Complex Analysis/Curve Integral

Note - Integrability

• ${\displaystyle f(\gamma (t_{i}))\cdot {\big (}\gamma (t_{i})-\gamma (t_{i-1}){\big )}}$ entspricht in reaL Analysis it corresponds to be oriented area of rectangles that approximate yhe integral in the Reimann integral.
• ${\displaystyle \sum _{i=1}^{n}f(\gamma (t_{i}))\cdot {\big (}\gamma (t_{i})-\gamma (t_{i-1}){\big )}}$ calculate the Complex numbersthe Riemann Sum of all individual terms a decomposition of the interval ${\displaystyle [a,b]}$ with ${\displaystyle a=t_{0}<\ldots <t_{n}=b}$
• ${\displaystyle \left|I-\sum _{i=1}^{n}f(\gamma (t_{i}))\cdot {\big (}\gamma (t_{i})-\gamma (t_{i-1}){\big )}\right|<\epsilon }$ means, that the Integral ${\displaystyle I}$ can be approximate with arbitrary precision using Riemann Sum.

Integration over chains

If ${\displaystyle \Gamma =\sum _{i=1}^{n}n_{i}\cdot \gamma _{i}}$ is a chain in ${\displaystyle \mathbb {C} }$, then a function ${\displaystyle f\colon \mathrm {Spur} (\Gamma )\to \mathbb {C} }$ is said to be integrable over ${\displaystyle \Gamma }$ if it is integrable over each ${\displaystyle \gamma _{i}}$, and we set

${\displaystyle \int _{\Gamma }f(z)\,dz:=\sum _{i=1}^{n}n_{i}\cdot \int _{\gamma _{i}}f(z)\,dz}$

Connection to real integration

If ${\displaystyle \gamma }$ is even piecewise differentiable, then the curve integral can be reduced to an integral over the parameter domain using the Mean Value Theorem, and we have in this case

${\displaystyle \int _{\gamma }f(z)\,dz=\int _{a}^{b}f(\gamma (t))\cdot \gamma '(t)\,dt}$

where a complex-valued function is integrated over a real interval, with the real and imaginary parts calculated separately.

Representation of the integral

Both examples together give us

${\displaystyle \int _{\gamma }z^{n}\,dz=2\pi i\cdot \delta _{n,-1}}$

where

$${\displaystyle {\begin{array}{rrcl}\delta _{n,-1}:&\mathbb {N} &\rightarrow &\{0,1\}\\&n&\mapsto &\delta _{n,-1}(n)={\begin{cases}1&,&n=-1\\0&,&n\not =-1\\\end{cases}}\end{array}}}$$

This fact plays an important role in the definition of the Residue and the proof of the Complex Analysis/Residue Theorem

Independence of parameterization

Let ${\displaystyle \gamma \colon [a,b]\to \mathbb {C} }$ be a piecewise ${\displaystyle C^{1}}$-path, ${\displaystyle \phi \colon [\alpha ,\beta ]\to [a,b]}$ a ${\displaystyle C^{1}}$-diffeomorphism that preserves orientation. Then ${\displaystyle \gamma \circ \phi \colon [\alpha ,\beta ]\to \mathbb {C} }$ is a piecewise ${\displaystyle C^{1}}$-path and we have

${\displaystyle \int _{\gamma }f(z)\,dz=\int _{\gamma \circ \phi }f(z)\,dz}$

i.e. the value of the integral is independent of the chosen parameterization of the path.

Proof

It is

${\displaystyle {\begin{array}{rl}\displaystyle \int _{\gamma \circ \phi }f(z)\,dz&=\displaystyle \int _{\alpha }^{\beta }f(\gamma (\phi (s)))(\gamma \circ \phi )'(s)\,ds\\&=\displaystyle \int _{\alpha }^{\beta }f(\gamma (\phi (s)))\gamma '(\phi (s))\phi '(s)\,ds\\&=\displaystyle \int _{\phi (\alpha )}^{\phi (\beta )}f(\gamma (t))\gamma '(t)\,dt\quad {\rm {Substitution}}\ t=\phi (s)\\&=\displaystyle \int _{a}^{b}f(\gamma (t))\gamma '(t)\,dt\\&=\int _{\gamma }f(z)\,dz\end{array}}}$

Linearity

Since the integral is defined over linear combinations of ${\displaystyle f}$, it is itself linear in the integrand, i.e. we have

${\displaystyle \int _{\gamma }(\alpha f+\beta g)(z)\,dz=\alpha \int _{\gamma }f(z)\,dz+\beta \int _{\gamma }g(z)\,dz}$

for rectifiable ${\displaystyle \gamma }$, ${\displaystyle \alpha ,\beta \in \mathbb {C} }$ and integrable ${\displaystyle f,g\colon \mathrm {spur} (\gamma )\to \mathbb {C} }$.

Orientation reversal

Let ${\displaystyle \gamma \colon [a,b]\to \mathbb {C} }$ be a rectifiable path, and let ${\displaystyle \gamma ^{-}\colon [a,b]\to \mathbb {C} }$ be the reversed path defined by ${\displaystyle \gamma ^{-}(s)=\gamma (a+b-s)}$. Then for integrable ${\displaystyle f\colon \mathrm {spur} (\gamma )\to \mathbb {C} }$

