$F(z)=\int_{{\mathbb R}^n}f(x)e^{2\pi x \cdot z-\pi x \cdot x-\frac{\pi}{2}z^2}dx$

Given $f\in L^2({\mathbb R}^n)$ is a radial function.

$z\in{\mathbb C}^n$ and $z^2$ denotes $z \cdot z$ (dot product).

I tried integrating in polar co-ordinates but I have difficulties simplifying the integral in higher dimensions.

  • $\begingroup$ This looks like a Bargmann transform in case you weren't aware. You might find some relevant material if you Google around. $\endgroup$ – Cameron Williams Jun 21 '16 at 19:58
  • $\begingroup$ Yes, I know but I don't think it requires anything more than some integral manipulations. Like the one dimensional case can be obtained by averaging $F(z)$ and $F(-z)$. $\endgroup$ – Mathew George Jun 21 '16 at 20:03
  • $\begingroup$ @Did How can you separate $z$ from a dot product like that. $\endgroup$ – Mathew George Jun 21 '16 at 20:32
  • $\begingroup$ $z$ is complex in the Bargmann transform @Did. $\endgroup$ – Cameron Williams Jun 21 '16 at 20:42
  • $\begingroup$ @CameronWilliams OK. $\endgroup$ – Did Jun 21 '16 at 20:50

Here is the solution for anyone interested.

In polar co-ordinates,

$$F(z) = e^{-{\pi\over 2}z^2}\int_0^\infty f(r)r^{n-1}e^{-\pi r^2}\int_{S^{n-1}} e^{2\pi r(x'\cdot z)}\,d\sigma(x')\,dr$$

Let $g(z)=\int_{S^{n-1}} e^{2\pi r(x'\cdot z)}\,d\sigma(x')$ and $h=g|_{{\mathbb R}^n}$ for a fixed $r$.

$h$ is a radial function since if $K\in SO(n)$,

$$h(Kx)=\int_{S^{n-1}} e^{2\pi r(x'.Kx)}\,d\sigma(x')=\int_{S^{n-1}} e^{2\pi r((K^*x').x)}\,d\sigma(x')=h(x)$$ using the rotational invariance of surface measure.

So we have $h(x)=h_0(x\cdot x)$.

$h_0$ can be analytically extended to ${\mathbb{C}}^n$ as, say, $H$. Now $H$ and $g$ are two entire functions and $H(z\cdot z)=g(z)$ on ${\mathbb{R}}^n$. So $H(z\cdot z)=g(z)$ everywhere.

Hence, $$F(z) = e^{-{\pi\over 2}z^2}\int_0^\infty f(r)r^{n-1}e^{-\pi r^2}K(r,z^2)\,dr$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.