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.


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 $C$.

  • $\begingroup$ Thank you very much. Let me ask something. What does "upon applying $f$ to $\phi$ and $\hat{f}$ to $\hat{\phi}$" mean? $\endgroup$ – Tom TJ Mar 17 '17 at 17:00
  • $\begingroup$ @TomTJ I meant $\langle f,\phi\rangle = \langle \hat f,\hat \phi\rangle$. $\endgroup$ – TZakrevskiy Mar 17 '17 at 17:20
  • $\begingroup$ Sorry, I am still confused. I got $\langle f',\phi\rangle=\langle\hat{f}',\phi\rangle \Rightarrow\langle i\xi\hat{f},\phi\rangle=\langle(i\xi/2)\hat{f},\phi\rangle \Rightarrow C=0$. $\endgroup$ – Tom TJ Mar 17 '17 at 21:06
  • $\begingroup$ No, without the derivative. Try to find explicitly $\langle f,\phi\rangle = \int_{\Bbb R} \exp\left(-(\frac 12 +i)x^2\right)dx$ and the corresponding integral for $\langle \hat f,\hat \phi\rangle$. $\endgroup$ – TZakrevskiy Mar 17 '17 at 21:25
  • 1
    $\begingroup$ You are right, a simple change of variables won't help you here, it is not allowed. The idea would be to show that the map $z\to \int_{\Bbb R}\exp(-zx^2/2)dx$ as a holomorphic function over $\{z\in\Bbb C:Re\ z>0 \}$. After that you say for real positive $z$ the value of the integral is known, $\frac{\sqrt {2\pi}}{\sqrt{z}}$. Finally, by identity theorem, you can extend this formula to $z$ with non-zero imaginary part. YOu might want to look at this question:(math.stackexchange.com/questions/1098902/…) $\endgroup$ – TZakrevskiy Mar 18 '17 at 23:11

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