復(fù)分析中最基本的結(jié)果當(dāng)屬Cauchy積分公式了,即若$D$是由可求長(zhǎng)簡(jiǎn)單閉曲線$\gamma$圍成的區(qū)域,并且$f\in H(D)\cap C(\overline{D})$,則$\forall z\in D$有$$f(z)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta$$

設(shè)$\gamma_0,\gamma_1,\cdots,\gamma_n$是$n+1$條可求長(zhǎng)簡(jiǎn)單閉曲線,$\gamma_1,\cdots,\gamma_n$都在$\gamma_0$內(nèi)部,而$\gamma_1,\cdots,\gamma_n$中每一條都在其他$n-1$條的外部,而$D$是由這$n+1$條曲線圍成的區(qū)域,用$\gamma$記$D$的邊界$\partial D$,如果$f\in C^1(\overline{D})$,那么對(duì)任意的$z\in D$有$$f(z)=\frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta+\frac{1}{2\pi i}\int_{D}\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\cdot\frac{1}{\zeta-z}{\rm d}\zeta\wedge{\rm d}\overline{\zeta}$$

證明   不妨設(shè)$D$為單連通區(qū)域（無(wú)本質(zhì)區(qū)別）,在點(diǎn)$z$附近取圓盤(pán)$B_r=B(z,r)$,如果記一次外微分形式$\omega=\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta$,那么根據(jù)Stokes公式$$\frac{1}{2\pi i}\int_{\partial D+\left(\partial B_r\right)^{-}}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta=\frac{1}{2\pi i}\int_{D\setminus \overline{B_r}}{\rm d}\omega=-\frac{1}{2\pi i}\int_{D\setminus\overline{B_r}}\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\cdot\frac{1}{\zeta-z}{\rm d}\zeta\wedge{\rm d}\overline{\zeta}$$這說(shuō)明$$\frac{1}{2\pi i}\int_{\partial D}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta=-\frac{1}{2\pi i}\int_{D\setminus\overline{B_r}}\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\cdot\frac{1}{\zeta-z}{\rm d}\zeta\wedge{\rm d}\overline{\zeta}+\frac{1}{2\pi i}\int_{\partial B_r}\frac{f(\zeta)}{\zeta-z}{\rm d}\zeta\tag{1}$$

僅需注意到\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$\overline{B_r}$上連續(xù),從而存在$M>0$使得$$\left|\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\right|\leq M,\forall \zeta\in \overline{B_r}$$于是$$\left|\frac{1}{2\pi}\int_{\overline{B_r}}\frac{\partial f(\zeta)}{\partial\overline{\zeta}}\cdot\frac{1}{\zeta-z}{\rm d}\zeta\wedge{\rm d}\overline{\zeta}\right|\leq\frac{M}{\pi}\int_{\overline{B_r}}\frac{1}{|\zeta-z|}{\rm d}\sigma=rM\to 0,(r\to0)$$

這樣在(1)式兩端令$r\to 0$便可得到欲證等式.