复值函数的积分是这样定义的.设有向曲线$gamma:z=z(t),tin[alpha,eta]$,并且$a=z(alpha)$为起点,$b=z(eta)$为终点.现沿着$gamma$方向任取分点
[a=a_{0},a_{1},cdots,a_{n}=b]
考虑和式
$$S_{n}=sum_{i=1}^{n}f(xi_{k})Delta_{k},xi_{k}in[a_{k-1},a_{k}]$$
当分点无限增多,而
$$max{Delta_{k}} o0$$
时,如果$limlimits_{n oinfty}S_{n}$存在且等于$I$,称$f(z)$沿$gamma$可积,记做
$$I=int_{gamma}f(z){ m d}z.$$
如果设$f(z)=u(x,y)+iv(x,y)$,则
[int_{gamma}f(z){ m d}z=int_{gamma}(u+iv)({ m d}x+i{ m d}y)]
[=int_{gamma}u{ m d}x-v{ m d}y+iint_{gamma}v{ m d}x+u{ m d}y]
如同实函数一样,同样的我们用外微分形式的观点来看复值函数的积分.为此我们将给定函数$f(z)$视作$z,overline{z}$的函数且二者独立.那么
[{ m d}overline{z}wedge{ m d}z=({ m d}x-i{ m d}y)wedge({ m d}x+i{ m d}y)]
[=2i{ m d}sigma]
这里${ m d}sigma$为二维的面积元素.并且我们再定义算子
[partial f=frac{partial f}{partial z}{ m d}z,overline{partial}f=frac{partial f}{partialoverline{z}}{ m d}overline{z}]
并且设$omega$是一个复的外微分形式,定义
[{ m d}omega=partialomega+overline{partial}omega]
显然${ m dd}omegaequiv0$.据此我们便可得到复形式的Green公式:设$omega=omega_{1}{ m d}z+omega_{2}{ m d}overline{z}$为区域$Omega$上的一次外微分形式,则
[int_{partialOmega}omega=int_{Omega}{ m d}omega]
他的证明也是容易的,只需设$omega_{1}=f_{1}+ig_{1},omega_{2}=f_{2}+ig_{2}$,其中$f,g$都是实函数,只需计算${ m d}omega$即可.
而由Green公式可得到所谓的Pompeiu公式：若$Omegasubsetmathbb C$为有界区域,且有$C^1$边界条件,设$f(z)in C^1(overline{Omega})$,那么我们有
[f(z)=frac{1}{2pi i}int_{partialOmega}frac{f(zeta)}{zeta-z}{ m d}zeta-frac{1}{pi}int_{Omega}frac{partial f(zeta)}{partialoverline{zeta}}cdotfrac{{ m d}sigma}{zeta-z}]
我们来证明一下,取$z$的邻域$B(z,varepsilon)subset Omega$,据Green公式有
[int_{partialOmega-partial B(z,varepsilon)}frac{f(zeta)}{zeta-z}{ m d}zeta=int_{Omegasetminus B(z,varepsilon)}{ m d}left(frac{f(zeta)}{zeta-z}{ m d}zeta ight)]
我们来计算外微分形式${ m d}left(frac{f(zeta)}{zeta-z}{ m d}zeta ight)$,由于
[{ m d}left(frac{f(zeta)}{zeta-z}{ m d}zeta ight)=frac{partial}{partialoverline{zeta}}left(frac{f(zeta)}{zeta-z} ight){ m d}overline{zeta}wedge{ m d}zeta]
[=frac{partial f}{partialoverline{zeta}}cdotfrac{{ m d}overline{zeta}wedge{ m d}zeta}{zeta-z}+f(zeta)frac{partial}{partialoverline{zeta}}left(frac{1}{zeta-z} ight){ m d}overline{zeta}wedge{ m d}zeta]
注意到函数$g(zeta)=frac{1}{zeta-z}$在$Omegasetminus B(z,varepsilon)$内全纯,从而
