A closed one-form on the punctured plane with nonzero period

Let $D={\mathbb{R}}^2\setminus \{(0,0)\}$ and $\omega =Pdx+Qdy$ with $P(x,y)=-y\mathrm{/}(x^2+y^2)$ and $Q(x,y)=x\mathrm{/}(x^2+y^2)$. The domain $D$ is open since each nonzero point has a ball avoiding the origin. The coefficient vector $(P,Q)=(-y,x)\mathrm{/}(x^2+y^2)$ is nonzero at every point of $D$. The domain is path-connected: if the segment joining $p,q\in D$ avoids $0$, use that segment; if it contains $0$, then $p,q,0$ are collinear. Writing $p=(p_1,p_2)$, take $r=(-p_2,p_1)$; it is nonzero and perpendicular to $p$, so $p,r$ and $q,r$ are linearly independent pairs and the segments $[p,r]$ and $[r,q]$ avoid $0$.

Write $r^2=x^2+y^2>0$. For sufficiently small $h$, direct difference quotients give ${\mathrm{\partial }}_{x}P=2xy\mathrm{/}{r}^4$, ${\mathrm{\partial }}_{y}Q=-2xy\mathrm{/}{r}^4$, and ${\mathrm{\partial }}_{x}Q={\mathrm{\partial }}_{y}P=(y^2-x^2)\mathrm{/}{r}^4$: for example, $(Q(x+h,y)-Q(x,y))\mathrm{/}h=(y^2-x^2-xh)\mathrm{/}(r^2((x+h{)}^2+y^2))$ and $(P(x,y+h)-P(x,y))\mathrm{/}h=(y^2-x^2+yh)\mathrm{/}(r^2(x^2+(y+h{)}^2))$, while the other two quotients are $y(2x+h)\mathrm{/}(r^2((x+h{)}^2+y^2))$ and $-x(2y+h)\mathrm{/}(r^2(x^2+(y+h{)}^2))$. These limits are continuous on $D$, so $P,Q\in C^1(D)$; the cross partials agree and the coordinate definition [F1] gives $d\omega =({\mathrm{\partial }}_{x}Q-{\mathrm{\partial }}_{y}P)dx\wedge dy=0$.

The unit circle $\gamma (t)=(\cos t,\sin t)$, $0\le t\le 2\pi$, lies in the domain and uses the form defined in step 1.1. Along it $(P,Q)=(-\sin t,\cos t)={\gamma }^{\mathrm{\prime }}(t)$, hence [F2] gives ${\int }_{\gamma }\omega ={\int }_0^{2\pi }1dt=2\pi$ by [F4].

The closedness is proved in step 2.1. If $\omega =dS$ globally on $D$, then $F=\mathrm{\nabla }S$ for the corresponding vector field, and [F3] would give ${\int }_{\gamma }\omega =S(\gamma (2\pi ))-S(\gamma (0))=0$, contradicting step 2.2. Thus $\omega$ is closed but not globally exact. This does not assert that every possible global integrating factor is absent.