Complex Analysis/Curve Integral

Introduction

The complex curve integral is the function-theoretic generalization of the integral from real analysis. Instead of an interval,rectifiable curve serves as the integration domain. The integral is taken over complex-valued functions instead of real-valued functions.

Definition - Rectifiable Curve

Let $\gamma :[a,b]\to ℂ$ be a rectifiable curve, $f:\mathrm{s}\mathrm{p}\mathrm{u}\mathrm{r}(\gamma )\to ℂ$ a mapping. $f$ is said to be integrable over $\gamma$ if there exists a complex number $I\in ℂ$ such that for every $ϵ>0$, there exists a $\delta >0$ such that for every partition $a=t_0<\dots <t_n=b$ of the interval $[a,b]$ with $t_i-t_{i-1}<\delta$ for all $1\le i\le n$, we have

$|I-\sum _{i=1}^{n}f(\gamma (t_i))(\gamma (t_i)-\gamma (t_{i-1})\left)\right|<ϵ$

The complex number $I$ is called the integral of $f$ over $\gamma$ and is denoted by $\underset{\gamma }{\int }f(z)dz:=I$ designated.

Note - Integrability

• $f(\gamma (t_i))\cdot (\gamma (t_i)-\gamma (t_{i-1}))$ entspricht in reaL Analysis it corresponds to be oriented area of rectangles that approximate yhe integral in the Reimann integral.
• $\sum _{i=1}^{n}f(\gamma (t_i))\cdot (\gamma (t_i)-\gamma (t_{i-1}))$ calculate the Complex numbersthe Riemann Sum of all individual terms a decomposition of the interval $[a,b]$ with $a=t_0<\dots <t_n=b$
• $|I-\sum _{i=1}^{n}f(\gamma (t_i))\cdot (\gamma (t_i)-\gamma (t_{i-1})\left)\right|<ϵ$ means, that the Integral $I$ can be approximate with arbitrary precision using Riemann Sum.

Integration over chains

If $\mathrm{\Gamma }=\sum _{i=1}^n{n}_i\cdot {\gamma }_i$ is a chain in $ℂ$, then a function $f:\mathrm{S}\mathrm{p}\mathrm{u}\mathrm{r}(\mathrm{\Gamma })\to ℂ$ is said to be integrable over $\mathrm{\Gamma }$ if it is integrable over each ${\gamma }_i$, and we set

$\underset{\mathrm{\Gamma }}{\int }f(z)dz:=\sum _{i=1}^n{n}_i\cdot \underset{{\gamma }_i}{\int }f(z)dz$

Connection to real integration

If $\gamma$ 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

$\underset{\gamma }{\int }f(z)dz=\underset{a}{\overset{b}{\int }}f(\gamma (t))\cdot {\gamma }^{\prime }(t)dt$

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

Example 1

We consider the curve $\gamma :[0,1]\to ℂ$, $\gamma (t):=\exp (2\pi it)$, and the function $f(z)=\frac{1}{z}$. Since the curve is differentiable, we have

$\begin{array}{cc}\underset{\gamma }{\int }f(z)dz& =\underset{\gamma }{\int }\frac{1}{z}dz\\ & =\underset{0}{\overset{1}{\int }}\frac{1}{\exp (2\pi it)}\cdot 2\pi i\cdot \exp (2\pi it)dt\\ & =\underset{0}{\overset{1}{\int }}2\pi idt\\ & =2\pi i\end{array}$

Example 2

We modify our first example slightly and consider the curve $\gamma :[0,1]\to ℂ$, $\gamma (t):=\exp (2\pi it)$, and the function $f(z)=z^n$ for $n\ne -1$. Since the curve is differentiable, we have

$\begin{array}{cc}\underset{\gamma }{\int }f(z)dz& =\underset{\gamma }{\int }{z}^{n}dz\\ & =\underset{0}{\overset{1}{\int }}\exp (2\pi it{)}^n\cdot 2\pi i\cdot \exp (2\pi it)dt\\ & =\underset{0}{\overset{1}{\int }}2\pi i\exp (2\pi i(n+1)t)dt\\ & =\frac{1}{n+1}\exp (2\pi i(n+1)t){|}_0^1\\ & =0\end{array}$

Representation of the integral

Both examples together give us

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

where

$$\begin{array}{cccc}{\delta }_{n,-1}:& \mathbb{N}& \to & \{0,1\}\\ & n& \mapsto & {\delta }_{n,-1}(n)=\{\begin{array}{ccc}1& ,& n=-1\\ 0& ,& n=-1\end{array}\end{array}$$

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

Properties

Independence of parameterization

Let $\gamma :[a,b]\to ℂ$ be a piecewise $C^1$-path, $\varphi :[\alpha ,\beta ]\to [a,b]$ a $C^1$-diffeomorphism that preserves orientation. Then $\gamma \circ \varphi :[\alpha ,\beta ]\to ℂ$ is a piecewise $C^1$-path and we have

$\underset{\gamma }{\int }f(z)dz=\underset{\gamma \circ \varphi }{\int }f(z)dz$

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

Proof

It is

$\begin{array}{cc}\underset{\gamma \circ \varphi }{\int }f(z)dz& =\underset{\alpha }{\overset{\beta }{\int }}f(\gamma (\varphi (s)))(\gamma \circ \varphi {)}^{\prime }(s)ds\\ & =\underset{\alpha }{\overset{\beta }{\int }}f(\gamma (\varphi (s))){\gamma }^{\prime }(\varphi (s)){\varphi }^{\prime }(s)ds\\ & =\underset{\varphi (\alpha )}{\overset{\varphi (\beta )}{\int }}f(\gamma (t)){\gamma }^{\prime }(t)dt\mathrm{S}\mathrm{u}\mathrm{b}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{t}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}t=\varphi (s)\\ & =\underset{a}{\overset{b}{\int }}f(\gamma (t)){\gamma }^{\prime }(t)dt\\ & =\underset{\gamma }{\int }f(z)dz\end{array}$

Linearity

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

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

for rectifiable $\gamma$, $\alpha ,\beta \in ℂ$ and integrable $f,g:\mathrm{s}\mathrm{p}\mathrm{u}\mathrm{r}(\gamma )\to ℂ$.

