TITLE: How to evaluate $\int_{0}^{1} \! \frac{-\ln(1-t)}{t} \, \mathrm{d}t$ without using taylor series expansion?
QUESTION [6 upvotes]: I want to prove the Basel Problem and I managed to turn it into an integral which I can't solve.
I am interested to know how it can be evaluated without using Taylor series expansions, perhaps with the use of special functions?
$$\int_{0}^{1} \! \frac{-\ln(1-t)}{t} \, \mathrm{d}t.$$
I tried using substitutions that can relate it to the Gamma or Beta function but I always hit a seemingly insurmountable roadblock.
I will highly appreciate a detailed answer. Thank you!

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\begin{align}
\int_{0}^{1}{-\ln\pars{1 - t} \over t}\,\dd t & =
-\int_{0}^{1}{\ln\pars{t} \over 1 - t}\,\dd t =
\left.\partiald{}{\nu}\int_{0}^{1}{1 - t^{\nu} \over 1 - t}\,\dd t
\,\right\vert_{\ \nu\ =\ 0}
\\[5mm] &=
\left.\partiald{\Psi\pars{\nu + 1}}{\nu}\,\right\vert_{\ \nu\ =\ 0}
=
\Psi\, '\pars{1} = \sum_{n = 0}^{\infty}{1 \over \pars{n + 1}^{2}}
\\[5mm] & =
\sum_{n = 1}^{\infty}{1 \over n^{2}} = \bbx{\pi^{2} \over 6\phantom{^{2}}} \\ &
\end{align}