A possible closed form?

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!

Solutions Collecting From Web of "A possible closed form?"

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\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}

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.