Prove that $\frac{1}{2\pi}\frac{xdy-ydx}{x^2+y^2}$ is closed

Question

I would like to prove that $\alpha = \frac{1}{2\pi} \frac{xdy-ydx}{x^2+y^2}$ is a closed differential form on $\mathbb{R}^2-\{0\}$ . However when I apply the external derivative to this expression (and ignore the $\frac{1}{2\pi}\cdot\frac{1}{x^2+y^2}$ factor ), I get:

\begin{equation} d \alpha = \frac{\partial x}{\partial x}dx\wedge dy + \frac{\partial x}{\partial y}dy\wedge dy - \frac{\partial y}{\partial x}dx\wedge dx - \frac{\partial y}{\partial y}dy\wedge dx \end{equation}

\begin{equation} d \alpha = dx\wedge dy - dy\wedge dx \end{equation}

\begin{equation} d \alpha = 2 dx\wedge dy \end{equation}

Which is not closed on $\mathbb{R}^2-\{0\}$. Where is my mistake ?

Answer 1

Well, let us write $\alpha=f\cdot \omega $ with $f(x,y)=\frac{1}{x^2+y^2}$ and $\omega=xdy-ydx$. Then \begin{align} d\alpha=df\wedge \omega+fd\omega&=-\frac{1}{(x^2+y^2)^2}\left(2xdx+2ydy\right)\wedge\omega+ \frac{1}{x^2+y^2}2dx\wedge dy=\\ &=-\frac{1}{(x^2+y^2)^2}\left(2x^2 dx\wedge dy-2y^2dy\wedge dx\right)+ \frac{1}{x^2+y^2}2dx\wedge dy=\\ &=-\frac{1}{(x^2+y^2)^2}\cdot 2\left(x^2+y^2\right) dx\wedge dy+ \frac{1}{x^2+y^2}2dx\wedge dy= \\ &=0. \end{align}

Answer 2

It may be helpful to point out that $\alpha$ is, locally and up to a constant, the $d\theta$ of the (polar) angle $\theta$. Then $d\alpha=dd\theta=0$ since one always has $d^2=0$.