Theorem 12.4.5 (Gaussian under Fourier Transform).label Let $f(x)=e^{-\pi a\abs{x}^2}, a>0$ then $\ha{f}(\xi)=a^{-n/2}e^{-\pi\abs{\xi}^2/a}$
Proof. We scale $a$ to $1$ by $x\mto a^{-1/2}x$ then
\begin{align*}\ha{f}(\xi)=\integral{\R^n}{}e^{-2\pi ix\cd \xi}e^{-\pi \abs{x}^2}dx=\prods{n}{j=1}\integral{\R}{}e^{-2\pi ix_j\xi_j-\pi x_j^2}dx_{j}=\prods{n}{j=1}e^{-\pi \xi_j^2}\end{align*}
where the last index we use complex analysis. First drop $j$ and let $z=x+i\xi$ since $e^{-\pi z^2}$ is entire and decay rapidly in modulus, for $\abs{\Im z}$ bounded and $\abs{\Re z}\to\infty$. By Cauchy and taking limit we deform contour to the real axis
\begin{align*}e^{-\pi \xi^2}\integral{}{}e^{-\pi z^2}dz=e^{-\pi \xi^2}\integral{}{}e^{-\pi x^2}dx=e^{-\pi \xi^2}=1\end{align*}
$\square$
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