Problem:
Let $P={-y \over x^2+y^2}$ and $Q={x \over x^2+y^2}$ for $(x,y)\ne(0,0)$.
Show that $\oint_{\partial \Omega}(Pdx + Qdy)=2\pi$ if $\Omega$ is any open set containing $(0,0)$ and with a sufficiently regular boundary.
Working:
Clearly, we cannot immediately apply Green's Theorem, because $P$ and $Q$ are not continuous at $(0,0)$. So, we can create a new region $\Omega_\epsilon$ which is $\Omega$ with a disc of radius $\epsilon$ centered at the origin excised from it.
We then note ${\partial Q \over \partial x} – {\partial P \over \partial y} = 0$ and apply Green's Theorem over $\Omega_\epsilon$. Furthermore, $\oint_C(Pdx + Qdy)=2\pi$ if $C$ is any positively oriented circle centered at the origin.
I get the general scheme of how to approach this problem, however I am unsure of how to argue it in a rigorous manner.
Best Answer
By the definition of open set, since $(0,0) \in \Omega$, there is an $r>0$ such that the disk of radius $r$ centred at $(0,0)$, $B_r(0,0)$ is contained in $\Omega$. Hence you can define $\Omega_r = \Omega \setminus B_r(0,0)$, which has two distinct boundaries, $\partial \Omega_r$ and $\partial B_r(0,0)$.
Now, as you say, we can use Green's Theorem on $\Omega_r$, so $$ 0 = \iint_{\Omega_r} (\partial_x Q - \partial_y P)\, dx\,dy = \int_{\partial\Omega_r} P \, dx + Q \, dy = \left( \int_{\partial\Omega} -\int_{x^2+y^2=r^2} \right) (P \, dx + Q \, dy), $$ because the boundary of $B_r(0,0)$ is traversed in the opposite direction because we have to keep the interior of $\Omega_r$ on the left.
Therefore we have to compute $\int_{x^2+y^2=r^2} (P \, dx + Q \, dy)$. Since this is just a circle, we parametrise it by $x=-r\cos{t}$, $y=r\sin{t}$, $dx=r\cos{t} \, dt$, $ dy=r\sin{t} \, dt $: $$\int_{x^2+y^2=r^2} (P \, dx + Q \, dy) = \int_0^{2\pi} \frac{-r\sin{t}}{r^2(\cos^2{t}+\sin^2{t})} (-r\sin{t}) \, dt + \frac{r\cos{t}}{r^2(\cos^2{t}+\sin^2{t})} (r\cos{t}) \, dt \\ = \int_0^{2\pi} 1 \, dt = 2\pi. $$