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:
$$
I \approx -0.3962906410900101751594101405188072631361627457\dots
$$
 A: $$\begin{align}
\int_0^{\infty} \frac{x\cos x-\sin x}{x(e^x-1)}dx=\int_0^{\infty} \frac{(x\cos x-\sin x)e^{-x}}{x(1-e^{-x})}dx
\\=\int_0^{\infty} \frac{x\cos x-\sin x}{x}\sum_{n=1}^{\infty}e^{-xn}\,\,dx
\\=\sum_{n=1}^{\infty}\left(\Re\int_0^{\infty}e^{-xn}e^{ix}dx-\tan^{-1}\left(\frac1{n}\right)\right)
\\=\sum_{n=1}^{\infty}\left(\Re\frac{1}{n-i}-\tan^{-1}\left(\frac1{n}\right)\right)
\\=\sum_{n=1}^{\infty}\left(\frac{n}{n^2+1}-\tan^{-1}\left(\frac1{n}\right)\right)
\\=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\tan^{-1}\left(\frac1{n}\right)-\frac{1}{n(n^2+1)}\right)
\\=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{(2m+1)n^{2m+1}}-\Re\sum_{n=1}^{\infty}\frac{i}{n(n+i)}
\\=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\zeta(2n+1)}{2n+1}-\Re{H_{i}}
\\=\Im\sum_{n=2}^{\infty} \frac{(-1)^n\zeta(n)}{n}i^n-\Re(H_{i}-\frac1{i})
\\=\Im\left(i\gamma+\ln(i)+\log(\Gamma(i))\right)-\Re H_{i-1}
\\=\gamma+\frac{\pi}{2}+\operatorname{arg}\Gamma(i)-\Re\left(\gamma+\psi(i)\right)
\\=\frac{\pi}{2}+\operatorname{arg}\Gamma(i)-\Re\psi(i).
\end{align}$$
Where I used $\,\,\,\,\,\displaystyle \sum_{n=2}^{\infty}\frac{(-1)^n\zeta(n)}{n}x^n=\gamma x+\log\Gamma(x+1)$ 
which follows from the logarithm of the Weierstrass-form Gamma function,
and $\,\,\,\,\,\displaystyle \int_0^{\infty} \frac{\sin x}{x}e^{-nx}dx=\tan^{-1}\left(\frac1{n}\right)\,\,\,\,$ is a standard exercise in differentiation under the integral sign.
A: We may write the original integral in the form:
$$ I = \text{Re}\int_{0}^{+\infty}\frac{(x+i)e^{ix}-i}{x(e^x-1)}\,dx $$
but since we have:
$$ \mathcal{L}\left(\frac{(x+i)e^{ix}-i}{x}\right)=\frac{1}{s-i}+i\left(\log(s)-\log(s-i)\right)$$
your claim follows from:
$$ \mathcal{L}^{-1}\left(\frac{1}{e^x-1}\right) = \delta(s-1)+\delta(s-2)+\ldots $$
and the Weierstrass product for the $\Gamma$ function, leading to the usual representations for $\log\Gamma$ and its derivative, $\psi$.
A: 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 $$\int_0^\infty\frac{1-\dfrac{\sin x}x}{e^x-1}~dx~=~\frac\pi2+\gamma+\arg\Big(\Gamma(i)\Big).$$
