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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)\,


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


\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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