In 3D, Coulomb function in Cartesian coordinates $(x, y, z)$ reads
$$\frac{1}{|\boldsymbol{x}-\boldsymbol{x'}|} = \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}}$$
which may be expanded in spherical coordinates $(r, \theta, \phi)$ with Legendre polynomials (see here) or spherical harmonics (see here) in Laplace expansion.
Consider function in 2D
$$\frac{1}{|\boldsymbol{x}-\boldsymbol{x'}|} = \frac{1}{\sqrt{(x-x')^2 + (y-y')^2}}$$
What forms of expansion in Cartesian coordinates $(x, y)$ or polar coordinates $(r, \theta)$?
What special functions used?
The Laplace expansion will work: since $$ \frac{1}{\lvert (r\cos{\theta},r\sin{\theta})-(r'\cos{\theta'},r'\sin{\theta'}) \rvert} = \frac{1}{\sqrt{r^2+r'^2-2rr'\cos{(\theta-\theta')}}}, $$ one has $$ \frac{1}{\sqrt{r^2+r'^2-2rr'\cos{(\theta-\theta')}}} = \frac{1}{r}\sum_{l \geq 0} \left( \frac{r'}{r}\right)^l P_{l}(\cos{(\theta-\theta')}), $$ since the left-hand side is always the generating function of the Legendre polynomials, whatever the dimension. That's fine, but while one can apply the formula $$P_l(\cos{\theta}) = \frac{4\pi}{2l+1} \sum_{m=-l}^l Y^{-m}_l(\theta,0)Y^{m}_l(\theta',0), $$ it's rather low on physical meaning: it seems to suggest we are working on a great circle on the sphere, rather than in the plane proper.
On the other hand, the coordinate-free expansion of $$ \frac{1}{\lvert \mathbf{x}-\mathbf{x}'\rvert} = \frac{1}{\lvert \mathbf{x}\rvert} \left(1 -\frac{2\mathbf{x} \cdot \mathbf{x}'}{\lvert \mathbf{x}\rvert^2} + \frac{\lvert \mathbf{x}'\rvert^2}{\lvert \mathbf{x}\rvert^2} \right)^{-1/2} $$ using the binomial expansion works in exactly the same way, except that the expansion is no longer has traceless tensor in the $r^{-5}$ term: $3r_ir_j-\delta_{ij}r^2$ has nonzero trace for $d \neq 3$.