[frac{partial g}{partialoverline{zeta}}=0]
因此
[int_{partialOmega}frac{f(zeta)}{zeta-z}{ m d}zeta=int_{partial B(z,varepsilon)}frac{f(zeta)}{zeta-z}{ m d}zeta+2iint_{Omegasetminus B(z,varepsilon)}frac{partial f}{partialoverline{zeta}}cdotfrac{{ m d}sigma}{zeta-z}]
[=int_{partial B(z,varepsilon)}frac{f(zeta)-f(z)}{zeta-z}{ m d}zeta+f(z)int_{partial B(z,varepsilon)}frac{{ m d}zeta}{zeta-z}+2iint_{Omegasetminus B(z,varepsilon)}frac{partial f}{partialoverline{zeta}}cdotfrac{{ m d}sigma}{zeta-z}]
(记上式为"*")先来看第一项,注意到$f(z)in C^1(overline{Omega})$,设
[f(z)=u(x,y)+iv(x,y)]
则$u,v$均在$overline{Omega}$上具有一阶连续偏导数,再设$zeta=x_{1}+iy_{1},z=a+ib$,则
[left|frac{f(zeta)-f(z)}{zeta-z} ight|=frac{1}{|zeta-z|}cdotleft|left.frac{partial u}{partial x} ight|_{(xi_{1},y_{1})}(x_{1}-a)+left.frac{partial u}{partial y} ight|_{(a,eta_{1})}(y_{1}-b)+ileft.frac{partial v}{partial x} ight|_{(xi_{2},y_{1})}(x_{1}-a)+ileft.frac{partial v}{partial y} ight|_{(a,eta_{2})}(y_{1}-b) ight|]
[leqleft|left.frac{partial u}{partial x} ight|_{(xi_{1},y)} ight|+cdotsleq M]
因此
[left|int_{partial B(zvarepsilon)}frac{f(zeta)-f(z)}{zeta-z}{ m d}zeta ight|leq 2Mpivarepsilon]
而第二项
[f(z)int_{partial B(z,varepsilon)}frac{{ m d}zeta}{zeta-z}=f(z)int_{0}^{2pi}frac{ie^{i heta}}{e^{i heta}}{ m d} heta]
[=2pi if(z)]
这样在"*"式中令$varepsilon o0^+$(此时$Omegasetminus B(z,varepsilon) oOmega$)即得欲证等式.
特别的,如果$f(z)$还在$Omega$内全纯,那么我们有所谓的Cauchy积分公式
[f(z)=frac{1}{2pi i}int_{partialOmega}frac{f(zeta)}{zeta-z}{ m d}zeta.]
这是因为如果$f(z)$全纯,那么
[frac{partial f}{partialoverline{zeta}}=0]
根据Pompeiu公式即可得到.
而由Cauchy积分公式即可得到Cauchy积分定理：设$Omegainmathbb C$为有界区域且有$C^1$边界条件,$f(z)$在$Omega$内全纯,且$f(z)in C^1(overline{Omega})$,那么我们有
[int_{partialOmega}f(zeta){ m d}zeta=0]
证明是简单的,任取$z_{0}inOmega$,则函数$g(zeta)=(zeta-z_{0})f(zeta)$在$Omega$内全纯,且$g(zeta)in C^1(overline{Omega})$,据Cauchy积分公式即得结果.
反过来,我们也可以用Cauchy积分定理得出Cauchy积分公式.简单证明一下
[int_{partialOmega}frac{f(zeta)}{zeta-z}{ m d}zeta=int_{partialOmega-partial B(z,varepsilon)}frac{f(zeta)}{zeta-z}{ m d}zeta+int_{partial B(z,varepsilon)}frac{f(zeta)}{zeta-z}{ m d}zeta]
据Cauchy积分定理知上式第一项为零,而第二项我们再证明Pompeiu公式时已经计算过,综合一下即可得到Cauchy积分公式.
所以说Cauchy积分公式和Cauchy积分定理是等价的.