TITLE: How to get the integral representation of $f'$ from that of $f$? (Cauchy's Integral Formula)
QUESTION [6 upvotes]: I know that if $f$ is holomorphic on an open set $\Omega$ containing a circle $C$ and it's interior, then for any $z_0$ in the interior of $C$ we have 
$$f(z_0) = \dfrac {1}{2 \pi i} \int_{C} \dfrac {f(w)}{w-z}dw$$
where $C$ has positive orientation. If we compute the difference quotient, it comes out to 
$$f'(z_0) = \dfrac {1}{2 \pi i}  \lim_{h \to 0} \int_C \dfrac {f(w)}{(w-z_0)(w-z_0 - h)}dw$$
Obviously I would like to interchange the limit and the integral. I guess that would be possible by bringing in some fancier theorems, but is there a way to get around this with just the basics and get to
$$f'(z_0) =\dfrac {1}{2 \pi i} \int_C \dfrac {f(w)}{(w-z_0)^2}dw$$

REPLY [3 votes]: Let $\epsilon=\min\limits_{w\in C}|w-z_0|>0$ and $|h|<\frac{\epsilon}2$ so that $|w-z_0-h|>\frac{\epsilon}2$ for all $w\in C$. We have
$$\begin{align*}
\frac{f(z_0+h)-f(z_0)}{h}-\frac1{2\pi i}\int_C \frac{f(w)}{(w-z_0)^2} dw&=\frac1{2\pi i}\int_C \frac{f(w)}{w-z_0}\left(\frac{1}{w-z_0-h}-\frac1{w-z_0}\right) dw\\&=\frac{h}{2\pi i}\int_C \frac{f(w)dw}{(w-z_0)^2(w-z_0-h)} 
\end{align*}$$, and
$$\begin{align*}
\left|\frac{f(z_0+h)-f(z_0)}{h}-\frac1{2\pi i}\int_C \frac{f(w)}{(w-z_0)^2} dw\right|&\le \frac{|h|}{2\pi}\cdot \int_C \frac{\max_{w\in C}|f(w)|}{\epsilon^3/2}\ |dw|\\&\le \frac{\text{len}(C)}{\pi} \frac{\max_{w\in C}|f(w)|}{\epsilon^3}\cdot |h|.
\end{align*}$$ If we let $|h|\to 0$, it follows that
$$
\lim_{h\to 0}\left|\frac{f(z_0+h)-f(z_0)}{h}-\frac1{2\pi i}\int_C \frac{f(w)}{(w-z_0)^2} dw\right|=0
$$, or equivalently,
$$
f'(z_0)=\lim_{h\to 0}\frac{f(z_0+h)-f(z_0)}{h}=\frac1{2\pi i}\int_C \frac{f(w)}{(w-z_0)^2} dw.
$$