Prove that $\int_0^\infty\frac{x\cos(x)-\sin(x)}{x\left({e^x}-1\right)}\,dx = \frac{\pi}2+\arg\left(\Gamma(i)\right)-\Re\left(\psi_0(i)\right)$ While I was working on this question, I've found that
$$
I=\int_0^\infty\frac{x\cos(x)-\sin(x)}{x\left({e^x}-1\right)}\,dx = \frac{\pi}2+\arg\left(\Gamma(i)\right)-\Re\left(\psi_0(i)\right),
$$
where $\arg$ is the complex argument, $\Re$ is the real part of a complex number, $\Gamma$ is the gamma function, $\psi_0$ is the digamma function.

How could we prove this? Are there any more simple closed-form?

A numerical approximation:
$$
I \approx -0.3962906410900101751594101405188072631361627457\dots
$$

Too long for a comment: After simplifying the integrand with x, write the $($new$)$ 
numerator as $\bigg(1-\dfrac{\sin x}x\bigg)-(1-\cos x),~$ then, for the latter, use Euler's formula 
in conjunction with an exponential substitution to $($formally$)$ express the second 
integral as a sum of harmonic numbers of complex argument, whose connection 
with the digamma function is well-known. Now “all” that's left to do is showing that $$\int_0^\infty\frac{1-\dfrac{\sin x}x}{e^x-1}~dx~=~\frac\pi2+\gamma+\arg\Big(\Gamma(i)\Big).$$
