For a function $f(x,y)$ on $\mathbb{R}^2,$ defined possibly outside the origin, write


$$\int\_\epsilon ' f \,dx\,dy : = \int\_{\mathbb{R}^2\setminus D\_\epsilon}f \, dx\,dy,$$


(the integral on the complement to the $\epsilon$-disk) and


$$\int' f \,dx \,dy : = \lim\_{\epsilon\to 0} \int'\_\epsilon f \,dx\,dy,$$ when defined.


We view $\mathbb{R}^2$ as the complex line with coordinate $z = x+iy$. Then I claim that the assignment $$\phi:f(x,y) \mapsto \int\_{\mathbb{R}^2}' f\cdot z^{-1} \,dx\,dy$$ makes sense as a functional on compactly supported, smooth functions (indeed, if $f$ is rotationally symmetric then $\phi(f) = 0$ even before taking the limit, and if $f(0) = 0$ then $f\cdot z^{-1}$ is bounded, hence integrable, at $0$; now in the space of smooth compactly supported functions, any function can be written as a rotationally symmetric function plus a function that vanishes at the origin). We can thus formally define a *distribution* $z^{-1}\in C^{-\infty}(\mathbb{R}^2)$ as $$z^{-1}: = \frac{\phi}{dx\,dy}.$$


Now write $\bar{\partial} = \partial\_x + i \partial\_y$ .Like any vector field, this acts on distributions, and so we have a distribution $\bar{\partial} z^{-1}.$ Since $z^{-1}$ is holomorphic where it is smooth, we must have $\bar{\partial} z^{-1}$ be a distribution supported at the origin. In fact, it is known to mathematical physicists that the result is a delta function at the origin: $$\bar{\partial} z^{-1} = 2\pi \delta\_0,$$ and this fact is useful in conformal field theory.


I would like to see a proof of this result (the physics sources I have seen do not prove this). In fact, I know how to give one using a direct calculation with polar coordinates: but this seems *ad hoc* and is not very satisfying to me.


I suspect there might be "nicer" proofs using one or more of the following three techniques, and I am hoping that more analytically literate MO users can provide them.


1. If one can show that $\bar{\partial} z^{-1}$ is determined by its values on holomorphic (near the origin) functions, the Cauchy residue formula would imply that $\bar{\partial} z^{-1} = 2\pi \delta\_0.$
2. I suspect there should be a proof using the stationary phase approximation.
3. This is probably overkill, but it would be nice if there were a proof using pseudodifferential operators.



Elegant proofs of $\bar{\partial}z^{-1} = 2\pi \delta_0$
The identification of $\partial\_{\bar{z}}z^{-1}$ with a delta function follows directly from the [Cauchy–Pompeiu formula](https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula#Smooth_functions)
$$f(\zeta) = \frac{1}{2\pi i}\int\_{\partial D} \frac{f(z) \,dz}{z-\zeta} - \frac{1}{\pi}\iint\_D \frac{\partial f(z)}{\partial \bar{z}} \frac{dx\wedge dy}{z-\zeta},\qquad\qquad(\ast)$$
for $f$ a complex-valued $C^1$ function on the closure of the disk $D$ in $\mathbb{C}$. (Note that $f$ need not be holomorphic, as in the usual Cauchy formula.) If the support of $f$ is within $D$ the boundary integral $\int\_{\partial D}$ does not contribute, while the area integral $\iint\_D$ states that$^\ast$
$$\frac{\partial }{\partial \bar{z}} \frac{1}{z-\zeta}=\pi\delta^{(2)}(z-\zeta).$$


*Historical note:* [Wikipedia](https://en.wikipedia.org/wiki/Cauchy%27s_integral_formula#cite_note-2) gives a [1905](http://archive.numdam.org/item/AFST_1905_2_7_3_265_0/) paper as the source of the formula ($\ast$), but I could not locate it there. I did find it in a [1913](https://doi.org/10.1007/BF03015607) paper by Pompeiu (equation 3):


![](https://i.stack.imgur.com/e5wGE.png)
The first term $h(z)$ is the boundary integral. The capital $S$ denotes the area integral, the function $\varphi$ is the derivative $\partial f/\partial \bar{\zeta}$.


---


$^\ast$ The OP has a factor $2\pi$ instead of $\pi$, because of a different definition of the [Wirtinger derivative](https://en.wikipedia.org/wiki/Wirtinger_derivatives). Here I follow the definition $\partial/\partial\bar{z}=\tfrac{1}{2}(\partial/\partial x+i\partial/\partial y)$, while in the OP there is no coefficient $\tfrac{1}{2}$.


