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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: $$ I \approx -0.3962906410900101751594101405188072631361627457\dots $$

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$$\begin{align} \int_0^{\infty} \frac{x\cos x-\sin x}{x(e^x-1)}dx=\int_0^{\infty} \frac{(x\cos x-\sin x)e^{-x}}{x(1-e^{-x})}dx \\=\int_0^{\infty} \frac{x\cos x-\sin x}{x}\sum_{n=1}^{\infty}e^{-xn}\,\,dx \\=\sum_{n=1}^{\infty}\left(\Re\int_0^{\infty}e^{-xn}e^{ix}dx-\tan^{-1}\left(\frac1{n}\right)\right) \\=\sum_{n=1}^{\infty}\left(\Re\frac{1}{n-i}-\tan^{-1}\left(\frac1{n}\right)\right) \\=\sum_{n=1}^{\infty}\left(\frac{n}{n^2+1}-\tan^{-1}\left(\frac1{n}\right)\right) \\=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\tan^{-1}\left(\frac1{n}\right)-\frac{1}{n(n^2+1)}\right) \\=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{(2m+1)n^{2m+1}}-\Re\sum_{n=1}^{\infty}\frac{i}{n(n+i)} \\=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\zeta(2n+1)}{2n+1}-\Re{H_{i}} \\=\Im\sum_{n=2}^{\infty} \frac{(-1)^n\zeta(n)}{n}i^n-\Re(H_{i}-\frac1{i}) \\=\Im\left(i\gamma+\ln(i)+\log(\Gamma(i))\right)-\Re H_{i-1} \\=\gamma+\frac{\pi}{2}+\operatorname{arg}\Gamma(i)-\Re\left(\gamma+\psi(i)\right) \\=\frac{\pi}{2}+\operatorname{arg}\Gamma(i)-\Re\psi(i). \end{align}$$

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.

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  • $\begingroup$ Could you provide a link or hint as to why $\frac{1}{n}-\tan^{-1}\left(\frac1{n}\right)=\sum_{m=1}^{\infty} \frac{(-1)^{m+1}}{(2m+1)n^{2m+1}}$ $\endgroup$ – user5389726598465 Jun 4 '17 at 19:00
  • $\begingroup$ Also, $\int_0^{\infty} \frac{\sin x}{x}e^{-nx}dx=-\tan^{-1}\left(n\right)$ (ex3.2:fy.chalmers.se/~tfkhj/FeynmanIntegration.pdf)I don't see how that can equal $tan^{-1}(1/n)$ if $\tan(x)+\tan(1/x)=\pi/2$. If you change the constant of integration to convert to cot, the original formula won't have the right value at infinity. $\endgroup$ – user5389726598465 Jun 4 '17 at 19:06
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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, $\psi$.

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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 $~\displaystyle\int_0^\infty\frac{1-\dfrac{\sin x}x}{e^x-1}~dx~=~\frac\pi2+\gamma+\arg\Big(\Gamma(i)\Big)$.

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