# CVRC6

A function $\phi(z)\,$ is zero when $z = 0\,$, and is real when $x\,$ is real, and is analytic when $|z| \leq 1\,$. If $f(x, y)\,$ is the imaginary part of $\phi(x + iy)\,$ prove that $\int_0^{2\pi} \frac{x\sin\theta}{1 - 2x\cos\theta + x^2}f(\cos\theta, \sin\theta)\ d\theta = \pi\phi(x)$ holds when $-1 < x < 1\,$.

Parametrizing with $z = e^{i\theta}\,$, the function $\phi(z) = u(\cos\theta, \sin\theta) + if(\cos\theta, \sin\theta)\,$. Consider

$\int_{\Gamma} \frac{x\left(z - \frac{1}{z}\right)\frac{1}{2i}}{1 - 2x\left(z + \frac{1}{z}\right)\frac{1}{2} + x^2}\phi(z)\ \frac{dz}{iz}$

on the unit circle. Then

 $= \frac{1}{2}\int_\Gamma \frac{x\left(z - \frac{1}{z}\right)}{1 - x\left(z + \frac{1}{z}\right) + x^2}\phi(z)\ \frac{dz}{-z}$ $= \frac{1}{2}\int_\Gamma \frac{x\left(z - \frac{1}{z}\right)}{(x - z)\left(x - \frac{1}{z}\right)}\phi(z)\ \frac{dz}{-z}$ $= \frac{1}{2}\int_\Gamma \frac{x\left(z - \frac{1}{z}\right)}{(z - x)(xz - 1)}\phi(z)\ dz$ $= \frac{1}{2}\int_\Gamma \frac{x(z^2 - 1)}{z(z - x)(xz - 1)}\phi(z)\ dz$

So the poles are at $z = 0\,$ and $z = x\,$, since $-1 < x < 1\,$, and the other singularity $1/x\,$ is outside the unit circle. Thus

 $= \frac{1}{2}\left(2\pi i \left[\left(\mathrm{Residue\ at\ } z = 0\right) + \left(\mathrm{Residue\ at\ } z = x\right)\right]\right)$ $= \pi i \left[\left(\lim_{z\to 0} \frac{x(z^2 - 1)}{(z - x)(xz - 1)}\phi(z)\right) + \left(\lim_{z\to x} \frac{x(z^2 - 1)}{z(xz - 1)}\phi(z)\right)\right]$ $= \pi i \left[\left(\frac{x(-1)}{(-x)(-1)}\phi(0)\right) + \left(\frac{x(x^2 - 1)}{x(xx - 1)}\phi(x)\right)\right]$ $= \pi i \left[0 + 1\phi(x)\right]$ $= \pi i\phi(x)\,$

Thus

 $i\pi\phi(x)\,$ $= \int_0^{2\pi}\frac{x\sin\theta}{1 - 2x\cos\theta + x^2}u(\cos\theta, \sin\theta)\ d\theta$ $+\ i\int_0^{2\pi}\frac{x\sin\theta}{1 - 2x\cos\theta + x^2}f(\cos\theta, \sin\theta)\ d\theta$

Therefore,

$\int_0^{2\pi}\frac{x\sin\theta}{1 - 2x\cos\theta + x^2}f(\cos\theta, \sin\theta)\ d\theta = \pi\phi(x)$

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