Prove $\int_0^1 \frac{x-1}{(x+1)\log{x}} \text{d}x = \log{\frac{\pi}{2}}$

Question

Prove $$\int_0^1 \frac{x-1}{(x+1)\log{x}} \text{d}x = \log{\frac{\pi}{2}}$$

Tried contouring but couldn't get anywhere with a keyhole contour.

Geometric Series Expansion does not look very promising either.

Answer 1

Hint. One may set $$ f(s):=\int_0^1 \frac{x^s-1}{(x+1)\log{x}}\: \text{d}x, \quad s>-1, \tag1 $$ then one is allowed to differentiate under the integral sign, getting $$ f'(s)=\int_{0}^{1}\frac{x^s}{x+1}\:dx=\frac12\psi\left(\frac{s}2+\frac12\right)-\frac12\psi\left(\frac{s}2+1\right), \quad s>-1, \tag2 $$where we have used a standard integral representation of the digamma function.

One may recall that $\psi:=\Gamma'/\Gamma$, then integrating $(2)$, observing that $f(0)=0$, one gets

$$ f(s)=\int_0^1 \frac{x^s-1}{(x+1)\log{x}}\: \text{d}x=\log \left(\frac{\sqrt{\pi}\cdot\Gamma\left(\frac{s}2+1\right)}{\Gamma\left(\frac{s}2+\frac12\right)}\right), \quad s>-1, \tag3 $$

from which one deduces the value of the initial integral by putting $s:=1$, recalling that $$ \Gamma\left(\frac12+1\right)=\frac12\Gamma\left(\frac12\right)=\frac{\sqrt{\pi}}2. $$

Edit. The result $(3)$ is more general than the given one.

Answer 2

We have $$I=\int_{0}^{1}\frac{x-1}{\log\left(x\right)\left(x+1\right)}dx=\sum_{k\geq0}\left(-1\right)^{k}\int_{0}^{1}\frac{x^{k+1}-x^{k}}{\log\left(x\right)}dx $$ $$\stackrel{x=e^{-u}}{=}\sum_{k\geq0}\left(-1\right)^{k+1}\int_{0}^{\infty}\frac{e^{-\left(k+2\right)u}-e^{-\left(k+1\right)u}}{x}du $$ and now we can apply the Frullani's theorem and get $$I=\sum_{k\geq1}\left(-1\right)^{k}\log\left(\frac{k}{k+1}\right)=\log\left(\prod_{k\geq1}\left(\frac{k}{k+1}\right)^{\left(-1\right)^{k}}\right) $$ and now note that $$\prod_{k=1}^{2N}\left(\frac{k}{k+1}\right)^{\left(-1\right)^{k}}=\prod_{k=1}^{N}\frac{2k}{2k+1}\prod_{k=1}^{N}\frac{2k}{2k-1}=\frac{\left(2N\right)!!^{2}}{\left(2N+1\right)!!\left(2N-1\right)!!}=\frac{\pi N!^{2}}{2\Gamma\left(N+\frac{1}{2}\right)\Gamma\left(N+\frac{3}{2}\right)} $$ where the last identity follows from the classic estimations for the double factorial. Hence it is sufficient to take the limit as $N\rightarrow\infty $ and get $$I=\color{red}{\log\left(\frac{\pi}{2}\right)}$$ as wanted. Maybe it is interesting to see that the we get a very famous product, known as the Wallis' product. See here for some proof of it.

Answer 3

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\,\mathcal{#1}} \newcommand{\mrm}[1]{\,\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{0}^{1}{x - 1 \over \pars{x + 1}\ln\pars{x}}\,\dd x} & = \int_{0}^{1}{1 \over 1 + x}\ \overbrace{\int_{0}^{1}x^{t}\,\dd t}^{\ds{x - 1 \over \ln\pars{x}}}\ \,\dd x = \int_{0}^{1}\int_{0}^{1}{x^{t} \over 1 + x}\,\dd x\,\dd t \\[5mm] & = \int_{0}^{1}\int_{0}^{1}{x^{t} - x^{t + 1} \over 1 - x^{2}}\,\dd x\,\dd t\,\,\, \stackrel{x\ \mapsto\ x^{1/2}}{=}\,\,\, \half\int_{0}^{1} \int_{0}^{1}{x^{t/2 - 1/2}\,\,\, -\,\,\, x^{t/2} \over 1 - x} \,\dd x\,\dd t \\[5mm] & = \half\int_{0}^{1}\bracks{\int_{0}^{1}{1 - x^{t/2} \over 1 - x}\,\,\,\dd x - \int_{0}^{1}{1 - x^{t/2 - 1/2} \over 1 - x}\,\,\,\dd x}\,\dd t \\[5mm] & = \half\int_{0}^{1}\bracks{\Psi\pars{{t \over 2} + 1} - \Psi\pars{{t \over 2} + \half}}\,\dd t \\[1mm] & \pars{~\Psi:\ Digamma\ Function~} \end{align} with the $\ds{\Psi}$ integral representation $\ds{\left.\Psi\pars{z} + \gamma = \int_{0}^{1}{1 - t^{z - 1} \over 1 - t} \,\dd t\,\right\vert_{\ \Re\pars{z}\ >\ 0}.\quad}$ $\ds{\gamma}$ is the Euler-Mascheroni Constant.

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Since $\ds{\Psi\pars{z} = \totald{\ln\pars{\Gamma\pars{z}}}{z}}$: \begin{align} \color{#f00}{\int_{0}^{1}{x - 1 \over \pars{x + 1}\ln\pars{x}}\,\dd x} & = \left.\ln\pars{\Gamma\pars{t/2 + 1} \over \Gamma\pars{t/2 + 1/2}}\right\vert_{\ 0}^{\ 1}\qquad \pars{~\Gamma:\ Gamma\ Function~} \\[5mm] & = \ln\pars{{\Gamma\pars{3/2} \over \Gamma\pars{1}}\, {\Gamma\pars{1/2} \over \Gamma\pars{1}}} = \ln\pars{\half\,\Gamma^{\, 2}\pars{\half}} = \color{#f00}{\ln\pars{\pi \over 2}} \end{align}

Above we used $\ds{\Gamma\pars{1} = 1}$, the $\ds{\Gamma}$-Recursion Formula $\ds{\Gamma\pars{z + 1} = z\,\Gamma\pars{z}}$ and $\ds{\Gamma\pars{\half} = \root{\pi}}$.