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$fin H(D)cap C(overline{D})$,则$forall zin D$有$$f(z)=frac{1}{2pi i}int_{gamma}frac{f(zeta)}{zeta-z}{ m d}zeta$$
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$fin C^1(overline{D})$,那么对任意的$zin D$有$$f(z)=frac{1}{2pi i}int_{partial D}frac{f(zeta)}{zeta-z}{ m d}zeta+frac{1}{2pi i}int_{D}frac{partial f(zeta)}{partialoverline{zeta}}cdotfrac{1}{zeta-z}{ m d}zetawedge{ m d}overline{zeta}$$
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$omega=frac{f(zeta)}{zeta-z}{ m d}zeta$,那么根据Stokes公式$$frac{1}{2pi i}int_{partial D+left(partial B_r ight)^{-}}frac{f(zeta)}{zeta-z}{ m d}zeta=frac{1}{2pi i}int_{Dsetminus overline{B_r}}{ m d}omega=-frac{1}{2pi i}int_{Dsetminusoverline{B_r}}frac{partial f(zeta)}{partialoverline{zeta}}cdotfrac{1}{zeta-z}{ m d}zetawedge{ m d}overline{zeta}$$这说明$$frac{1}{2pi i}int_{partial D}frac{f(zeta)}{zeta-z}{ m d}zeta=-frac{1}{2pi i}int_{Dsetminusoverline{B_r}}frac{partial f(zeta)}{partialoverline{zeta}}cdotfrac{1}{zeta-z}{ m d}zetawedge{ m d}overline{zeta}+frac{1}{2pi i}int_{partial B_r}frac{f(zeta)}{zeta-z}{ m d}zeta ag{1}$$
仅需注意到egin{align*}frac{1}{2pi i}int_{partial B_r}frac{f(zeta)}{zeta-z}{ m d}zeta&=frac{1}{2pi}int_{0}^{2pi}f(z+re^{i heta}){ m d} heta o f(z),(r o 0)end{align*}
此外由于$frac{partial f(zeta)}{partialoverline{zeta}}$在$overline{B_r}$上连续,从而存在$M>0$使得$$left|frac{partial f(zeta)}{partialoverline{zeta}} ight|leq M,forall zetain overline{B_r}$$于是$$left|frac{1}{2pi}int_{overline{B_r}}frac{partial f(zeta)}{partialoverline{zeta}}cdotfrac{1}{zeta-z}{ m d}zetawedge{ m d}overline{zeta} ight|leqfrac{M}{pi}int_{overline{B_r}}frac{1}{|zeta-z|}{ m d}sigma=rM o 0,(r o0)$$
这样在(1)式两端令$r o 0$便可得到欲证等式.