A boundary version of Cauchy's theorem I am looking for a reference for the following theorem (or something like it) that is not Kodaira's book.


Let $D$ be a domain and $\overline{D}$ be it's closure. Suppose that $f:\overline{D} \rightarrow \mathbb{C}$ is holomorphic in $D$ and continuous on $\overline{D}$. If the boundary of $D$, denoted $\partial D$ is composed of piecewise $C^1$ curves then
    $$f(w) = \frac{1}{2\pi i}\oint_{\partial D}\frac{f(z)}{z-w}dz$$
    for all $w\in D$.


This is stronger than the classical Cauchy integral theorem, because here we can allow the path we integrate over to be the boundary. This appears in Kodaira's book, but the book seems sloppily written and I found an egregious error in the first few pages, so I'd like another reference. 
This does not appear in Ahlfors or any other standard text I looked at. I am looking for any proof that extends to usual Cauchy's theorem to allow paths that lie on the boundary of a region, instead of requiring them to lie entirely in the open region like the usual version does. 
I managed to find a paper that sketches a high-powered proof when the boundary is a rectifiable Jordan arc (or some multiply connected region where each boundary component is) and gives a reference to a more elementary proof of the same result. I think a simpler proof should be possible in this case, where the boundary is assumed to be $C^1$. 
This is closely related to an old question that did not get much attention.
 A: I guess here that you mean that the boundary of $D$ is a finite number of pairwise disjoint curves. In this case, this Cauchy formula follows directly from Mergelyan's Theorem about uniform rational approximation of holomorphic functions. Indeed, the formula holds for rational functions with poles outside $\overline{D}$, and Mergelyan's Theorem says that any $f$ continuous on $\overline{D}$, holomorphic in $D$, can be approximated uniformly on $\overline{D}$ by rational functions with poles outside $\overline{D}$.
It also works under the weaker hypothesis that the curves are rectifiable.
A: For domains with (piecewise) $C^1$ boundary, there is a simple proof using Stokes' theorem, at least if you assume that $f$ extends to $C^1$ on $\bar D$. A little computation will show that for $f \in C^1(\bar D)$,
$$2\pi i f(w) = \int_{\partial D} \frac{f(z)}{z-w}\,dz + \iint_D \frac{\frac{\partial f}{\partial \bar z}}{z-w}\,dz\wedge d\bar z,$$
from which your version of Cauchy's theorem follows.
I'll try to dig up a reference for the $C(\bar D)$ case.
A: My apologies on my misreading the first time. Knowing $f$ is continuous on $\overline D$, for $\delta>0$ with $|w|<1-\delta$, we will know that the Cauchy integrand
$$\frac{f(z)}{z-w}$$
is uniformly continuous on $\{z: 1-\delta\le z\le 1\}$, and so the integral over the circle of radius $1-\delta$ converges to the integral over $\partial D$.
