Closed form for ${\large\int}_0^\infty\frac{x-\sin x}{\left(e^x-1\right)x^2}\,dx$ 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.
 A: This is another solution just for reference. 
The laplace transform of $\displaystyle\frac{x-\sin{x}}{x^2}$ is given by
\begin{align}
\mathcal{L}_s\left(\frac{x-\sin{x}}{x^2}\right)
&=\int^s_\infty\int^t_\infty\frac{1}{u^2}-\frac{1}{1+u^2}\ du\ dt\tag1\\
&=\int^s_\infty-\frac{1}{t}-\arctan{t}+\frac{\pi}{2}\ dt\\
&=-\ln{s}+\frac{1}{2}\ln(1+s^2)-s\arctan{s}+\frac{\pi}{2}s-1\\
&=\frac{1}{2}\ln\left(1+\frac{1}{s^2}\right)+s\arctan\left(\frac{1}{s}\right)-1
\end{align}
Thus
\begin{align}
\int^\infty_0\frac{x-\sin{x}}{x^2(e^x-1)}dx
&=\sum^\infty_{n=1}\mathcal{L}_n\left(\frac{x-\sin{x}}{x^2}\right)\tag2\\
&=\frac{1}{2}\sum^\infty_{n=1}\ln\left(1+\frac{1}{n^2}\right)+\sum^\infty_{n=1}\left(n\arctan\left(\frac{1}{n}\right)-1\right)\\
&=\frac{1}{2}\ln\left.\frac{\pi z}{\pi}\prod^\infty_{n=1}\left(1+\frac{z^2}{n^2}\right)\right|_{z=1}+\sum^\infty_{n=1}\sum^\infty_{k=1}\frac{(-1)^k}{(2k+1)n^{2k}}\tag{3}\\
&=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\sum^\infty_{k=1}\frac{(-1)^k\zeta(2k)}{(2k+1)}\tag4\\
&=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2i}\int^{i}_0\left(1-\pi z\cot(\pi z)\right)\ dz\tag5\\
&=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2}-\frac{1}{8\pi}\int^{\exp(-2\pi)}_1\frac{\ln{u}(1+u)}{u(1-u)}du\tag6\\
&=\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)+\frac{1}{2}-\frac{1}{8\pi}\left[2\mathrm{Li}_2(1-u)+\frac{\ln^2{u}}{2}\right]^{\exp(-2\pi)}_1\tag7\\
&=\color{#E2062C}{\frac{1}{2}-\frac{\pi}{4}+\frac{1}{2}\ln\left(\frac{\sinh(\pi)}{\pi}\right)-\frac{1}{4\pi}\mathrm{Li}_2\left(1-e^{-2\pi}\right)}
\end{align}
Explanation:
$(1)$: Differentiated under the integral twice. 
$(2)$: Expanded $(e^{x}-1)^{-1}$. 
$(3)$: Expanded $\arctan\left(n^{-1}\right)$. 
$(4)$: Recognised the Weierstrass product for $\sinh$, summed in $n$. 
$(5)$: Used the fact that $\displaystyle\pi z\cot(\pi z)=1-2\sum^\infty_{k=1}\zeta(2k)z^{2k}$. 
$(6)$: Substitued $u=e^{2\pi iz}$. 
$(7)$: $\displaystyle \frac{\ln{u}(1+u)}{u(1-u)}=\frac{2\ln{u}}{1-u}+\frac{\ln{u}}{u}$.
A: 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 $a=1$, 
$$\int_0^{\infty} \frac{x-\sin(x)}{x^2(e^x-1)}\,dx=\frac12+\frac{5}{24}\pi-\frac12\ln(2\pi)+\frac1{4\pi}\operatorname{Li}_2(e^{-2\pi }).$$
A: $$\frac{x-\sin x}{x^2}=\sum_{n\geq 1}\frac{(-1)^{n+1}}{(2n+1)!}x^{2n-1},$$
and since:
$$ \int_{0}^{+\infty}\frac{x^{2n-1}}{e^x-1}\,dx = (2n-1)!\cdot \zeta(2n),$$
we have:
$$\begin{eqnarray*} &&\int_{0}^{+\infty}\frac{x-\sin x}{x^2(e^x-1)}\,dx = \sum_{n\geq 1}\frac{(-1)^{n+1}}{2n(2n+1)}\zeta(2n)\\&=&\color{red}{\sum_{n\geq 1}\left(-1+n\arctan\frac{1}{n}+\frac{1}{2}\,\log\left(1+\frac{1}{n^2}\right)\right)}\\&=&\color{blue}{\log\sqrt{\frac{\sinh \pi}{\pi}}+\sum_{n\geq 1}\left(-1+n\arctan\frac{1}{n}\right)}.\tag{1}\end{eqnarray*} $$
Combining this identity with the robjonh's answer to another question, we finally get:
$$\color{purple}{\int_{0}^{+\infty}\frac{x-\sin x}{x^2(e^x-1)}\,dx=\frac{1}{2}+\frac{5\pi}{24}-\log\sqrt{2\pi}+\frac{1}{4\pi}\operatorname{Li}_2(e^{-2\pi})}.\tag{2}$$

On the other hand, the identity claimed by user111187, 
$$ \int_{0}^{+\infty}\frac{x-\sin x}{x(e^x-1)} = \gamma+\Im\log\Gamma(1+i)\tag{3} $$
follows from the integral representation for the $\log\Gamma$ function and for the Euler-Mascheroni constant. By considering the Weierstrass product for the $\Gamma$ function,
$$\Gamma(z+1) = e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{\frac{z}{n}}$$
we have:
$$ \log\Gamma(z+1) = -\gamma z + \sum_{n\geq 1}\left(\frac{z}{n}-\log\left(1+\frac{z}{n}\right)\right)$$
so:
$$ \int_{0}^{+\infty}\frac{x-\sin x}{x(e^x-1)}\,dx = \sum_{n\geq 1}\left(\frac{1}{n}-\arctan\frac{1}{n}\right).\tag{4}$$
