledge of the Gamma function, that: $$\int_0^\infty u\cdot e^{-u}\d u=1$$So we are left with: $$\sum_{n\ge0}\frac{1}{(2n+1)^2}$$

We know that: $$\zeta(2)=\sum_{n\ge1}\frac{1}{n^2}=\frac{\pi^2}{6}$$Some algebra gives: $$\underset{\text{sum of all squares}}{\underbrace{\zeta(2)}}-\underset{\text{sum of even squares}}{\underbrace{\frac{1}{4}\zeta(2)}}=\underset{\text{sum of odd squares}}{\underbrace{\sum_{n\ge0}\frac{1}{(2n+1)^2}}}$$

Giving: $$\int_0^\infty\frac{u}{e^u-e^{-u}}\d u=\frac{3}{4}\cdot\frac{\pi^2}{6}=\frac{\pi^2}{8}$$As desired.

FShrike
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Proceed as follows \begin{align} \int_0^\infty{\frac{\ln^2x}{\sqrt{x}(1-x)^2}}dx = &\>8\int_0^\infty{\frac{\ln^2 t}{(1-t^2)^2}}dt \overset{t\to 1/t}= 4\int_0^\infty {\frac{(1+t^2)\ln^2 t}{(1-t^2)^2}}dt\\ = &\>4\int_0^\infty \ln^2 t\>d\left( \frac t{1-t^2}\right) \overset{ibp}=-8 \int_0^\infty \frac {\ln t}{1-t^2}dt\\ =&-8\left( -\frac{\pi^2}4\right)=2\pi^2 \end{align} where $\int_0^\infty \frac {\ln t}{1-t^2}dt=-\frac{\pi^2}4 $

Quanto
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