How find this integral $I=\int_{0}^{1}\frac{1}{2-x}\ln{\frac{1}{x}}dx$

Question

Find this integral $$I=\int_{0}^{1}{1 \over 2 - x}\,\ln\left(1 \over x\right)\,{\rm d}x$$

My idea: let $1-x=t$, then $$I=\int_{0}^{1}{\ln\left(1 - t\right) \over 1 + t}\,{\rm d}t$$

Answer 1

We have $$I=\int_0^1\sum_{n=0}^\infty\frac{x^n}{2^{n+1}}\ln (1/x)\,dx =\sum_{n=0}^\infty\frac{1}{2^{n+1}}\int_0^1x^n\ln(1/x)dx =\sum_{n=0}^\infty\frac{1}{2^{n+1}(n+1)^2}$$ That is $$I=\sum_{n=1}^\infty\frac{1}{2^nn^2}=\hbox{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{12} - \frac{\ln^22}{2}.$$

For more information on the Dilogarithm $\hbox{Li}_2$ see here.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I \equiv \int_{0}^{1}{1 \over 2 - x}\ln\pars{1 \over x}\,\dd x:\ {\large ?}}$

\begin{align} \color{#00f}{\large I}&=-\,\half\int_{0}^{1}{\ln\pars{x} \over 1 - x/2}\,\dd x =-\,\half\int_{0}^{1}\ln\pars{x}\sum_{n = 0}^{\infty}\pars{x/2}^{n}\,\dd x =-\,\half\sum_{n = 0}^{\infty}{1 \over 2^{n}}\int_{0}^{1}\ln\pars{x}x^{n}\,\dd x \\[3mm]&=\left.-\,\half\sum_{n = 0}^{\infty}{1 \over 2^{n}} \partiald{}{\mu}\int_{0}^{1}x^{n + \mu}\,\dd x\right\vert_{\mu = 0} =\half\sum_{n = 0}^{\infty}{1 \over 2^{n}\pars{n + 1}^{2}} \\[3mm]&=\half\ \overbrace{\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}^{2}}} ^{\ds{{\pi^{2} \over 6}}}\ -\ \half\ \overbrace{\sum_{n = 0}^{\infty}{1 - 2^{-n} \over \pars{n + 1}^{2}}} ^{\ds{\ln^{2}\pars{2}}} = \color{#00f}{\large{\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}} \end{align}

ADDENDA

Also, it's equivalent to

\begin{align} \color{#00f}{\large I}&=-\,\half\int_{0}^{1}{\ln\pars{x} \over 1 - x/2}\,\dd x =-\int_{0}^{1}{\ln\pars{2\bracks{x/2}} \over 1 - x/2}\,{\dd x \over 2} =-\int_{0}^{1/2}{\ln\pars{2x} \over 1 - x}\,\dd x \\[3mm]&=-\int_{0}^{1/2}\ln\pars{1 - x}\,\pars{{1 \over 2x}\,2}\,\dd x =\int_{0}^{1/2}{\rm Li}_{2}'\pars{x}\,\dd x ={\rm Li}_{2}\pars{\half} \\[3mm]&=\color{#00f}{\large{\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}} \end{align}