# Fourier transform of $e^{-i\pi x^2}$

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 $\pi/4$.
• Thank you so much! Then it seems that my guess for the answer was wrong, and we cannot treat $i$ as if it were real, right? – Singfook Sangwood Nov 8 '17 at 6:59
• @ThomasYang The formula $$\int_{-\infty}^{\infty} e^{-a\pi x^2}e^{-2\pi i x \xi} dx = \frac{1}{\sqrt{a}} e^{-\pi \xi^2/a}$$ only holds when $a>0$. If you treat $a$ as a complex number, the path of integration will no longer be the real line, this causes problems. – pisco Nov 8 '17 at 7:06