I want to represent the function:
\begin{equation} f(x)=e^{-a(x-b)^{2}} \end{equation}
where, $0<a<1$, $x\in\mathbb{R}$, and $b\in\mathbb{R}$.
As a power series for an integral I am working on. First, I re-wrote $f(x)$ using the power series definition of the exponential to get,
\begin{equation} f(x) = \sum_{n=0}^{\infty} \frac{(-a)^{n}}{n!}(x-b)^{2n} \end{equation}
Using binomial expansion on the $(x-b)^{2n}$ yields
\begin{equation} f(x) = \sum_{n=0}^{\infty} \frac{(-a)^{n}}{n!}\sum_{k=0}^{2n}\binom{2n}{k}(-b)^{2n-k}\,x^{k} \end{equation}
I want to then simplify this expression by switching the order of summation. Here is what I did:
\begin{equation} \begin{aligned} f(x) &= \sum_{k=0}^{\infty}x^{k} \sum_{n=k/2}^{\infty}\binom{2n}{k}\frac{(-a)^{n}}{n!}(-b)^{2n-k}\\ &=\sum_{k=0}^{\infty}\frac{(-\sqrt{-a}x)^{k}}{\frac{k}{2}!}\,{_{1}}F_{1}\left[\frac{k+1}{2},\frac{1}{2},-ab^{2}\right] \end{aligned} \end{equation}
where, ${_{1}}F_{1}(a,b,z)$ is the confluent hypergeometric function of the 1st kind.
At this point the form of $f(x)$ makes the solution to my integral much simpler. However, I am not sure if switching the order of summation like this is allowed under these circumstances. Looks like the sum has an imaginary component that one would not expect. Any thoughts?