Question : Can you tell if this proof of the Cauchy integral theorem/formula is correct, and if it is, is there a good reason for not using it in complex analysis courses ?
Let $\gamma$ be a closed rectifiable piecewise $C^1$ contour parametrized by $t \in [a,b]$, and consider the dilatation of the contour around a point : $\gamma_{r,s}(t) = s + r(\gamma(t)-s)$. Let $$g(r) = \frac{1}{r}\int_{\textstyle \gamma_{r,0}} f(z) dz = \int_a^b f(r \gamma(t))\gamma'(t)dt$$ If $f(z)$ is continuously complex differentiable on $U$ a convex open containing $\gamma$ and the origin, then $$r g'(r) = \int_a^b \gamma(t) f'(r \gamma(t)) r\gamma'(t)dt = \gamma(t) f(r\gamma(t))|_a^b - \int_a^b \gamma'(t) f(r \gamma(t))dt = -g(r)$$
so with $h(r) = r\ g(r)$ we have $h'(r) = g(r) + r g'(r) = 0$ $$ \implies \int_\gamma f(z) dz = h(1) = \lim_{r \to 0} h(r) = \lim_{r \to 0}r\int_a^b f(r \gamma(t)\gamma'(t)dt = 0$$
If $f(z)$ is non-continuously complex differentiable, then as before we have $h'(r) = 0$. Letting $u = f(r\gamma(t)), v = \gamma(t)$ : $$r \frac{\partial}{\partial r} [u \frac{\partial}{\partial t}v] = v \frac{\partial}{\partial t}u = \frac{\partial}{\partial t}[uv]-u\frac{\partial}{\partial t}v $$ Also note that $\int_a^b \frac{\partial}{\partial t}[uv] dt = uv|_a^b = 0$, so that $$rg'(r) = \int_a^b r \frac{\partial}{\partial r} [u \frac{\partial}{\partial t}v]dt = \int_a^b \left(\frac{\partial}{\partial t}[uv]-u\frac{\partial}{\partial t}v\right)dt$$ $$ = uv|_a^b - \int_a^b f(r\gamma(t))\gamma'(t)dt= -g(r)$$ and we have the Cauchy integral theorem for convex regions containing the origin.
Corollaries :
Shifting what was done before, we have that if $f(z)$ is holomorphic on any convex region $U \ni s$ then $\int_\gamma f(z) dz = \lim_{ r \to 0} \int_{\gamma_{r,s}} f(z) dz =0$
if $f(z)$ is holomorphic on a simply connected bounded open $U$ containing a closed rectifiable piecewise $C^1$ contour $\gamma$, then $U$ can be decomposed as an union of convex sub-regions $U_k$, and $\gamma = {\bigcup^{\bigoplus}}_{k=1}^K \gamma_k$ where each $\gamma_k \in U_k$ is a closed contour and the symbol $\bigcup^{+}$ means that the union of two curves traversed in opposite direction cancel. Hence we have $$\int_\gamma f(z) dz = \sum_{k=1}^K \int_{\gamma_k} f(z) dz = \sum_{k=1}^K 0 = 0$$
if homotopically $\gamma \cong \gamma'$ on $U$ where $f(z)$ is holomorphic then $\gamma \cup \gamma'_{\ominus}$ ($\ominus$ means traversed in opposite direction) can be completed to yield $\lambda = \gamma \cup^+ \delta \cup^+ \gamma'_{\ominus} \cup^+ \delta_{\ominus}$ a closed contour supported on a simply connected region, so that $\int_\lambda f(z)dz = \int_\gamma f(z) dz + \int_{\gamma'} f(z) dz = 0$
With $s \in U$ simply connected, by definition $\eta(\gamma,s)$ is the number of small circles $|z-s| = \epsilon$ to which a closed contour $\gamma \subset U$ is homotopically equivalent on $U \setminus \{s\}$, so that : $$\int_\gamma \frac{f(z)}{z-s} dz = \lim_{ r \to 0} \int_{\gamma}\frac{f(z)}{z-s} dz= \eta(\gamma,s) \int_{|z-s| = \epsilon}\frac{f(z)}{z-s} dz$$ $$ = \eta(\gamma,s)\ f(s) \lim_{\epsilon\to 0}\int_{|z-s| = \epsilon}\frac{1}{z-s} dz = 2i \pi \ \eta(\gamma,s) \ f(s)$$