Fourier transform of $e^{-ax^2}$ in $\mathbb{R}^2$

Question

I'm trying to solve the following Fourier transform, or at least I'm trying to simplify it a bit

$$ \mathcal{F}(e^{-ax^2})(\xi,\eta) = \int_{\mathbb{R^2}} e^{-i(x\xi+y\eta)}e^{-ax^2}dxdy $$

for $a > 0$. One of the problems I'm finding is that the integrand function is not summable in $\mathbb{R}^2$, hence integrating is not that immediate. Thank you for any help.

Answer 1

As you pointed out, the integral you wrote is not well defined. You have to think of $f(x,y) = e^{-ax^2}$ as a tempered distribution. Let $f_1(x) = e^{-ax^2}$ and $f_2(y) = 1$. The function $f$ can be written as $f_1 \otimes f_2$. Therefore we have $$\mathcal F(f)(\xi, \eta) = \mathcal F(f_1 \otimes f_2)(\xi, \eta) = \mathcal F(f_1)\otimes\mathcal F(f_2)(\xi, \eta).$$

The Fourier transform of $f_1(x) = e^{-ax^2}$ is $\sqrt{\frac{\pi}{a}}e^{-\frac{\xi^2}{4a}}$ and the Fourier transform of $f_2(y) = 1$ is $2\pi \delta(\eta)$.

So, we obtain the formula $$\mathcal F(f) =2\pi \sqrt{\frac{\pi}{a}}e^{-\frac{\xi^2}{4a}}\delta(\eta).$$