# Integral involving Hermite polynomials

## Homework Statement

The Hermite polynomials $$H_n(x)$$ may be defined by the generating function
$$e^{2hx-h^2} = \sum_{n=0}^{\infty}H_n(x)\frac{h^n}{n!}$$

Evaluate
$$\int^{\infty}_{-\infty} e^{-x^2/2}H_n(x) dx$$
(this should be from -infinity to infinity, but for some reason the latex won't work!)

Given.

## The Attempt at a Solution

I know that for odd $$n$$, this integral is 0, but I have no idea how to evaluate it. I know that the Hermite polynomials are orthogonal with respect to the given weight function, but I don't think I can use that for this integral (also, that's not in my book, I just found it on wikipedia). I'm assuming I need to use the generating function to derive my answer somehow, but I can't imagine how I would do that.

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vela
Staff Emeritus
Homework Helper
I haven't worked this problem out yet, but I'd try looking at

$$\int^{\infty}_{-\infty} e^{-x^2/2}e^{2hx-h^2}\,dx= \int^{\infty}_{-\infty} e^{-x^2/2}\sum_{n=0}^{\infty}H_n(x)\frac{h^n}{n!}\,dx$$

See if you can work out the lefthand side.

I haven't worked this problem out yet, but I'd try looking at

$$\int^{\infty}_{-\infty} e^{-x^2/2}e^{2hx-h^2}\,dx= \int^{\infty}_{-\infty} e^{-x^2/2}\sum_{n=0}^{\infty}H_n(x)\frac{h^n}{n!}\,dx$$

See if you can work out the lefthand side.
Huh, I didn't even think of that. I'll try it and report back. Thank you so much!

I can't get anywhere with that integral. The best alternatives I've come up with have been repeatedly integrating by parts or using a recursion relation derived from the generating function, but I can't get either of those to work out. I'm almost positive I'm supposed to use the recursion relation ($$H_{n+1} = 2xH_n - 2nH_{n-1}$$), since I derived it in an earlier problem and the book has basically no other information on Hermite polynomials. I thought I could mash the integral into the gamma function by reducing the integral to powers of x times the exponential (and in fact I still think I can, and that this is roughly the right way to do it), but I can't make it work.

fzero
Homework Helper
Gold Member
The integral on the LHS is Gaussian if you complete the square in the exponent.

Thanks. I've never heard of a gaussian integral before. So now I've got

$$\sqrt{2\pi}e^{h^2} = \sum_{n}^{\infty}\frac{h^n}{n!}I$$
where I is the integral I'm trying to get.
Expanding the exponential on the LHS I get a series in powers of $$h^{2n}$$. Is it kosher to simply rewrite that as $$\sum_{n=0}{\infty}\frac{h^n}{(n/2)!}$$? If it is, equating coefficients gives $$I = \frac{\sqrt{2\pi}}{(n/2)!}$$
Does that look right?

fzero