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.

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 $u=e^{2\pi iz}$. 
$(7)$: $\displaystyle \frac{\ln{u}(1+u)}{u(1-u)}=\frac{2\ln{u}}{1-u}+\frac{\ln{u}}{u}$.
