TITLE: Closed form for ${\large\int}_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$
QUESTION [14 upvotes]: I'm interested in a closed form for this simple looking integral:
$$I=\int_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$$
Numerically,
$$I\approx0.235708612100161734103782517656481953570915076546754616988...$$
Note that if we try to split the integral into two parts, each with only one term in the numerator, then both parts will be divergent.

REPLY [3 votes]: Here is a bit of a generalization, and what I think an easier solution.
Start with $\displaystyle \int_0^{\infty} \sin(ax) \,e^{-xn}\,dx=\frac{a}{a^2+n^2}$. (Easily established through integration by parts.)
Integrate both sides w.r.t. $a$ from $0$ to $t$, to get $\displaystyle \int_0^{\infty} \frac{1-\cos(tx)}{x} \,e^{-xn}\,dx=\frac12 \ln\left(1+\frac{t^2}{n^2}\right).$ 
Sum both sides from $n=1$ to $\infty$, to get $\displaystyle \int_0^{\infty} \frac{1-\cos(tx)}{x(e^x-1)}\,dx=\frac12 \ln \dfrac{\sinh(\pi t)}{\pi t} $
Finally, integrate both sides w.r.t. $t$ from $0$ to $a$, to get:
$$\int_0^{\infty} \frac{xa-\sin(xa)}{x^2(e^x-1)}\,dx=\frac12 \int_0^a \ln \frac{\sinh(\pi t)}{\pi t}\,dt
\\\\=-\frac12 a \ln\pi-\frac12(a\ln a-a)+\frac12\int_0^a (\ln(1-e^{-2\pi t})-\ln2+\pi t)\,dt
\\\\=-\frac12 a \ln(2\pi)-\frac12 a\ln a+\frac{a}{2}+\frac{\pi}{4}a^2-\frac12\int_0^a \sum_{n=1}^{\infty} \frac{e^{-2\pi t n}}{n}\,\,dt
\\\\=\frac{\pi}{24}(6a^2-1)+\frac{a}{2}(1-\ln(2\pi))-\frac12 a\ln a+\frac1{4\pi}\operatorname{Li}_2(e^{-2\pi a})
$$
In particular, when $a=1$, 
$$\int_0^{\infty} \frac{x-\sin(x)}{x^2(e^x-1)}\,dx=\frac12+\frac{5}{24}\pi-\frac12\ln(2\pi)+\frac1{4\pi}\operatorname{Li}_2(e^{-2\pi }).$$