Integrals of Hermite polynomials over $(-\infty, 0)$

Question

Does there exist a simple expression for integrals of the form,

$I = \int_{-\infty}^0 H_n(u) H_m(u)\, \mathrm{e}^{-u^2}\,du$,

where $m$ and $n$ are nonnegative integers and $H_n$ is the $n$'th (physicists') Hermite polynomial?

When $n+m$ is even, the symmetry of the integrand and the orthogonality of $H_n$ imply,

$I = \sqrt{\pi} \,2^{n-1} n! \,\delta_{n,\,m}$ (for $n+m$ even).

For $n+m$ odd, $I$ is nonzero and increases in magnitude with $n+m$, but I have been unable to find a general formula.

Answer 1

It looks to me like we have exponential generating functions

$$\sum_{n=0}^\infty I(n,n+2k+1) t^n/n! = \dfrac{(-1)^{k+1}(2k)!}{k! (1-2t)^{k+3/2} (1+2t)^{k+1/2}}$$

EDIT: Hmm, these can be combined into a bivariate exponential generating function

$$ \sum_{n=0}^\infty \sum_{k=0}^\infty I(n,n+2k+1) \frac{s^k t^n}{k! n!} = \frac{1}{(-1+2t) \sqrt{1+4s-4t^2}}$$

Answer 2

Although this is a very old question, I think it has another answer that may be useful to people.

I prefer the probabilists definition $$ He_{\alpha}(x) = 2^{-\frac{\alpha}{2}}H_\alpha\left(\frac{x}{\sqrt{2}}\right) $$ so the integral becomes $$ I = 2^{\frac{n+m-1}{2}}\int_{-\infty}^0He_n(x)He_m(x)\omega(x)dx $$ where $\omega(x) = e^{-\frac{x^2}{2}}$.

Using the linearization of Hermite polynomials $$ He_\alpha(x)He_\beta(x)=\sum_{k=0}^{\min(\alpha,\beta)}{\alpha \choose k}{\beta \choose k}k!He_{\alpha+\beta-2k}(x) $$ we can rewrite the integral as a single Hermite polynomial, since we know this indefinite integral $$ I = 2^{\frac{n+m-1}{2}}\sum_{k=0}^{\min(n,m)}{n \choose k}{m \choose k}k!\int_{-\infty}^0 He_{n+m-2k}(x)\omega(x)dx. $$

The indefinite integral can be calculated by noting $$ \frac{d}{dx}\left[He_n(x)\omega(x)\right]=\frac{d^{n+1}}{dx^{n+1}}\omega(x)\\ = (-1)He_{n+1}(x)\omega(x) $$ and therefore $$ \int He_{n+1}(x)\omega(x)dx = -He_n(x)\omega(x) $$ and in particular $$ \int_{-\infty}^0 He_{n+m-2k}(x)\omega(x)dx = -He_{n+m-2k-1}(0). $$ These are known as the Hermite numbers and are zero for odd indices, therefore $n+m-2k-1$ is even, or has zero modulus $2$. Since $2k$ also has zero modulo $2$, then the parity of $n$ and $m$ must be opposite, ie $\operatorname{mod}(n,2) + \operatorname{mod}(m,2) = 1$.

Assume $n>m$ and $n$ is odd, so it can be written $n=2l+1$ and $m$ is even, so it can be written $m=2s$. The integral becomes $$ I = -2^{l+s}\sum_{k=0}^{2s}{2l+1 \choose k}{2s \choose k}k! He_{2s+2l-2k}(0). $$ We can re-index the sum by $\alpha=s+l-k$ $$ I = -2^{l+s}\sum_{\alpha=l-s}^{l+s}{2l+1 \choose l+s-\alpha}{2s \choose l+s-\alpha}(l+s-\alpha)! He_{2\alpha}(0). $$ The Hermite numbers are $He_{2\alpha}(0) = \frac{(-1)^{\alpha}(2\alpha)!}{2^\alpha\alpha!}$ leading to $$ I = 2^{l+s}\sum_{\alpha=l-s}^{l+s}{2l+1 \choose l+s-\alpha}{2s \choose l+s-\alpha}(l+s-\alpha)! \frac{(-1)^{\alpha+1}(2\alpha)!}{2^\alpha \alpha!}. $$ Although complicated, it has your desired properties that for $n+m$ odd it is zero, and it increases as $n,m$ increase.