Sum of Hermite functions with factorials

Question

I'm trying to evaluate the following quantity for $n \in \mathbb{N}$, $x\in \mathbb{R}$ and $\alpha_1, \alpha_2, \beta_1$ scalars.

$$I(x)=\sum_{k=0}^{n-1}\frac{\alpha_1^k \alpha_2^{n-k}}{k!(n-k)!}H_k(x)H_{n-1-k}(\beta_1 x)$$ where $H_n$ is the n-th "physicist" Hermite polynomial.

I know the following result about a related sum :

$$ \sum_{k=0}^{n}\frac{\alpha_1^k \alpha_2^{n-k}}{k!(n-k)!}H_k(x)H_{n-k}(\beta_1 x)=\frac{(\alpha_1^2+\alpha_2^2)^{n/2}}{n!}H_n\left(\frac{\alpha_1 + \alpha_2 \beta_1 }{(\alpha_1^2+\alpha_2^2)^{1/2}}x\right)$$

but I cannot manage to use this result to evaluate the former one because of the indices of the Hermite polynomials...

I could try to integrate or derivate with respect to $\alpha_1$ or $\alpha_2$ but it does not seem to work...

Thanks a lot !

Answer 1

It is unlikely that there is a nice expression. Write the generating function: $$ \sum_{n=1}^\infty I_n(x)z^{n-1}=\sum_{n=1}^\infty \sum_{k=0}^{n-1}\frac{(\alpha_1z)^k (\alpha_2z)^{n-k}}{k!(n-k)!}H_k(x)H_{n-1-k}(\beta_1 x) \\ = \sum_{k=0}^\infty \frac{(\alpha_1z)^k}{k!}H_k(x)\sum_{l=1}^\infty\frac{ (\alpha_2z)^{l}}{l!}H_{l-1}(\beta_1 x) \\= \exp\{2\alpha_1xz - \alpha_1^2 z^2\}\int_0^z\exp\{2\alpha_2\beta x y - \alpha_2^2 y^2\}dy. $$ The shift in index resulted in appearance of a nasty integral. Without this shift, we could collect the exponents and expand the expression, but here it seems hopeless.