TITLE: A stupid question on Stokes' theorem
QUESTION [7 upvotes]: Suppose I want to do the simple integral 
$I=\int_{ R^2} \frac{\mathrm{d}r\wedge\mathrm{d}\phi}{(1+r)^2}$.
Just evaluating the integral one quickly gets $\left. -2\pi\frac{1}{1+r} \right |^\infty_0=2\pi$.
However, since $ \frac{\mathrm{d}r\wedge\mathrm{d}\phi}{(1+r)^2} =\mathrm{d} \left(- \frac{\mathrm{d}\phi}{(1+r)} \right) $ I could also use Stokes' theorem.
The form I am integrating is not compactly supported, but I can integrate over a 2-ball of radius $R$ and then take the limit $R \rightarrow \infty$:
$I=\lim_{R \rightarrow \infty} \int_{B^2_R} \mathrm{d} \left(- \frac{\mathrm{d}\phi}{(1+r)} \right)=
-\lim_{R \rightarrow \infty} \int_{S^1_R}  \frac{\mathrm{d}\phi}{(1+r)} 
=-\lim_{R \rightarrow \infty} 2\pi  \frac{1}{(1+R)}=0 $ which is clearly wrong.
So where is the mistake? The function $1/(1+r)$ is regular all the way from $0$ to $\infty$. I think that the problem is the fact that $\mathrm{d}\phi$ is not well defined at $r=0$, is that right? Would it be correct then to integrate over the domain $\{(x,y)\in R^2\;|\; \epsilon\leq x^2+y^2\leq R \}$, and then take both the limit $R\rightarrow \infty $ and $\epsilon\rightarrow 0$?

REPLY [4 votes]: Yes, you are right, the problem is that $d\phi$ isn't well-behaved at the origin. If we compute it in cartesian coordinates, we find
$$d\phi = \frac{-ydx + xdy}{x^2+y^2},$$
so the "magnitude" of $d\phi$ is proportional to $r^{-1}$, thus you can't apply Stokes ignoring that singularity.
If you consider the region $\varepsilon < \sqrt{x^2+y^2} < R$, the integral over the inner boundary converges to the value $2\pi$ that is the value of the area integral.