Compute $\int_{-\infty}^\infty |H_n( a x)|^2 \exp(-\frac{x^2}{2} )$ where $H_n$ is the Hermite polynomial

Question

Is there a closed-form expression (or tight upper bound) for the integral $$\int_{-\infty}^\infty |H_n( a x)|^2 \exp(-\frac{x^2}{2} ) dx$$ for $a>0$. Here $H_n$ is a Hermite polynomial (physicists').

If $a=1/\sqrt{2} $, we can compute the above as (e.g., see here) \begin{align} 2^{n}\int_{-\infty}^\infty |H_{e_n}( x)|^2 \exp(-\frac{x^2}{2} ) dx= \sqrt{2 \pi} 2^{n} n! \end{align}

The question is how to do this for a general $a$? It doesn't appear to be a simple change of variable. I am also sure that his has probably been looked at, but I couldn't find anything from the literature review.

I think the correct answer must be of the form $$\int_{-\infty}^\infty |H_n( a x)|^2 \exp(-\frac{x^2}{2} ) dx \propto a^n \sqrt{2 \pi} 2^{n} n!$$

Here are the first five computed in wolfram:

$$ n=2: 4 \sqrt{2 \pi} (12*a^4-4a^2-1) $$ $$n=3: 48 \sqrt{2 \pi} a^2 (3 - 12 a^2 + 20 a^4) $$ $$n=4: 48 \sqrt{2 \pi} (560 a^8 - 480 a^6 + 168 a^4 - 24 a^2 + 3)$$ $$ n=5: \sqrt{2 \pi}960 a^2 (15 - 120 a^2 + 520 a^4 - 1120 a^6 + 1008 a^8)$$

Answer 1

For the even polynomial $P(x) = H_n(x)^2$, you are trying to find a closed form for $$F(a^2) = \int_{-\infty}^\infty P(ax)e^{-x^2/2}\,dx.$$ Differentiating with respect to $a$, $$\begin{align*}2aF'(a^2) &= \int_{-\infty}^\infty P'(ax)\,xe^{-x^2/2}\,dx \\ &\mathop{=}^{\text{I.B.P.}}-P'(ax)\,e^{-x^2/2}\left. \right]_{-\infty}^\infty + \int_{-\infty}^\infty aP''(ax)\,e^{-x^2/2}\,dx\\ &= \int_{-\infty}^\infty aP''(ax)\,e^{-x^2/2}\,dx \end{align*}$$ so $$F'(a^2) = \frac{1}{2}\int_{-\infty}^\infty P''(ax)\,e^{-x^2/2}\,dx.$$ Repeating this process, we find $$F^{(k)}(a^2) = \frac{1}{2^k}\int_{-\infty}^\infty P^{(2k)}(ax)\,e^{-x^2/2}\,dx. \tag{1}$$ so $$\frac{F^{(k)}(0)}{k!} = \frac{\sqrt{2\pi}}{2^k k!}P^{(2k)}(0);$$ in other words the coefficient of $a^{2k}$ in $F(a^2)$ is $\sqrt{2\pi}\frac{(2k)!}{2^k k!}$ times the coefficient of $x^{2k}$ in $P(x)$.

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For $P(x) = H_n(x)^2$ we have $$\frac{P^{(2k)}(x)}{(2k)!} = \sum_{i+j = 2k}\frac{H_n^{(i)}(x)}{i!}\frac{H_n^{(j)}(x)}{j!}.$$ From the identity $H_n^{(m)}(x) = 2^m \frac{n!}{(n-m)!}H_{n-m}(x)$, we get $$\frac{P^{(2k)}(x)}{(2k)!} = 2^{2k}\sum_{i+j = 2k}\binom{n}{i}\binom{n}{j}H_{n-i}(x)H_{n-j}(x). \tag{2}$$ From here we could plug in $x=0$ to derive an explicit (albeit messy) expression for $F(a^2)$.

Alternatively, we can exploit orthogonality by substituting equation $(2)$ into equation $(1)$ and setting $a = \frac{1}{\sqrt{2}}$. Then all of the terms disappear except for the $i=j=k$ term, and we are just left with

$$ F^{(k)}(1/2) = (2k)!\,2^k\binom{n}{k}^2\int_{-\infty}^\infty \left[H_{n-k}\left(\frac{x}{\sqrt{2}}\right)\right]^2 e^{-x^2/2}\,dx.$$ By the identity in the original post, this is equal to $$F^{(k)}(1/2) =(2k)!\binom{n}{k}^2 2^{n} \sqrt{2\pi} (n-k)!.$$ Dividing by $k!$ and simplifying gives us $$\frac{F^{(k)}(1/2)}{k!} = \sqrt{2\pi}\, 2^{n} n!\,\binom{2k}{k}\binom{n}{k}.$$ These are the coefficients of the expansion of $F$ centered at $a^2 = \tfrac{1}{2}$, so we arrive at $$\int_{-\infty}^\infty |H_n(ax)|^2e^{-x^2/2}\,dx = F(a^2) = \sqrt{2\pi}\,2^n\,n!\cdot\sum_{k=0}^n \binom{2k}{k}\binom{n}{k}\left(a^2-\tfrac{1}{2}\right)^k.$$