Is there a way to calculate the improper integral $\int_0^\infty \big(\frac{\ln x}{x - 1}\big)^2 dx$?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \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}

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