I want to solve the integral:
$$ f(x) = \int_{-\infty}^{\infty} \frac{ e^{ -\frac{1}{2} \frac{\rho}{\beta} (t - j \alpha \frac{\rho}{\beta})^{2} }}{(t-j \rho)^{M} (t+j \rho)^{M}} dt $$
where $j$ is the imaginary unity, with $ \beta, \rho \in \mathbb{R}_{++}$ and $\alpha \in \mathbb{R}_{+}$. The numerator is simply a complex Gaussian, with a peak at $t = j \alpha \frac{\rho}{\beta}$, which is analytic. If we take a contour in the upper half plane, encircling $t = j \rho$, then we know as the radius of our contour $R \rightarrow \infty$, we have that the numerator $e^{ -\frac{1}{2} \frac{\rho}{\beta} (t - j \alpha \frac{\rho}{\beta})^{2} } \rightarrow 0$.
Our integral therefore reduces to: $$ f(x) =j \cdot \text{Res}(z = j \rho) $$
The residue is:
$$ \text{Res}(z = j \rho) = \underset{z \rightarrow j \rho}{\text{lim}} \frac{1}{(M-1)!} \frac{d^{M-1}}{dz^{M-1}} \left[ (z-j\rho)^{M} \frac{ e^{ -\frac{1}{2} \frac{\rho}{\beta} (z - j \alpha \frac{\rho}{\beta})^{2} }}{(z-j \rho)^{M} (z+j \rho)^{M}} \right] $$
$$ \text{Res}(z = j \rho) = \underset{z \rightarrow j \rho}{\text{lim}} \frac{1}{(M-1)!} \frac{d^{M-1}}{dz^{M-1}} \left[\frac{ e^{ -\frac{1}{2} \frac{\rho}{\beta} (z - j \alpha \frac{\rho}{\beta})^{2} }}{ (z+j \rho)^{M}} \right] $$
We can compute this residue using the Generalized Leibniz rule:
$$ \frac{d^{M-1}}{dz^{M-1}} [ p(z) \cdot q(z)] = \sum_{k=0}^{M-1} \binom{M-1}{k} \frac{d^{M-1-k}}{dz^{M-1-k}} p(z) \cdot \frac{d^{k}}{dz^{k}} q(z) $$
with $p(z) = (z+j \rho)^{M}$ and $q(z) = e^{ -\frac{1}{2} \frac{\rho}{\beta} (z - j \alpha \frac{\rho}{\beta})^{2} }$
Question:
Is there a simpler way? I know I can evaluate $\frac{d^{k}}{dz^{k}} e^{ -\frac{1}{2} \frac{\rho}{\beta} (z - j \alpha \frac{\rho}{\beta})^{2}}$ via Faa di Bruno's formula (https://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula) or via the method presented here (http://www.m-hikari.com/ams/ams-2012/ams-73-76-2012/oliveiraAMS73-76-2012-1.pdf).
Doing either gives me a complicated mess though. Is there a simpler way? Is this a standard integral of any type? Am I missing an obvious simplification?
Thanks!