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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – Procyon lotor Sep 17 '19 at 15:10
  • $\begingroup$ You are welcome! I corrected the typo you noticed. $\endgroup$ – Paul Enta Sep 17 '19 at 15:57

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