For $k ∈ \mathbb{N}$ let $\gamma_k$ be the curve traced by a point travelling, at constant speed 1, along a circle of radius $k$ centered in $O ∈ \mathbb R^2$. The point starts at $(k, 0)$ and completes a single loop. For odd $k$ the orientation is counter-clockwise, for even $k$ it’s clockwise. Write down a parametrisation of $γ_k$ and discuss to what extent is it unique. With $$g(x,y)=(\frac {-y} {x^2 +y^2},\frac {x} {x^2 +y^2})$$
find $$\sum_{k=1}^n \int_{γ_k} g$$ for every $n \in \mathbb N$ as well as $$\sum_{k=1}^\infty \int_{γ_k}^* g$$ where $$\int_{γ_k}^* g= \frac 1 {\text{length}(γ_k)} \int_{γ_k} g$$ note $\int^*$ is a special sign.
I am trying to find the parametrisation but am not sure where to start and why would it be unique? For the sum of integrals, do I integrate scalar (from $0$ to $k$) for the scalar of $g$ and add them up, is there a shortcut? Thanks.
$\frac{\partial}{\partial x}\frac{x}{x^2+y^2}=\frac{y^2-x^2}{(x^2+y^2)^2}=\frac{\partial}{\partial y}\frac{-y}{x^2+y^2}=\frac{y^2-x^2}{(x^2+y^2)^2}$. Yes, and as Maxim pointed out, $\vec{g}$ is not a conservative vector field but this is only because the partial derivatives have a unique point of discontinuity at $(0,0)$. Therefore, the line integral must vanish whenever we take it over a simple closed curve that does not enclose $(0,0)$ ; the line integral is invariant when taken over those simple closed curves both enclosed by $\gamma_k$ and intersecting $\gamma_k$ at least twice. For odd $k$ we have $$\int_{\gamma_k} \vec{g}(x,y)\;\cdot d\vec{r}=\int_0^{2\pi}\frac{1}{k^2}\langle-\sin(t),\cos(t)\rangle\cdot k\langle-\sin(t),\cos(t)\rangle\;dt=\frac{2\pi}{k}$$ with $\;\;\sum_{k=1}^n\int_{\gamma_k}^* \vec{g}(x,y)\;dr=\sum_{k=1}^n \frac{1}{k^2}(-1)^{k+1}=\frac{\pi^2}{12}$ as changing the curve orientation only changes the sign of the line integral yielding an alternating Basel sum.