Getting a Hermite polynomial expansion of Gaussian with given variance. I am trying to find an expansion of centered Gaussian - $\frac{1}{\sqrt{2\pi}\sigma}\exp({-\frac{x^2}{2\sigma^2})}$ in terms of Hermite polynomials. 
Namely to calculate $a_n=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}{\exp({-\frac{x^2}{2\sigma^2}})}H_{n}(x)\exp({-\frac{x^2}{2})}dx$ 
Any comments are welcome.

Edited later:  
"""
Equivalently, I am looking for the value of - 
$a_n=\int_{-\infty}^{\infty}{\exp({-\frac{x^2}{\alpha}})}H_{n}(x)dx$
for some arbitrary $\alpha$
"""
 A: If $\hat{H}_{n}$ are the normalized Hermite functions, and $-1 < r < 1$, then
$$
\begin{align}
     \sum_{n=0}^{\infty}r^{n}\hat{H}_{n}(x)^{2} & =\frac{1}{\sqrt{\pi(1-r^{2})}}\exp\left(-\frac{1-r}{1+r}x^{2}+x^{2}\right) \\
          & = \frac{1}{\sqrt{\pi(1-r^{2})}}\exp\left(\frac{2r}{1+r}x^{2}\right).
\end{align}
$$
The normalized $\hat{H}_{n}$ are chosen so that $\int_{-\infty}^{\infty}\hat{H}_{n}(x)e^{-x^{2}}dx = 1$. By choosing $r$ appropriately, you can get what you want. Look for Mehler's kernel on this Wikipedia page: http://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions . You'll find more information about how one derives the above special case of Mehler's kernel $\sum_{n=0}^{\infty}r^{n}H_{n}(x)H_{n}(y)$ where $x=y$.
A: The Hermite expansion of the probability distribution function for $\mathcal{N}(0,\sigma^2)$
$$
 \omega_\sigma(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}
$$
is
$$
 \omega_\sigma(x)=\sum_{m=0}^\infty \frac{(-1)^m}{m!2^m \sqrt{2\pi\left(\sigma^2+1\right)^{2m+1}}}H_{2m}(x).
$$
The starting point is the expression for the general coefficient for any function $f(x)=\sum_n d_n H_n(x)$ is given by the expression
$$
 d_n = \frac{1}{n!}\int_{-\infty}^\infty \frac{\partial^n f(x)}{\partial x^n} \omega(x) dx,
$$
which, by the definition of the stretched Hermite polynomials $ H_n\left(\frac{x}{\sigma}\right) = \frac{(-\sigma)^n}{\omega_\sigma(x)}\frac{\partial^n}{\partial x^n}\omega_\sigma(x)$,
reduces to
$$
  d_n = \frac{1}{(-\sigma)^nn!\sqrt{2\pi\sigma^2}}\int_{-\infty}^\infty H_n\left(\frac{x}{\sigma}\right) \omega_\sigma(x)\omega(x)dx.
$$
The product $\omega_\sigma(x)\omega(x)$ can be transformed into $\omega(y)$ by the change of variables $y = \frac{x}{\sigma}\sqrt{\sigma^2+1}$.
The integral of a general stretched Hermite polynomial
$$
 I_\alpha =\int_{-\infty}^\infty H_\alpha(\gamma x)\omega(x)dx
$$
can be done via a Taylor expansion
$$
 I_\alpha = \sum_{\beta=0}^{\alpha} {\alpha \choose\beta} H_{\alpha-\beta}(0)\gamma^\beta \int_{-\infty}^\infty x^\beta\omega(x)dx
$$
with result
$$
 I_{\alpha} = (\alpha-1)!!\left(\gamma^2-1\right)^{\frac{\alpha}{2}}.  
$$
Putting this into the formula for $d_n$, we get
$$
  d_n = \begin{cases}
   \frac{1}{(\sigma)^nn!\sqrt{2\pi(\sigma^2+1)}}(n-1)!!\left(\frac{-\sigma^2}{\sigma^2+1}\right)^{\frac{n}{2}} & n\textrm{ even}\\
   0 & n \textrm{ odd},
   \end{cases}
$$
which, after a reindex of the summation leads to the expansion given at the start of this answer.
Please read my recently submitted manuscript https://www.researchgate.net/publication/352374514_A_GENERAL_EXPRESSION_FOR_HERMITE_EXPANSIONS_WITH_APPLICATIONS
for all of the details.
