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

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.

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

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