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I want to perform Fourier Transform of $$\frac{\exp(jkr)}{r},$$ where $k=\frac{2 \pi}{\lambda}$ and $r=\sqrt{x^2+y^2+z^2}$. The result should be $\exp\left(jkz \sqrt{1-(\lambda u)^2 - (\lambda v)^2} \right).$

My below code returns the function itself after a long time of running

FourierTransform[Exp[I k Sqrt[x^2 + y^2 + z^2]]/Sqrt[x^2 + y^2 + z^2], 
{x, y}, {u, v}, FourierParameters -> {0, 2 Pi}]

Can anyone tell me what's the problem with my code? Thanks a lot.

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    $\begingroup$ You need to use correct syntax in Mathematica for any chance of getting an answer. so e is Exp and i should be I and e^(...) should be Exp[....]. But after doing these changes, it does not look like it can do it. But maybe if you wait long time it can. $\endgroup$
    – Nasser
    Commented Feb 1, 2022 at 7:51
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    $\begingroup$ Going to a cylindrical coordinate system might help (there seems to be such a symmetry in the integral). If I have time later, I might give it a shot. $\endgroup$
    – Hans Olo
    Commented Feb 1, 2022 at 10:15
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    $\begingroup$ Your LATEX formula $\frac{\exp(jkr)}{r}$ differs from your code Exp[I k Sqrt[x^2 + y^2 + z^2]]/(x^2 + y^2 + z^2). What should we trust? In both cases the usual Fourier transform does not exist for real values of k,x,y,z. $\endgroup$
    – user64494
    Commented Feb 3, 2022 at 13:16
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    $\begingroup$ 1) Why are you converting to Cartesian before doing the transform? 2) Why is there a z in your expected result? $\endgroup$
    – Rudy Potter
    Commented Feb 3, 2022 at 20:51
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    $\begingroup$ Even FourierTransform[Exp[I Sqrt[r1^2 + 1]]/Sqrt[r1^2 + 1], r1, s] returns the input. $\endgroup$
    – user64494
    Commented Feb 4, 2022 at 7:51

1 Answer 1

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So we have

f[r_] = E^(-I k r)/r /. k -> (2 \[Pi])/\[Lambda]

enter image description here
Taking the Fourier transform in spherical coordinates

FourierTransform[f[r], r, p, FourierParameters -> {0, 2 Pi}]

give a nice compact but complex result:
enter image description here
but suppose we convert to cylindrical coordinates:

fcy[\[Rho]_, z_] = 
 TransformedField["Spherical" -> "Cylindrical", 
  f[r], {r, \[Theta], \[Phi]} -> {\[Rho], \[Phi]2, z}]

enter image description here
and then take the Fourier transform

Simplify[FourierTransform[
  fcy[\[Rho], z], {\[Rho], z}, {\[CapitalRho], \[CapitalZeta]}, 
  FourierParameters -> {0, 2 Pi}]]

we get
enter image description here
Which is less compact but is real and looks oddly similar to your expected result.

I'm still not able to get the transform of the expression in Cartesian coordinates. Interestingly, I am also unable to get the inverse Fourier transform of your expected result.

We can transfer to Cartesian coordinates with

fxyz[x_, y_, z_] = 
 TransformedField["Spherical" -> "Cartesian", 
  f[r], {r, \[Theta], \[Phi]} -> {x, y, z}]

enter image description here
And try to take the transform in just X and Y leaving Z constant.

FourierTransform[fxyz[x, y, z], {x, y}, {u, v}, 
 FourierParameters -> {0, 2 Pi}]

Which is basically the code you have in your question. This ran for about 4 hours on my PC before Mathematica gave up and returned the input.

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    $\begingroup$ If I am not mistaken, the usual Fourier transform of $$\frac{e^{-\frac{2 i \pi \sqrt{\rho ^2+z^2}}{\lambda }}}{\sqrt{\rho ^2+z^2}}$$ does not exist so it should result in terms of generalized functions. $\endgroup$
    – user64494
    Commented Feb 4, 2022 at 7:56
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    $\begingroup$ If I correctly understand it, z is treated as a constant in the question. $\endgroup$
    – user64494
    Commented Feb 4, 2022 at 8:04
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    $\begingroup$ @user64494 Yes, z should not be in the Fourier Transform. $\endgroup$
    – Neysa
    Commented Feb 4, 2022 at 8:14

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