TITLE: A proof of $\int_{0}^{1}\left( \frac{\ln t}{1-t}\right)^2\,\mathrm{d}t=\frac{\pi^2}{3}$
QUESTION [26 upvotes]: What is the proof of the following:
$$\int_{0}^{1} \left(\frac{\ln t}{1-t}\right)^2 \,\mathrm{d}t=\frac{\pi^2}{3} \>?$$

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{1}\bracks{\ln\pars{t} \over 1 - t}^{2}\,\dd t
     ={\pi^{2} \over 3}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large\int_{0}^{1}%
\bracks{\ln\pars{t} \over 1 - t}^{2}\,\dd t}
=\lim_{\epsilon\ \to\ 0^{+}}\int_{\epsilon}^{1}%
{\ln^{2}\pars{t} \over \pars{1 - t}^{2}}\,\dd t
\\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{%
-\,{\ln^{2}\pars{\epsilon} \over 1 - \epsilon}
-\int_{\epsilon}^{1}{1 \over 1 - t}\,2\ln\pars{t}\,{1 \over t}\,\dd t}
\\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{%
-\,{\ln^{2}\pars{\epsilon} \over 1 - \epsilon}
-2\int_{\epsilon}^{1}{\ln\pars{t} \over 1 - t}\,\dd t
-2\int_{\epsilon}^{1}{\ln\pars{t} \over t}\,\dd t}
\\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{%
-\,{\ln^{2}\pars{\epsilon} \over 1 - \epsilon}
-2\int_{\epsilon}^{1}{\ln\pars{t} \over 1 - t}\,\dd t
+2\ln^{2}{\epsilon}}
=-2\int_{0}^{1}{\ln\pars{1 - t} \over t}\,\dd t
\\[3mm]&=2\int_{0}^{1}{\rm Li}_{2}'\pars{t}\,\dd t
=2\,{\rm Li}_{2}\pars{1}=2\
\underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}
_{\ds{=\ \color{#c00000}{\pi^{2} \over 6}}}\ =\
\color{#66f}{\Large{\pi^{2} \over 3}} \approx {\tt 3.2899}
\end{align}

$\ds{{\rm Li_{s}}\pars{z}}$ is the PolyLogarithm Function and we used well known properties of them as reported in the above cited link.