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I would like to know the equivalent in Fourier space of an double integral over a circular domain:

$$\int\int_C f(x,y) =\int_0^l \rho d\rho\int_0^{2\pi}d\theta f(\rho,\theta)\rightarrow ????????$$

I know the $\hat{f}(k)$ but there's no closed form for its inverse Fourier transform.

Thanks

JFNJr
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1 Answers1

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If your function $f$ is radially symmetric, then the equivalent of the FT is

$$\hat{f}(k) = 2 \pi \int_0^{\infty} dr \, r \, f(r)\, J_0(k r) $$

where $J_0$ is the Bessel function of the first kind of zeroth order and is equal to

$$J_0(x) = \frac{1}{2 \pi} \int_0^{2 \pi} d\theta \, e^{i x \cos{\theta}}$$

If not, then you may write $f(r,\theta)$ as a Fourier series in $\theta$

$$f(r,\theta) = \sum_{m=-\infty}^{\infty} c_m(r) e^{i k \theta}$$

and

express in terms of Bessels of higher order.

Ron Gordon
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  • Probably my question was mistakenly written. Since I only know the FT, I would like to know how to perform the integral over the circular domain in the fourier space. I could then go back doing IFT, if then it will be possible in a closed form, otherwise I keep it in F space. – JFNJr May 13 '13 at 17:42
  • Perhaps if you stated what $\hat{f}$ was exactly, I would have a better idea of what you want. – Ron Gordon May 13 '13 at 17:43
  • $\hat{f}$ is the one that I posted before and you were answering about the incapability to have a closed form. – JFNJr May 13 '13 at 17:45
  • @StefanoCoppola: oh, I see. – Ron Gordon May 13 '13 at 17:55
  • So, any clue for my problem? – JFNJr May 13 '13 at 18:41