Hermite Polynomials Integrals

  • #1
Hi. I'm off to solve this integral and I'm not seeing how

[itex]\int dx Hm(x)Hm(x)e^{-2x^2}[/itex]

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

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

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


Thank you.
 

Answers and Replies

  • #2
vela
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I haven't worked this out completely, but I'll suggest you try using the generating function
$$g(x,t) = \exp(-t^2+2tx) = \sum_{n=0}^\infty H_n(x)\frac{t^n}{n!}$$
 
  • #3
vela
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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.
 

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