# Function series involving Hermite Polynomials

I am wondering if there is a simpler form of the following function series involving the (physicists') even Hermite polynomials: $$\begin{equation} f(x) = e^{-\frac{x^2}{2}} \cdot \sum_{n = 0}^{\infty} \, \frac{1}{4^n n!} \cdot \frac{H_{2n}\left(x\right)}{a + n} \end{equation}$$ where $$a>0$$.

The series seems to converge. It would be great to have a closed-form for it. I expect a hump-shaped function with Cauchy-like fat-tails.

Not a close form, but an integral representation of the series which helps finding the asymptotic behavior. Using the representation of the even Hermite polynomials in terms of the Laguerre polynomials here, $$\begin{equation} H_{2n}\left( x \right)=(-1)^n4^nL_n^{(-1/2)}\left( x^2 \right) \end{equation}$$ the function reads $$\begin{equation} f(x)=e^{-x^2/2}\sum_{n\ge0}(-1)^n\frac{L_n^{(-1/2)}\left( x^2 \right)}{n+a} \end{equation}$$ The generating function for the generalized Laguerre polynomials $$\begin{equation} \sum_{n\ge0}t ^nL_n^{(-1/2)}\left(X \right)=\frac{e^{-\frac{t X}{1-t }}}{\sqrt{1-t }} \end{equation}$$ can be modified by changing $$t\to -t$$: $$\begin{equation} \frac{e^{\frac{t X}{1+t }}}{\sqrt{1+t }}=\sum_{n\ge0}(-1)^nt ^nL_n^{(-1/2)}\left(X \right) \end{equation}$$ Now, multiplying this expression by $$t^{a-1}$$ and integrating between 0 and 1, we obtain $$\begin{equation} \int_0^1 \frac{e^{\frac{t X}{1+t }}}{\sqrt{1+t }}t^{a-1}\,dt=\sum_{n\ge0}(-1)^n\frac{L_n^{(-1/2)}\left(X \right)}{n+a} \end{equation}$$ Then \begin{align} f(x)&=e^{-x^2/2}\int_0^1 \frac{e^{\frac{t x^2}{1+t }}}{\sqrt{1+t }}t^{a-1}\,dt\\ &=\int_0^1 \frac{e^{-x^2\frac{1-t}{2(1+t) }}}{\sqrt{1+t }}t^{a-1}\,dt\\ &=\sqrt{2}\int_0^1\left( 1-u \right)^{a-1}\left( 1+u \right)^{a-1/2}e^{-ux^2/2}\,du \end{align} An explicit expression of the latter representation in terms of two-variables hypergeometric sum can be found in Ederlyi I, 4.3.24 p.154. Apart from $$a$$ integer or half-integer where a CAS gives an explicit expressions in terms of the error function, this integral representation can be use to find the asymptotic behavior of the function.
Near $$x=0$$, Expanding the exponential term and using an integral representation of the hypergeometric function, we obtain $$\begin{equation} f(x)\sim \frac{\sqrt{2}}{a}\,_2F_1\left(1, \frac{1}{2}-a;1+a;-1 \right)-\frac{x^2}{\sqrt2 a(1+a)}\,_2F_1\left(2, \frac{1}{2}-a;2+a;-1 \right)+O(x^4) \end{equation}$$
For $$x\to\infty$$, the major contribution to the integral is near $$u=0$$, the end integration point can be taken infinite, as $$\begin{equation} \left( 1-u \right)^{a-1}\left( 1+u \right)^{a-1/2}\sim 1+\left( \frac{1}{2}-2a \right)u+O(u^2) \end{equation}$$ Watson lemma gives $$\begin{equation} f(x)\sim \frac{2\sqrt{2}}{x^2}+\frac{2\sqrt{2}\left( 1-4a \right)}{x^4}+O\left( x^{-6} \right) \end{equation}$$ Higher order terms for both developments can easily be obtained.
• This is great, thank you! The two asymptotic developments are really helpful. For the one near $x=0$, I assume there is a missing $x^2$ factor in the second term? – Procyon lotor Sep 17 '19 at 15:10