# Deriving an alternate form of the Cauchy Integral Formula

There is a well known formula in complex analysis called the Cauchy Integral Formula:

$$f(z) = \frac{1}{2\pi i} \int_{C} \frac{f(p)}{p-z} \, dp$$

which holds for the circle of integration $C$ when $f$ is holomorphic in an open region containing the disk defined by $C$ and any $z$ strictly inside the disk outlined by $C$.

There is an alternate formula which I am trying to derive for the special case when $C$ is a circle of radius $R$ around the origin:

$$f(z) = \frac{1}{2 \pi} \int_{0}^{2\pi} f(Re^{i \theta}) \mathbb{R}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right) \, d\theta$$ where the symbol $\mathbb{R}(\cdot)$ is meant to mean the real part of some complex number.

Here is a hint from the text: note that if $w = \frac{R^{2}}{\overline{z}}$, then the integral of $\frac{f(p)}{p -w}$ around $C$ is $0$, which I'm pretty sure follows from holomorphicity of said integrand. However, this has not been much help to me.

What I've tried:
It is necessary and sufficient to show:

$$f(z) + \frac{1}{2 \pi} \int_0^{2\pi} f(Re^{i \theta}) \mathbb{I}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right)i \, d\theta$$ $$= \frac{1}{2 \pi} \int_{0}^{2\pi} f(Re^{i \theta}) \mathbb{R}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right) \, d\theta + \frac{1}{2 \pi} \int_0^{2\pi} f(Re^{i \theta}) \mathbb{I}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right)i \, d\theta$$

*Here I've only added the "imaginary" version of the formula to both sides. Now working with the right hand side:

\begin{align} & = \frac{1}{2 \pi} \int_0^{2\pi} f(Re^{i \theta})\frac{Re^{i\theta} + z}{Re^{i\theta} - z}d\theta \\[8pt] & = \frac{1}{2 \pi} \int_0^{2\pi} f(Re^{i \theta})\frac{Re^{i\theta} + z}{Re^{i\theta} - z}d\theta \\[8pt] & = \frac{1}{2 \pi} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}(Re^{i\theta} + z) \, d\theta \\[8pt] & = \frac{1}{2 \pi} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}Re^{i\theta}d\theta + \frac{1}{2 \pi} \int_{0}^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta \\[8pt] & = \frac{1}{2 \pi i} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}Rie^{i\theta}d\theta + \frac{1}{2 \pi} \int_{0}^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta \\[8pt] & = \frac{1}{2\pi i} \int_C \frac{f(p)}{p-z}dp + \frac{1}{2 \pi} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta \\[8pt] & = f(z) + \frac{1}{2 \pi} \int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta \end{align}

And canceling $f(z)$ from both sides yields: $$\int_0^{2\pi} f(Re^{i \theta}) \mathbb{I}\left(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\right) \, d\theta =\int_0^{2\pi}\frac{f(Re^{i \theta})}{Re^{i\theta} - z}z \, d\theta$$

But I don't know where to take it from here.

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Just a small typo: You should probably include a factor if $i = \sqrt{-1}$ in each term which has the operator $\mathbb I$ since, somewhat confusingly, the imaginary part of a complex number is real. –  Morgan Sherman Aug 26 '12 at 19:19
Oh sorry I didn't realize $\mathbb{I}$ was a real thing. In my version it includes the $i$ term. I'll fix it. –  Mark Aug 26 '12 at 21:17

This is called the Poisson integral formula, by the way.

$$\dfrac{1}{2\pi}\int_0^{2 \pi} f(Re^{i\theta})\ \text{Re}\left(\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\right)\ d\theta = \frac{1}{4\pi} \int_0^{2\pi} f(Re^{i\theta}) \left( \frac{Re^{i\theta}+z}{Re^{i\theta}-z} + \frac{Re^{-i\theta} + \overline{z}}{Re^{-i\theta}-\overline{z}}\right)\ d\theta$$

Break this up into four terms. Taking $\zeta = R e^{i\theta}$ on the circle,

$$\frac{1}{4\pi}\int_0^{2\pi} f(Re^{i\theta}) \frac{Re^{i\theta}}{Re^{i\theta} - z}\ d\theta = \frac{1}{4\pi i} \oint_C f(\zeta) \dfrac{d\zeta}{\zeta - z} = \frac{f(z)}{2}$$

\eqalign{\frac{1}{4\pi}\int_0^{2\pi} f(Re^{i\theta}) \frac{z}{Re^{i\theta} - z}\ d\theta &= \frac{1}{4\pi i} \oint_C f(\zeta) \frac{z}{\zeta-z} \frac{d\zeta}{\zeta}\cr &= \frac{1}{4\pi i} \oint_C \left( \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta} \right) \ d\zeta = \frac{f(z)}{2} - \frac{f(0)}{2}\cr}

\eqalign{\frac{1}{4\pi}\int_0^{2\pi} f(Re^{i\theta}) \frac{Re^{-i\theta}}{Re^{-i\theta} - \overline{z}}\ d\theta &= \frac{1}{4 \pi i} \oint_C f(\zeta) \frac{R^2 \zeta^{-1}}{R^2 \zeta^{-1} - \overline{z}} \frac{d\zeta}{\zeta}\cr &= \frac{1}{4\pi i} \oint_C \left( \frac{f(\zeta)}{\zeta} + \frac{\overline{z} f(\zeta)}{R^2 - \overline{z} \zeta}\right) d\zeta = \frac{f(0)}{2} } (note that if $\overline{z} \ne 0$, $|R^2/\overline{z}| > R$). I'll let you do the last term.

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Can you elaborate why the first line holds? –  Mark Aug 26 '12 at 19:51
$\text{Re}(w) = \frac{w + \overline{w}}2$ –  Robert Israel Aug 26 '12 at 19:53
What led you to take the integrands like $f \frac{z}{p-z} \frac {1}{p}$ which were products and turn them into sums? It seems like these steps are crucial to the solution. Thanks! –  Mark Aug 26 '12 at 21:10
Partial fractions is a very useful technique, both in first-year calculus and here. –  Robert Israel Aug 26 '12 at 21:52
And the last integrand is holomorphic. –  Mark Aug 27 '12 at 23:58