What is the Fourier transform of $f(x)e^{2\pi ix^2}$ with $f\in L^2({\mathbb{R}})$?
Attempt.
Suppose $f\in L^2$ and let $g(x):=e^{2\pi ix^2}$. Clearly, $fg$ is in $L^2(\mathbb R)$, so its Fourier transform is well defined and in $L^2(\mathbb R)$, but can it be written in terms of the Fourier transform of $f$?
The Gaussian is its own Fourier transform $$ \int_{-\infty}^\infty e^{-\pi x^2} e^{-2i \pi \xi x}dx = e^{-\pi \xi^2}$$
(complete the square assuming $2i \pi \xi \in \mathbb{R}$ and extend the obtained result to $2i \pi \xi \in \mathbb{C}$ by analytic continuation)
Thus for $a > 0$
$$H_a(\xi) = \int_{-\infty}^\infty e^{-\pi a x^2} e^{-2i \pi \xi x}dx =\frac{1}{a^{1/2}} e^{-\pi \xi^2/a}$$
Again we can use analytic continuation to extend the result to $a \in \mathbb{C}, \Re(a) > 0$.
Therefore $$\mathcal{F}[f e^{- \pi a x^2}](\xi)=\widehat{f} \ast \frac{e^{-\pi \xi^2/a}}{a^{1/2}} $$ And by continuity in the $L^2$ sense $$\mathcal{F}[f e^{- \pi (-2i) x^2}]=\lim_{b \to 0^+}\mathcal{F}[f e^{- \pi (b-2i) x^2}] =\lim_{b \to 0^+} \widehat{f} \ast \frac{e^{-\pi \xi^2/(b-2i)}}{(b-2i)^{1/2}} $$ If $\widehat{f} \in L^1$ then the limits goes out and this is $$\mathcal{F}[f e^{- \pi (-2i) x^2}](\xi)=\widehat{f} \ast \frac{e^{-\pi \xi^2/(-2i)}}{(-2i)^{1/2}} $$