# Fourier transform integral

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.

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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
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

Inserting the Jacobian $r^2\sin\theta$ and $\sqrt{x^2+y^2+z^2}=r$ in polar coordinates gives
$$\int_0^\infty r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \frac{1}{r} e^{i\mathbf{k}\cdot \mathbf{r}}$$
$$= \int_0^\infty r^2 dr \int_0^{2\pi} d\phi \int_0^\pi \sin\theta d\theta \frac{1}{r} e^{ikr\cos\theta}$$
and with $z=\cos\theta$, $dz=-\sin\theta d\theta$ $$= 2\pi \int_0^\infty r dr \int_0^\pi \sin\theta d\theta e^{ikr\cos\theta} = -2\pi \int_0^\infty r dr \int_{1}^{-1} dz e^{ikrz} = 2\pi \int_0^\infty r dr \int_{-1}^{1} dz e^{ikrz}$$ and with $t=ikrz$, $dz=dt/(ikr)$ $$= 2\pi \int_0^\infty r dr \frac{1}{ikr} \int_{-ikr}^{ikr} dt e^t$$ $$= 2\pi \int_0^\infty r dr \frac{1}{ikr} [e^{ikr}-e^{-ikr}] = 4\pi \int_0^\infty r dr \frac{1}{kr} \sin(kr) = \frac{4\pi}{k^2} \int_0^\infty kr d(kr) \frac{1}{kr} \sin(kr)$$ $$= \frac{4\pi}{k^2} \int_0^\infty d(kr) \sin(kr) = \frac{4\pi}{k^2} \int_0^\infty dz \sin z$$ and this exists only in the theory of distributions.
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. – Sangchul Lee Mar 26 '14 at 11:09