I have a problem trying to compute a derivative... (it is from Nakahara's book on Geometry, Topology an physics, the part of the Dirac monopoles).
He defines the vector potential as
$$ \vec A^N(x,y,z)=\frac{g}{r(r+z)}(-y,x,0) $$
where $r=\sqrt{x^2+y^2+z^2}$, and then he claims the curl is given by
$$ \nabla \times \vec A^N= g\frac{\vec r}{r^3}+4\pi g\delta(x)\delta(y)\theta(-z) $$
where $\delta$ is the dirac delta and $\theta$ the Heaviside function. It is easy to find that for $r\neq 0,-z$, the result holds. But exactly at line where the second term is important I couldn't do it. First of all I think it is kind of weird that the second term is just a scalar.
The followup analysis highly depends of this divergent term, so I wanted to now how to do it. But the way I could try to pass to spherical coordinates where the vector $\vec A^N$ has quite a simple form, but the coordinate singularity lies at the very same point of the divergence hence I wouldn't want to.
Edit (solution)
I found a solution if everyone else is looking for it. (Thanks for the useful coments) I found particularly useful the book on topological solitons by Manton and Sutcliffe.
Leting $f=\nabla\times\vec A-g\vec r/r^3$ and $\phi$ some test function.
$$ \left< f,\phi \right>=\int_{-\infty}^0dz\int_{B_\epsilon} \, dx \, dy \, f\phi $$
where $B_R$ is the ball of radius $R$. And $\epsilon$ is any positive number and we will take the limit $\epsilon\to 0$. The integrals of $(g\vec r/r^3)\phi$ yield $0$ by rotational symmetry. (it is not divergent, even at $z=0$, because the area of $B_\epsilon$ grows as $\epsilon^2$ and $1/r^2$ diverges as $1/\epsilon^2$ for $z=0$). Then
$$ \left< f,\phi \right>=\int_{-\infty}^0\int_{B_\epsilon} \, dx \, dy \, \nabla\times \vec A \phi $$
For the $x$ component we use $$ \int_{B_\epsilon} \, dx \, dy \, (\nabla\times \vec A)_x \phi=\int_{B_\epsilon} \, dx \, dy - \partial_z\left(\frac{gx}{r(r+z)}\right)\phi=-\partial_z\left(\int_{B_\epsilon} \, dx \, dy \frac{gx}{r(r+z)}\phi\right) $$
And the term inside parenthesis is 0 by similar arguments. For the $y$ component we have exactly the same
Lastly. for the z component we use $\nabla\times(\phi \vec A)=\phi \vec A+\nabla \phi\times \vec A$. $$ \int_{B_\epsilon} \, dx \, dy(\nabla\times \vec A)_z\phi=\int_{B_\epsilon}\vec da\cdot \phi\nabla \times \vec A=\int_{B_\epsilon}\vec da\cdot \nabla\times(\phi \vec A)+\int_{B_\epsilon}\vec da \cdot \nabla \phi\times \vec A $$
The second integral can be shown to be $0$ exactely in the same fashion as for the $x$ an $y$ coordinates. And for the first one we use Stokes theorem to get
$$ \int_{B_\epsilon}\vec da\cdot \nabla\times(\phi \vec A)=\int_{S_e}\vec dl\cdot\phi\vec A $$
where $S_R$ is the sphere of radius $\epsilon$. The line element is $\vec dl=\epsilon \hat\varphi$ and when $\epsilon\to 0$, $\vec A=\frac{2g}{\epsilon}\hat \varphi$. Hence
$$ \left< f,\phi \right>=\int_{-\infty}^04\pi g\phi(0,0,z)=\int_{\mathbb R^3}dV4\pi g\theta(-z)\delta(x)\delta(y)\phi\hat z $$
Then
$$ \nabla \times \vec A^N= g\frac{\vec r}{r^3}+4\pi g\delta(x)\delta(y)\theta(-z)\hat z $$