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

1. May 31, 2015

### 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?

2. Jun 1, 2015

### dongsh2

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

3. Jun 3, 2015

4. Jun 3, 2015

### dongsh2

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.

5. Jun 4, 2015

### 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.

6. Jun 4, 2015

### 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.

7. Jun 5, 2015

### Svein

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