Editing Cauchy's integral formula (section)

== Proof sketch ==

By using the [[Cauchy integral theorem]], one can show that the integral over {{math|''C''}} (or the closed rectifiable curve) is equal to the same integral taken over an arbitrarily small circle around {{math|''a''}}. Since {{math|''f''(''z'')}} is continuous, we can choose a circle small enough on which {{math|''f''(''z'')}} is arbitrarily close to {{math|''f''(''a'')}}. On the other hand, the integral
<math display="block">\oint_C  \frac{1}{z-a} \,dz = 2 \pi i,</math>
over any circle {{math|''C''}} centered at {{math|''a''}}.  This can be calculated directly via a parametrization ([[integration by substitution]]) {{math|''z''(''t'') {{=}} ''a'' + ''εe<sup>it</sup>''}} where {{math|0 ≤ ''t'' ≤ 2π}} and {{math|''ε''}} is the radius of the circle.

Letting {{math|''ε'' → 0}} gives the desired estimate
<math display="block">\begin{align}
\left | \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-a} \,dz  - f(a) \right |
&= \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)-f(a)}{z-a} \,dz \right | \\[1ex]
&= \left | \frac{1}{2\pi i}\int_0^{2\pi}\left(\frac{f\bigl(z(t)\bigr)-f(a)}{\varepsilon e^{it}}\cdot\varepsilon e^{it} i\right )\,dt\right | \\[1ex]
&\leq \frac{1}{2 \pi} \int_0^{2\pi} \frac{ \left|f\bigl(z(t)\bigr) - f(a)\right| } {\varepsilon} \,\varepsilon\,dt \\[1ex]
&\leq \max_{|z-a|=\varepsilon} \left|f(z) - f(a)\right|
~~ \xrightarrow[\varepsilon\to 0]{} ~~ 0.
\end{align}</math>