Understand this Fourier transform $\int \frac{1}{|x|}e^{ikx} d^3 x = \frac{4 \pi}{k^2}$

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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.

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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?