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Contour Integral Representation of a Function
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[QUOTE="julian, post: 6809546, member: 142346"] I have partial result, in that I might have proved the integral for functions specified in the Schwarz Reflection Principle. -------- [HEADING=2]Schwarz Reflection Principle:[/HEADING] Let ##D## be an open domain in the complex plane who's reflection about the real axis is symmetric. Define ##D^+ = \{ z \in D : Im(z) >0 \}##. Suppose that ##f## is an analytic function which is defined in ##D^+##. Further suppose that ##f## extends to a continuous function on the real axis, and takes on real values on the real axis. Then ##f## can be extended to an analytic function on ##D## by the formula $$ f(\overline{z}) = \overline{f(z)} . $$ and the values for ##z## reflected across the real axis are the reflections of ##f(z)## across the real axis. -------- We assume that ##D## contains the disk ##\overline{\mathbb{D}} = \{ z \in \mathbb{C} : |z| \leq 1 \}##. In the following we only consider functions as described above. First note: ##Im (f(0)) =0##. Now \begin{align} \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+z}{w-z} dw & = \frac{1}{2 \pi i} \int_0^{2 \pi} \frac{1}{2} [f (e^{i \theta}) + \overline{f (e^{i \theta})}] \dfrac{e^{i \theta} + z}{e^{i \theta} - z} i d \theta \nonumber \\ & = \frac{1}{2 \pi i} \int_0^{2 \pi} \frac{1}{2} [f (e^{i \theta}) + f (e^{-i \theta})] \dfrac{e^{i \theta} + z}{e^{i \theta} - z} i d \theta \nonumber \\ & = \frac{1}{4 \pi i} \int_0^{2 \pi} \left[ f (e^{i \theta}) \dfrac{e^{i \theta} + z}{e^{i \theta} - z} + f (e^{-i \theta}) \dfrac{e^{i \theta} + z}{e^{i \theta} - z} \right] i d \theta \nonumber \\ & = \frac{1}{4 \pi i} \int_0^{2 \pi} f (e^{i \theta}) \left[ \dfrac{e^{i \theta} + z}{e^{i \theta} - z} + \dfrac{e^{-i \theta} + z}{e^{-i \theta} - z} \right] i d \theta \nonumber \\ & = \frac{1}{4 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} \left[ \dfrac{w+z}{w-z} + \dfrac{1+wz}{1-wz} \right] dw \quad (*) \nonumber \end{align} First, using ##(*)## when we have ##z=0##: $$ \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+0}{w-0} dw = \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} dw = f(0) . $$ Now take ##z \not=0##. Using ##(*)##, \begin{align} \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{Re (f(w))}{w} \dfrac{w+z}{w-z} dw & = \frac{1}{4 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w} \left[ \dfrac{w+z}{w-z} - \dfrac{w+z^{-1}}{w-z^{-1}} \right] dw \nonumber \\ & = \frac{1}{4 \pi i} \oint_{|w|=1} f(w) \left[ \dfrac{2}{w-z} - \frac{1}{w} - \dfrac{2}{w-z^{-1}} + \frac{1}{w} \right] dw \nonumber \\ & = \frac{1}{2 \pi i} \oint_{|w|=1} \dfrac{f(w)}{w-z}dw = f(z) . \nonumber \end{align} [/QUOTE]
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Contour Integral Representation of a Function
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