Inverse Fourier transform of $\exp(4\pi^2i|\xi|^2t)$ I would like to compute the inverse Fourier transform of $\exp(4\pi^2i|\xi|^2t)$.
\begin{equation}
f(x,t) = \int_{\mathbb{R}^n} e^{2\pi i x\cdot\xi} e^{4\pi^2i|\xi|^2t} \,\mathrm{d}\xi
\end{equation}
Can I use the fact that the Fourier transform of $e^{-\pi|x|^2}$ is $e^{-\pi|\xi|^2}$, and then apply to the function above with the change of variable (dilation)? I wonder if it works for complex numbers.
Thank you.
 A: Before editing the question you were asking the Inverse Fourier transform of
$\xi\mapsto \exp(4\pi^2i|\xi|^2)$
\begin{eqnarray}
f(x)&=&\int_{\mathbb{R}^n}e^{2\pi ix\cdot\xi}e^{4\pi^2i|\xi|^2}\,d\xi\stackrel{y=2\pi\xi}{=}\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i(x\cdot y+|y|^2)}\,dy=\frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i(|y+x/2|^2-|x|^2/4)}\,dy\\
&=&\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i|y+x/2|^2}\,dy\stackrel{z=y+x/2}{=}\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\int_{\mathbb{R}^n}e^{i|z|^2}\,dz=\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\left(\int_{\mathbb{R}}e^{it^2}\,dt\right)^n\\
&=&\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\left(\int_{\mathbb{R}}\cos(t^2)\,dt+i\int_{\mathbb{R}}\sin(t^2)\,dt\right)^n=\frac{e^{-i\frac{|x|^2}{4}}}{(2\pi)^n}\left[\sqrt{\frac\pi2}(1+i)\right]^n\\
&=&\left(\frac{1+i}{\sqrt{2}}\right)^n\cdot\left(\frac{\sqrt{\pi}}{2\pi}\right)^ne^{-i\frac{|x|^2}{4}}=\frac{e^{in\frac\pi4}}{(2\sqrt{\pi})^n}e^{-i\frac{|x|^2}{4}}=\frac{1}{(2\sqrt{\pi})^n}\exp\left(-i\frac{|x|^2-n}{4}\right).
\end{eqnarray}

After editing the question you are asking the Inverse Fourier transform of $(\xi,t)\mapsto \exp(4\pi^2i|\xi|^2t)$ (w.r.t. $\xi$). For $t\ne 0$, let us denote by $s$ the sign of $t$. Then
\begin{eqnarray}
f(x,t)&=&\int_{\mathbb{R}^n}e^{2\pi ix\cdot\xi}e^{4\pi^2i|\xi|^2t}\,d\xi\stackrel{y=2\pi\sqrt{|t|}\xi}{=}\frac{1}{(2\pi\sqrt{|t|})^n}\int_{\mathbb{R}^n}\exp\left(is(s|t|^{-1/2}x\cdot y+|y|^2)\right)\,dy\\
&=&\frac{1}{(2\pi\sqrt{|t|})^n}\int_{\mathbb{R}^n}\exp\left[is\left(\left|y+\frac{s|t|^{-1/2}}{2}x\right|^2-\frac{|x|^2}{4|t|}\right)\right]\,dy\\
&=&\frac{\exp\left(-is\frac{|x|^2}{4|t|}\right)}{(2\pi\sqrt{|t|})^n}\int_{\mathbb{R}^n}\exp\left(is\left|y+\frac{s|t|^{-1/2}}{2}x\right|^2\right)\,dy\\
&\stackrel{z=y+\frac{s|t|^{-1/2}}{2}x}{=}&\frac{e^{-it\frac{|x|^2}{4}}}{(2\pi\sqrt{|t|})^n}\int_{\mathbb{R}^n}e^{is|z|^2}\,dz=\frac{e^{-it\frac{|x|^2}{4}}}{(2\pi\sqrt{|t|})^n}\left(\int_{\mathbb{R}}e^{isu^2}\,du\right)^n\\
&=&\frac{e^{-it\frac{|x|^2}{4}}}{(2\pi\sqrt{|t|})^n}\left(\int_{\mathbb{R}}\cos(su^2)\,du+i\int_{\mathbb{R}}\sin(su^2)\,du\right)^n=\frac{e^{-it\frac{|x|^2}{4}}}{(2\pi\sqrt{|t|})^n}\left[\sqrt{\frac\pi2}(1+si)\right]^n\\
&=&\left(\frac{1+si}{\sqrt{2}}\right)^n\cdot\left(\frac{\sqrt{\pi}}{2\pi\sqrt{|t|}}\right)^ne^{-it\frac{|x|^2}{4}}=\frac{e^{isn\frac\pi4}}{(2\sqrt{\pi|t|})^n}e^{-it\frac{|x|^2}{4}}\\
&=&\frac{1}{(2\sqrt{\pi|t|})^n}\exp\left(-it\frac{|x|^2-sn}{4}\right)=\frac{1}{(2\sqrt{\pi|t|})^n}\exp\left(-i\frac{t|x|^2-|t|n}{4}\right).
\end{eqnarray}
