Classic example of a non exact form Let $\dfrac{xdy-ydx}{x^2+y^2}$ be a 1-form defined in $\mathbb{R}^2\backslash\{0\}$. 
Where can I find a detailed proof that it is not exact? I would prefer a proof that doesn't use results about conservative vector fields (I know that way). I have searched Jefree Lee, Do Carmo & some others, but they all skip this part.  Thanks a lot.
 A: I'm not sure whether this is acceptable or not, but the simplest thing to do (since the form is closed) is to attempt to integrate it to find supposed $\phi$ such that your form $\omega$ is $\mathrm d \phi$.
One can derive a contradiction most straightforwardly by going around a circle, say $x^2+y^2=1$. We use the path $x(t)=\cos t,y(t)=\sin t$.
$$\int_0^{2\pi} \omega = \int_0^{2\pi} \frac{\cos^2 t +\sin^2 t}{1}\mathrm d t = 2\pi$$
But if $\omega=\mathrm d \phi$ then the expression on the left is $\phi((1,0))-\phi((1,0))=0$, a contradiction.

In response to your complaints about this method: The definition of exact is that $\omega=\mathrm d \phi$ for some $\phi$. The only result we've used is that integrals of derivatives are given by boundary values (a very mini version of Stokes's theorem), i.e. 
$$\int_{\mathbf x}^{\mathbf y} \mathrm d \phi = \left[\int_{t_x}^{t_y} \frac{\mathrm d \phi}{\mathrm d t} \mathrm d t = \right]\phi({\mathbf y})-\phi({\mathbf x})$$
which I don't think is really a mysterious result. If there's something you're not happy with do just say.
A: Yes, everyone's doing the standard FTC proof we all know. Instead, let's try to integrate to find a potential function $f$. In the right half-plane you get $f(x,y) = \arctan(y/x)$. Now, to get a continuous continuation of this into the upper left half-plane, we need to use
$g(x,y) = \arctan(x/y)+\pi/2$. To continue this into the lower left half-plane, we need
$h(x,y) = \arctan(y/x)+\pi$, and, finally, into the lower right half-plane, we need $k(x,y)=\arctan(x,y)+3\pi/2$. So, piecing, these together, we have a continuous potential function defined on $\mathbb R^2 - \{(x,0): x>0\}$. But $f(1,0) = 0$ and $\lim_{(x,y)\to (1,0)} k(x,y) = 2\pi$, so there's no continuous potential function on $\mathbb R^2-\{(0,0)\}$. 
Of course, this $2\pi$ is the same thing we got by integrating the closed $1$-form around the circle. It represents the fact that the polar angle has made one full revolution.
Now, how do I know my construction is essentially unique and that there is no cleverer way around it?  Potential functions on a connected set are unique up to constants. So, up to shifting by a constant, my attempt is the only possible function.
A: If you integrate it along a circle containing the origin in its interior, you get a nonzero value (try it! put $x=\cos \theta, y = \sin \theta$). If the form were exact, the integral would vanish.
I don't know if that's what you mean by a "conservative result", but this is really the reason why this form isn't exact.
A: You can prove this using the Green's function for the vector derivative.
There exists a Green's function $G$ for the vector derivative $\partial$:
$$\partial \cdot G = \delta \implies G(r) = \frac{r}{2\pi |r|^2}$$
The Hodge dual of this field is what you're looking for.
$$\partial \cdot G = \delta \implies \partial \wedge (Gi) = i \delta$$
All we have to do then is compute the Hodge dual $Gi$:
$$\omega = Gi = \frac{ri}{2\pi |r|^2} = \frac{x \sigma^y - y \sigma^x}{2 \pi |r|^2}$$
where $\sigma^x, \sigma^y$ are basis covectors.  Thus, $\partial \wedge \omega = i\delta$, the (dual of the) Dirac delta function, and $\omega$ is therefore not exact.
