# Cauchy's integral formula with high tech methods

Using Stokes theorem there is particularly nice way to prove Cauchys theorem. If $U$ is a domain in $\mathbb{C}$ with (piecewise) smooth boundary $\Gamma$, then for all differentiable functions $g: \bar U \rightarrow \mathbb{C}$ we have $$\int_\Gamma g dz = \int_Ud(gdz)=\int_U \frac{\partial g}{\partial \bar z} d\bar z \wedge dz = 2i \int_U \frac{\partial g}{\partial \bar z} d\mathcal{L}^2$$ (where $\mathcal{L}^2$ is the two dimensional Lebesgue measure on $\mathbb{C}$). Consequently, if $g$ is holomorphic, then the line integral over $\Gamma$ vanishes.

I want to know if this reasoning also gives rise to Cauchy's integral formula. How do I want to do this? For simplicity we assume that $0 \in U$ and want to prove that $$f(0) = \frac{1}{2\pi i} \int_\Gamma \frac{f(z)}{z} dz.$$ The function $z \mapsto \frac{1}{z}$ is locally integrable over $\mathbb{C}$ and homogeneous of degree $-1$, hence it can be considered as a tempered distribution which is also homogeneous of degree $-1$. Since this distribution is holomorphic away from $0$, we conclude that $\frac{\partial}{\partial \bar z}\frac{1}{z}$ has its support in $\{0\}$ and is homogeneous of degree $-2$. Consequently it is a multiple of the Dirac distribution at $0$, which we denote by $\delta$. Either by a computation or in anticipation of Cauchy's formula we may assume that $$\frac{\partial}{\partial \bar z}\frac{1}{z} = \pi \delta$$ To prove Cauchy's integral formula I want to perform the following computation: Let $f$ be holomorphic in a neighbourhood of $\bar U$, then (trusting our inner physicist) we see $$\frac{1}{2\pi i} \int_\Gamma \frac{f(z)}{z} dz = \frac{1}{\pi} \int_U\frac{\partial }{\partial \bar z}\left(\frac{f(z)}{z}\right) d\mathcal{L}^2 = \frac{1}{\pi} \int_U f(z) \frac{\partial}{\partial \bar z}\frac{1}{z} d\mathcal{L}^2 = \int_U f(z) \delta d\mathcal{L}^2 = f(0)$$ Of course this is not rigorous, the main problem being that we are dealing with a tempered distribution, which we can only apply to Schwartz functions and not to holomorphic functions.

My question is: Can we do this in a rigorous way, somehow using holomorphic functions as test functions? (In light of this question I read about hyperfunctions, which somehow seem to be related, but at a first glance the theory seemed to be quite extensive - too much to answer the question myself right now. So maybe if anyone is familiar with these concepts my question might be very easy to answer. Any ideas are welcome!)