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.
 A: 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.
