Weird Integral Involving Hermite Polynomials
The answer follows from Gradshteyn and Ryzhik, 7.376, $$ I_n(y):=\int_{-\infty}^\infty e^{i\,x\,y}e^{-x^2/2} \, H_n(x) \, dx = \sqrt{2 \pi} e^{-y^2/2} H_n(y) i^n .$$ Just set $n=m$ and $y=m.$ I'd attempt a proof, but I need to what is allowed as a starting point.
Formula above is easily derived from generating function $$ \sum_{n=0}^\infty \frac{t^n}{n!} H_n(x) = \exp{(-t^2+2x\,t)} $$ Make a generating formula for $I_n(y),$ interchange integral and sum, and use the previous formula like so: $$\sum_{n=0}^\infty \frac{t^n}{n!} I_n(y) = \int_{-\infty}^\infty e^{i\,x\,y} e^{-x^2/2} \exp{(-t^2+2x\,t)}dx = e^{-t^2} \exp{(\frac{1}{2}(2t+iy)^2 )}\sqrt{2\pi}$$ where the well-known Gaussian integral has been used in the last step. Arrange the exponentials as follows: $$e^{-t^2} \exp{(\frac{1}{2}(2t+iy)^2 )}= e^{-y^2/2} \exp{( -(it)^2 +2(it)y )} $$ Use the generating function again, but now the argument is $it.$ Equate coefficients of $t$ to complete the proof.