Doubts Regarding Proof Of Cauchy Integral Formula

Question

I was trying to understand the proof of Cauchy Integral formula from the book "Complex Analysis" By Stein and Shakarchi. I understand most steps of that proof, but I have one doubt:

I understand that due to Cauchy Integral Theorem, the integral of the domain bounded by the curve $C$ is 0, and this irrespectively of the value around the $|\zeta|=\epsilon$ part. However in book they state that in the limit $\delta \to0$ the integral of $f(\zeta)/{(\zeta -z)}$ equals the one around the exterior circle part of $C$ plus the integral around the circle $|\zeta|=\epsilon$, and I do not get this part.

Any Help will be appreciated

Answer 1

It is simply a matter of what is the limiting value of $$ \oint_{\Gamma_{\epsilon,\delta}}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta,\tag{1}\label{1} $$ calculated while keeping $\epsilon$ constant and letting $\delta\to 0$. The value of the curvilinear integral on the two parallel segments of $\Gamma_{\epsilon,\delta}$ has the following form $$ \int_{z_1}^{z_2}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta + \int_{z_2-\delta \xi_2}^{z_1-\delta \xi_1}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta=\int_{z_1}^{z_2}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta-\int_{z_1-\delta \xi_1}^{z_2-\delta \xi_2}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta\tag{2}\label{2} $$ where $z_1,z_1-\delta\xi_1\in C_\epsilon$ and $z_2,z_2-\delta\xi_2\in C$: when $\delta\to 0$, \eqref{2} goes to $0$, therefore $$ \lim_{\delta\to 0}\oint_{\Gamma_{\epsilon,\delta}}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta=\oint_C\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta- \oint_{C_\epsilon}\frac{f(\zeta)}{\zeta-z}\mathrm{d}\zeta\tag{3}\label{3} $$ Edit: in order to explain the concept in a better way, I added the following picture:

For $\delta\to 0$, the two parallel segment parts of the keyhole domain $\Gamma_{\epsilon,\delta}$, having as endpoints respectively $z_1,z_2$ and $z_1-\delta\xi_1,z_2-\delta\xi_2$, tend to the same segment: since these two paths are traveled in opposite direction during the evaluation of the curvilinear integral \eqref{1} (as made explicit by equation \eqref{2}), their joint contribution tend to vanish and equation \eqref{3} holds.

Answer 2

$$\begin{align} 0&=\lim_{\delta,\epsilon \to 0}\int_{\Gamma_{\delta,\epsilon}}F(\zeta)\,d\zeta\\ &= \int_{C}F(\zeta)\,d\zeta +\lim_{\epsilon \to 0}\int_{C_\epsilon}F(\zeta)\,d\zeta\\ &= \int_{C}F(\zeta)\,d\zeta +\lim_{\epsilon \to 0}\int_{C_\epsilon}\frac{f(\zeta)-f(z)}{\zeta - z}\,d\zeta +\lim_{\epsilon \to 0}\int_{C_\epsilon}\frac{f(\zeta)}{\zeta - z}\,d\zeta\\ &= \int_{C}F(\zeta)\,d\zeta + 0 -f(z)2\pi i\\ \end{align}$$ We thus conclude that $$2\pi i f(z) = \int_{C}F(\zeta)\,d\zeta= \int_{C}\frac{f(\zeta)}{\zeta - z}\,d\zeta$$ Which is the Cauchy Integral Formula.