Cauchy积分公式的一个推广形式
复分析中最基本的结果当属Cauchy积分公式了,即若$D$是由可求长简单闭曲线$\gamma$围成的区域,并且$f\in H(D)\cap C(\overline{D})$,则$\forall z\in D$有$$f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta$$
设$\gamma_0,\gamma_1,\cdots,\gamma_n$是$n+1$条可求长简单闭曲线,$\gamma_1,\cdots,\gamma_n$都在$\gamma_0$内部,而$\gamma_1,\cdots,\gamma_n$中每一条都在其他$n-1$条的外部,而$D$是由这$n+1$条曲线围成的区域,用$\gamma$记$D$的边界$\partial D$,如果$f\in C^1(\overline{D})$,那么对任意的$z\in D$有$$f(z)=\frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta+\frac{1}{2\pi i}\int_{D}\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\cdot\frac{1}{\zeta-z}{\rm d}\zeta\wedge{\rm d}\overline{\zeta}$$
证明 不妨设$D$为单连通区域（无本质区别）,在点$z$附近取圆盘$B_r=B(z,r)$,如果记一次外微分形式$\omega=\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta$,那么根据Stokes公式$$\frac{1}{2\pi i}\int_{\partial D+\left(\partial B_r\right)^{-}}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta=\frac{1}{2\pi i}\int_{D\setminus \overline{B_r}}{\rm d}\omega=-\frac{1}{2\pi i}\int_{D\setminus\overline{B_r}}\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\cdot\frac{1}{\zeta-z}{\rm d}\zeta\wedge{\rm d}\overline{\zeta}$$这说明$$\frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta=-\frac{1}{2\pi i}\int_{D\setminus\overline{B_r}}\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\cdot\frac{1}{\zeta-z}{\rm d}\zeta\wedge{\rm d}\overline{\zeta}+\frac{1}{2\pi i}\int_{\partial B_r}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta\tag{1}$$
仅需注意到\begin{align*}\frac{1}{2\pi i}\int_{\partial B_r}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta&=\frac{1}{2\pi}\int_{0}^{2\pi}f(z+re^{i\theta}){\rm d}\theta\to f(z),(r\to 0)\end{align*}
此外由于$\frac{\partial f(\zeta)}{\partial\overline{\zeta}}$在$\overline{B_r}$上连续,从而存在$M>0$使得$$\left|\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\right|\leq M,\forall \zeta\in \overline{B_r}$$于是$$\left|\frac{1}{2\pi}\int_{\overline{B_r}}\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\cdot\frac{1}{\zeta-z}{\rm d}\zeta\wedge{\rm d}\overline{\zeta}\right|\leq\frac{M}{\pi}\int_{\overline{B_r}}\frac{1}{|\zeta-z|}{\rm d}\sigma=rM\to 0,(r\to0)$$
这样在(1)式两端令$r\to 0$便可得到欲证等式.
