Can you help me prove the integral for Hermite polynomials?

[Gabriel Maia]

Hi. I'm off to solve this integral and I'm not seeing how

\int dx Hm(x)Hm(x)e^{-2x^2}

Where Hm(x) is the hermite polynomial of m-th order. I know the hermite polynomials are a orthogonal set under the distribution exp(-x^2) but this is not the case here.

Using Hm(x)=(-1)^m e^{x^2}\frac{d^m}{dx^m}e^{-x^2} I was able to rewrite the integral as

\int \left(\frac{d^m}{dx^m}e^{-x^2}\right)^2 dx

I have calculated this integral for m=0,1,2,3,4,5 with mathematica and the result seems to be (2m-1)!\sqrt{\frac{\pi}{2}} but I need to prove it formally. Can you help me?


Thank you.

 

[vela]

I haven't worked this out completely, but I'll suggest you try using the generating function
$$g(x,t) = \exp(-t^2+2tx) = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}$$

 

[vela]

Using the generating function, it was pretty straightforward to show that
$$\int H_m(x)H_m(x) e^{-2x^2}\,dx = \sum_{m=2p+q} \frac{1}{q!}\left(\frac{m!}{p!2^p}\right)^2,$$ which appears to reproduce the result you found. I don't see how to simplify that sum to $(2m-1)!$ however.