TITLE: Evaluate $\int_0^{\infty}\frac{e^x-1}{xe^x(e^x+1)}dx.$
QUESTION [7 upvotes]: Evaluate the integral
  $$
\int_{0}^{\infty}{\mathrm{e}^{x} - 1 \over x\,\mathrm{e}^{x}\left(\mathrm{e}^x+1\right)}\,\mathrm{d}x\,.
$$

I have no idea how to approach integrals like this and I can't get any valuable result. Any hints will be appreciated.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \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}^{\infty}{\expo{x} - 1 \over x\expo{x}\pars{\expo{x} + 1}}\,\dd x & =
\int_{0}^{\infty}{\expo{-x} - \expo{-2x} \over x\pars{1 + \expo{-x}}}\,\dd x
\,\,\,\stackrel{\expo{-x}\ =\ t}{=}\,\,\,
\int_{1}^{0}{t - t^{2} \over \bracks{-\ln\pars{t}}\pars{1 + t}}
\,\pars{-\,{\dd t \over t}}
\\[5mm] & =
\int_{0}^{1}{1 \over 1 + t}\,\
\overbrace{{t - 1 \over \ln\pars{t}}}^{\ds{\int_{0}^{1}t^{x}\,\dd x}}\
\,\dd t =
\int_{0}^{1}\int_{0}^{1}{t^{x} - t^{x + 1} \over 1 - t^{2}}\,\dd t\,\dd x
\\[5mm] & \stackrel{t^{2}\ \mapsto\ t}{=}\,\,\,
{1 \over 2}\int_{0}^{1}
\int_{0}^{1}{t^{\pars{x - 1}/2}\,\,\, -\,\,\, t^{x/2} \over 1 - t}\,\dd t\,\dd x \\[5mm] & =
{1 \over 2}\int_{0}^{1}\bracks{%
\int_{0}^{1}{1 - t^{x/2} \over 1 - t}\,\dd t -
\int_{0}^{1}{1 - t^{\pars{x - 1}/2} \over 1 - t}\,\dd t}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{1}
\bracks{\Psi\pars{{x \over 2} + 1} - \Psi\pars{x + 1 \over 2}}\,\dd x
\end{align}
where
$\ds{\Psi}$ is the Digamma Function and we used the well known identity
$$
\left.\vphantom{\LARGE A}\Psi\pars{z}\right\vert_{\,\Re\pars{z}\ >\ 0} = -\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t
$$
$\ds{\gamma}$ is the Euler-Mascheroni Constant. Then,
\begin{align}
\int_{0}^{\infty}{\expo{x} - 1 \over x\expo{x}\pars{\expo{x} + 1}}\,\dd x & =
\left.\ln\pars{\Gamma\pars{x/2 + 1} \over
\Gamma\pars{\bracks{x + 1}/2}}\right\vert_{\ 0}^{\ 1} =
\ln\pars{{\Gamma\pars{3/2} \over \Gamma\pars{1}}\,
{\Gamma\pars{1/2} \over \Gamma\pars{1}}} =
\ln\pars{{1 \over 2}\,\Gamma^{2}\pars{1 \over 2}}
\\[5mm] & =
\bbox[15px,#ffe,border:2px dotted navy]{\ds{\ln\pars{\pi \over 2}}}
\end{align}