I am trying to solve the contour integral
$$\frac{1}{2\pi i}\int_C\frac{e^{t}}{t(2at+x^2)}{\rm{exp}}\left(\frac{ax^2}{2(2at+x^2)}\right)\,{\rm d}t $$
where the path of integration $C$ starts at $-\infty-i0$ on the real axis, goes to $-\varepsilon-i0$, circles the origin in the counterclockwise direction with radius $\varepsilon$ to the point $-\varepsilon+i0$ and returns to the point $-\infty+i0$ (I got such path from Hankel's contour integral of reciprocal Gamma function $1/\Gamma(z)$).
I concluded that a given function has two poles: First one is $t=0$ and I also calculated the residual in that pole and it is equal to $\frac{e^{a/2}}{x^2}$; second one is at $t=-\frac{x^2}{2a}$ and I need to express a given function as a Laurent series and find a coefficient $c_{-1}$ which will be a residual... In order to do that first I used substitution $2at+x^2=y$ so then I can express my function around $t=0$: \begin{align*}\frac{2a*{\rm{exp}}\left(\frac{-x^2}{2a}\right)*{\rm{exp}}\left(\frac{y}{2a}\right)*{\rm{exp}}\left(\frac{ax^2}{2y}\right)}{-x^2y\left(1-\frac{y}{x^2}\right)}&= \frac{-2a*{\rm{exp}}\left(\frac{-x^2}{2a}\right)}{x^2y}\sum_{n\ge0}\left(\frac{y}{x^2}\right)^n\sum_{m\ge0}\frac{\left(\frac{y}{2a}\right)^m}{m!} \sum_{k\ge0}\frac{\left(\frac{ax^2}{2y}\right)^k}{k!}\\&=\frac{-2a*{\rm{exp}}\left(\frac{-x^2}{2a}\right)}{x^2}\sum_{n,m,k\ge0}\dfrac{y^{n+m-k-1}(ax^2)^k}{x^{2n}(2a)^m2^km!k!}\end{align*}
Please can you help me to find the desired coefficient...