Let $x,y$ be the standard coordinates on $\mathbb{R}^2 \setminus \{0\}$ and $r, \theta$ the polar coordinates. Show that the Poincaré dual of the ray $\{(x,0) \mid x > 0 \}$ in $\mathbb{R}^2 \setminus \{0\}$ is $d\theta/2\pi$ in $H^1(\mathbb{R}^2 \setminus \{0\})$.
I found this answer on the site and there is part which I don't understand. They show the following
Now this implies $$\frac{d}{d\theta} \int_{R_\theta} \omega = \frac{d}{d\theta}\int_0^\infty f(r, \theta) \, dr = \int_0^\infty \frac{\partial f(r, \theta)}{\partial \theta} \, dr = \int_0^\infty \frac{\partial g(r, \theta)}{\partial r} \, dr = 0$$
and conclude that
So indeed, $\int_{R_\theta} \omega$ is independent of $\theta$.
But I have no idea what is happening with the chain of equalities in $$\frac{d}{d\theta} \int_{R_\theta} \omega = \frac{d}{d\theta}\int_0^\infty f(r, \theta) \, dr = \int_0^\infty \frac{\partial f(r, \theta)}{\partial \theta} \, dr = \int_0^\infty \frac{\partial g(r, \theta)}{\partial r} \, dr = 0.$$ Why are they differentiating the integral over the form with respect to $\theta$ and why does this imply that it is independent of $\theta$?