How to understand the singularity of Green's Theorem? I'm learning complex analysis right now where I encountered the line integral of $\frac{1}{z}$ over a circle containing the origin, e.g. $\int_{\Gamma}{\frac{1}{z}}dz$, where $\Gamma: z(t) = e^{it}, t \in [0,2\pi]$. This is just the integral $\int_{\Gamma}\frac{-ydx+xdy}{x^2+y^2}$that one should encounter when studying Green's Theorem. 
Specifically,$$\int_{\Gamma}{\frac{1}{z}}dz = \int_{\Gamma}{\frac{x-iy}{x^2+y^2}(dx+idy) = (\int_{\Gamma}{\frac{x}{x^2+y^2}}dx+\frac{y}{x^2+y^2}dy}) + i(\int_{\Gamma}{\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy}) = i(\int_{\Gamma}{\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy}) = i2\pi.$$
How do I interpret this singularity at the origin? Physically, we know the Green's theorem is a description of circulation/curl. This singularity seems to say that there is a "source of curl" at the origin in some sense such that even though we have the curl vanishing everywhere except for the origion, we still get nonzero circulation. It smells somewhat like when we integrate the delta function $\delta(x)$ in distribution sense. Is there any connection to this?
This is just my blurry understanding of this scenario. I hope someone could help me completely understand this recurrent singularity in mathematics.
 A: Sorry for the length, hopefully it is somewhat insightful.
This isn't answering the question exactly as you've posed it, but this is some motivation for studying de Rham Cohomology and differential forms. One interpretation of the importance of this integral is that it detects a topological property of $\mathbb{R}^2\setminus \{(0,0)\}$, namely the puncture at the origin. The absence of a point allows us to define the differential form 
$$ \omega=\frac{-y}{r^2}dx+\frac{x}{r^2}dy$$
where $r^2=x^2+y^2$. You can just think of this (for now) as a thing that you can integrate along a path $\gamma$ in the obvious way. Let's say $\gamma$ is the unit circle path around the origin oriented counterclockwise. If there were a function $f$ defined on $\mathbb{R}^2\setminus \{(0,0)\}$ so that $df=\omega$, i.e. $\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=\omega$, then by Stokes' Theorem we would get that 
$$ \int_\gamma\omega=\int_\gamma df=\int_{\partial \gamma}f=0.$$
Indeed, we can see that this is not the case since we can compute the integral and see that it is nonzero. So, we see that there is no function $f$ with $df=\omega$.
Now, let's say that a differential $1-$form is an expression $fdx+gdy$ for $f,g$ smooth functions on $M=\mathbb{R}^2\setminus \{(0,0)\}$. Let's say that a differential $2-$form is an expression of the form $hdx\wedge dy$ where $h$ is a smooth function, and $\wedge$ is what is known as the wedge product of forms. $\wedge$ has the following properties $dx\wedge dx=dy\wedge dy=0$ and $dx\wedge dy=-dy\wedge dx$. It is also bilinear with respect to smooth functions in the sense that 
$$ fdx\wedge gdy=fgdx\wedge dy.$$
Denote the $\mathbb{R}-$vector space of $1-$forms $\Omega^1(M)$ and the $\mathbb{R}-$vector space of $2-$forms by $\Omega^2(M)$. We can define a map $d:\Omega^1(M)\to\Omega^2(M)$ called the exterior derivative using the definition of $df$ above for a smooth function (note the double use of $d$). For a $1-$form $\omega=fdx+gdy$, define 
$$ d\omega=df\wedge dx+dg\wedge dy.$$
$d\omega$ is a $2-$form. So, we get a sequence of maps 
$$ \mathscr{C}^\infty(M)\xrightarrow{d} \Omega^1(M)\xrightarrow{d}\Omega^2(M).$$
You can check by hand that $d\circ d=0$, in the sense that for any $f\in \mathscr{C}^\infty(M)$, $d(df)=0$. It follows that 
$$ \operatorname{im}(d:\mathscr{C}^\infty(M)\to \Omega^1(M))\subseteq \ker(d:\Omega^1(M)\to \Omega^2(M)).$$
In particular, $\operatorname{im}(d)\subseteq \ker(d)$ as a subspace (abbreviating the above notation). So, it makes sense to define 
$$ H^1_{\text{dR}}(M)=\frac{\ker(d:\Omega^1(M)\to \Omega^2(M))}{\operatorname{im}(d:\mathscr{C}^\infty(M)\to \Omega^1(M))}$$ 
to be the first de Rham Cohomology group of $M$. It is nothing more than a quotient of vector spaces (hence a vector space). Recalling what we did before, the form $\omega=-\frac{y}{r^2}dx+\frac{x}{r^2}dy$ does not lie in $\operatorname{im} d$, but it does lie in $\ker d$ as you can check by hand. It follows that 
$$ \dim_{\mathbb{R}} H^1_{\text{dR}}(M)\ge 1.$$
So, we have shown that the first de Rham cohomology of the punctured plane is nonzero. Actually, $H^1_{\text{dR}}(M)\cong \mathbb{R}$, but no matter. By the Poincaré Lemma from multivariable calculus of forms (which you will probably see soon) we get that $H^1_{\text{dR}}(\mathbb{R}^2)=0$. A nice property of de Rham Cohomology is that it is preserved under homeomorphism (in fact homotopy equivalence), so that if $M$ and $N$ are appropriate topological spaces, then $M\cong N$ implies $H^1_{\text{dR}}(M)\cong H^1_{\text{dR}}(N)$. As a corollary of this, we can see that $\mathbb{R}^2\setminus \{(0,0)\}$ is not homeomorphic to $\mathbb{R}^2$, since they have distinct first de Rham cohomologies. 
The elements of $H^1_{\text{dR}}(M)$ are called cohomology classes. One reason the form $\omega$ in question is significant is that it represents a nontrivial cohomology class generating the first de Rham cohomology of the punctured plane. In particular, if someone asked you to explain why the punctured plane and the plane are distinct intuitively, you would say that the former has a puncture, while the latter does not. At a basic level, the observation that this form integrates to a nonzero value around a closed loop indicates that there is a nontrivial first cohomology class. So, this allows us to formalize the notion of having a puncture. There are other ways to formalize this, and this falls naturally under the heading of algebraic topology.

