0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.