Proving that a function is analytic I'm struggling with the following problem:

Problem: Suppose that $h$ is a continuous function on a simple closed curve $\gamma$. Define
$$
H(w) = \oint_{\gamma} \frac{h(z)}{z - w} \, dz.
$$
Show that $H$ is analytic on $\mathbb{C} \setminus \gamma$.

I feel like the solution should fall very easy from the Cauchy integral formula, but I'm having trouble seeing how to connect the two. Could anyone lend a helping hand?
 A: Let $w_1, w_2 \in \Bbb C - \gamma$. We have
\begin{align}
\left|H(w_2) - H(w_1)\right| &= \left|\int_\gamma \frac{h(z)(z - w_1) - h(z)(z - w_2)}{(z - w_2)(z - w_1)}dz\right| \\
&= \left|\int_\gamma \frac{h(z)(w_2 - w_1)}{(z - w_2)(z - w_1)}dz\right| \\
&= \left|\int_\gamma \frac{h(z)}{(z - w_2)(z - w_1)}dz\right| \left|w_2 - w_1\right|. \tag{*}
\end{align}
By the ML lemma, it follows that $H$ is continuous.
Now we can use Morera's theorem. Let $\Delta \subset \Bbb C - \gamma$ be a closed triangle. We have
\begin{align}
\int_{\partial \Delta} H(w)dw &= \int_{\partial \Delta} \int_\gamma \frac{h(z)}{z - w} dz dw \\
&= \int_\gamma \int_{\partial \Delta} \frac{h(z)}{z - w} dw dz \\
&= 0.
\end{align}
We are able to change the order of integration since we are integrating a continuous function over compact sets. The last equality follows from Cauchy's integration theorem since $h(z)/(z - w)$ is holomorphic over $\Bbb C - \{z\}$ as a function of $w$.

Alternatively, as Christian Blatter suggests, we can use (*) to get
\begin{align}
\lim_{w_2 \to w_1} \frac{H(w_2) - H(w_1)}{w_2 - w_1} &= \lim_{w_2 \to w_1} \int_\gamma \frac{h(z)}{(z - w_2)(z - w_1)} dz \\
&= \int_\gamma \frac{h(z)}{(z - w_1)^2} dz.
\end{align}
Swapping the limit and integral is justified for the same reason as above. This gives an expression for $H'(w_1)$.
