Sum of Hermite polynomials

Question

I am trying to find a closed form (or the tightest upper bound possible) for the following sum: $$ \sum_{n=0}^{\infty} \frac{H_{2n}(x)w^n}{(2n+1)!} $$

From the equalities $ \sum_{n=0}^{\infty} \frac{H_{n}(x)w^n}{n!}=e^{2xw-w^2} $ and $H(-x)=(-1)^nH(x)$ I can obtain $$ \sum_{n=0}^{\infty} \frac{H_{2n}(x)w^n}{(2n)!}=\frac{e^{2xw-w^2}+e^{-2xw-w^2}}{2} $$ Which is close but not what I want. I thought of using the recurrence relation on the hermite polynomials or the identity $H_n'(x)=2nH_{n-1}(x)$ but I couldn't find a solution. This sum looks like it should have a closed form but I am currently stuck, any help would be appreciated.

Answer 1

With Mathematica:

$$\sum _{n=0}^{\infty } \frac{w^n H_{2 n}(x)}{(1+2 n)!}=\frac{e^{x^2} \sqrt{\pi } \left(\text{erf}\left(\sqrt{w}-x\right)+\text{erf}\left(\sqrt{w}+x\right)\right)}{4 \sqrt{w}}$$

Answer 2

Well, as the previous answer is just CAS, I will offer a detailed proof as an alternative. Posit: $$f(x, w) = \sum_{n=0}^{\infty} \frac{H_{2n}(x)w^n}{(2n+1)!} \tag{1}$$ $$g(x, w) = w f(x, w^2) \tag{2}$$ Then observe that for $w$ in any given compact of the complex plane, the series of derivarives in $w$ is dominated in norm by a geometric series, so that $g(x, w)$ can be derived term-by-term. Now:

\begin{align} \frac{\partial{g}}{\partial{w}}(x,w) &= \sum_{n=0}^\infty \frac{H_{2n}(x) w^{2n}}{(2n)!}\\ &= \frac{e^{- (w - x)^2} + e^{- (w + x)^2}}{2}\,e^{x^2} \end{align}

by use of the classic generating function $e^{2xt-t^2} = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}$.

whence, by an obvious change of variables ($v=w \pm x$):

\begin{align}g(x,w)=g(x,w) -g(x,0) &= \frac{e^{x^2}}{2} \left(\int_{-x}^{w-x} e^{-t^2} dt + \int_{x}^{w+x} e^{-t^2}dt\right) \\ &= \frac{e^{x^2}}{4} \sqrt{\pi } \left((\text{erf}(w-x)+\text{erf}(x)) + (\text{erf}(w+x)-\text{erf}(x))\right)\\ &= \frac{e^{x^2}}{4} \sqrt{\pi } \left(\text{erf}(w-x) + \text{erf}(w+x)\right) \tag{3} \end{align}

hence, choosing $w>0$:

$$f(x, w) = \frac{g(x, \sqrt{w})}{\sqrt{w}} = \frac{e^{x^2}}{4} \sqrt{\frac{\pi}{w} } \left(\text{erf}(\sqrt{w}-x) + \text{erf}(\sqrt{w}+x)\right)$$

which is the CAS answer.

(Edit) For $w<0$, $(2)-(3)$ being true for any complex $w$, we can write:

$$f(x, w) = f(x, (i\sqrt{-w})^2) = -i\frac{g(x, i\sqrt{-w})}{\sqrt{-w}} = -i \frac{e^{x^2}}{4} \sqrt{\frac{\pi}{-w} } \left(\text{erf}(i\sqrt{-w}-x) + \text{erf}(i\sqrt{-w}+x)\right)$$

which can also be directly inferred from $(3)$ by analytic continuation.

Finally for $w=0$, the answer is obvious: $f(x, 0) = H_0(x)=1$.