A Polar Fourier transform of derivatives

hunt_mat
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TL;DR Summary
Computing Fourier transforms is simple in Cartesian co-ordinates but how do you do it for polar co-ordinates?
The 2D Fourier transform is given by: \hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy
In terms of polar co-ordinates: \hat{f}(\rho,\phi)=\int_{0}^{\infty}\int_{-\pi}^{\pi}rf(r,\theta)e^{-ir\rho\cos(\theta-\phi)}drd\theta

For Fourier transforms in cartesian co-ordinates, relating the Fourier transform of a derivative of a function to the Fourier transform of the function. However, what happens with the polar Fourier transform? I've done a simple calculation for the r-derivative and I get:
\widehat{\frac{\partial f}{\partial r}}=-\rho\sum_{n\in\mathbb{Z}}\int_{0}^{\infty}rf_{n}(r)J_{n-1}(r\rho)dr
This is after I expanded f(r,\theta) as a Fourier series:f(r,\theta)=\sum_{n\in\mathbb{Z}}f_{n}(r)e^{in\theta}
I feel a little lost at this point. Can anyone suggest anything?
 
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hunt_mat said:
TL;DR Summary: Computing Fourier transforms is simple in Cartesian co-ordinates but how do you do it for polar co-ordinates?

The 2D Fourier transform is given by: \hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy[\tex]<br />
<br /> Did you mean to type this? <br /> The 2D Fourier transform is given by: ##\hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy##
 
DaveE said:
Did you mean to type this?
The 2D Fourier transform is given by: ##\hat{f}(k,l)=\int_{\mathbb{R}^{2}}f(x,y)e^{-ikx-ily}dxdy##
I was correcting my memory on how to include LaTeX here, you'll see that it's corrected now.
 
Search for "Hankel transform". The Wikipedia article doesn't give an expression for the transform of a derivative, but differentiating the inverse transform and using Bessel function identities should work.
 
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