How to evaluate $\int_{0}^{1} \! \frac{-\ln(1-t)}{t} \, \mathrm{d}t$ without using taylor series expansion? 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!
 A: A similar problem and solution can be found here. Proposed by Khalef and solved by Sujee.
Since $\int_0^1 \frac{dx}{1+x^2}=\frac{\pi}{4}$, we have
$$\frac{\pi^2}{16}=\int_0^1\int_0^1\frac{dydx}{(1+x^2)(1+y^2)}\overset{t=xy}{=}\int_0^1\int_0^x\frac{dtdx}{x(1+x^2)(1+t^2/x^2)}$$
$$=\frac12\int_0^1\int_t^1\frac{dxdt}{x(1+x^2)(1+t^2/x^2)}\overset{x^2\to x}{=}\frac12\int_0^1\left(\int_{t^2}^1\frac{dx}{(1+x)(x+t^2)}\right)dt$$
$$=-\frac12\int_0^1\frac{\ln\left(\frac{4t^2}{(1+t^2)^2}\right)}{1-t^2}dt\overset{t=\frac{1-x}{1+x}}{=}-\frac12\int_0^1\frac{\ln\left(\frac{1-x^2}{1+x^2}\right)}{x}dx$$
$$\overset{x^2\to x}{=}-\frac14\int_0^1\frac{\ln\left(\frac{1-x}{1+x}\right)}{x}dx=-\frac14\int_0^1\frac{\ln\left(\frac{(1-x)^2}{1-x^2}\right)}{x}dx$$
$$=-\frac12\int_0^1\frac{\ln(1-x)}{x}dx+\frac14\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{x^2\to x}$$
$$=-\frac38\int_0^1\frac{\ln(1-x)}{x}dx\Longrightarrow \int_0^1\frac{-\ln(1-x)}{x}dx=\frac{\pi^2}{6}$$

Remark:
This solution can be considered a proof that $\zeta(2)=\frac{\pi^2}{6}$ as we have $\int_0^1\frac{-\ln(1-x)}{x}dx=\text{Li}_2(x)|_0^1=\text{Li}_2(1)=\zeta(2)$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}
