Let $A$ be the two-dimensional area integral, over the complex plane, of a gaussian function:
$$A = \int_\text{plane} \frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2i} \ \exp[-z\bar{z}].$$
Of course one could evaluate this integral by converting to Cartesian (or alternatively to polar) coordinates, where the answer is evident,
$$A = \int_\text{plane} \mathrm{d}x \wedge \mathrm{d}y \ \exp[-(x^2+y^2)] = \pi.$$
But, is there a way to do the complex area integration directly in the complex coordinates $z,\bar{z}$?
In particular, let us define the (non-holomorphic) function $f$, $$f=-\frac{1}{z}\exp[-z\bar{z}], \\ \frac{\partial f}{\partial \bar{z}} = \exp[-z\bar{z}].$$
Then I wonder whether one can proceed with the following manipulations: first "integrate" over $\bar{z}$,
$$A = \int_\text{plane} \frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2i} \ \exp[-z\bar{z}]
= \oint_\text{contour} \frac{\mathrm{d}z}{2i}\ f(\bar{z},z) .$$
and then evaluate the positive definite Gaussian part at the $z=0$ pole, to use the Residue theorem,
$$
= \oint_\text{contour} \frac{\mathrm{d}z}{2i} \ \frac{1}{z} = \pi ,$$
which produces the correct value of the integral. Is there a sense in which these manipulations are correct?
Can one perform the complex plane area integral by first integrating over $\bar{z}$, and then over $z$?
Best Answer
Well, there is certainly various manipulations one can perform in the $z$ and $\bar{z}$ variables:
There is a complex Stoke's theorem $$ \int_R\! \mathrm{d}\eta ~=~\oint_{\partial R} \!\eta, \qquad R ~\subseteq~ \mathbb{C}, \qquad \mathrm{d}~=~\partial+\bar{\partial}~=~\mathrm{d}z\frac{\partial}{\partial z} + \mathrm{d}\bar{z}\frac{\partial}{\partial \bar{z}}, \tag{1} $$ where$^1$ $$\eta~=~ f(z,\bar{z})~ \mathrm{d}z +g(z,\bar{z}) ~ \mathrm{d}\bar{z} \tag{2}$$ is a 1-form in the Dolbeault double complex.
Often one can argue that an integration over the complex plane $$ \int_{\mathbb{C}} \!\mathrm{d}\bar{z} \wedge \mathrm{d}z~f(z,\bar{z}) ~=~\lim_{R\to \infty} \int_{B(z_0,R)}\!\mathrm{d}\bar{z}\wedge \mathrm{d}z ~f(z,\bar{z}) \tag{3}$$ is a limit of integrations over disks $B(z_0,R)$.
One may prove that $$\forall a,b~\in~\mathbb{C}:~~\int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i}~\exp\left[-(z-a)(\bar{z}-b)\right] ~=~1, \tag{4}$$ for instance by translating $z=x+iy$ into real and imaginary parts.
If e.g. $P(z)$ is a holomorphic polynomial, then $$\fbox{$\forall a~\in~\mathbb{C}:~~ I~:=~\int_{\mathbb{C}} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i}~P(z)~\exp\left[-\bar{z}(z-a)\right]~=~P(a).$}\tag{5}$$ Sketched proof of eq. (5): $$\begin{align} &\quad P(a)-I \quad\text{for}\quad R~\to~\infty \cr\cr &\quad\uparrow(3)+(4) \cr\cr &\int_{B(z_0,R)} \!\frac{\mathrm{d}\bar{z} \wedge \mathrm{d}z}{2 \pi i}~\left\{P(a)-P(z)\right\}~\exp\left[-\bar{z}(z-a)\right] \cr ~=~&\int_{B(z_0,R)} \!\mathrm{d}\bar{z} \wedge \frac{\partial}{\partial \bar{z}} \left\{\frac{P(z)-P(a)}{z-a}\exp\left[-\bar{z}(z-a)\right]\frac{\mathrm{d}z}{2 \pi i}\right\} \cr ~=~&\int_{B(z_0,R)} \!\mathrm{d}\left\{\frac{P(z)-P(a)}{z-a}\exp\left[-\bar{z}(z-a)\right]\frac{\mathrm{d}z}{2 \pi i}\right\} \cr ~\stackrel{(1)}{=}~&\oint_{\partial B(z_0,R)} \!\frac{\mathrm{d}z}{2 \pi i} \frac{P(z)-P(a)}{z-a}\exp\left[-\bar{z}(z-a)\right]\cr\cr &\quad\downarrow \cr\cr &\quad 0 \quad\text{for}\quad R~\to~\infty. \end{align}\tag{6}$$ $\Box$
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$^1$ Note that $\bar{z}=x-iy$ denotes the complex conjugate variable. It is not an independent complex variable. In particular,
$$ \frac{\partial}{\partial z} ~:=~\frac{1}{2}\left(\frac{\partial}{\partial x}-i \frac{\partial}{\partial y}\right), \qquad \frac{\partial}{\partial \bar{z}} ~:=~\frac{1}{2}\left(\frac{\partial}{\partial x}+i \frac{\partial}{\partial y}\right).\tag{7}$$
The double argument notation $f(z,\bar{z})$ is traditionally used to indicate that $f(z,\bar{z})$ is not necessarily a holomorphic function. A holomorhic function $\bar{\partial}g(z) / \partial \bar{z}=0$ is in turn written with only a single argument $g(z)$.