Consider the following integral for any $a \in \mathbb{R}$ and $b>0$: $$ I(a,b) \ = \ \int_{-\infty}^{\infty} \frac{ \exp\left( i a e^{u}\right) }{ e^{b \cosh(u)} - 1 } du $$
Is there any way to do this integral? My thoughts are to try and use the Residue theorem since the function falls off to 0 very quickly as you get further out from the origin.
However, the issue is then that there are infinitely poles (when $\cosh(z) = 2\pi i n$ for $n \in \mathbb{Z}$).
Is there a way to do this integral?
We have $$ \text{PV}\int_{0}^{+\infty}e^{iat}\exp\left[-\frac{b}{2}\left(t+\frac{1}{t}\right)\right]\frac{dt}{t} = 2\cdot K_0\left(\sqrt{b^2-2iab}\right) $$ where $K_0$ is a modified Bessel function of the second kind (appearing, for instance, in the normal product distribution); hence by enforcing the substitution $u=\log t$ and by exploiting a geometric series expansion we have $$ I(a,b) = 2\sum_{k\geq 1} K_0\left(\sqrt{k^2 b^2-2ikab}\right) $$ where the RHS is simple to approximate numerically due to the rapid decay of $K_0$ on $\mathbb{R}^+$.
I do not believe we may get any simpler "closed" form.