Integration for holomorphic function: a homotopy view

Let ${\gamma }_1$$\gamma_1$ and ${\gamma }_2$$\gamma_2$ be two paths in $\mathbb{C}$$\mathbb C$, which connect $a$$a$ and $b$$b$. $H(t,s)$$H(t,s)$ is $C^1$$C^1$-homotopy between ${\gamma }_1$$\gamma_1$ and ${\gamma }_2$$\gamma_2$. $f(z)$$f(z)$ is a holomorphic function.

Then we have

$$\begin{array}{}\text{(1)}& {\int }_{{\gamma }_1}f(z)dz={\int }_{{\gamma }_2}f(z)dz\end{array}$$

Proof

Let

$$\begin{array}{}\text{(2)}& I(t)={\int }_{H(t,s)}f(z)dz\end{array}$$

What we need to prove is

$$\begin{array}{}\text{(3)}& \frac{d}{dt}I(t)=0\end{array}$$

i.e.

$$\begin{array}{}\text{(4)}& \frac{d}{dt}I(t)=\frac{d}{dt}{\int }_a^{b}f(H)\frac{\partial H}{\partial s}ds={\int }_a^b\frac{d}{dt}f(H)\frac{\partial H}{\partial s}ds={\int }_a^b{f}^{\prime }(H)\frac{\partial H}{\partial t}\frac{\partial H}{\partial s}+f(H)\frac{{\partial }^{2}H}{\partial t\partial s}ds\end{array}$$

It is not hard to see that $S_2$$S_2$ is the automorphism of this integration.

i.e. if you change the order of $s,t$$s,t$, the integration will not change.

Therefore we have

$$\begin{array}{}\text{(5)}& {\int }_a^b\frac{d}{dt}f(H)\frac{\partial H}{\partial s}ds={\int }_a^b\frac{d}{ds}f(H)\frac{\partial H}{\partial t}ds\end{array}$$

i.e.

$$\begin{array}{}\text{(6)}& \frac{d}{dt}I(t)=f(H)\frac{\partial H}{\partial t}{|}_{s=a}^{s=b}\end{array}$$

But $H(t,b)$$H(t,b)$ and $H(t,a)$$H(t,a)$ is constant respect to $t$$t$.

Thus

$$\begin{array}{}\text{(7)}& \frac{d}{dt}I(t)=0\end{array}$$

Then we can prove Cauchy integral formula.(The previous one is not a proof, since we can not see holomorphic means analytic without Cauchy integral formula!,but it could gives you some idea and understanding)

$$\begin{array}{}\text{(10)}& {\oint }_{\gamma }\frac{1}{z-a}dz=2\pi i\end{array}$$

Thus

$$\begin{array}{}\text{(11)}& \frac{1}{2\pi i}({\oint }_{\gamma }\frac{f(z)}{(z-a)}dz-f(a))=\frac{1}{2\pi i}({\oint }_{\gamma }\frac{f(z)}{(z-a)}-\frac{f(a)}{(z-a)}dz)=\frac{1}{2\pi i}\left({\oint }_{\gamma }\frac{f(z)-f(a)}{(z-a)}dz\right)\end{array}$$

Let $\gamma =a+ϵe^{it}$$\gamma=a+\epsilon e^{i t}$.

Then

$$\begin{array}{}\text{(12)}& \frac{1}{2\pi i}\left({\oint }_{\gamma }\frac{f(z)-f(a)}{(z-a)}dz\right)=\frac{1}{2\pi i}\left({\int }_0^{2\pi }\frac{f(a+ϵe^{it})-f(a)}{ϵe^{it}}ϵi{e}^{it}dt\right)=\frac{1}{2\pi }({\int }_0^{2\pi }f(a+ϵe^{it})-f(a))\end{array}$$

Since it is homotopy invarience, we could let $ϵ⟶0$$\epsilon\longrightarrow 0$.

Thus

$$\begin{array}{}\text{(13)}& \begin{array}{c}\frac{1}{2\pi }({\int }_0^{2\pi }f(a+ϵe^{it})-f(a))=0\end{array}\end{array}$$

i.e.

$$\begin{array}{}\text{(14)}& f(z)=\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(\xi )}{\xi -z}d\xi \end{array}$$

$$\begin{array}{}\text{(15)}& f(z)=\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(\xi )}{\xi -z}d\xi =\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(\xi )}{(\xi -a)-(z-a)}d\xi =\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(\xi )}{\xi -a}\frac{1}{1-\frac{z-a}{\xi -a}}d\xi \end{array}$$

$$\begin{array}{}\text{(16)}& \frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(\xi )}{\xi -a}\frac{1}{1-\frac{z-a}{\xi -a}}d\xi =\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(\xi )}{\xi -a}\sum _{n=0}^{\mathrm{\infty }}{\left(\frac{z-a}{\xi -a}\right)}^{n}d\xi =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(\xi )}{(\xi -a{)}^{n+1}}d\xi (z-a{)}^n\end{array}$$

i.e.

$$\begin{array}{}\text{(17)}& f(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(\xi )}{(\xi -a{)}^{n+1}}d\xi (z-a{)}^n\end{array}$$

Thus holomorphic function is analytic as well.

The integration can be viewed as a dual basis as well,

$$\begin{array}{}\text{(18)}& \frac{1}{2\pi i}{\oint }_{\gamma }\frac{(\xi -a{)}^i}{(\xi -a{)}^{j+1)}}={\delta }_{ij}\end{array}$$

thus we have

$$\begin{array}{}\text{(19)}& \frac{D^n}{n!}=\frac{1}{2\pi i}{\oint }_{\gamma }\frac{(-)}{(\xi -a{)}^{n+1)}}\end{array}$$