An expression involving Hermite polynomials at imaginary values Let $H_n(x)$ be the $n$-th (probabilists') Hermite polynomial. I need to bound from above the expression:
$$
\frac{1}{d}\sum_{n=0}^{d-1}\exp(-t^2/2)\frac{\vert H_n(it)\vert^2}{n!}
$$
where $t>0$ is real. Preferably I'd like to show that this is goes to zero as $t\to\infty$ uniformly in $d$, with explicit bounds on convergence. I wasn't able to find asymptotic formulae for $H_n$ evaluated at imaginary times.
 A: An upper bound, valid for $t\geq 1$ uniformly in $d$, can be obtained as follows.
Using the known formulas for (probabilists') Hermite polynomials
$$
H'_n(x)=nH_{n-1}(x),\qquad xH_n(x)=H_{n+1}(x)+nH_{n-1}(x)
$$
it follows that
$$
(x^{-n}H_n(x))'=n(n-1)x^{-n-1}H_{n-2}(x).
$$
Since the Hermite polynomials $H_n(x)$ are even/odd with nonzero coefficients alternating in sign, it follows that $t^{-n}|H_n(it)|$ is a decreasing function of $t>0$, for all $n$.
Hence, for all $t\geq 1$,
$$
\frac 1de^{-\frac{t^2}2}\sum_{n=0}^d\frac{|H_n(it)|^2}{n!}\leq
\frac 1de^{-\frac{t^2}2}t^{2(d-1)}\sum_{n=0}^d\frac{|H_n(i)|^2}{n!}
$$
i.e., denoting $C_d:=\frac 1d\sum_{n=0}^d\frac{|H_n(i)|^2}{n!}$,
$$
\frac 1de^{-\frac{t^2}2}\sum_{n=0}^d\frac{|H_n(it)|^2}{n!}\leq C_d e^{-\frac{t^2}2}t^{2(d-1)},\quad \forall t>1.
$$
The integers $|H_n(i)|$ are the so-called telephone numbers (https://en.wikipedia.org/wiki/Telephone_number_(mathematics)), and their large $n$ asymptotics is known. In particular (cf. Wikipedia link),
$$
|H_n(i)|\sim \biggl(\frac ne\biggr)^{n/2}\frac{\exp(\sqrt n)}{(4e)^{1/4}}
$$
so that by Stirling's formula
$$
\frac{|H_n(i)|^2}{n!}\sim\frac{\exp(2\sqrt n)}{\sqrt{8\pi e n}},\ \ n\to+\infty
$$
we obtain the heuristic asymptotics
$$
C_d\approx \frac 1d\int_0^d\frac{\exp(2\sqrt n)}{\sqrt{8\pi e n}}\mathrm{d}n \sim
\frac{\exp(2\sqrt d)}{d\sqrt{8\pi e}},\quad d\to+\infty,
$$
which is corroborated by numerics and might be not too hard to prove rigorously.
