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Does it have a closed form?

$$\int _{0}^{\infty }\!{\frac {x\cos \left( x \right) -\sin \left( x \right) }{{x} \left( {{\rm e}^{x}}-1 \right) }}{dx}$$

EDIT: no need for answer, I just found the closed form. Thanks!

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  • $\begingroup$ maybe apply fraction division? divide both side by $x*(e^{x}-1)$ $\endgroup$ – dato datuashvili Feb 1 '14 at 14:08
  • $\begingroup$ Perhaps convert $\sin$ and $\cos$ to exponentials? I haven't thought it out, but that might be an approach... $\endgroup$ – apnorton Feb 1 '14 at 14:17
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    $\begingroup$ Please, post the closed form (my result seems to be close to -1/pi). $\endgroup$ – Claude Leibovici Feb 1 '14 at 14:22
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\int _{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x\pars{\expo{x} - 1}}\,\dd x = \Re\int _{0}^{\infty} \bracks{\expo{\ic x} - {\sin\pars{x} \over x}}\,{1 \over \expo{x} - 1}\,\dd x \\[3mm]&= \Re\int _{0}^{\infty} \bracks{\expo{\pars{\ic - 1}x} - {\sin\pars{x}\expo{-x} \over x}}\, {\dd x \over 1 - \expo{x}} = \Re\int _{0}^{\infty} \bracks{\expo{-\pars{1 - \ic}x} - {\sin\pars{x}\expo{-x} \over x}}\, \sum_{\ell = 0}^{\infty}\expo{-x}\,\dd x \\[3mm]&= \Re\sum_{\ell = 0}^{\infty}\int _{0}^{\infty} \bracks{\expo{-\pars{\ell + 1 - \ic}x} - {\sin\pars{x}\expo{-\pars{\ell + 1}x} \over x}}\,\dd x \end{align}

Also $$ \totald{}{\mu}\int_{0}^{\infty}{\sin\pars{x}\expo{-\mu x} \over x}\,\dd x = -\Im\int_{0}^{\infty}\expo{\pars{\ic - \mu}x}\,\dd x =-\Im\pars{1 \over \mu - \ic} = -\,{1 \over \mu^{2} + 1} $$ $$ \int_{0}^{\infty}{\sin\pars{x}\expo{-\pars{\ell + 1}x} \over x}\,\dd x = {\pi \over 2} - \int_{0}^{\ell + 1}{\dd\mu \over \mu^{2} + 1} ={\pi \over 2} - \arctan\pars{\ell + 1} = \arctan\pars{1 \over \ell + 1} $$

\begin{align} &\color{#00f}{\large\int _{0}^{\infty}{x\cos\pars{x} - \sin\pars{x} \over x\pars{\expo{x} - 1}}\,\dd x = \Re\sum_{\ell = 0}^{\infty} \bracks{{1 \over \ell + 1 - \ic} - \arctan\pars{1 \over \ell + 1}}} \end{align}

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Is this what you have found?

$$\int _{0}^{\infty }\!{\frac {x\cos \left( x \right) -\sin \left( x \right) }{{x} \left( {{\rm e}^{x}}-1 \right) }}{dx} = \frac{\pi}{2}+\arg\left(\Gamma\left(i\right)\right)-\Re\left(\psi_0\left(i\right)\right).$$

Here $\arg$ is the complex argument, $\Re$ is the real part of a complex number, $\Gamma$ is the gamma function, $\psi_0$ is the digamma function, $i$ is the imaginary unit and $\pi$ is also a famous constant.

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