I need to evaluate: $$F_{n,m} = \int_{-\infty}^\infty e^{-cz^2+bz} H_n(z)H_m(z) dz$$
I tried to use the generating functions for the Hermite Polynomials of form:
$$e^{zt - \frac{t^2}{2}} = \sum_{n=0}^\infty\frac{t^n}{n!}H_n(z)$$ and, $$e^{zr - \frac{r^2}{2}} = \sum_{m=0}^\infty\frac{r^m}{m!}H_m(z)$$
Thus, I have obtained the following equation:
$$\sum_{n=0,m=0}^\infty\frac{t^n}{n!}\frac{r^m}{m!}F_{n,m} =e^{\frac{-1}{2}(t^2+r^2)} \int_{-\infty}^\infty e^{-c z^2 + (b+t+r)z}dz$$ Performing the Gaussian integral in RHS, the above equation takes form of:
$$\sum_{n=0,m=0}^\infty\frac{t^n}{n!}\frac{r^m}{m!}F_{n,m} =\frac{\sqrt\pi}{\sqrt c}e^{\frac{-1}{2}(t^2+r^2)}e^{\frac{(b+t+r)^2}{4c}}$$
I then rearranged the RHS in form:
$$\sum_{n=0,m=0}^\infty\frac{t^n}{n!}\frac{r^m}{m!}F_{n,m} =\frac{\sqrt\pi}{\sqrt c}e^{\frac{b^2}{4c}}e^{t^2(\frac{1}{4c}-\frac{1}{2})}e^{t(\frac{b}{2c})}e^{ r^2(\frac{1}{4c}-\frac{1}{2})}e^{r(\frac{b}{2c})}e^{\frac{tr}{2c}}$$
This is where I have got stuck and haven't been able to move forward to extract the integral $F_{n,m}$ in LHS. Any help would be much appreciated. Thanks
When $c=1$ this has a solution that is relatively straight-forward. The steps are
For arbitrary $c$ there is an extra step to scale the variable.
Completing the square $$ F_{nm}=e^{\frac{b^2}{4}}\int_{-\infty}^\infty e^{-(z-\frac{b}{2})^2}\operatorname{H}_n(z)\operatorname{He}_m(z)dz. $$ Change to probabilists $$ F_{nm}=e^{\frac{b^2}{4}}2^{\frac{n+m-1}{2}}\int_{-\infty}^\infty e^{-(z-\frac{b}{2})^2}\operatorname{He}_n(\sqrt{2}z)\operatorname{He}_m(\sqrt{2}z)dz. $$ Linearize $$ F_{nm}=e^{\frac{b^2}{4}}2^{\frac{n+m-1}{2}}\sum_{k=0}^{\operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\int_{-\infty}^\infty e^{-(z-\frac{b}{2})^2}\operatorname{He}_{n+m-2k}(\sqrt{2}z)dz. $$ Make the substitution $\frac{x}{\sqrt{2}}=\left(z-\frac{b}{2}\right)$ $$ F_{nm}=e^{\frac{b^2}{4}}2^{\frac{n+m-1}{2}}\sum_{k=0}^{\operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\int_{-\infty}^\infty e^{-\frac{x^2}{2}}\operatorname{He}_{n+m-2k}\left(x + \frac{b}{\sqrt{2}}\right)dz. $$ Taylor expand the Hermite polynomial $$ F_{nm}=e^{\frac{b^2}{4}}2^{\frac{n+m-1}{2}}\sum_{k=0}^{\operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\sum_{l=0}^{n+m-2k}{n+m-2k\choose l}\left(\frac{b}{\sqrt{2}}\right)^l\\ \times\int_{-\infty}^\infty e^{-\frac{x^2}{2}}\operatorname{He}_{n+m-2k-l}(x)dz. $$ Use orthogonality to obtain $$ F_{nm}=e^{\frac{b^2}{4}}2^{\frac{n+m}{2}}\sqrt{\pi}\sum_{k=0}^{\operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\left(\frac{b}{\sqrt{2}}\right)^{n+m-2k}, $$ which can be rewritten as $$ F_{nm}=e^{\frac{b^2}{4}}n!m!\sqrt{\pi}\sum_{k=0}^{\operatorname{min}(n,m)}\frac{2^kb^{n+m-2k}}{k!(n-k)!(m-k)!}. $$
When $c\ne 1$ the steps are the same up until the actual integral, where instead the integral that needs to be done is $$ F_{nm}=2^{\frac{n+m-1}{2}}e^{-\frac{b^2}{4c}}\frac{1}{\sqrt{c}}\sum_{k=0}^{ \operatorname{min}(n,m)}{n\choose k}{m\choose k}k!\sum_{l=0}^{n+m-2k}{n+m-2k \choose l}\left(\frac{b}{\sqrt{2}c}\right)^l\\ \times\int_{-\infty}^\infty e^{-\frac{x^2}{2}}\operatorname{He}_{n+m-2k-l}\left(\frac{x}{\sqrt{c}}\right)dx $$ This now needs the scaling of Hermite polynomials $$ \operatorname{He}_n(\gamma x) = n!\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\frac{1}{2^kk!(n-2k)!}\gamma^{n-2k}\left(\gamma^2-1\right)^k \operatorname{He}_{n-2k}(x). $$ Denoting the summation variable as $s$, the integral is then over the Hermite polynomial with index $n+m-2k-l-2s$ which constrains the $s$ sum to its highest value $s=\frac{n+m-l}{2}-k$. This also puts a restriction on the $l$ sum, as when $n+m-2k-l$ is odd, the scaling series never includes a constant term (since it is an odd polynomial, and therefore has no even components). We can rewrite this condition as $\operatorname{mod}(n+m,2)=\operatorname{mod}(l,2)$ to see that $l$ must have the same parity as $n+m$, so we enforce this by writing $l=2a+p$ where $p=\operatorname{mod}(n+m,2)$.
The integral now becomes $$ F_{nm}=e^{-\frac{b^2}{4c}}\sqrt{\frac{\pi}{c}}n!m!\sum_{k=0}^{ \operatorname{min}(n,m)}\frac{2^k}{k!(n-k)!(m-k)!}\sum_{a=0}^{\frac{n+m-p}{2}-k}{n+m-2k \choose 2a+p}\left(\frac{b}{c}\right)^{2a+p}\\ \times \left(\frac{1}{c}-1\right)^{\frac{n+m-2a-p}{2}-k}\frac{(n+m-2k-2a-p)!}{\left(\frac{n+m-2a-p}{2}-k\right)!} $$ This simplifies a bit, but not much, to the final answer $$ F_{nm}=e^{-\frac{b^2}{4c}}\sqrt{\frac{\pi}{c}}n!m!\sum_{k=0}^{ \operatorname{min}(n,m)}\frac{2^k}{k!(n-k)!(m-k)!}\sum_{a=0}^{\frac{n+m-p}{2}-k}{n+m-2k \choose 2a+p}\frac{(n+m-2k-2a-p)!}{\left(\frac{n+m-2a-p}{2}-k\right)!}\\ \times b^{2a+p}\frac{(1-c)^{\frac{n+m-2a-p}{2}-k}}{c^{\frac{n+m+2a+p}{2}-k}} $$
This does reduce to the case when $c=1$ by virtue of the only term in the $a$ sum left is when $(1-c)^{\frac{n+m-2a-p}{2}-k}$ is $1$, ie $\frac{n+m-2a-p}{2}-k=0$, in which case the above answer is obtained.