Ytukyg

Let $A$ be the two-dimensional area integral, over the complex plane, of a gaussian function:
$$A = int_text{plane} frac{mathrm{d}bar{z} wedge mathrm{d}z}{2i} exp[-zbar{z}].$$

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 $f$, $$f=-frac{1}{z}exp[-zbar{z}], \ frac{partial f}{partial bar{z}} = exp[-zbar{z}].$$
Then I wonder whether one can proceed with the following manipulations: first "integrate" over $bar{z}$,
$$A = int_text{plane} frac{mathrm{d}bar{z} wedge mathrm{d}z}{2i} exp[-zbar{z}]
= oint_text{contour} frac{mathrm{d}z}{2i} f(bar{z},z) .$$
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 $z$?

Well, there is certainly various manipulations one can perform in the $z$ and $bar{z}$ variables:

1. There is a complex Stoke's theorem
   $$ int_R! mathrm{d}eta ~=~oint_{partial R} !eta, qquad R ~subseteq~ mathbb{C}, qquad mathrm{d}~=~partial+bar{partial}~=~mathrm{d}zfrac{partial}{partial z} + mathrm{d}bar{z}frac{partial}{partial bar{z}}, tag{1} $$
   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.




2. Often one can argue that an integration over the complex plane
   $$
   int_{mathbb{C}} !mathrm{d}bar{z} wedge mathrm{d}z~f(z,bar{z}) ~=~lim_{Rto infty} int_{B(z_0,R)}!mathrm{d}bar{z}wedge mathrm{d}z ~f(z,bar{z})
   tag{3}$$
   is a limit of integrations over disks $B(z_0,R)$.




3. One may prove that
   $$forall a,b~in~mathbb{C}:~~int_{mathbb{C}} !frac{mathrm{d}bar{z} wedge mathrm{d}z}{2 pi i}~expleft[-(z-a)(bar{z}-b)right] ~=~1, tag{4}$$
   for instance by translating $z=x+iy$ into real and imaginary parts.




4. If e.g. $P(z)$ is a holomorphic polynomial, then
   $$fbox{$forall a~in~mathbb{C}:~~ I~:=~int_{mathbb{C}} !frac{mathrm{d}bar{z} wedge mathrm{d}z}{2 pi i}~P(z)~expleft[-bar{z}(z-a)right]~=~P(a).$}tag{5}$$
   Sketched proof of eq. (5):
   $$begin{align}
   int_{B(z_0,R)} !frac{mathrm{d}bar{z} wedge mathrm{d}z}{2 pi i}~left{P(a)-P(z)right}~expleft[-bar{z}(z-a)right]&quadstackrel{(3)+(4)}{longrightarrow}quad P(a)-I quadtext{for}quad R~to~infty cr
   ~=~int_{B(z_0,R)} !mathrm{d}bar{z} wedge frac{partial}{partial bar{z}} left{frac{P(z)-P(a)}{z-a}expleft[-bar{z}(z-a)right]frac{mathrm{d}z}{2 pi i}right}& cr
   ~=~int_{B(z_0,R)} !mathrm{d}left{frac{P(z)-P(a)}{z-a}expleft[-bar{z}(z-a)right]frac{mathrm{d}z}{2 pi i}right}& cr
   ~stackrel{(1)}{=}~oint_{partial B(z_0,R)} !frac{mathrm{d}z}{2 pi i} frac{P(z)-P(a)}{z-a}expleft[-bar{z}(z-a)right]&quadlongrightarrowquad 0 quadtext{for}quad R~to~infty. tag{6}end{align}$$
   $Box$

--

$^1$ 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 $g(z)$.