What would be a counterexample to Cauchy's integral formula or Cauchy's theorem?

Question

Here is the Cauchy's theorem.

Let $G$ be open in $\mathbb{C}$. (Not necessarily connected)

Let $f:G\rightarrow \mathbb{C}$ be a holomorphic function.

Let $\gamma_k$ be closed rectifiable curves in $G$ for $1\leq k \leq n$.

If $\sum Wnd(\gamma_k,z)=0$ for all $z\in \mathbb{C}\setminus G=0$, then $\sum \int_{\gamma_k} f(z) dz = 0$.

As you can see, it requires curves to lie inside $G$.

Now, let's consider this case:

Let $\gamma$ be a simple closed rectifiable curve in the plane.

Let $G$ be the interior of $\gamma$ and $f:G\cup\{\gamma\} \rightarrow \mathbb{C}$ be a function holomorphic on $G$.

Assume that $f$ is complex-differentiable on $\{\gamma\}$. (I mean complex differentiablity at points, not holomorphy)

Then, is $\int_\gamma f(z) dz=0$?

What would be a counterexample?

Answer 1

This is true, even under slightly weaker assumptions, but is fairly technical to prove.

In fact, we have

Theorem If $\gamma$ is a rectifiable Jordan curve and $f$ is holomorphic on the interior $G$ of $\gamma$ and continuous on $\bar G = G \cup \gamma$, then $$ \int_\gamma f(z)\,dz = 0. $$

The proof is apparently due to Denjoy and appeared in Compt. Rend., 196, 29-33 (1933). I also managed to find a reference to a different proof by Walsh from 1933.

Answer 2

Partial result: Suppose $\gamma$ encloses a convex region $G.$ For convenience, assume $0\in G.$ Then for $0<r<1,$ $r\gamma$ is a closed contour in $G.$ Since $G$ is simply connected, $\int_{r\gamma} f(z)\,dz = 0$ by Cauchy. As $r\to 1^-,$ the uniform continuity of $f$ on $\overline G$ shows

$$\int_{r\gamma} f(z)\,dz \to \int_{\gamma} f(z)\,dz.$$

Therefore $\int_{\gamma} f(z)\,dz =0.$

Here's the thing: The same result would be true if we merely assumed $f$ is continuous on $\overline G$ and $ f\in H(G).$ So at this point I can't tell if this "complex differentiability" at points of $\gamma$ hypothesis gives any advantage over continuity on $\overline G.$