The Fundamental Theorem of Calculus

 

Def 1

Let $\Omega$ be an area and $F : \Omega \to \mathbb{C}$ be differentiable on $\Omega$

If $F' = f$,  $F$ is called primitive function of $f$

 

 

 

Thm 1

Let $\Omega$ be an area  and  $f : \Omega \to \mathbb{C}$ be continuous on $\Omega$

Let $F$ be a primitive function of $f$  and  $\Gamma : [a, b] \to \Omega$ be a curve. Then

$\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = F(\Gamma (b)) - F(\Gamma (a))$

 

  Since $\dfrac{\mathrm{d} }{\mathrm{d} t} F(\Gamma (t)) = F'(\Gamma(t)) \Gamma'(t)$,

  $\displaystyle \int_{\Gamma} f \; \mathrm{d}z = \int_{\Gamma} F' \; \mathrm{d}z$
                  $= \displaystyle \int_{a}^{b} F'(\Gamma(t)) \Gamma'(t) \; \mathrm{d}t$
                  $= \displaystyle \int_{a}^{b} \frac{\mathrm{d} }{\mathrm{d} t} F(\Gamma (t)) \; \mathrm{d}t$
                  $= F(\Gamma(b)) - F(\Gamma(a)) $

 

 

 

Def 2

Let $\Gamma : [a,b] \to \mathbb{C}$ be a curve such that $\Gamma(a) = \Gamma(b)$ .

$\Gamma$ is called closed curve. If $\Gamma$ is 1-1 on $[a, b)$, it is called simple closed curve.

 

# When performing a line integral along a simple closed curve, the curve is assumed to be oriented counterclockwise.

 

 

 

Cor 1

Let $\Gamma$ be a closed curve. Suppose the primitive of $f$ exists. Then

$\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = 0$

 

 

 

Cor 2

Let $\Gamma$ be a closed curve. Then

$\displaystyle \int_{\Gamma} z^{n} \; \mathrm{d}z = 0$     ( $n \in \mathbb{Z} \setminus \left\{-1 \right\}$ )

 

 

 

Cor 3

Let $\Gamma$ be a closed curve and $f(z)$ be power series. Then

$\displaystyle \int_{\Gamma} f(z) \; \mathrm{d}z = 0$

 

Thm 2

Let $z_{0} \in \mathbb{C}$ . For $r \in \mathbb{R}$, $C_{r}$ : $\left|z - z_{0} \right| = r $ . Then

$\displaystyle \int_{C_{r}} \dfrac{1}{z - z_{0}} \; \mathrm{d}z = 2 \pi i$ 

 

  Let $C_{r}(t) = e^{it} + z_{0}$  ( $0 \leq t \leq 2\pi$ )

  $\displaystyle \int_{C_{r}} \dfrac{1}{z - z_{0}} \; \mathrm{d}z = \int_{0}^{2\pi} \dfrac{1}{e^{it}} i e^{it} \; \mathrm{d}t = 2 \pi i$

 

 

 

Def 3

Let $\Gamma : [a, b] \to \mathbb{C}$ be a curve and $\Gamma(t) = x(t) + i y(t)$ .

The length of the curve is defined as follows.

$\displaystyle \int_{a}^{b} \sqrt{x'(t)^{2} + y'(t)^{2}} \; \mathrm{d}t = \int_{a}^{b} \left|x'(t) + iy'(t) \right| \; \mathrm{d}t = \int_{a}^{b} \left|\Gamma'(t) \right| \; \mathrm{d}t = \int_{\Gamma} \left|\mathrm{d}z \right| $

( We write $\left|\mathrm{d}z \right| = \left|\Gamma'(t) \right| \mathrm{d}t$ )

 

 

 

Thm 3

Let $\Omega$ be an area,  $\Gamma : [a, b] \to \Omega$  and  $f : \Omega \to \mathbb{C}$

Suppose length of $\Gamma$ is $L$  and  $\left|f(z) \right| \leq M$ on $\Gamma$ . Then

$\displaystyle \left|\int_{\Gamma} f(z) \; \mathrm{d}z \right| \leq \int_{\Gamma} \left|f(z) \right| \left|\mathrm{d}z \right| \leq ML$

 

  $\displaystyle \left|\int_{\Gamma} f(z) \; \mathrm{d}z \right| = \left|\int_{a}^{b} f(\Gamma(t)) \Gamma'(t) \; \mathrm{d}t \right|$
                            $\leq \displaystyle \int_{a}^{b} \left|f(\Gamma(t)) \right| \left|\Gamma'(t) \right| \; \mathrm{d}t$
                            $= \displaystyle \int_{\Gamma} \left|f(z) \right| \left|\mathrm{d}z \right|$
                            $\leq \displaystyle M \int_{\Gamma} \left|\mathrm{d}z \right|$

                            $= ML$