TITLE: Integration (Fourier transform)
QUESTION [8 upvotes]: $\mathrm{Re}(i) = 0$, but the fourier transform of $f(x) = e^{-ix^2}$ is $g(\alpha) = \sqrt{\pi\over i}\times e^{i\alpha^2 \over 4}$, is it not? 
Is there an easy to show that it is so, knowing the fourier transform of $h(x) = e^{-x^2}$? (It is $k(\alpha) = \sqrt\pi\times e^{-\alpha^2 \over 4}$.)
The non-unitary, angular frequency Fourier transform is taken into consideration.

REPLY [7 votes]: The Fourier transform of $f(x) = \exp(-ix^2)$ doesn't exist in the usual sense. Note that $f \notin L^1$.
However, the function $f$ is bounded and continuous and, as such, can be identified with a tempered distribution, so $\hat f$ exists, also as a tempered distribution. Working out the details, you do in fact end up with
$$\hat f(\omega) = \sqrt{\frac{\pi}{2}} (1-i) e^{i\omega^2/4}$$
which is the same as what you have written, with a suitable interpretation of $\sqrt{i}$. The approach outlined by rlgordonma is probably the easiest way to get to the result.