For $R>0$ let the domain $\Omega \subset \mathbb{R}^2$ be defined by $\Omega := \bar{B}_R(0)\setminus\{0\}$. Consider a unit vector field $w \in C^1(\Omega, \mathbb{S}^1)$ whose index is zero at each point in the interior of $\Omega$
$$ 0 = \oint\limits_C w_1dw_2 -w_2dw_1 = \oint\limits_C (w_1\nabla w_2 - w_2\nabla w_1)\cdot d\underline{S}$$ for all sufficently smooth curves $C \subset \Omega$. Given this show that the Dirichlet integral $$E=\int\limits_\Omega (|\nabla w_1|^2 + |\nabla w_2|^2 )d\underline{x}$$ is finite.
I had a few ideas on how to prove this. Firstly, have been able to use polar co-ordinates and the unit nature of the vector field to show that \begin{align} E &= \int\limits_0^R \int\limits_0^{2\pi}\left( \left|\frac{\partial w}{\partial r}\right|^2 + \frac{1}{r^2}\left|\frac{\partial w}{\partial \theta}\right|^2\right)d\theta rdr, \\ &= \int\limits_0^{2\pi}\int\limits_0^Rr\left(w_1\frac{\partial w_2}{\partial r} - w_2 \frac{\partial w_1}{\partial r} \right)^2drd\theta + \int\limits_0^R\frac{1}{r}\int\limits_0^{2\pi}\left(w_1\frac{\partial w_2}{\partial \theta} - w_2 \frac{\partial w_1}{\partial \theta} \right)^2d\theta dr \end{align} The zero index property implies that $$0 = \int\limits_0^{2\pi}\left(w_1\frac{\partial w_2}{\partial \theta} - w_2 \frac{\partial w_1}{\partial \theta} \right)d\theta. $$ I thought that I could eliminate the log term and then bound the other one. However I was unable to continue this further as Cauchy-Schwartz takes the absolute value. The second idea that I had was to prove that $w$ is $C^1$ at the origin and then bound the integral from above by bounding the maxmimum and minimum gradient value, but I am having problems proving the continuity at the origin as the zero index condition implies that $w_1\nabla w_2 - w_2\nabla w_1$ is a conservative vector field and hence $$\nabla \phi = w_1\nabla w_2 - w_2\nabla w_1, \quad \phi = \arctan\left(\frac{w_2}{w_1} \right) +C$$ for $C \in \mathbb{R}$ an arbitrary constant, and I am unsure how to proceed.