Fourier transform in the tempered distributions space

Question

I'm trying to prove that the following formula holds in the tempered distributions space $S'(\mathbb{R}^n)$: $$\widehat{\exp(-a|x|^2)}=\left(\frac{\pi}{a}\right)^{\frac{n}{2}}\exp\left(-\frac{\pi^2|\xi|^2}{a}\right)\hspace{0.1cm};\hspace{0.1cm} Re(a)\geq0\hspace{0.1cm};\hspace{0.1cm} a\neq0,$$ where $\sqrt{a}$ is defined as the branch with $Re(a)>0$.

I have no problem when I consider $a\in\mathbb{R}$ because in that case $\exp(-a|x|^2)\in L^1(\mathbb{R}^n)$ and its Fourier transform (given by that formula) coincides with the transform in the tempered distributions sense. But I don't know what to do in the general case ($a\in \mathbb{C}$). The book (Introduction to Nonlinear Dispersive Equations - Linares,Ponce) suggest to use an analytic continuation argument, but I don't really see how to do it. Can anyone help me with this, please? Thanks.

Answer 1

You can do this as follows: fix $t\in\mathbb R^n$, and set $$F(\alpha)=\int_{\mathbb R^n}e^{-\alpha|x|^2}e^{-2\pi itx}\,dx,\quad G(\alpha)=\left(\frac{\pi}{a}\right)^{\frac{n}{2}}\exp\left(-\frac{\pi^2|t|^2}{a}\right).$$ If $\alpha>0$, then $F(\alpha)=G(\alpha)$. You then show that $F$ and $G$ are holomorphic on ${\rm Re}(\alpha)>0$ (using, for example, the Cauchy-Riemann equations), and since $F$ and $G$ coincide on the positive real axis, they should coincide on $\{\alpha\in\mathbb C|{\rm Re}(\alpha)>0\}$.

To extend this to the $\alpha$ with ${\rm Re}(\alpha)=0$, you could use continuity.