I'm dealing with the Hermitian operator, and I've been asked to prove that all $f(x) = x^n e^{\alpha x}$ belong to $L^2(-\infty,\infty;e^{-x^2/2})$ by showing that:
$$\int_{-\infty}^{\infty}x^m e^{2\alpha x}e^{-x^2/2}dx = e^{2\alpha^2}\int_{-\infty}^{\infty}(x+2\alpha)^me^{-x^2/2}dx$$
Of course, after proving the identity, it is clear that all $f(x)$ of this particular form will belong, because you've simplified this integral to an $m^{th}$ degree polynomial, and all polynomials will belong to $L^2(-\infty,\infty;e^{-x^2/2})$.
I'm not sure if it is of any use to recognize that this is just $<x^m,e^{\alpha x}>$ (Hermite inner product), or if I just need to start integrating by parts a whole bunch.. Any help please :)
In order to prove the identity, we have to complete the squares; more precisely: $$-\frac{x^2}2+2\alpha x=-\frac 12\left(x^2 -2\cdot 2\alpha x\right)= -\frac 12\left((x-2\alpha)^2-4\alpha^2\right).$$ Therefore, $$\int_{-\infty}^{\infty}x^m e^{2\alpha x}e^{-x^2/2}\mathrm dx =e^{2\alpha^2}\int_{-\infty}^{\infty}x^m\exp\left(-\frac{\left(x-2\alpha\right)^2} 2\right)\mathrm dx.$$ To conclude, use the substitution $t:=x-2\alpha$ (the bounds are unchanged and $\mathrm dx=\mathrm dt$).