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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?

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$$\begin{aligned} \int_{-\infty}^\infty e^{-i\pi x^2} e^{-2\pi i x \xi} dx &= \int_{-\infty}^\infty e^{-i\pi [(x+\xi)^2-\xi^2]} dx \\ &= e^{i\pi \xi^2} \int_{-\infty}^\infty e^{-i\pi x^2} dx \\ &= \frac{e^{i\pi \xi^2}}{\sqrt{2}} \end{aligned}$$

The last integral (Fresnel function) is easily evaluated using a circular arc contour subtending an angle of $\pi/4$.

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  • $\begingroup$ 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? $\endgroup$ – Singfook Sangwood Nov 8 '17 at 6:59
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    $\begingroup$ @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. $\endgroup$ – pisco Nov 8 '17 at 7:06

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