How can I prove that $\mathcal{F} (e^{-\pi \alpha x^2}) =\frac 1{\sqrt{\alpha}} e^{-\pi \xi^2 /\alpha}$ for $\alpha \neq 0$ purely imaginary, in the sense of tempered distributions? (Here, $\sqrt {a}$ denotes the standard analytic branch of the complex square root.)
I tried to generalize the proof for the case $\mathfrak{Re} \alpha >0$. In this case, we can conclude that $(e^{\pi \xi^2 /a } \hat f )' =0$, so that $ \hat f (\xi ) = e^{-\pi \xi^2 /a } \hat f (0)$. But for the case $\alpha \neq 0 $ purely imaginary, $\hat f$ may not be a $C^1$ function (a priori, it is just a tempered distribution) so that this method does not seem to work. What should I do?
Thank you for any help!
Let $$g(a,b)=\int_{-\infty}^\infty e^{-a t^2+bt}dt$$
For $a > 0,b\in \Bbb{R}$ by completing the square $$g(a,b)-e^{-b^2/(4a)} (\pi/a)^{1/2}=0$$ With the branch of $z^{1/2}$ analytic on $\Re(z) > 0$ and positive on $z > 0$ the LHS extends to a complex analytic function of $\Re(a)> 0,b\in \Bbb{C}$, that it vanishes on $(0,\infty) \times \Bbb{R}$ implies it vanishes everywhere.
Finally for $c\in \Bbb{R}^*$ we have the limit in the sense of tempered distributions $$\mathcal{F}[e^{-ic t^2}](\xi) = \lim_{\epsilon \to 0^+}\mathcal{F}[e^{-(ic+\epsilon) t^2}](\xi) =\lim_{\epsilon \to 0^+} e^{\pi \xi^2/(ic+\epsilon)} (\pi/(ic+\epsilon))^{1/2}= e^{\pi \xi^2/(ic)} (\pi/(ic))^{1/2}$$