An integral over $\mathbb C$: I would like to know is that this equality is just:
$$\int_{\mathbb C}e^{-z}dz=\int_{\mathbb R}e^{-x}dx \times \int_{\mathbb R}e^{-iy}dy$$
for all $z=x+iy\in \mathbb C$. Or in general:
$$\int_{\mathbb C}f(z)dz=\int_{\mathbb R}g(x)dx \times \int_{\mathbb R}g(y)dy,$$ where $f(z)=g(x) g(y)$ ?
Thank you in advance.
 A: Notation $\int(..){\rm d}z$ usualy denotes integration along a curve, not over a surface. As such, it has no relation to $\iint(..){\rm d}x {\rm d}y$.
For $f(x+{\rm i}y)=u(x,y)+{\rm i}v(x,y)$ integral $\int_\gamma f(z) {\rm d}z$, where $\gamma$ is a curve in $\mathbb{C}\equiv\mathbb{R^2}$ is defined as 
$$ \int_\gamma f(z) {\rm d}z = \int_\gamma (u+{\rm i}v)({\rm d}x+{\rm i\,d}y)= \int_\gamma (u \,{\rm d}x - v\,{\rm d}y) + {\rm i}\int_\gamma (v \,{\rm d}x + u \,{\rm d}y)$$
You can define a surface integral in a way that you propose, but then it's usually denoted differently. For ${\rm d}z = {\rm d}x+{\rm i\,d}y$, ${\rm d}z^* = {\rm d}x-{\rm i\,d}y$ you have ${\rm d}z\wedge{\rm d}z^*=-2{\rm i\,d}x\wedge{\rm d}y$, so you define
$$ \int_\Omega f(z) {\rm d}z{\rm d}z^* = -2{\rm i} \int_\Omega f(x+{\rm i}y){\rm d}x {\rm d}y$$
A: You are trying to compare two unrelated integrals:
$$\int e^{-(x+iy)}d(x+iy)$$ and $$\iint e^{-(x+iy)}\,dx\,dy.$$
In addition, your integration domains do not make sense.
