I am looking for ways to do this integration analytically
\begin{equation}
\int\_0^\infty \mathrm{d}p\frac{e^{-p \sin (\phi )} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)}
\end{equation}


For context:


I encountered an integral of the form
$$
E\_b=\pi+\int \frac{i\left(e^{c|p|}+e^{2|p| \sin (\phi)}\right) e^{-|p|(c+\sin (\phi))+i p \cos (\phi)}}{p\left(e^{c|p|}+1\right)} \mathrm{d} p-2 i \tanh ^{-1}\left(1+\frac{2 i e^{i \phi}}{c}\right)+2 i \tanh ^{-1}\left(1-\frac{2 i e^{-i \phi}}{c}\right)-2 \pi
$$
as an energy of a particle in a quantum field theory model Hamiltonian that I am looking at.


The curious property of the above expression that I can verify *Numerically* is that for $c>0, 0<\phi<\pi/2$ and $3c<2\sin(\phi)$, the expression is negative for $c<\sin(\phi)$ and positive for $c>\sin(\phi)$.


I would like to find an analytic expression of the energy.


To do so, I wrote the integral as
$$\int \frac{i \left(e^{c | p| }+e^{2 | p| \sin (\phi )}\right) e^{-| p| (c+\sin (\phi ))+i p \cos (\phi )}}{p \left(e^{c | p| }+1\right)}\mathrm{d}p=-2\int\_0^{\infty } \frac{e^{-p \sin (\phi )} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, dp-2\int\_0^{\infty } \frac{e^{p \sin (\phi )-c p} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)} \, dp$$


But I failed to compute this Laplace transform of $\frac{\sin(ax)}{x(e^{bx}+1)}$.


Any ideas would be greatly appreciated.



Analytic expression for $\int_0^\infty \mathrm{d}p\frac{e^{-p \sin (\phi )} \sin (p \cos (\phi ))}{p \left(e^{c p}+1\right)}$
We seek \begin{align}I&=\int\_0^\infty\frac{e^{-st}\sin t}{t(e^{at}+1)}\,dt\\&=\sum\_{n\ge0}(-1)^n\int\_0^\infty\frac{e^{-(s+a(n+1))t}\sin t}t\,dt=-\sum\_{n\ge1}(-1)^n\arctan\frac1{s+an}\end{align} which has a "closed form" involving complex log-gamma; for instance, [see the case $(s,a)=(0,1)$](https://math.stackexchange.com/questions/776182/a-closed-form-for-the-infinite-series-sum-n-1-infty-1n1-arctan-left?rq=1) which is readily extendable to our case.


Explicitly, $$I=\Im\log\frac{\Gamma\left(1+\frac{s+i}{2a}\right)\Gamma\left(\frac12+\frac s{2a}\right)}{\Gamma\left(1+\frac s{2a}\right)\Gamma\left(\frac12+\frac{s+i}{2a}\right)}$$


