TL;DR I'm doing an integral and have reason to believe it can be further simplified, but cannot for the life of me (nor that of Mathematica) manage to massage it any further.
I'm working on a research problem in physics (computing a two-point function in a certain toy model of low-dimensional gravity, for those curious about the context) and have reduced my expression to a linear combination of integrals of the form \begin{equation} I_n(\alpha) := \int_0 ^\pi d\theta e^{- \alpha \cos(\theta) + i n \theta} \ \ \ (n \in \mathbb{Z}, \alpha \in \mathbb{R}). \end{equation} My life would be immensely improved if I had a nice expression for this integral. Here's what I've managed so far: For $n \geq 0$ we can apply De Moivre, expand as a binominal in $\sin(\theta)$ and $\cos(\theta)$, and then apply Feynman's trick to differentiate under the integral sign. \begin{align*} \int_0^\pi d\theta e^{- \alpha \cos(\theta)} e^{i n \theta} & = \int_0^\pi d\theta e^{- \alpha \cos(\theta)} (\cos(\theta) + i \sin(\theta))^n \\ \ & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \int_0^\pi d\theta e^{- \alpha \cos(\theta)} \cos^{n-k}(\theta) \sin^k(\theta) \\ \ & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \left( - \frac{d}{d \alpha} \right)^{n-k} \int_0^\pi d\theta e^{- \alpha \cos(\theta)} \sin^k(\theta). \end{align*} Substituting $y \equiv \cos (\theta) \implies d \theta = - (1 - y^2)^{-1/2} dy$ then gives \begin{align*} \int_0^\pi d\theta e^{- \alpha \cos(\theta)} e^{i n \theta} & = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \left( - \frac{d}{d \alpha} \right)^{n-k} \int_{-1}^1 dy e^{- \alpha y} (1 - y^2)^{\frac{k-1}{2}} \end{align*} This inner integral can be done in Mathematica, and we find that \begin{equation*} \therefore \int_0^\pi d\theta e^{- \alpha \cos(\theta)} e^{i n \theta} = - \frac{\alpha \sqrt{\pi}}{2} \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \Gamma \left( \frac{1+k}{2} \right) \left( - \frac{d}{d \alpha} \right)^{n-k} F \left( 1 + \frac{k}{2}, \frac{\alpha^2}{4} \right) \end{equation*} where where $F$ is the regularized confluent hypergeometric function $_0 F_1(; a; z)/\Gamma(a)$ (implemented in Mathematica as Hypergeometric0F1Regularized).
Unfortunately, while technically I've done the integral (in the sense that there is no longer an $\int$ sign on the RHS), the end result isn't much more clarifying than what I started with.
If I opt not to differentiate under the integral sign, the resulting form \begin{equation*} I_n(\alpha) = \sum_{k=0}^n \begin{pmatrix} n \\ k \end{pmatrix} \int_{-1}^1 dy e^{- \alpha y} (1 - y^2)^{\frac{k-1}{2}} y^{n-k} \end{equation*} has an inner integral with an exact expression, but the latter is built from regularized generalized hypergeometric functions and gamma functions sufficiently messy that it doesn't help much, and certainly doesn't admit a neat sum over $k$. It would be great if, for example, I had a nice simple expression for
\begin{equation*} \int_0^\pi d\theta e^{- \alpha \cos (\theta)} \cos(n \theta) \mbox{ and } \int_0^\pi d\theta e^{- \alpha \cos (\theta)} \sin(n \theta) \end{equation*}
so that I could avoid the binomial expansion.
The integral $I_n(\alpha)$ smells like it ought to be doable by a clever substitution, given that it's built so simply, so any help would be appreciated.
Edit: for further context if it's helpful or interesting, at one level zoomed out I am trying to compute \begin{equation*} \int_0 ^\pi d \theta e^{- \alpha \cos (\theta)} \frac{(e^{\pm i 2 \theta}; q)_\infty}{(\tilde{q} e^{\pm i \theta}; q)_\infty} \end{equation*} where $(a;q)_\infty$ is the $q$-Pochhammer symbol, each $\pm$ is shorthand for the product $(a_\pm;q)_\infty := (a_+;q)_\infty (a_-;q)_\infty$, and $q,\tilde{q}$ are positive constants less than 1. I've attacked this by essentially expanding as a series in $e^{i \theta}$ using the identities found at https://en.wikipedia.org/wiki/Q-Pochhammer_symbol, which gives a linear combination of integrals of the form $I_n(\alpha)$ for various $n \in \mathbb{Z}$. I could be barking up the wrong tree with that approach though.