How can a curl lead to delta distributions? My professor defined the following vector field:
$$
\vec{A}=\left(\frac{-gy}{r(r+z)},\frac{gx}{r(r+1)},0\right)
$$
($g$ being a constant, $r=\sqrt{x^2+y^2+z^2}$ and told us to derive the 3-dimensional curl:
$$
\mathrm{curl}\vec{A}=\vec\nabla\times\vec{A}=g\frac{\vec{r}}{r^3}+4\pi\delta(x)\delta(y)\theta(-z)
$$
where $\theta$ is the Heaviside function, and $\delta$ a Dirac distribution.
I have absolutely no clue how to derive that. I have never seen a curl leading to Dirac delta distributions.
Any hints are welcome.
 A: Take the function
$$f(x,y,z) = \frac{x}{\sqrt{x^2+y^2+z^2}}$$
And say we want to find $\frac{\partial f}{\partial x}$. If either of $y$ or $z$ are nonzero, then this is straightforward, as the function is differentiable; you get $\frac{y^2+z^2}{r^3}$. However, if $y = z = 0$, then
$$f(x, 0, 0) = \frac{x}{\sqrt{x^2}} = \textrm{sign}(x)$$
Strictly speaking, thus function is not differentiable at $x = 0$. However, we can informally say 
$$\frac{d}{dx} \textrm{sign}(x) = 2\delta(x)$$
because 
$$\int_a^b 2\delta(x) dx = \textrm{sign}(b)-\textrm{sign}(b)$$
since both the LHS and RHS are equal to $2$ when $a < 0 < b$, and $0$ when $a < b < 0$ or $0 < a < b$. Notice that $2$ is the "surface area" of a unit $0$-sphere; similar to how you have a $4\pi$ in your epression, the surface area of a unit $2$-sphere. This isn't a coincidence. 
Anyway, putting these together, we see that 
$$\frac{\partial f}{\partial x} = \frac{y^2+z^2}{r^3} + 2\delta(x) \chi(x,y,z)$$
where $\chi$ is the indicator function for the set $\{(x,0,0) : x\in\Bbb{R}\}$. 
I hope this makes more sense now. Again, everything here was completely informal/heuristic. 
