What is the charge density of a pure electric dipole? I've only found two definitions of the charge density of a pure electric dipole:
\begin{equation}
\rho(\vec{r})=\vec{p}\cdot\nabla\delta(\vec{r}-\vec{r}_0)
\end{equation}
and
\begin{equation}
\rho(\vec{r})=-\vec{p}\cdot\nabla\delta(\vec{r}-\vec{r}_0).
\end{equation}
Neither of these came with reasons or derivations, so I'm not really sure which to use?
 A: In electrostatics, when you write the multipole expansion of the potential you find
\begin{equation}
\Phi(\vec{x})=\frac{1}{4 \pi \epsilon_0} \left[ \frac{q}{r} + \frac{\vec{p} \cdot \vec{x}}{r^3} + ... \right] \, ,
\end{equation}
where $r=|\vec{x}|$ and $...$ indicates the higher order multipole terms. We can find the effective charge distribution with
\begin{equation}
\nabla^2 \Phi = - \frac{\rho}{\epsilon_0} \, .
\end{equation}
Now we just need appropriate expressions for the laplacian of the terms we have. Remember that
\begin{equation}
\frac{\vec{x}}{r^3} = -  \nabla \frac{1}{r} \quad \mathrm{and} \quad \nabla^2 \frac{1}{r} = - 4 \pi \, \delta(\vec{x}) \, .
\end{equation}
Now we will be able to reproduce the result if we notice that
\begin{eqnarray}
\nabla^2 \left[ \frac{\vec{p} \cdot \vec{x}}{r^3}\right]  &=& \nabla \cdot \left[  \nabla \left( \frac{\vec{p} \cdot \vec{x}}{r^3}\right)  \right] = \nabla \cdot \left[  (\vec{p} \cdot \nabla) \left( \frac{\vec{x}}{r^3}\right)  \right] = \vec{p} \cdot \nabla \left[ \nabla \cdot \left(\frac{\vec{x}}{r^3}\right) \right]
\\
&=& - \vec{p} \cdot \nabla \left[ \nabla \cdot \left( \nabla \frac{1}{r}\right) \right]
= - \vec{p} \cdot \nabla \left[ \nabla^2 \frac{1}{r} \right] = 4 \pi \, \vec{p} \cdot \nabla \delta (\vec{x}) \, .
\end{eqnarray} 
Check each equality carefully. Hence we find the effective charge density 
\begin{equation}
\rho(\vec{x}) = q \, \delta(\vec{x}) - \vec{p} \cdot \nabla \delta(\vec{x}) \, + \, ... \, .
\end{equation}
EDIT:
Alternatively you can say
\begin{eqnarray}
-4\pi\nabla \delta(\vec{x}) &=&\nabla \left[ \nabla \cdot \left(\nabla \frac{1}{r}\right)\right]=\nabla \left[ \nabla \cdot \left(- \frac{\vec{x}}{r^3}\right)\right] = \nabla \times \left[ \nabla \times \left(- \frac{\vec{x}}{r^3}\right) \right] + \nabla^2 \left(- \frac{\vec{x}}{r^3}\right) 
\\
&=& - \nabla^2 \left( \frac{\vec{x}}{r^3}\right) \, .
\end{eqnarray}
Now I think everything is consistent, sorry for the confusion.
