Consider the one-form $\alpha = \frac{xdy-ydx}{x^2 + y^2}$ on $\mathbb{R}^2 \setminus \{(0,0)\}$. Let $i: S^1 \to \mathbb{R}^2 \setminus \{(0,0)\}$ be the inclusion.
Compute $\int_{S^1} i^*\alpha$. Does there exist $f \in C^\infty(\mathbb{R}^2\setminus \{(0,0)\})$ such that $df = \alpha$. Show that $d \alpha = 0$, and explain why combining this with Stokes' theorem does not give a contradiction.
Attempt:
Denote the unit sphere with boundary by $D$. Then
$$\int_{S^1} i^*\alpha = \int_{S^1}i^{*}(xdy-ydx)= \int_D dx \land dy - dy \land dx = 2 \int_D dx \land dy = 2 \pi$$ where Stokes' theorem was used. There does not exist an $f$ with $df = \alpha$. I'm not entirely sure how I can formally show this. This would imply that $\partial f/\partial x = 1/2\ln(x^2 + y^2)$ and integrating we get something like $f(x,y) = 1/2\ln(x^2 + y^2)c(y)+ d(y)$ and I gueuss the $\ln$ is problematic.
Further, I was able to show that $d \alpha = 0$ with a direct computation but I can't see why this doesn't Stokes' theorem.
EDIT: Is the problem the following? If one is naive, one would say that
$$\int_{S^1} i^* \alpha = \int_Dd \alpha$$
but this doesn't make sense because $\alpha$ is not defined at the origin!
Thanks in advance for any help!
Based on the first part of the problem, I'm guessing they want you to say: If there was such an $f$, then you would have $$\int_{S^1} i^*\alpha = \int_{S^1} i^* df = \int_{S^1} di^*f = \int_{\partial S^1 =\varnothing} i^*f = 0$$ by Stokes' theorem, but you just computed that this integral is not 0.