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Is there any closed-form expression for the integral \begin{equation} I(n,m) = \int_0^{+\infty} \mathcal{H}_n(u) \mathcal{H}_m(u)\exp(-u^2) du \end{equation} where $\mathcal{H}_n(x)$ is the Hermite polynomial, defined as \begin{equation} \mathcal{H}_n(x) = \dfrac{(-1)^n}{\omega(x)} \dfrac{d^n \omega(x)}{dx^n} \end{equation} with $\omega(x)=\exp(-x^2)$. I have calculated this integral for a few $(n,m)$ values, and cannot find any pattern in the solution, apart from \begin{equation} I(n,n) = 2^{n-1} n! \sqrt{\pi} \,. \end{equation}

Maziar Noei
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    For $n,m\in2\mathbb{N}$ or $n,m\in2\mathbb{N}+1$ with $n\neq m$ this vanishes. This is because the integral is then half the integral over $\left(-\infty,\infty\right)$ and the latter vanishes because of orthogonality. – eranreches Jun 12 '19 at 09:10

2 Answers2

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Maybe this helps:

For $i$ and $j$ having different parities: $$ I(i,j) = \int\limits_0^\infty H_i H_j e^{-x^2}dx = \int\limits_0^\infty (2xH_{i-1}-2(i-1)H_{i-2})H_je^{-x^2}\\ = -2(i-1)I(i-2,j)-(H_{i-1}H_j)_{|0}+\int\limits_0^\infty\partial_x (H_{i-1}H_j)e^{-x^2}dx\\ = -2(i-1)I(i-2,j)-(H_{i-1}H_j)_{|0}+2(i-1)I(i-2,j)+2jI(i-1,j-1)\\ =2jI(i-1,j-1)-(H_{i-1}H_j)_{|0} $$ where we used $H_i =2xH_{i-1}-2(i-1)H_{i-2}$, $\partial_xH_i = 2iH_{i-1}$ and the product rule.

denklo
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  • The 2nd term of your 2nd line should be $(e^{-x^2}H_{i-1}H_j)$ not just $H_{i-1}H_j$ evaluated at $0$. – jack Jul 18 '22 at 02:52
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Transforming the desired integral to use the probabilists Hermite polynomials yields $$ I_{nm}=2^\frac{n+m-1}{2}\int_{0}^\infty\operatorname{He}_n(x)\operatorname{He}_m(x)e^{-\frac{1}{2}x^2}dx. $$ This can be turned into a sum of Hermite polynomials using the linearization formula $$ \operatorname{He}_n(x)\operatorname{He}_m(x)=\sum_{l=0}^{\operatorname{min}(n,m)}{n \choose l}{m \choose l}l!\operatorname{He}_{n+m-2l}(x) $$ to obtain $$ I_{nm}=2^\frac{n+m-1}{2}\sum_{l=0}^{\operatorname{min}(n,m)}{n \choose l}{m \choose l}l!\int_0^\infty\operatorname{He}_{n+m-2l}(x)e^{-\frac{1}{2}x^2}dx. $$ The indefinite integral can be done by using the definition of the Hermite polynomial and noticing $$ \frac{d}{dx}[\operatorname{He}_n(x)\omega(x)] = (-1)\omega(x)\frac{d^{n+1}}{dx^{n+1}}\omega(x) = -\operatorname{He}_{n+1}(x)\omega(x), $$ and therefore $$ \int_0^\infty \operatorname{He}_n(x)\omega(x)dx = \operatorname{He}_{n+1}(0)\omega(0). $$ However care must be taken for the integral of $\operatorname{He}_0(x)=1$, this will result in the error function (or extending the Hermite to negative indices). The index of the Hermite polynomial in the linearization sum is between ${n+m - 2\operatorname{min}(n,m), n+m}$ which is either $[n-m,n+m]$ or $[m-n,m+n]$ depending on which is bigger. Therefore $0$ is only present when $n=m$, in which case $$ I_{nm}=2^{n-\frac{1}{2}}n!\int_0^\infty e^{-\frac{1}{2}x^2}dx=2^{n-1}n!\sqrt{\pi}. $$

In the other cases, the index of the Hermite polynomial never reaches zero, so we don't have to concern ourselves with the error function integral. In this case, the integral becomes $$ I_{nm}=2^\frac{n+m-1}{2}\sum_{l=0}^{\operatorname{min}(n,m)}{n \choose l}{m \choose l}l!\operatorname{He}_{n+m-2l-1}(0). $$ The Hermite polynomial evaluated at zero is only non-zero when the index is even. Therefore $\operatorname{mod}(n+m-2l-1,2)=0$, or $\operatorname{mod}(n,2)+\operatorname{mod}(m,2)=1$, $n$ and $m$ must have opposite parity for the integral to be non-zero.

Tom Davis
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