Now, from a complex analytic perspective, a theorem that will be presented soon in your complex analysis studies (if not already) is the following: 

If a holomorphic function $f$ defined on a domain $U\subseteq \mathbb{C}$ has a primitive $F$, then $\oint_\gamma fdz=0$ for all closed contours $\gamma$ contained in $U$. 

A primitive is a function $F$ such that $F'=f$.
Now, let's examine $f(z)=\frac{1}{z}$, which is a holomorphic function defined on $\mathbb{C}\setminus \{0\}$ (which is topologically the same as a punctured plane). We actually know, essentially, that its primitive is a "function" called $\log z$. There is an issue here, though. To define $\log z$ in any sensible way, we need to make a branch cut. This means we need to remove a ray emanating from the origin in $\mathbb{C}$ to define $\log z$ in a reasonable way. What this really means is: there is no global primitive for $f(z)=\frac{1}{z}$ on $\mathbb{C}\setminus \{0\}$. This is related to the topological problem we discussed above, since using $z=x+iy$ and $dz=dx+idy$, you can compute that $\frac{1}{z}dz$ is a slightly modified version of our form $\omega$ from before.
Actually, the fact that the de Rham cohomology is $1-$dimensional means that in some sense this is the "only" problem function up to equivalence. This is somewhat illustrated by the fact that $\frac{1}{z^n}$ poses no problem for $n\ge 2$. Indeed, $\frac{1}{z^n}$ has primitive
$$ -\frac{1}{n-1}\cdot \frac{1}{z^{n-1}}$$
valid for $n\ge 2$.
