<s>[INST] Prove that $\frac{1}{2\pi}\frac{xdy-ydx}{x^2+y^2}$ is closed 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 ?
 [/INST] 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}
</s>