The closed angular form on the punctured plane is not exact

Given: The witness $\alpha =(-ydx+xdy)\mathrm{/}(x^2+y^2)$ on ${\mathbb{R}}^2\setminus \{0\}$.

Closed and exact differential forms: For the complex def-de-rham-cochain-complex, put $Z^k(M)=\ker (d:{\mathrm{\Omega }}^k(M)\to {\mathrm{\Omega }}^{k+1}(M))$ and $B^k(M)=\mathrm{im}(d:{\mathrm{\Omega }}^{k-1}(M)\to {\mathrm{\Omega }}^k(M))$. A form is closed if it belongs to $Z^k$ and exact if it belongs to $B^k$. If $\omega =d\eta$, then $d\omega =d^2\eta =0$ by thm-the-exterior-derivative-squares-to-zero, so $B^k\subseteq Z^k$. In particular $B^0=0$, since ${\mathrm{\Omega }}^{-1}=0$. The zero form is both closed and exact in every degree.

The local coordinate formula for the exterior derivative: Let $(U,x^1,\dots ,x^n)$ be a smooth chart on a smooth manifold and $\omega$ a smooth $k$-form on $U$, with $k\ge 0$. Summing over increasing $k$-tuples $I$, and writing $d{x}^I=d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$, if $\omega ={\sum }_I{\omega }_{I}d{x}^I$, then $d\omega ={\sum }_{I}d{\omega }_I\wedge d{x}^I\mathrm{.}$

Newton–Leibniz needs only continuity on $[a,b]$, differentiability on $(a,b)$, and a Riemann-integrable extension of the interior derivative: Let $a<b$. Suppose $G:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f:[a,b]\to \mathbb{R}$ is Riemann integrable and $f(x)=G^{\mathrm{\prime }}(x)(a<x<b),$ then ${\int }_a^{b}f=G(b)-G(a)\mathrm{.}$ No derivative of $G$ at either endpoint is assumed, and the two endpoint values assigned to the integrable extension $f$ do not enter the conclusion.