Prove holomorphism of expression

I have to prove that given $\gamma$ contour of $G \subset \mathbb{C}$ and $f$ continuous function on $G$. Then the function $g(z) = \int_{\gamma} \frac{f(w)}{w-z}dw$ is holomorphic on $\mathbb{C} - \bar{G}$

My attempt:

I started by using that given any parametrization $\alpha :[a,b] \rightarrow \gamma$ we can rewrite $g(z)$ as:

$g(z) = \int_a^b \frac{f(\alpha(t))}{\alpha(t)-z}\alpha'(t)dt$

Now using the fundamental theorem I derivate to see where $g'(z)$ is defined and I get:

$g'(z) = \frac{f(\alpha(b))}{\alpha(b)-z}\alpha'(b) - \frac{f(\alpha(a))}{\alpha(a)-z}\alpha'(a) $

Here I get stuck, I fail so see how this links back to $g$ being holomorphic at $\mathbb{C} - \bar{G}$.

Any hints on how to proceed? Thanks!

Solution 1:

The difference quotient can be computed via $$ g(z_2) - g(z_1) = \int_\gamma \left(\frac{1}{w-z_2} - \frac{1}{w-z_1} \right) f(w)\, dw \\ = (z_2 - z_1) \int_\gamma \frac{f(w)}{(w-z_2)(w-z_1)} \, dw \, . $$ Now show that the integral has a limit for $z_2 \to z_1$. Conclude that $g$ is differentiable in $\Bbb C \setminus \overline G$, and $$ g'(z) = \int_\gamma \frac{f(w)}{(w-z)^2} \, dw \, . $$