Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$ PartII The orginal problem is "Calculating $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$, where L is a smooth, simple closed, and postively oriented curve that does not pass through the orgin".
But what if I modify the hypothesis and allow non-simple closed curve? I mean is there something like green formula that allows us calculuating "non-simple closed curve integral"? 
EDIT:
It seems to me that this integral should also resemble the simple closed one. Because common interior line integral should be canceled only remaining the outter curve. Am I right?
 A: HINT: Suppose $L$ intersects with it self at finitely many points. Then, if we consider a parametrization of $L$, we deduce $L=L_1\cup L_2\cup\cdots L_n$ where $L_k$ is simple. 
Now consider
$$ \oint_L f = \sum_k\oint_{L_k} f$$
A: You should parameterize your curve using polar coordinates:
write the typical point $(x,y)=(r(t)\cos\theta(t),r(t)\sin\theta(t))$
where $r$ and $\theta$ are functions of $t$ on an interval $[a,b]$.
Also note that as the curve "closes up" then $r(b)=r(a)$
and $\theta(b)=\theta(a)+2m\pi$ for some integer $m$.
A: First you need to prove that $\oint_{L} \frac{xdy - ydx}{x^2 + y^2}$ is the same as calculating the integral over a circle of radius '$r$' such that the circle of radius '$r$' lies completely within the domain enclosed by the curve $L$.
Once you prove that the problem now becomes simple. All you need is to evaluate the integral over a circle. This can be done by parameterizing the circle as $x = r \cos(\theta)$ and $y = r \sin(\theta)$. Note that $dx = - r \sin(\theta)d\theta$ and $dy = r cos(\theta)d\theta$.
Also, $x^2 + y^2 = r^2$.
The integral then becomes
$$ \int_{0}^{2 \pi} \frac{r\cos(\theta) \times r \cos(\theta) d\theta - r\sin(\theta) \times -r \sin(\theta) d\theta}{r^2} = \int_{0}^{2 \pi} d\theta =2\pi$$
Hence, we get $$\oint_{L} \frac{xdy - ydx}{x^2 + y^2} = 2\pi$$
A: In the punctuated plane $\dot{\mathbb R}^2$ we have the arg-"function" taking values in $\mathbb R/(2\pi\mathbb Z)$. Its gradient is a well defined vector field, namely $\nabla$arg$=(-{y\over x^2+y^2}, {x\over x^2+y^2})$. It follows that  for any nice curve $L$, closed or not, simple or not (but not going through the origin), your integral sums up the ongoing argument increases/decreases along $L$. If $L$ is closed then these arg-increases/decreases sum up to $2\pi$ times the winding number of $L$ around the origin.
