I just wanted to check my result.
Let's define the Fourier transform as (the integral over whole real line):
$$g(k)=\frac{1}{\sqrt{2 \pi}} \int e^{-i k x} f(x) dx$$
We have the following function:
$$f(\mathbf{r}_1,\mathbf{r}_2)=\frac{1}{|\mathbf{r}_1-\mathbf{r}_2|}$$
I want to Fourier it in all the Cartesian coordinates, which should be:
$$g(\mathbf{k}_1,\mathbf{k}_2)=\frac{1}{(2 \pi)^3} \int \int e^{-i \mathbf{k}_1 \mathbf{r}_1-i \mathbf{k}_2 \mathbf{r}_2} \frac{d\mathbf{r}_1 d\mathbf{r}_2}{|\mathbf{r}_1-\mathbf{r}_2|}=$$
Substituting:
$$\mathbf{r}_1=\mathbf{r}+\mathbf{r}_2$$
$$=\frac{1}{(2 \pi)^3} \int \int e^{-i (\mathbf{k}_1+\mathbf{k}_2) \mathbf{r}_2-i \mathbf{k}_1 \mathbf{r}} \frac{d\mathbf{r}_2 d\mathbf{r}}{|\mathbf{r}|}=\delta(\mathbf{k}_1+\mathbf{k}_2) \int e^{-i \mathbf{k}_1 \mathbf{r}} \frac{d\mathbf{r}}{|\mathbf{r}|}=\frac{4 \pi \delta(\mathbf{k}_1+\mathbf{k}_2)}{\mathbf{k}_1^2}$$
The last integral is evaluated using spherical coordinates and the properties of Bessel functions.
What's confusing to me is that the transform should be symmetric in $1 \leftrightarrow 2$, meaning that if we exchange the indices, the result should be the same. But it doesn't seem to be symmetric. Where's my mistake?
Appendix:
$$\int e^{-i \mathbf{k} \mathbf{r}} \frac{d\mathbf{r}}{|\mathbf{r}|}=\int_{0}^{2 \pi} \int_{0}^{\pi} \int_{0}^{\infty} e^{-i \rho ( k_{x} \sin \theta \cos \phi + k_{y} \sin \theta \sin \phi+k_{z} \cos \theta)} \rho \sin \theta d \rho d \theta d \phi=$$
$$=2 \pi \int_{0}^{\pi} \int_{0}^{\infty} e^{-i \rho k_{z} \cos \theta} J_0 (\sqrt{k_x^2+k_y^2} \rho \sin \theta ) \rho \sin \theta d \rho d \theta=$$
$$=2 \pi \int_{0}^{\pi} \int_{0}^{\infty} e^{-i \rho k_{z} \cot \theta} J_0 (\sqrt{k_x^2+k_y^2} \rho ) \rho \csc \theta d \rho d \theta=$$
$$=4 \pi \int_{0}^{\infty} K_0 (k_z \rho) J_0 (\sqrt{k_x^2+k_y^2} \rho ) \rho d \rho=\frac{4 \pi}{k_x^2+k_y^2+k_z^2}$$
Since $\delta\ne0\implies\mathbf{k}_1=-\mathbf{k}_2$, an equivalent way to write the result would be the manifestly symmetric $\dfrac{4\pi\delta(\mathbf{k}_1+\mathbf{k}_2)}{|\mathbf{k}_1||\mathbf{k}_2|}$.