向量分析

Suppose L is a curve on the complex plain. Take dividing points $z_0,z_1\cdots z_n$, where $z_0$ is the start of L and $z_n$ is the end. Let $f:\mathbb{C}\to\mathbb{R}^n,g:\mathbb{C}\to\mathbb{C}.$

definition Let \(F(z)= \left[ \begin{array}{c} F_1(z )\\ F_2(z )\\\cdots \end{array} \right].\) Define

\[F'(z_i)=\frac{\mathrm d F}{\mathrm dx}=\lim_{\lambda\to 0}\frac{F(z_{i+1})-F(z_i)}{z_{i+1}-z_i}= \left[ \begin{array}{c} F_1'(z )\\ F_2'(z )\\\cdots \end{array} \right].\] \[\int_{L}f(z)\mathrm dz =\lim_{\lambda\to 0}\sum_{i=0}^{n-1}f(z_i)(z_{i+1}-z_i).\]

Let \(\displaystyle{f(z)= \left[ \begin{array}{c} z \\ z ^{ 2 } \end{array} \right],g(z)=z }\) , we have

\[\displaystyle \begin{align*} \int_{L}f(z)\mathrm dz =&\int_{L}\left[ \begin{array}{c} z \\ z ^{ 2 } \end{array} \right]\mathrm dz\\ =&\lim_{\lambda\to 0}\sum_{i=0}^{n-1}\left[ \begin{array}{c} z_i \\ z_i ^{ 2 } \end{array} \right](z_{i+1}-z_i)\\ =&\left[ \begin{array}{c}\lim\limits_{\lambda\to 0}\sum\limits_{i=0}^{n-1} z_i (z_{i+1}-z_i)\\ \lim\limits_{\lambda\to 0}\sum\limits_{i=0}^{n-1}z_i ^{ 2 }(z_{i+1}-z_i) \end{array} \right]\\ =&\left[ \begin{array}{c} \int_{L}z\mathrm dz \\ \int_{L}z ^{ 2 } \mathrm dz\end{array} \right].\\ \end{align*}\]

Generally, we have

theorem 1 Let \(\displaystyle{f(z)= \left[ \begin{array}{c} f_1(z )\\ f_2(z )\\\cdots \end{array} \right]}\) . Then

\[\displaystyle \begin{align*} \int_{L}f(z)\mathrm dz =&\left[ \begin{array}{c} \int_{L} f_1(z )\mathrm dz \\ \int_{L} f_2(z )\mathrm dz\\\cdots\end{array} \right].\\ \end{align*}\tag1\]

which implies the Newton-Leibniz formula.

theorem 2 (Newton-Leibniz formula) let \(\displaystyle{f(z)= \left[ \begin{array}{c} f_1(z )\\ f_2(z )\\\cdots \end{array} \right]}\) , $F’(z)=f(z)$. then

\[\displaystyle \begin{align*} \int_{L}f(z)\mathrm dz =&F(b)-F(a).\\ \end{align*}\tag2\]

proof 1 From theorem 1 we have

\[\displaystyle \begin{align*} \int_{L}f(z)\mathrm dz =&\left[ \begin{array}{c} \int_{L} f_1(z )\mathrm dz \\ \int_{L} f_2(z )\mathrm dz\\\cdots\end{array} \right]\\ =&\left[ \begin{array}{c} F_1(b)-F_1(a) \\ F_2(b)-F_2(a)\\\cdots\end{array} \right]\tag{Newton-Leibniz}\\ =&F(b)-F(a). \end{align*}\]

proof 2

\[\begin{align*} \int_{L}f(z)\mathrm dz =&\lim_{\lambda\to 0}\sum_{i=0}^{n-1}\frac{F(z_{i+1})-F(z_i)}{z_{i+1}-z_i}(z_{i+1}-z_i)\\ =&\lim_{\lambda\to 0}\sum_{i=0}^{n-1}F(z_{i+1})-F(z_i)\\ =&F(z_{n})-F(z_0)\\ \end{align*}\]

example Let \(\displaystyle{f(z)= \left[ \begin{array}{c} z \\ z ^{ 2 } \end{array} \right],g(z)=z }\) . Then

\[\displaystyle \begin{align*} \int_{0}^1f(z)\mathrm dz =&F(1)-F(0)\\ =&\left[ \begin{array}{c} \frac{ 1 }{ 2 } \\ \frac{ 1 }{ 3 }\end{array} \right]-\left[ \begin{array}{c} 0 \\0\end{array} \right]\\ =&\left[ \begin{array}{c} \frac{ 1 }{ 2 } \\ \frac{ 1 }{ 3 }\end{array} \right].\\ \end{align*}\]