$$\int_0^\infty \text{d}x \frac{\sin ax}{e^{2\pi x}-1}=\frac{1}{4}\coth \frac{a}{2}-\frac{1}{2a}$$
Is there a closed form solution to this, slightly different integral (in terms of $a$)?
$$\int_{\color\red{-\infty}}^\infty \text{d}x \frac{\sin ax}{e^{2\pi x}-\color\red{x}}$$
The manipulation is too difficult for me, but I have tried generating an inverse Mellin transform by substituting $x=iu$
That actually got me kind of far, but failed because there was still the $iu$ term left. Thanks in advance!