Let $\alpha$ a simple closed curve of class $C^1$, and traverse (counterclockwise). We define. $g: \mathbb{C}-\{\alpha\} \to \mathbb{C}$ as:
\begin{align} g(z) = \int_{\alpha} \frac{w^3 + 2w}{(w-z)^3}dw \end{align}
I want to find the function rule of $g$.
Here is my work:
Consider $f(w) = w^3+2w$. Then $f'(w) = 3w^{2} + 2$; $f''(w)=6w$.
Using the Cauchy Integral it follows: \begin{align} f''(z) &= \frac{2!}{2 \pi i} \int_{\alpha} \frac{w^3+2w}{(w-z)^{2+1}}\\ \pi i f''(z) &= \int_{\alpha} \frac{w^3+2w}{(w-z)^{3}}\\ 6 \pi i z &= \int_{\alpha} \frac{w^3+2w}{(w-z)^{3}}\\ \end{align}
$\therefore g(z) = 6 \pi i z$
I know this only works inside the $\alpha$ region, but $g$ is also defined outside. What should I do to find the function rule outside $\alpha$ region? Any suggestion?