I have been re-reading through my complex analysis text and wanted to try something different. Cauchy's Integral Theorem is typically proved using an application of Green's Theorem and then by virtue of the Cauchy-Riemann Equations the integral vanishes. I have been trying to do the same to Cauchy's Integral Formula. That is, starting with $\int_{\gamma}\frac{f(z)}{z-a}dz$, with $a$ inside the region defined by the curve $\gamma$ I want to get back $f(a) 2 \pi i$. However, whenever I try Green's theorem on the integral, I get that it vanishes. Here is my work so far (mostly based on the proof of Cauchy's Integral Theorem):
$\begin{align} & \int_\gamma \frac{f(z)}{z-a}dz \\[10pt] & = \int_\gamma \frac{u(x,y)+iv(x,y)}{z-a}(dx+i\,dy) \end{align} $
Now let $l(x,y)=\dfrac1{x+iy+a}$
So we have $ \begin{align} & = \int_\gamma \frac{u(x,y)+iv(x,y)}{z-a}(dx+i\,dy) \\ & = \int_\gamma ul\;dx -vl\; dy + i \int_{\gamma}vl\; dx+ ul\; dy \\ & = \int\int_D -\frac{\partial vl }{\partial x} -\frac{\partial ul }{\partial y} dx\,dy + i \int\int_D \frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \end{align} $
$\dfrac{\partial ul }{\partial x}=\dfrac{\partial l }{\partial x}u+\dfrac{\partial u }{\partial x}l$ and we know $\frac{\partial l }{\partial x}=\frac{\partial }{\partial x} \frac1{x+iy+a}=\frac{-1}{(x+iy+a)^2} $
and similarly
$\frac{\partial }{\partial x}\frac1{x+iy+a}=\frac{-i}{(x+iy+a)^2} $
Thus we have $ \begin{align} & {} \quad \int\int_{D}-\frac{\partial vl }{\partial x} -\frac{\partial ul }{\partial y} dx\,dy + i \int\int_{D}\frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \\[8pt] & =-\int\int_{D}\frac{\partial vl }{\partial x} + \frac{\partial ul }{\partial y} dx\,dy + i \int\int_{D}\frac{\partial ul }{\partial x}-\frac{\partial vl }{\partial y} dx\,dy \\[8pt] & =-\int\int_{D}\frac{\partial l }{\partial x}v+\frac{\partial v }{\partial x}l + \frac{\partial l }{\partial y}u+\frac{\partial u }{\partial y}l \, dx\,dy + i \int\int_{D}\frac{\partial l }{\partial x}u+\frac{\partial u }{\partial x}l-\frac{\partial l }{\partial y}v-\frac{\partial v }{\partial y}l \, dx\,dy \\[8pt] & =-\int\int_{D}\frac{-1}{(x+iy+a)^2}v+\frac{\partial v }{\partial x}l + \frac{-i}{(x+iy+a)^2}u+\frac{\partial u }{\partial y}l\; dx\,dy + i \int\int_{D}\frac{-1}{(x+iy+a)^2}u+\frac{\partial u }{\partial x}l-\frac{-i}{(x+iy+a)^2}v-\frac{\partial v }{\partial y}l\;dx\,dy \end{align} $
edit: Taking out a small area around the singularity gives the correct answer. Thank you to froggie for pointing this out!
This also gives:
If $g(z)$ is a holomorphic function that is bounded for $| z| < 1$ then for all $|a|< 1$ we have $\displaystyle\pi g(a)=\int\int_{|z|<1} \dfrac{g(z)}{(1-a\bar z)^2}\,dx\,dy$ where $z=x+iy$.