Surface integral of complex function using residues I am stuck with a math problem that I thought should be straightforward. Maybe I'm missing something here and you can help me. 
The key idea is that I have to integrate this function over a volume:
$ I = \int_V \frac{\partial u}{\partial x} dV $
where the volume V is in the x-y plane. I thought of using complex analysis since I do not have a closed form for u, but instead I have an expression for F(z) with
$ F(z) = u + iv \quad \text{with} \quad z=x+iy$
I therefore thought of using complex analysis. 
I'll explain my first approach, which I believe was incorrect:
First, I thought of the integral "I" as the real part of the whole thing:
$ I = Re\left(\int_V \frac{d F}{d z} dV \right)$
and then use the divergence theorem to cast this not as a volume integral, but as a contour integral.
As you can see already, the approach is wrong in many levels: mostly because dF/dz does not correspond to the divergence (does it?).
I am sure there should be an easy way of converting this integral into a contour integral, but I have been unsuccessful finding it. 
Anyone has dealt with this type of problem before?
Many thanks in advance!
Ignacio
Edit:
Ok, after some research I found out that, by defining J as
$J = \displaystyle \frac{i}{2} \int_V  \frac{dF}{dz} dz \wedge dz^*  $
it is shown that I = Re(J), since $dz\wedge dz^* = -2idxdy$
I believe that J can be calculated using Stokes theorem... work in progress...
 A: If you are still interested now--
This is the correct formula you can use:
$$\int_S \frac{\partial F(z,\bar{z})}{\partial z}dz d\bar{z}=i\oint_{\partial S}F(z,\bar{z})d\bar{z},$$
where $\bar{z}=z^*=x-iy$, $dzd\bar{z}=2dxdy$, $S$ is the region we are integrating over, and $\partial S$ is the boundary of $S$ (assuming the contour to be anti-clockwise). Here I have reserved the possibility that $F$ might not be meromorphic, i.e. it may depend explicitly on $\bar{z}$.
$\mathbf{Proof}$: 
\begin{eqnarray}
\int_S \frac{\partial F(z,\bar{z})}{\partial z}dz d\bar{z}=\int_{S}2\partial_z F dxdy=\int_S (\partial_x-i\partial_y)F dx dy=\oint_{\partial S}(F dy+i Fdx)=i\oint_{\partial S} F d\bar{z},
\end{eqnarray}
where in the third step I have used Green's formula $\int_S (\partial_x M+\partial_y L)dx dy=\oint_{\partial S}(Mdy-Ldx)$.
Now returning to your question, we have to calculate $\oint_{\partial S} F(z)d\bar{z}$ or its complex conjugate $\oint_{\partial S} F^*(\bar{z})dz$. Because $F^*(\bar{z})$ is not analytic, the integral would depend on the shape of the contour--not just the topology. 
Let's see an example here: Take $S$ to be a circular region of radius $R$ centered on $z=0$ and suppose $F(z)$ can be Taylor expanded in $S$. We have
$$\oint_{\partial S} F^*(\bar{z}) dz=\oint_{\partial S}\sum_{n\geq 0} \frac{\bar{z}^n}{n!}F^{(n)}(0)^* dz=\oint_{\partial S}\sum_{n\geq 0} \frac{R^{2n}}{z^n n!}F^{(n)}(0)^* dz=2\pi i R^2 F'(0)^*.$$
Here is a specific application
$$\int_S 2\cos z dS=\int_S \partial_z \sin z dz d\bar{z}=i\oint_{\partial S} \sin z d\bar{z}=i [2\pi i R^2 \cos(0)]^*=2\pi R^2.$$
Basically with circular geometry, all functions $F(z,\bar{z})$ can be integrated this way, as long as it can be Taylor expanded.
