Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
Possible DUplicate: math.stackexchange.com/questions/55419/… –  JavaMan Oct 10 '11 at 0:24
    
@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 0:37
    
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 0:53
3  
The downvoter should perhaps explain the reason for the downvote. –  Srivatsan Oct 10 '11 at 0:59
    
Have you tried spherical coordinates? –  rcollyer Oct 10 '11 at 3:01

1 Answer 1

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.

share|improve this answer
1  
For OP: Heuristically, $\int_{0}^{\infty} \sin z \, dz = 1$ and hence the Fourier transform of $\frac{1}{|x|}$ is $\frac{4\pi}{|\xi|^{2}}$. Of course, this is justified in distribution sense. –  sos440 Mar 26 at 11:09
    
For some rigorous treatment, one can refer to Proposition 4.1 of Lectures on Harmonic Analysis by Thomas H. Wolff. –  sos440 Mar 26 at 11:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.