I would like to find a closed form for this series:$$ \sum_{n=0}^\infty\frac{(-1)^n(2n+1)!!\cdot H_{2n+1}(y)}{(2n+1)!\cdot(2n+1-2m)}v^n,$$ where $H_n$ are the Hermite polynomials. One can rewrite this expression as $$ \sum_{n=0}^\infty\frac{ H_{2n}(0)H_{2n+1}(y)}{(2n)!\cdot(2n+1-2m)}v'^n,$$ and this again can be solved for the case $m=0$. The solution for this case is here and uses the recurrence formula $H'_n(x)=2nH_{n-1}(x)$ and that the $H_{2n+1}(0)$ vanish, so that you can bring it into a form of Mehler's formula:
$$\sum_{n=0}^\infty\frac{H_n(x)H_n(y)}{n!}(\frac{u}{2})^n=\frac{1}{\sqrt{1-u^2}}\ e^{\frac{2xyu-(x^2+y^2)u^2}{1-u^2}}$$
Does someone have an idea how to approach this? Any idea would be appreciated!
I guess one thing you could do is put a function $$f(c)=c^{1+2m/(2n+1)}$$ into the series, differentiate with respect to $c$ and get the case for $m=0$ and thus a closed expression for the derivative. Then integrate and put $c=1$. The resulting integral looks more than painful, though.