<s>[INST] 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.
 [/INST] 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 $u=e^{2\pi iz}$. 
$(7)$: $\displaystyle \frac{\ln{u}(1+u)}{u(1-u)}=\frac{2\ln{u}}{1-u}+\frac{\ln{u}}{u}$.
</s>