Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Additional Information:
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.

share|improve this question
    
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
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.