Curl and circulation of a vector field that is ill-defined at the origin: any interesting physical effects? In the cylindrical polar $(\rho,\phi,z)$ coordinate, suppose the velocity field in a liquid is given by $$\vec{v}=\frac{K}{\rho}\hat{e}_{\phi}, \qquad K=\text{constant}.\tag{1}$$ It can be easily checked that $$\vec\nabla\times\vec{v}=0, \qquad \rho\neq 0.\tag{2} $$ Is it possible to write down a compact expression of $\vec\nabla\times\vec{v}$ valid at the points including $\rho=0$? 
Next, I want to use the Stokes theorem to calculate the value of $\oint\vec{v}\cdot\vec{dl}$. I understand that this looks like a math question. However, how would a physicist interpret this singularity in the sense that how physical it is? Will it have any important physical consequence in the sense that this singularity cannot be ignored?
 A: *

*Physically, OP's vector field $\vec{v}$ is like a magnetic $\vec{B}$-field created by an electric current in a thin wire along the $z$-axis, cf. Ampere's circuital law.   

*Mathematically, it is possible to lift OP's eq. (2) to distribution theory
$$ \vec{\nabla}\times\vec{v}~=~2\pi K ~\delta(x)~\delta(y)~\hat{e}_z, \tag{A}$$
so that Kelvin–Stokes theorem works. 
Sketched proof of eq. (A). The simplest is probably to regularize OP's vector field (1) as as a smooth vector field $$ \vec{v}~=~\frac{K\rho}{\rho^2+\varepsilon}\hat{e}_{\phi},\tag{B}$$
where $\varepsilon>0$ is a regularization parameter. Then the curl becomes
$$ \vec{\nabla}\times\vec{v}~=~ K ~\frac{2\varepsilon}{(\rho^2+\varepsilon)^2} ~\hat{e}_z. \tag{C}$$
Finally use the following representation of the 2D Dirac delta distribution:
$$\delta(x)\delta(y)~=~\frac{1}{\pi}\lim_{\varepsilon\to 0^+}\frac{\varepsilon}{(\rho^2+\varepsilon)^2}, \qquad \rho~=~\sqrt{x^2+y^2}.  \tag{D}$$
$\Box$
A: Mathematical answer
Yes, one can carefully re-define field theory so that it holds in the sense 
of distributions, mainly you will end up with equations containing Dirac deltas in the right hand side (a standard textbook exercise is the calculation of the potentials generated by a moving charge).
Physical answer
Your example describes cases that very often do occur in physics. Take for instance
the electric field generated by a point particle $q$: it is stated that
$$
\textbf{E}(\textbf{r})=\frac{q}{r^2}\textbf{u}_r \tag{1}
$$
up to scaling constants. Now one can easily see that most quantities derived thereupon will not 
be defined (let alone continuous) in $r=0$ as a consequence of the fact that the field itself is not defined at the origin. This per se' is not really a problem because $(1)$ holds (even if rarely stated in physics textbooks) only far away from the charge, namely as we approach $r\to 0$ (without even the need to reach the origin) such formula ceases to hold true.
This is proven in exercises when you calculate the potential (or equivalently the field) of what textbooks authors call the infinite plate, the infinite conductor, the infinite solenoid or so forth, 
namely whenever you use charge densities rather than charges themselves: this is because as you get closer to a point charge its dimensions increase and the only quantity that you can measure is the density. For those cases you can still perfectly apply Maxwell's equations, differentiate and integrate without boundaries and domain problems because, by definition, you are not in the origin
anymore. 
To directly answer your questions:


*

*if you have an equation that is undefined for some values of the coordinates (say $r=0$) this means
that by definition you are far away from the origin, the domain of such equation is $\mathbb{R^3}-
\left\{0\right\}$ and you need not know what happens in the origin, as it is not part of the domain.

*if you are close to the origin, then arguably the equations must change (fields will be replaced
by densities) and you will not have discontinuities anymore.
If your equations describe dynamics (as I already mentioned in the example of the moving charge) and the points in the space-time where divergences occur may change (i. e. not just the origin anymore) then you must pay the price and move to distributions.
