04 - Fundamental Theorem of Calculus

Fundamental Theorem of Calculus

Let $U\subseteq \mathbb{C}$ be open
Let $f:U\to \mathbb{C}$ be an continuous function
Let $\gamma :[a,b]\to U$ be a piecewise $C^1$ path in $U$

If $F:U\to \mathbb{C}$ is a primitive for $f$ then

$${\int }_{\gamma }f(z)dz=F(\gamma (b))-F(\gamma (a))$$

If $\gamma$ is a closed path then the integral is then $0$

Proof

Suppose that $\gamma$ is $C^1$ then

$$\begin{array}{rl}{\int }_{\gamma }f(z)dz& ={\int }_{\gamma }{F}^{\prime }(z)dz\\ & ={\int }_a^b{F}^{\prime }(\gamma (t)){\gamma }^{\prime }(t)dt(\text{by chain rule})\\ & ={\int }_a^b\frac{d}{dt}(F\circ \gamma )(t)dt\\ & =F(\gamma (b))-F(\gamma (a))\end{array}$$

However if $\gamma$ is piecewise $C^1$ then there exists
Partition $a=a_0<a_1<\cdots <a_k=b$ such that $\gamma$ is $C^1$ on $[a_i,a_{i+1}]$ for $i=0,1,\cdots ,k-1$
Hence obtain telescoping sum

$$\begin{array}{rl}{\int }_{\gamma }f(z)dz& ={\int }_a^{b}f(\gamma (t)){\gamma }^{\prime }(t)dt\\ & =\sum _{i=0}^{k-1}{\int }_{a_i}^{a_{i+1}}f(\gamma (t)){\gamma }^{\prime }(t)dt\\ & =\sum _{i=0}^{k-1}(F(\gamma (a_{i+1}))-F(\gamma (a_i)))\\ & =F(\gamma (b))-F(\gamma (a))\end{array}$$

If $\gamma$ is closed $\gamma (a)=\gamma (b)$ hence the integral of $f$ along the closed path is $0$