Understand this Fourier transform $\int \frac{1}{|x|}e^{ikx} d^3 x = \frac{4 \pi}{k^2}$ I found the equation 
$$\int \frac{1}{|x|}e^{ikx} d^3 x = \frac{4 \pi}{k^2}$$
in a 'physics' textbook and I just don't understand what this equation tries to tell me. Is there anybody who understands this calculation? I just don't see it, cause I interpreted this as a 3d- integral over the norm, but I guess that this integral would not converge. Especially, I don't know the limits, though this $4 \pi$ factor looks very much like the angular integral in spherical coordinates.
If anything is unclear, please let me know.
 A: Just a rough way to go:
Denote \begin{align}
I(k)=\int_{R^3}\frac{1}{|x|}e^{i \vec{k}\vec{x}}d^3x
\end{align}
1.) Introduce spherical coordinates $(r, \phi, \theta)$, with $|\det(J)|=r^2 sin(\phi)$. 
2.) Choose the Orientation of $\vec{k}$ as along the $z$-Axis: $\vec{k}\vec{x}=k r \cos(\phi) $
3.)  $kr \rightarrow r$ which gives us a nice $1/k^2$ overall factor
3.)  Do the trivial $\theta$ integral, so $2\pi$ shows up
4.)  Carry out the nearly trivial $\phi$ integral. You should have something like \begin{align}I(k)=\frac{2\pi}{i k^2}\int_0^{\infty}\left(e^{-ir }-e^{ir }\right)dr
\end{align}
5.) the crucial step is now to interpret this integral. As a comment suggest this has to be interpreted in a distributional sense. For computational purposes it is sufficient to add a small imaginary part to $r$ so that the integrals converge: $k\rightarrow r\pm i\delta $
6.) Take the limit $\delta \rightarrow 0  $ and you're done:
\begin{align}
I(k)=\frac{4\pi}{k^2}
\end{align}
Is this enough, so you can fill in the rest for yourself?
