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.

- 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 $\bar{\partial} z^{-1} = 2\pi \delta_0.$
- I suspect there should be a proof using the stationary phase approximation.
- This is probably overkill, but it would be nice if there were a proof using pseudodifferential operators.