Factor before Dirac delta in magnetic dipole field formula I bumped into this formula for the magnetic induction field generated by a dipole, containing Dirac's delta, while studying hyperfine splitting: $$\textbf{B}(\textbf{r}) = \frac{2}{3}\mu_0 \textbf{m}\delta(\textbf{r}) - \mu_0 \nabla\frac{1}{4\pi} \frac{\textbf{m}\cdot \textbf{r}} {|\textbf{r}|^3}.$$
If I try to compute the curl of the vector potential of a dipole, which should be $$\textbf{A}=-\frac{\mu_{0}}{4\pi}\cdot\textbf{m}×\nabla\frac{1}{r}$$ to obtain the $B$ field, I end up getting the formula quoted here -> Equation for the field of a magnetic dipole. In this formula, the delta hasn't got the $\frac{2}{3}$ factor on its side because it comes from the Laplacian of $\frac{1}{r}$. However every book and article reporting the equation says the 2/3 factor has to be there. Is there a way to reconcile the two formulae or one of the two is wrong? Why?
 A: It is somewhat problematic to rigorously define $\partial_i\partial_j\frac{1}{r}$ in 3D distribution theory, cf. e.g. this Math.SE post. Nevertheless, due to the identity
$$ \nabla^2\frac{1}{r}~=~-4\pi\delta^3(\vec{ r}),\tag{A}$$
it makes heuristic/physical sense to assign
$$ \partial_i\partial_j\frac{1}{r}~=~-\frac{4\pi}{3}\delta_{ij}\delta^3(\vec{ r})  ~+~ {\rm P.V.}\left(\frac{3x_ix_j}{r^5} -\frac{\delta_{ij} }{r^3}\right), \tag{B}$$
where ${\rm P.V.}$ stands for the Cauchy principal value.
Using
$$ \vec{A}~=~-\frac{\mu_{0}}{4\pi}\cdot\vec{m}\times \vec{\nabla}\frac{1}{r}, \tag{C} $$
we can calculate
$$\begin{align} \vec{B}
~=~&\vec{\nabla}\times\vec{A} \cr
~\stackrel{(C)}{=}~&-\frac{\mu_{0}}{4\pi}\left(\vec{m}\nabla^2\frac{1}{r}  ~-~\vec{\nabla}(\vec{m}\cdot \vec{\nabla}\frac{1}{r})  \right)  \cr
~\stackrel{(A)}{=}~&\mu_{0} \vec{m}\delta^3(\vec{ r})~+~ \frac{\mu_{0}}{4\pi}\vec{\nabla}(\vec{m}\cdot \vec{\nabla}\frac{1}{r})  \cr
~\stackrel{(B)}{=}~&\frac{2}{3}\mu_{0} \vec{m}\delta^3(\vec{ r}) ~+~ {\rm P.V.}~\frac{\mu_{0}}{4\pi}\frac{3\vec{m}\cdot\vec{r}-\vec{m}r^2}{r^5}\cr 
~=~&\frac{2}{3}\mu_{0} \vec{m}\delta^3(\vec{ r}) ~+~ {\rm P.V.}~\frac{\mu_{0}}{4\pi}\vec{\nabla}(\vec{m}\cdot \vec{\nabla}\frac{1}{r}) 
.\end{align}\tag{D}$$
