$\int_{S^1} \beta = 0 \Rightarrow \beta$ is the differential of a function. - Is this proof legit?

Question

Let $\beta$ be a smooth $1$-form on $S^1$, and $\int_{S^1} \beta = 0.$ Prove $\beta$ is the differential of a function.

I don't really have a clue for this question..

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I am trying to follow Anthony Carapetis' hint:

According to Change of Variable Formula in $\mathbb{R^k}$, $$\int_{S^1} \beta = \int_0^{2\pi} \beta d\theta = 0.$$

I am trying to follow Anthony Carapetis' hint invoking Fundamental Theorem of Algebra, however, I got stuck with Change of Variable Formula because I am not sure if all forms are integrable?

Change of Variable in $\mathbb{R^k}$. Assume that $f: V \to U$ is a diffeomorphism of open sets in $\mathbb{R^k}$ and $a$ is an integrable function on $U$. Then $$\int_U a dx_1 \cdots dx_k = \int_V (a \circ f) |\det(df)|dy_1 \cdots dy_k.$$

Answer 1

Integration of smooth $1$-forms (automatically closed for dimension reasons) on the circle induces an $\mathbb R$-linear surjective map $$I:\Omega^1(S^1)=Z^1(S^1)\to \mathbb R:\beta\mapsto \int_{S^1}\beta$$.
Passing to the quotient by the vector space $B^1(S^1)$ of exact $1$-forms yields an isomorphism for the first De Rham cohomology vector space: $$i:Z^1(S^1)/B^1(S^1)=H^1_{DR}(S^1)\stackrel {\cong}{\to} \mathbb R:[\beta] \mapsto \int_{S^1}\beta$$.
Since $i$ is in particular injective, we see that $\int_{S^1}\beta=0$ implies $[\beta] =0$ i.e. $\beta$ is exact.

All this is explained in exquisite detail by Loring Tu in his extraordinary book An Introduction to Manifolds, one of the most user-friendly and beautiful didactical textbooks I have ever seen.

Answer 2

Hint: functions on $S^1$ are just functions on $[0,1]$ that "match up" at the endpoints. The equivalent statement for $[0,1]$ (without the integral condition) is just the Fundamental Theorem of Calculus...