Understanding the Pole Singularity in Gradient of A

[alejandrito29]

in a text a read that

"$$\oint \nabla A \cdot dl = 2 \pi n$$

which implies that the gradient of A has a pole singularity"

why there is a singularity?

I thing that this is a contidion to integral is nonzero but ¿what is the theorem used?

 

[lurflurf]

by stokes theorem

$$\oint \! \bf{\nabla A} \cdot \mathrm{dl}=\iint \! \bf{\nabla \times (\nabla A)} \cdot \mathrm{ds}$$

clearly the curl of the gradient is zero so the integral is only nonzero if there is a singularity.