Special Integrals Hermite(2n+1,x)*Cos (bx) and e^(-x^2/2)

[dongsh2]

Do some one know how to integrate the
Hermite(2n+1,x)*Cos (bx) and e^(-x^2/2), x from 0 to infinity?

 

[dongsh2]

Do some one know how to integrate the
Integrate [Hermite(2n+1,x)*Cos (bx)*e^(-x^2/2), {x,0, \infinity}]?

 

[Svein]

Check out http://en.wikipedia.org/wiki/Gaussian_integral. It tells you most of what you need to know about that integral.

 

[dongsh2]

Check out http://en.wikipedia.org/wiki/Gaussian_integral. It tells you most of what you need to know about that integral.

Thanks. However, what I want to calculate is : Integrate[Exp[-x^2/2]Cos[ b x] HermiteH[2n+1,x],{x,0,\infinity}]. I have checked some references but I cannot find the result.

 

[Svein]

If we leave the Hermite polynoms for a moment, we can transform the rest: e^{-\frac{x^{2}}{2}}\cos(bx) is the real part of e^{-\frac{x^{2}}{2}+ibx}. The exponent can be further transformed: -\frac{x^{2}}{2}+ibx=-\frac{1}{2}(x^{2}-2ibx)=-\frac{1}{2}(x-ib)^{2}-\frac{1}{2}b^{2}. Thus you end up with e^{-\frac{1}{2}(x-ib)^{2}}\cdot e^{-\frac{1}{2}b^{2}}, where the last part is constant. Now, put z = (x-ib), then dz = dx. Use that with the contents of the link I gave you and see where you end up.

 

[dongsh2]

Thanks. However, it is not easy as what you thought. The problem is: x \in (0,\infinity). If we take z=x-i b and we have dz=dx, but the Hermite polynomial becomes
H[2n+1, z+ib] and integral interval becomes (-i b,\infinity). Using the mathematica, it still cannot find the solutions.

 

[Svein]

However, it is not easy as what you thought.

I did not say it was easy, I just gave you an idea of how to attack it.