${\displaystyle \int _{\gamma ^{-}}f(z)\,dz=-\int _{\gamma }f(z)\,dz}$

Proof

It is

${\displaystyle {\begin{array}{rl}\displaystyle \int _{\gamma ^{-}}f(z)\,dz&=\displaystyle \int _{a}^{b}f(\gamma ^{-}(s))(\gamma ^{-})'(s)\,ds\\&=\displaystyle \int _{a}^{b}f(\gamma (a+b-s))\gamma '(a+b-s)(-1)\,ds\\&=\displaystyle \int _{b}^{a}f(\gamma (t))\gamma '(t)\,dt\quad {\rm {Substitution}}\ t=a+b-s\\&=-\displaystyle \int _{a}^{b}f(\gamma (t))\gamma '(t)\,dt\\&=-\displaystyle \int _{\gamma }f(z)\,dz\end{array}}}$

Statement

Let ${\displaystyle G\subseteq \mathbb {C} }$ be a region, ${\displaystyle \gamma \colon [a,b]\to \mathbb {C} }$ a rectifiable path, ${\displaystyle f\colon G\to \mathbb {C} }$ continuous, and ${\displaystyle \epsilon >0}$. Then there exists a polygonal chain ${\displaystyle {\hat {\gamma }}\colon [a,b]\to \mathbb {C} }$ with ${\displaystyle \gamma (a)={\hat {\gamma }}(a)}$, ${\displaystyle \gamma (b)={\hat {\gamma }}(b)}$ and ${\displaystyle \left|\int _{\hat {\gamma }}f(z)\,dz-\int _{\gamma }f(z)\,dz\right|<\epsilon }$.

Proof

First of all let ${\displaystyle G=B_{R}(z_{0})}$ be a disk. Since ${\displaystyle \mathrm {spur} (\gamma )}$ is compact, there exists a ${\displaystyle r>0}$ with ${\displaystyle \mathrm {spur} (\gamma )\subseteq {\bar {B}}_{r}(z_{0})\subseteq G}$. On ${\displaystyle {\bar {B}}_{r}(z_{0})}$, ${\displaystyle f}$ is uniformly continuous, so we can choose a ${\displaystyle \delta >0}$ such that ${\displaystyle |f(z)-f(w)|<\epsilon }$ for ${\displaystyle z,w\in B_{r}(z_{0})}$ with ${\displaystyle |z-w|<\delta }$ holds.

Step 1 - Partition of Interval

Now choose, according to the definition of the integral, a partition ${\displaystyle a=t_{0}<\ldots <t_{n}=b}$ of ${\displaystyle [a,b]}$ such that ${\displaystyle |\gamma (s)-\gamma (t)|<\delta }$ for ${\displaystyle s,t\in [t_{i-1},t_{i}]}$ and

${\displaystyle \left|\int _{\gamma }f(z)\,dz-\sum _{i=1}^{n}f{\big (}\gamma (t_{i}){\big )}{\big (}\gamma (t_{i})-\gamma (t_{i-1}){\big )}\right|<\epsilon }$

holds.

Step 2 - Convex Combination

Define a convex combination with ${\displaystyle {\hat {\gamma }}}$ that connects ${\displaystyle \gamma (t_{i-1}}$ and ${\displaystyle \gamma (t_{i-1})}$ and ${\displaystyle \gamma (t_{i})}$ with ${\displaystyle \lambda _{t}\in [0,1]}$:

${\displaystyle (1-\lambda _{t})\cdot \gamma (t_{i-1})+\lambda _{t}\cdot \gamma (t_{i})}$

Step 3 - Convex Combination

With ${\displaystyle 1-\lambda _{t}={\frac {t_{i}-t}{t_{i}-t_{i-1}}}}$ and ${\displaystyle \lambda _{t}:={\frac {t-t_{i-1}}{t_{i}-t_{i-1}}}}$ the path ${\displaystyle {\widehat {\gamma }}(t)}$ is defined as:

${\displaystyle {\widehat {\gamma }}(t):={\frac {1}{t_{i}-t_{i-1}}}{\big (}\gamma (t_{i-1})(t_{i}-t)+\gamma (t_{i})(t-t_{i-1}){\big )},\qquad t\in [t_{i-1},t_{i}]}$

Step 5 - Inequality

${\displaystyle {\begin{array}{rl}\displaystyle \left|\int _{\gamma }f(z)\,dz-\int _{\hat {\gamma }}f(z)\,dz\right|&=\displaystyle \left|\int _{\gamma }f(z)\,dz-\int _{a}^{b}f({\hat {\gamma }}(t)){\hat {\gamma }}'(t)\,dt\right|\\&\leq \displaystyle \epsilon +\left|\sum _{i=1}^{n}f{\big (}\gamma (t_{i}){\big )}{\big (}\gamma (t_{i})-\gamma (t_{i-1}){\big )}-\sum _{i=1}^{n}{\frac {\gamma (t_{i})-\gamma (t_{i-1})}{t_{i}-t_{i-1}}}\int _{t_{i-1}}^{t_{i}}f({\hat {\gamma }}(t))\,dt\right|\\&\leq \displaystyle \epsilon +\sum _{i=1}^{n}|\gamma (t_{i})-\gamma (t_{i-1})|{\frac {1}{t_{i}-t_{i-1}}}\int _{t_{i-1}}^{t_{i}}|f({\hat {\gamma }}(t))-f(\gamma (t_{i}))|\\&\leq \displaystyle \epsilon +\sum _{i=1}^{n}|\gamma (t_{i})-\gamma (t_{i-1})|{\frac {1}{t_{i}-t_{i-1}}}\int _{t_{i-1}}^{t_{i}}\epsilon \\&=\epsilon +\epsilon \cdot {\mathcal {L}}(\gamma )=\epsilon \cdot (1+{\mathcal {L}}(\gamma ))\end{array}}}$

This implies the claim.

Wiki2Reveal

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Translation and Version Control

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• Source: Kurs:Funktionentheorie/Kurvenintegral - URL:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Kurvenintegral

• Date: 12/12/2024