Is there a way to calculate the improper integral $\int_0^\infty \big(\frac{\ln x}{x - 1}\big)^2 dx$? Is there a way to calculate the improper integral $\int_0^\infty \big(\frac{\ln x}{x - 1}\big)^2 dx$?
What have I tried:
$$\int_0^\infty \Big(\frac{\ln x}{x - 1}\Big)^2 dx = \int_0^1 \Big(\frac{\ln x}{x - 1}\Big)^2 dx + \int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2 dx = \int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2dx + \int_1^\infty\Big(\frac{\ln t}{\big(\frac{1}{t} - 1\big)t}\Big)^2dt = 2\int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2dx$$
From the above statement we can conclude, that the integral does indeed converge, as $0 \leq (\frac{\ln x}{x - 1})^2 \leq \frac{C}{(x - 1)^{\frac{3}{2}}}$, for some constant $C$. However, this new form of the integral is not much helpful in finding its exact value.
I also tried substituting $z = \ln x$ , then $dz = \frac{dx}{x}$. Then the integral becomes reduced to $2\int_{0}^{\infty} \frac{z^2}{e^z(e^z - 1)^2}dz$, which, unfortunately, also does not seem to be any easier.
 A: Starting in the same way one obtains:
$$\begin{align}
\int_0^\infty \Big(\frac{\log x}{x - 1}\Big)^2 dx &= 2 \int_0^1 \Big(\frac{\log x}{x - 1}\Big)^2 dx\\
&=2\int_0^1 \Big(\frac{\log(1-x)}{x}\Big)^2 dx\\
&=4\zeta(2)=\frac{2\pi^2}{3},
\end{align}$$
where the general expression
$$
\int_0^1\frac{\log^n(1-u)}{u^{m+1}}du=\frac{(-1)^n n!}{m!}\sum_{i=0}^{m}{m \brack i}\zeta(n+1-i)
$$
was used to obtain the second to the last equality.
A: I too was interested in this type of integrals a while ago. See here and here, but there's no need to bring a cannon to a gunfight.$$I=\int_0^\infty \frac{\ln^2 x}{(x-1)^2}dx=2\int_0^1 \frac{\ln^2 x}{(1-x)^2}dx$$
This is what you've shown too, now we can integrate by parts, however if we just take $\left(\frac{1}{1-x}\right)'=\frac{1}{(1-x)^2}$ as the derivative of the denominator we run into divergence issues, that's why we're gonna take:
$$\left(\frac{1}{1-x}-1\right)'=\frac{1}{(1-x)^2}\Rightarrow I=2\int_0^1 \left(\color{blue}{\frac{1}{1-x}-1}\right)'\color{red}{\ln^2 x} dx$$
$$=\underbrace{\left(\color{blue}{\frac{1}{1-x}-1}\right)\color{red}{\ln^2 x}}_{=0}\bigg|_0^1-2\int_0^1 \color{blue}{\frac{x}{1-x}}\color{red}{\frac{2\ln x}{x}}=4\int_0^1 \frac{\ln x}{x-1}dx=\frac{2\pi^2}{3}$$
The last integral can be found here.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\left. 2\int_{0}^{1}{\ln^{2}\pars{x} \over a - x}\,\dd x
\right\vert_{\ a\ \not\in\ \bracks{0,1}} & =
2\int_{0}^{1/a}{\ln^{2}\pars{ax} \over 1 - x}\,\dd x
\\[5mm] & =
-4\int_{0}^{1/a}\overbrace{\bracks{-\,{\ln\pars{1 - x} \over x}}}^{\ds{\mrm{Li}_{2}'\pars{x}}}\ \ln\pars{ax}\,\dd x
\\[5mm] & =
4\int_{0}^{1/a}\overbrace{\mrm{Li}_{2}\pars{x} \over x}^{\ds{\mrm{Li}_{3}'\pars{x}}}\,\dd x = 4\,\mrm{Li}_{3}\pars{{1 \over a}}
\\[1cm]
-2\int_{0}^{1}{\ln^{2}\pars{x} \over \pars{a - x}^{2}}\,\dd x & =
4\,\mrm{Li}_{3}'\pars{1 \over a}\pars{-\,{1 \over a^{2}}} =
4\,{\mrm{Li}_{2}\pars{1/a} \over 1/a}\,\pars{-\,{1 \over a^{2}}}
\\[5mm] & =
-4\,{\mrm{Li}_{2}\pars{1/a} \over a}
\end{align}

$$
a \to 1^{+} \implies 
2\int_{0}^{1}{\ln^{2}\pars{x} \over \pars{1 - x}^{2}}\,\dd x =
4\,\mrm{Li}_{2}\pars{1} = 4\sum_{n = 1}^{\infty}{1^{n} \over n^{2}} =
\bbx{2\pi^{2} \over 3}\ \approx\ 6.5797
$$
