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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.

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Check this problem. –  Mhenni Benghorbal Feb 1 '13 at 9:18
@MhenniBenghorbal, the question you linked to was concerned with integrals from $-\infty$ to $\infty$, while my question is about integrals from $-\infty$ to $0$. What I stated about the $m+n$ even case follows from considering the $-\infty$ to $\infty$ integrals. But I don't see how to infer something about the $m+n$ odd case. –  Ted Pudlik Feb 6 '13 at 4:56

1 Answer 1

up vote 1 down vote accepted

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

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

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I can't quite connect the dots here: what does $I(n,n+2k+1)$ stand for? How is $k$ related to $m$ and $n$? –  Ted Pudlik Feb 6 '13 at 5:11
$I(n,m)$ is your integral $\int_{-\infty}^0 H_n(u) H_m(u)\ e^{-u^2}\ du$. So I'm taking $m=n+2k+1$ where $k$ is a nonnegative integer. –  Robert Israel Feb 6 '13 at 19:28
For example, with $k = 3$ we have $I(0,7) = 120, I(1,8) = 240, I(2,9) = 4320, I(3,10) = 25920, I(4,11) = 570240$, corresponding to $$ \frac{6!}{3! (1-2t)^{9/2} (1+2t)^{7/2}} = \frac{120}{0!} + \frac{240}{1!} t + \frac{4320}{2!} t^2 + \frac{25920}{3!} t^3 + \frac{570240}{4!} t^4 + \ldots $$ –  Robert Israel Feb 6 '13 at 19:37
Now I see---a very neat relationship, thank you! –  Ted Pudlik Feb 8 '13 at 2:56

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