I need to take the residue of the following function at infinity: $$ f(z) = \oint_\infty \left(\frac{e^{-\alpha/z}e^{-\alpha z}}{z}\right)dz $$ Which, up to a sign, is invariant under inversions $z\rightarrow 1/w$, since $dz \rightarrow -\frac{1}{w^2} dw$.
How can I proceed to take a Laurent expansion of this function at $z = \infty$ (or $z = 0$)?
We have $e^{-\frac \alpha z}=\sum_{n=0}^\infty \left(-\frac \alpha z\right)^n$, so $$\frac{e^{-\frac \alpha z}}{z}=\frac1z-\frac \alpha{z^2}+\cdots.$$ Also $\lim_{z\to 0} e^{-\alpha z}=1$, hence $$\text{Res}_{z=0}\frac{e^{-\frac \alpha z}e^{-\alpha z}}{z}=1.$$ If you want to calculate the residue at $\infty$ directly, you may expand $e^{-\alpha z}$ and you will get the same answer.
Around $z = 0$, the exponential is the generating function for the Modified Bessel function, e.g. $$ e^{\frac{t}{2}(z + 1/z)} = \sum_{n=-\infty}^\infty I_n(t)z^n. $$
Thus, we find
$$ f(z) = \oint_\infty \left(\frac{e^{-\alpha/z}e^{-\alpha z}}{z}\right)dz = \oint_\infty \frac{dz}{z}\sum_{n=-\infty}^\infty I_n(2\alpha)z^n = I_0(2\alpha). $$ Which is the correct answer.