Fourier transform of $xe^{-x^2/2}$

Question

I'm trying to find the Fourier transform of $xe^{-\frac{x^2}{2}}$. I was given the hint that when $g(x) = e^{-\pi x^2}$, $\mathcal{F}(g) = g(x)$.

This is how far I have come. So we start off with the usual Fourier transform: $$ \int_{\mathbb{R}} e^{-2 \pi i x \xi} xe^{-x^2/2} \mathrm{d}x $$

Next we do a variable change, letting $x = \sqrt{2 \pi} y$ and therefore $\mathrm{d}x = \sqrt{2 \pi}\mathrm{d}y$.

$$ \int_{\mathbb{R}} e^{-2 \pi i \sqrt{2 \pi} y \xi} \sqrt{2 \pi} ye^{-\pi y^2} \sqrt{2 \pi} \mathrm{d}y = 2 \pi \int_{\mathbb{R}} e^{-2 \pi i \sqrt{2 \pi} y \xi} ye^{-\pi y^2} \mathrm{d}y $$

Now we have something in the form $e^{-\pi x^2}$ inside the integral, but I'm not sure how to use this to my advantage.

Answer 1

Hint: Note that $$xe^{-x^2/2}=-\frac{d}{dx}(e^{-x^2/2}).$$

So $$\int e^{-2\pi i xy} xe^{-x^2/2}dx= -\int e^{-2\pi i xy}\left(\frac{d}{dx}e^{-x^2/2}\right)dx.$$ Now apply integration by parts to the right hand side of this equation and use the fact that the Fourier transform of $e^{-\pi x^2}$ is itself.

Hope this helps!