Fourier transform of $e^{-i|x|^2}$

Question

I am trying to calculate the Fourier transform of $f(x)=e^{-i|x|^2}$ for $x\in\mathbb{R}^n$. Roughly speaking, the Fresnel integral implies that $$\hat{f}(\xi)=(2i)^{-n/2}e^{i|\xi|^2/4}$$ where the Fourier transform of a function $g$ is defined as $$\hat{g}(\xi)=(2\pi)^{-n/2}\int_{\mathbb{R}^n}g(x)e^{-ix\cdot\xi}dx,\ \ \xi\in\mathbb{R}^n.$$ However, as this post says, $f$ is not in $L^1$ and the Fresnel integral holds in the sense of improper Riemann integral not in the sense of Lebesgue integral. I think this equality $\hat{f}(\xi)=(2i)^{-n/2}e^{i|\xi|^2/4}$ is true in the sense of distributions and the proof of this calculation is based on distribution theory, but I cannot go further. How can we prove this? I appreciate any advice.

Answer 1

You can study the differential equation satisfied by $f$:

$$f'=-2ixf,$$ therefore, up to some normalization constants, $$i\xi \hat f = 2\hat f',$$ and hence $\hat f(\xi)=C\exp(i\xi^2/4)$

Now you need to find some conditions in order to define the arbitrary constant $C$ arising from the solution of the above equation.

We can try to apply our distribution to a test function $\phi\in S$ with good properties (here "good" means that we know its Fourier transform and we know how to apply $f$ to $\phi$). Take $\phi(x)=\exp(-x^2/2)$ so that $\phi = \hat \phi$. Upon applying $f$ to $\phi$ and $\hat f$ to $\hat \phi$ you will obtain the necessary constant $C$.