$\int_0^1 {\frac{{\ln (1 - x)}}{x}}$ without power series

Question

Is there a way to calculate $$\int_0^1{ \ln (1 - x)\over x}\;dx$$ without using power series?

Answer 1

\begin{align} &\int_0^1 {\frac{{\ln (1 - x)}}{x}}dx\\ =& \ \frac43 \bigg( \int_0^1 {\frac{{\ln (1 - x)}}{x}} \overset{x\to \frac{1-x}{1+x}} {dx} - \frac14\int_0^1 {\frac{{\ln (1 - x)}}{x}} \overset{x\to (\frac{1-x}{1+x} )^2} {dx}\bigg)\\ =& \ \frac43 \int_0^1 \frac{\ln x}{1-x^2}dx =\frac43 \int_0^1 \int_0^\infty \frac{-y}{(1+y^2)(1+x^2y^2)}dy\ dx \\ =& -\frac43\int_0^\infty\frac{\tan^{-1}y}{1+y^2}dy =-\frac43\cdot \frac12 (\tan^{-1}y)^2\bigg|_0^\infty=-\frac{\pi^2}6 \end{align}

Answer 2

A related problem. Using the change of variables $x=1-e^{-t}$ and taking advatage of the fact that

$$\Gamma(s)\zeta(s) = \int_{0}^{\infty} \frac{t^{s-1}}{e^{t}-1}\,, $$

the value of the integral follows

$$ -\int_{0}^{\infty} \frac{t}{e^{t}-1} \,dt = -\zeta(2) = -\frac{\pi^2}{6} \,.$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} & \color{#44f}{\int_{0}^{1}{\ln\pars{1 - x} \over x} \dd x} = \left.\partiald{}{\mu}\int_{0}^{1}x^{\epsilon\, - 1} \bracks{\pars{1 - x}^{\mu} - 1}\dd x \right\vert_{\substack{\epsilon\ =\ 0^{+}\\[1mm] \mu\ =\ 0\,\,}} \\[5mm] = & \ \partiald{}{\mu}\bracks{{\Gamma\pars{\epsilon}\Gamma\pars{\mu + 1} \over \Gamma\pars{\epsilon + \mu + 1}} - {1 \over \epsilon}} _{\substack{\epsilon\ =\ 0^{+}\\[1mm] \mu\ =\ 0\,\,}} \\[5mm] = & \ \partiald{}{\mu}\braces{{1 \over \epsilon}\bracks{{\Gamma\pars{1 + \epsilon}\Gamma\pars{\mu + 1} \over \Gamma\pars{\epsilon + \mu + 1}} - 1}} _{\substack{\epsilon\ =\ 0^{+}\\[1mm] \mu\ =\ 0\,\,}} \\[5mm] = & \ \left.-\,\partiald{H_{\mu}}{\mu} \right\vert_{\mu\ =\ 0} = \bbx{\color{#44f}{-\,{\pi^{2} \over 6}}} \approx -1.6449 \\ & \end{align}