Evaluate $\int_0^\infty{\frac{(\ln(x))^2}{\sqrt{x}(1-x)^2}}dx$ $$I=\int_0^\infty{\frac{(\ln(x))^2}{\sqrt{x}(1-x)^2}}dx$$
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 $2\pi^2$
 A: Evaluate the Mellin transform
$$\int_{0}^{\infty} \frac{x^{s-1}}{(1-x)^2} \, dx = -\pi (s-1)(\cot (\pi s)-i)$$
with Ramanujan’s master theorem. Then take the partial derivative with respect to $s$ twice and then take the limit $s\to\frac{1}{2}$ giving:
$$\int_{0}^{\infty}\frac{(\ln x)^2}{\sqrt{x}(1-x)^2} \, dx=2\pi^2$$
A: 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 $\zeta(2)= \pi^2/6$.
A: $\newcommand{\d}{\mathrm{d}}$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.
A: 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 $
