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I'm trying to calculate the 3D fourier transform of this function:

$$\frac{1}{(x^2+y^2+z^2)^{1/2}}$$

Any help would be appreciated, thanks.

user14192
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    Possible DUplicate: http://math.stackexchange.com/questions/55419/2-dimensional-fourier-transform-integral – JavaMan Oct 10 '11 at 00:24
  • @DJC Not an exact duplicate, as the question you linked to has $3/2$ in the denominator rather than $1/2$, but similar solution methods will probably work – Chris Taylor Oct 10 '11 at 00:37
  • Hi, if a similar method would work, how would you implement it? The method used before was 2D (it relied on using cylindrical bessel functions) and I was unable to adapt it to this question. – user14192 Oct 10 '11 at 00:53
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    The downvoter should perhaps explain the reason for the downvote. – Srivatsan Oct 10 '11 at 00:59
  • Have you tried spherical coordinates? – rcollyer Oct 10 '11 at 03:01

1 Answers1

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Inserting the Jacobian $r^2\sin\theta$ and $\sqrt{x^2+y^2+z^2}=r$ in polar coordinates gives

\begin{equation} \int_0^\infty r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \frac{1}{r} e^{i\mathbf{k}\cdot \mathbf{r}} \end{equation}

\begin{equation} = \int_0^\infty r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \frac{1}{r} e^{ikr\cos\theta} \end{equation}

and with $z=\cos\theta$, $dz=-\sin\theta d\theta$ \begin{equation} = 2\pi \int_0^\infty r dr \int_0^\pi \sin\theta d\theta e^{ikr\cos\theta} = -2\pi \int_0^\infty r dr \int_{1}^{-1} dz e^{ikrz} = 2\pi \int_0^\infty r dr \int_{-1}^{1} dz e^{ikrz} \end{equation} and with $t=ikrz$, $dz=dt/(ikr)$ \begin{equation} = 2\pi \int_0^\infty r dr \frac{1}{ikr} \int_{-ikr}^{ikr} dt e^t \end{equation} \begin{equation} = 2\pi \int_0^\infty r dr \frac{1}{ikr} [e^{ikr}-e^{-ikr}] = 4\pi \int_0^\infty r dr \frac{1}{kr} \sin(kr) = \frac{4\pi}{k^2} \int_0^\infty kr d(kr) \frac{1}{kr} \sin(kr) \end{equation} \begin{equation} = \frac{4\pi}{k^2} \int_0^\infty d(kr) \sin(kr) = \frac{4\pi}{k^2} \int_0^\infty dz \sin z \end{equation} and this exists only in the theory of distributions.

R. J. Mathar
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