Integrals of Hermite polynomials over $(-\infty, 0)$ 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.
 A: 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
$$ \sum_{n=0}^\infty \sum_{k=0}^\infty I(n,n+2k+1) \frac{s^k t^n}{k! n!}
= \frac{1}{(-1+2t) \sqrt{1+4s-4t^2}}$$
A: 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.
A: This question is getting a little old now, but I feel I can add something here, for my own conscience, if nothing else.
My take on this problem is - simply define $u = \sqrt{v}$ in the integral and take advantage of the relation between the Hermite and Laguerre polynomials. i.e.
$$
H_{2n}(\sqrt{v}) = (-1)^n 2^{2n}n! L^{\left(-\frac{1}{2}\right)}_n(v)\\
H_{2n+1}(\sqrt{v}) = (-1)^n 2^{2n+1}n! \sqrt{v} L^{\left(\frac{1}{2}\right)}_n(v)
$$
So, assuming that $m$ and $n$ are both even we get:
\begin{eqnarray}
\int^0_{-\infty}du e^{-u^2} H_n(u) H_m(u)= (-1)^{n+m}\int^{\infty}_0 dx e^{-x^2} H_n(x) H_m(x) \;\; \textrm{changing order of integration and changing variables $u \rightarrow -x$}\\
= \frac{1}{2} \int^{\infty}_0 \left(dv v^{-\frac{1}{2}} \right)e^{-v} H_n(\sqrt{v}) H_m(\sqrt{v}) \;\; \textrm{changing variables $x = \sqrt{v}$}\\
= \frac{1}{2} \int^{\infty}_0 \left(dv v^{-\frac{1}{2}} \right)e^{-v} (-1)^{\frac{n}{2}} 2^{n}\left(\frac{n}{2}\right)! L^{\left(-\frac{1}{2}\right)}_{\frac{n}{2}}(v)(-1)^{\frac{m}{2}} 2^{m}\left(\frac{m}{2}\right)! L^{\left(-\frac{1}{2}\right)}_{\frac{m}{2}}(v) \;\; \textrm{applying the above relation between Hermite and Laguerre polys. - the even case}\\
=\frac{(-1)^{\frac{n}{2}+\frac{m}{2}}2^{m+n}}{2} \left(\frac{n}{2}\right)!\left(\frac{m}{2}\right)!\int^{\infty}_0  dv v^{-\frac{1}{2}}  e^{-v}L^{\left(-\frac{1}{2}\right)}_{\frac{n}{2}}(v)L^{\left(-\frac{1}{2}\right)}_{\frac{m}{2}}(v) \;\; \textrm{decluttering algebra}\\
=(-1)^{\frac{n}{2}+\frac{m}{2}}2^{m+n-1} \left(\frac{n}{2}\right)!\left(\frac{m}{2}\right)! \frac{\Gamma\left( \frac{n+1}{2}\right)}{\left(\frac{n}{2}\right)!}\delta_{n,m} \;\; \textrm{applying Laguerre orthogonality}\\
=(-1)^{n}2^{2n-1} \left(\frac{n}{2}\right)! \Gamma\left( \frac{n+1}{2}\right)\delta_{n,m}
\end{eqnarray}
Since $n$ is even the expression is positive. Equivalent expressions exist for all other odd/even combinations of $m$ and $n$. I have probably screwed up the algebra in places, but I think the general argument is valid.
