Does there exist a direct way of proving Cauchy's integral formula?

Question

Cauchy's integral formula $:$

If $D$ be a simply connected domain and $\gamma$ is a closed contour in $D$, then for any analytic function $f$ in $D$ and $a \in D \setminus \{\gamma \}$, $$f(a)n(\gamma ; a) = \frac {1} {2 \pi i} \int_{\gamma} \frac {f(\zeta)} {\zeta - a}\ d{\zeta}.$$

Now my question is "Does there any process which directly proves the above formula?" i.e. a process which does not involve Cauchy-Goursat theorem, Cauchy's theorem for disk etc. Actually if it exists then that would be time preserving in exams, I guess. Otherwise we have to deal with lots of lemmas or results which actually wastes lots of time and exhausts us in exams. So I and some of my friends are feared off thinking that how could we manage time in exam hall if these kind of laborious questions happen to be in the question paper. Please give me some suggestions about overcoming such problems.

Thank you in advance.

Answer 1

For analytic functions the Cauchy integral theorem is not hard to prove : start with $f$ given by a power series on a disk

$f(z) =\sum_{n=0}^\infty c_n (z-a)^n $ for $|z-a| < r$ so that $F(z) = \sum_{n=0}^\infty \frac{c_n}{n+1} (z-a)^{n+1}$ is an (analytic) primitive. Therefore $\int_\gamma f(z)dz = F(\gamma(1))-F(\gamma(0))$

then extend to finite ($\scriptstyle \text{and simply connected}$) union of such disks, splitting $\gamma$ into parts contained in each disk.

Finally if $f$ is analytic then $$\int_\gamma \frac{f(z)}{z-s}dz = \int_\gamma \frac{f(s)}{z-s}dz+\int_\gamma \frac{f(z)-f(s)}{z-s}dz$$ where $\frac{f(z)-f(s)}{z-s}$ is analytic and $\int_\gamma \frac{f(s)}{z-s}dz = f(s)\, 2i\pi \, \eta(\gamma,s)$

---

To make it more clear : by definition, analytic functions have locally a primitive. And those assemble to a global primitive on simply connected open sets. Once this is achieved, the Cauchy integral theorem and formula become trivial.