# 调和函数的积分表示，泊松（Poison）积分公式

\begin{align*}f(z_0)&=\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)}{z-z_0}dz\\ &=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(Re^{i\theta})}{Re^{i\theta}-re^{i\theta}}iRe^{i\theta}d\theta\\ &=\frac{1}{2\pi}\int_0^{2\pi}\frac{f(Re^{i\theta})Re^{i\theta}}{Re^{i\theta}-re^{i\theta}}d\theta\end{align*}

\begin{align*}0&=\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)}{z-z^*}dz\\&=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(Re^{i\theta})\cdot iRe^{i\theta}}{Re^{i\theta}-\frac{R^2}{re^{-i\varphi}}}d\theta\\ &=\frac{1}{2\pi}\int_0^{2\pi}\frac{f(Re^{i\theta})re^{i\theta}}{re^{i\theta}-Re^{i\varphi}}d\theta\end{align*}

\begin{align*}f(z_0)&=\frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\left[\frac{Re^{i\theta}}{Re^{i\theta}-re^{i\theta}}-\frac{re^{i\theta}}{re^{i\theta}-Re^{i\varphi}}\right]d\theta\\&=\frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\frac{Re^{i\theta}(re^{i\theta}-Re^{i\varphi})-re^{i\theta}(Re^{i\theta}-re^{i\theta})}{(Re^{i\theta}-re^{i\theta})(re^{i\theta}-Re^{i\varphi})}\\&=\frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\frac{Rre^{2i\theta}-R^2e^{i(\theta+\varphi)}-Rre^{2i\theta}+r^2e^{i(\theta+\varphi)}}{Rre^{2i\theta}-r^2e^{i(\theta+\varphi)}-R^2e^{i(\theta+\varphi)}+Rre^{2i\varphi}}d\theta\\ &=\frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\frac{(r^2-R^2)e^{i(\theta+\varphi)}}{Rr(e^{2i\theta}+e^{2i\varphi})-(R^2+r^2)e^{i(\theta+\varphi)}}d\theta\\&=\frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\frac{R^2-r^2}{R^2+r^2-Rr(e^{2i\theta}+e^{2i\varphi})e^{-i(\theta+\varphi)}}d\theta\\ &=\frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\frac{R^2-r^2}{R^2+r^2-Rr(e^{i(\theta-\varphi)}+e^{-i(\theta-\varphi)})}d\theta\\&=\frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\frac{R^2-r^2}{R^2+r^2-2Rr\cos(\theta-\varphi)}d\theta\end{align*}