Analytic function outside a bounded domain Suppose $D\subseteq\mathbb C$ is a bounded domain, with $\partial D$ smooth. Then suppose $g:\mathbb C\backslash\overline D\to\mathbb C$ is analytic and extends continuously to $\partial D$. If $\lim_{z\to\infty}g(z)$ is finite, show that
$$
g(u)=\lim_{z\to\infty}(g(z))-\frac1{2\pi i}\int_{\partial D}\frac{g(z)dz}{z-u}
$$
for all $u\in\mathbb C\backslash\overline D$, assuming that $\partial D$ is oriented as the boundary of $D$.
This problem is confusing to me, because the integral term looks like the result of the Cauchy Integral Formula. But we are considering points outside $\overline D$. Would anyone be able to give some intuition for how to solve this problem?
 A: Recall the Cauchy integral formula:

Let $\Omega$ be a bounded domain with smooth boundary. Then for any $h \in \mathcal{O}(\Omega)$ which extends continuously to $\partial{\Omega}$, we have
$$ \forall u \in \Omega,\ h(u) = \frac{1}{2\pi i} \int_{\partial{\Omega}} \frac{h(z)}{z-u} dz. $$

We can assume that $D$ is contained in the unit disk $\Delta$.
Set $\Omega$ the image of $\mathbb{C}\backslash \overline{D}$ by $z \mapsto \frac{1}{z}$ and for all $z \in \Omega$, set $h(z) = g(\frac{1}{z})$. Notice that $\Omega$ is a bounded domain since $D \subset \Delta$. Alas, $h$ is not holomorphic on $\Omega$ but it is holomorphic on $\Omega\backslash\{0\}$. Since the boundary of this latter domain is not smooth, we replace it by $\Omega\backslash \Delta_{\varepsilon}$ where $\Delta_{\varepsilon}$ stands for the disc centered at the origin with radius $\varepsilon$, which is chosen small enough so that $\Delta_{\varepsilon} \subset \Omega$. Now you can apply the Cauchy integral formula above and let $\varepsilon$ go to zero to conclude.
