# 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!

• maybe apply fraction division? divide both side by $x*(e^{x}-1)$ – dato datuashvili Feb 1 '14 at 14:08
• Perhaps convert $\sin$ and $\cos$ to exponentials? I haven't thought it out, but that might be an approach... – apnorton Feb 1 '14 at 14:17
• Please, post the closed form (my result seems to be close to -1/pi). – Claude Leibovici Feb 1 '14 at 14:22

Also $$\totald{}{\mu}\int_{0}^{\infty}{\sin\pars{x}\expo{-\mu x} \over x}\,\dd x = -\Im\int_{0}^{\infty}\expo{\pars{\ic - \mu}x}\,\dd x =-\Im\pars{1 \over \mu - \ic} = -\,{1 \over \mu^{2} + 1}$$ $$\int_{0}^{\infty}{\sin\pars{x}\expo{-\pars{\ell + 1}x} \over x}\,\dd x = {\pi \over 2} - \int_{0}^{\ell + 1}{\dd\mu \over \mu^{2} + 1} ={\pi \over 2} - \arctan\pars{\ell + 1} = \arctan\pars{1 \over \ell + 1}$$
$$\int _{0}^{\infty }\!{\frac {x\cos \left( x \right) -\sin \left( x \right) }{{x} \left( {{\rm e}^{x}}-1 \right) }}{dx} = \frac{\pi}{2}+\arg\left(\Gamma\left(i\right)\right)-\Re\left(\psi_0\left(i\right)\right).$$
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.