- #1
Gabriel Maia
- 72
- 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.
[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.