I am looking for the Fourier transform of the function $$ f(x):=e^{-i\pi x^2}, $$ where we define $$ \hat f(\xi):=\int_{\mathbb R}f(x)e^{-2\pi i x\xi}dx. $$ First of all, the function $f$ is not integrable, but it is bounded and hence it can be regarded as a tempered distribution on $\mathcal S$, the Schwartz function space.
Completing squares as usual, it strongly suggests that its Fourier transform, in the sense of tempered distributions, is given by $$ \frac 1 {\sqrt i} e^{ i\pi \xi^2}. $$ The only problem is that I dont know how to determine the branch of $\sqrt i$. Any suggestions?
$$\begin{aligned} \int_{-\infty}^\infty e^{-i\pi x^2} e^{-2\pi i x \xi} dx &= \int_{-\infty}^\infty e^{-i\pi [(x+\xi)^2-\xi^2]} dx \\ &= e^{i\pi \xi^2} \int_{-\infty}^\infty e^{-i\pi x^2} dx \\ &= \frac{e^{i\pi \xi^2}}{\sqrt{2}} \end{aligned}$$
The last integral (Fresnel function) is easily evaluated using a circular arc contour subtending an angle of $\pi/4$.