It is intuitively clear that the Poincaré dual of a ray $\{(r,0);r>0\}$ in $\mathbb{R}^2-\{0\}$ is the form $\frac{1}{2\pi} d\theta$.
For some reason I do not understand, I failed to prove this rigorously. Can somebody help me?
(this example comes from the book of Bott & Tu, Differential Forms in Algebraic Topology)
More precisely, I failed to prove the equality of the integral of a closed form $\omega$ on the ray with the integral of the same form wedge $d\theta$ on the whole punctured plane.
Write $\omega = f(r,\theta) \, dr + g(r,\theta) \, d\theta$ in polar coordinates. The crucial observation is the following:
The integral of $\omega$ along a ray emanating from $0$ does not depend on the particular choice of the ray!
For the ray $R_\theta = \{(r\cos\theta,r\sin \theta)\mid r>0\}$ we have $\int_{R_\theta} \omega = \int_0^\infty f(r,\theta) \, dr$. By closedness of $\omega$, we have $\frac{\partial f}{\partial \theta} = \frac{\partial g}{\partial r}$. Now this implies
$$\frac{d}{d\theta} \int_{R_\theta} \omega = \frac{d}{d\theta}\int_0^\infty f(r, \theta) \, dr = \int_0^\infty \frac{\partial f(r, \theta)}{\partial \theta} \, dr = \int_0^\infty \frac{\partial g(r, \theta)}{\partial r} \, dr = 0$$
where the compact support of $\omega$ was used in the last equality. So indeed, $\int_{R_\theta} \omega$ is independent of $\theta$. Therefore we conclude
$$\int_{\mathbb R^2\setminus \{0\}} \omega \wedge \frac{d\theta}{2\pi} = \frac{1}{2\pi} \int_0^{2\pi} \int_0^\infty f(r,\theta) \, dr \, d\theta = \int_0^\infty f(r,0)\, dr = \int_{\{(r,0)\mid r>0\}} \omega$$ i.e. $\frac{d\theta}{2\pi}$ is the Poincaré dual of $\{(r,0)\mid r>0\}$ (and of any other ray emanating from $0$).
