I have been fighting with a contour integral with the following integrand
\begin{align*} f(z)=\frac{1}{1-az} e^{-\frac{i t}{2}\left(z+z^{-1}\right)} \end{align*} for $a,t>0$ real constants. I think this has a simple pole at $z=a^{-1}$ and two essential essential singularities at $z=0$ and $z=\infty$. I find the residue for the simple pole to be \begin{align*} \text{Res}\left(f(z),z=a^{-1}\right) = -\frac{1}{a} e^{-\frac{i t}{2}\left(a+a^{-1}\right)} \end{align*} For the essential singularity at $z=0$, I use the Laurent series \begin{align*} e^{-\frac{i t}{2}\left(z+z^{-1}\right)} = \sum\limits_{n\in\mathbb{Z}} z^n I_{n}(-it) = \sum\limits_{n=0}^{\infty} \left(z^n +\frac{1}{z^n}\right) I_{n}(-it) - I_{0}(-it) \end{align*} Where in the second equality I used $I_{-n}(t)=I_{n}(t)$ and subtracted the double counting. Similarly, around $z=0$ I use \begin{align*} (1-az)^{-1} = \sum\limits_{n=0}^{\infty} a^n z^n \end{align*} And so, \begin{align*} \frac{1}{1-az} e^{-\frac{i t}{2}\left(z+z^{-1}\right)} = \sum\limits_{n=0}^{\infty} \sum\limits_{k=0}^{\infty}\left(z^{2n-k}+z^{-k}\right) a^{n-k}I_{n}(-it) -I_{0}(-it)\sum\limits_{n=0}^{\infty}a^n z^n \end{align*} And the term with power $z^{-1}$ comes from $k=1$, \begin{align*} \text{Res}\left(f(z),z=0\right) = \sum\limits_{n=0}^{\infty} a^{n-1}I_{n}(-it) = \sum\limits_{n=0}^{\infty} a^{n-1}I_{n}(-it) \end{align*} Is this sensible? I am unsure about the convergence and about the product of the series. In the neighbourhood of $\infty$, the Laurent series of the exponential remains the same (symmetry under $z\to z^{-1}$), while $(1-ax)^{-1} = -\sum (a z)^{-(n+1)}$?
You are correct. In a punctured neighbourhood of $z=0$ we have
$$ f(z)=\frac{1}{1-az}e^{\frac{it}{2}\left(z+\frac{1}{z}\right)}=\sum_{l\geq 0}a^l z^l \sum_{m\geq 0}\frac{1}{m!}\left(\frac{it}{2}\right)^m z^m \sum_{n\geq 0}\frac{1}{n!}\left(\frac{it}{2}\right)^n \frac{1}{z^n}$$ hence $$ \text{Res}\left(f(z),z=0\right) = \sum_{l\geq 0}a^l\sum_{m\geq 0}\frac{1}{m!}\left(\frac{it}{2}\right)^m\frac{1}{(m+l+1)!}\left(\frac{it}{2}\right)^{l+m+1}$$ can be written as $$ \sum_{l\geq 0}a^l i^{l+1} J_{l+1}(t) $$ with $J_{l+1}$ being a Bessel function of the first kind. $J_{l+1}(t)$ is rapidly decreasing to zero as $l\to +\infty$, hence the previous series is absolutely convergent for any $a$. The residue at infinity can be computed in a similar way, by expanding $\frac{1}{1-az}$ as mentioned.