I want to show that $I=\displaystyle \oint_C \overline{v} \cdot d \overline{r}$ is multivalued where $\overline{v} = (-y/(y^2+x^2), ~x/(y^2+x^2))$ and $C$ is the circle $x^2+y^2=1$. I know we can't use Green's theorem because $\overline{v}$ is not defined at the origin, which is the centre of the circle. But mindlessly applying Green's theorem gives $I=0$. Meanwhile, the parametrisation $x = \cos{\theta}, ~ y= \sin{\theta}$ gives $I= 2\pi$. But I can't provide an actual proof for this.
How do you prove that it's multivalued/it can take either of these values? I'm not sure I can say that when we take away the origin, Green's theorem is valid so $I=0$ and when we include the origin, the parametrisation gives the correc answer $I=2\pi$ since it avoids this problem. Thanks for your help!
Green's Theorem doesn't apply, because its hypotheses include continuity of the partial derivatives of the integrand inside the entire region bounded by $C$. As for computing the integral directly, that looks relatively straight-forward. With the parametrization you've given, we have $$\mathbf{v}(\mathbf{r})=(-\sin(\theta),\cos(\theta)),\quad \mathbf{r}=(\cos(\theta),\sin(\theta)), \quad d\mathbf{r}=\mathbf{r}'\,d\theta=(-\sin(\theta),\cos(\theta))\,d\theta,$$ so that your integral becomes $$\oint\mathbf{v}\cdot d\mathbf{r}=\int_0^{2\pi}\mathbf{v}(\mathbf{r})\cdot\mathbf{r}'\,d\theta=\int_0^{2\pi}(-\sin(\theta),\cos(\theta))\cdot (-\sin(\theta),\cos(\theta))\,d\theta=\int_0^{2\pi}d\theta=2\pi,$$ as required.