In spherical coordinates, I believe that the laplacian of $1/r$ is zero everywhere except at $r = 0$ or
\begin{align} \nabla^2 \dfrac{1}{r} = -4\pi \delta^{(3)}({\vec{r}}). \end{align}
where $r$ is defined as the distance from the origin.
Is there a similar delta type function at $\rho = 0$ in cylindrical coordinates where $\rho$ the radial coordinate? When I write out the terms I get
\begin{align} \nabla^2 \dfrac{1}{\rho} = \left( \dfrac{\partial^2}{\partial \rho^2} + \dfrac{1}{\rho}\dfrac{\partial}{\partial \rho}\right)\dfrac{1}{\rho} = \dfrac{2}{\rho^3}-\dfrac{1}{\rho^3} = \dfrac{1}{\rho^3} \end{align}
Therefore I expect that the $\nabla^2 \dfrac{1}{\rho}$ should equal $\dfrac{1}{\rho^3}$ when $\rho \neq 0$ and perhaps some delta function when $\rho = 0$.
Is there a similar relation for $\nabla^2 \dfrac{1}{\rho^2}$ or $\nabla^2 \dfrac{1}{\rho^n}$?
I am asking because I am considering the biharmonic equation $\nabla^4f = 0$ in cylindrical coordinates where $\nabla^2 \dfrac{1}{\rho}$ and $\nabla^2\dfrac{1}{\rho^2}$ appear. I would be interested in any published papers on this topic as well.
EDIT: I have redefined the radial coordinate in the cylindrical coordinate system as $\rho$ instead of $r$
No, you're confusing the two meanings of $r$. In cylindrical coordinates, you need to be differentiating $(r^2+z^2)^{-1/2}$. And the Laplacian in $\Bbb R^3$ will of course have the $\partial^2/\partial z^2$ term.