Question: I am looking at the following series:
$$\sum_{n=0}^\infty t^n \cdot \frac{(2n-1)!!}{(2n+1)!}\cdot H_{2n+1}(x), $$ which can also be written as $$\sum_{n=0}^\infty u^n \cdot \frac{1}{(2n+1)!}\cdot H_{2n+1}(x)H_{2n}(0)$$ where $H_n$ are the Hermite polynomials and $t,u$ are some complex numbers. My goal is to somehow simplify it. Ideally, it should look similar to the usual type of genereting functions of the Hermite polynomials. So far, I tried using the Mehler Hermite polynomial formula See Wolfram Mathworld along with some recurrence relations but I cannot seem to avoid the problem that I only sum over the odd (or only over the even) polynomials.
Does someone have a better idea how to approach this problem?
Let's try and get a closed form for $$ F(x,y,z) = \sum_{k=1}^{\infty} \frac{(z/2)^k}{k!} H_{k}(x)H_{k-1}(y). $$ Since $H_{k-1}(0)$ is zero if $n$ is even, this reduces to your sum in the case $y=0$. Differentiating with respect to $x$ gives, via $H'_{k}(x) = 2k H_{k-1}(x)$, $$ F_x = 2\sum_{k=1}^{\infty} \frac{(z/2)^k}{(k-1)!} H_{k-1}(x)H_{k-1}(y) = z \sum_{k=0}^{\infty} \frac{(z/2)^k}{k!} H_{k}(x)H_{k}(y) \\ = \frac{z}{\sqrt{1-z^2}} \exp{\left( \frac{2xyz-(x^2+y^2)z^2}{1-z^2} \right)} $$ from Mehler's formula. It remains to put $y=0$ and integrate. So, $$ F_x(x,0,z) = \frac{z}{\sqrt{1-z^2}} \exp{\left( -x^2\frac{z^2}{1-z^2} \right)}. $$ We know all about this: it's just a Gaussian. Therefore its integral is related to the error function. $F(0,0,z)=0$ since one or the other polynomial is zero for each integer. Indeed, from here it follows that $$ F(x,0,z) = \frac{\sqrt{\pi}}{2} \operatorname{erf}{\left( \frac{xz}{\sqrt{1-z^2}} \right)}, $$ where of course $$ \operatorname{erf}{(w)} = \frac{2}{\sqrt{\pi}} \int_0^{w} e^{-t^2}\, dt. $$ That is probably about as simple as it gets, I'm afraid.