Orientation reversal

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

$\underset{{\gamma }^{-}}{\int }f(z)dz=-\underset{\gamma }{\int }f(z)dz$

Proof

It is

$\begin{array}{cc}\underset{{\gamma }^{-}}{\int }f(z)dz& =\underset{a}{\overset{b}{\int }}f({\gamma }^{-}(s))({\gamma }^{-}{)}^{\prime }(s)ds\\ & =\underset{a}{\overset{b}{\int }}f(\gamma (a+b-s)){\gamma }^{\prime }(a+b-s)(-1)ds\\ & =\underset{b}{\overset{a}{\int }}f(\gamma (t)){\gamma }^{\prime }(t)dt\mathrm{S}\mathrm{u}\mathrm{b}\mathrm{s}\mathrm{t}\mathrm{i}\mathrm{t}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}t=a+b-s\\ & =-\underset{a}{\overset{b}{\int }}f(\gamma (t)){\gamma }^{\prime }(t)dt\\ & =-\underset{\gamma }{\int }f(z)dz\end{array}$

Approximation by polygonal chains

The presented version of the integration path seems very general, but most integration paths that occur in practice are piecewise continuously differentiable. Since it is easier to work with piecewise continuously differentiable paths, we want to show in the following how an arbitrary integration path for continuous integrands can be approximated by polygonal chains. This can be used to transfer statements about general rectifiable paths to polygonal chains.

Statement

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

Proof

First of all let $G=B_R(z_0)$ be a disk. Since $\mathrm{s}\mathrm{p}\mathrm{u}\mathrm{r}(\gamma )$ is compact, there exists a $r>0$ with $\mathrm{s}\mathrm{p}\mathrm{u}\mathrm{r}(\gamma )\subseteq {\overline{B}}_r(z_0)\subseteq G$. On ${\overline{B}}_r(z_0)$, $f$ is uniformly continuous, so we can choose a $\delta >0$ such that $|f(z)-f(w)|<ϵ$ for $z,w\in B_r(z_0)$ with $|z-w|<\delta$ holds.

Step 1 - Partition of Interval

Now choose, according to the definition of the integral, a partition $a=t_0<\dots <t_n=b$ of $[a,b]$ such that $|\gamma (s)-\gamma (t)|<\delta$ for $s,t\in [t_{i-1},t_i]$ and

$|\underset{\gamma }{\int }f(z)dz-\sum _{i=1}^{n}f(\gamma (t_i)\left)\right(\gamma (t_i)-\gamma (t_{i-1})\left)\right|<ϵ$

holds.

Step 2 - Convex Combination

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

$(1-{\lambda }_t)\cdot \gamma (t_{i-1})+{\lambda }_t\cdot \gamma (t_i)$

Step 3 - Convex Combination

With $1-{\lambda }_t=\frac{t_i-t}{t_i-t_{i-1}}$ and ${\lambda }_t:=\frac{t-t_{i-1}}{t_i-t_{i-1}}$ the path $\hat{\gamma }(t)$ is defined as:

$\hat{\gamma }(t):=\frac{1}{t_i-t_{i-1}}(\gamma (t_{i-1})(t_i-t)+\gamma (t_i)(t-t_{i-1})),t\in [t_{i-1},t_i]$

Step 4 - Partition of Interval

So $\hat{\gamma }$ is the polygonal chain that connects the points $\gamma (t_i)$ by straight lines. In particular, $\hat{\gamma }$ lies in $B_r(z_0)$. By construction, we also have $|\hat{\gamma }(t)-\gamma (t_i)|<\delta$ for $t\in [t_{i-1},t_i]$. It follows that

Step 5 - Inequality

$\begin{array}{cc}|\underset{\gamma }{\int }f(z)dz-\underset{\hat{\gamma }}{\int }f(z)dz|& =|\underset{\gamma }{\int }f(z)dz-\underset{a}{\overset{b}{\int }}f(\hat{\gamma }(t)){\hat{\gamma }}^{\prime }(t)dt|\\ & \le ϵ+\left|\sum _{i=1}^{n}f\right(\gamma (t_i)\left)\right(\gamma (t_i)-\gamma (t_{i-1}))-\sum _{i=1}^n\frac{\gamma (t_i)-\gamma (t_{i-1})}{t_i-t_{i-1}}\underset{t_{i-1}}{\overset{t_i}{\int }}f(\hat{\gamma }(t))dt|\\ & \le ϵ+\sum _{i=1}^n|\gamma (t_i)-\gamma (t_{i-1})|\frac{1}{t_i-t_{i-1}}\underset{t_{i-1}}{\overset{t_i}{\int }}|f(\hat{\gamma }(t))-f(\gamma (t_i))|\\ & \le ϵ+\sum _{i=1}^n|\gamma (t_i)-\gamma (t_{i-1})|\frac{1}{t_i-t_{i-1}}\underset{t_{i-1}}{\overset{t_i}{\int }}ϵ\\ & =ϵ+ϵ\cdot ℒ(\gamma )=ϵ\cdot (1+ℒ(\gamma ))\end{array}$

This implies the claim.

Step 6 - Generalization

If $G$ is not a disk, we cover $\mathrm{S}\mathrm{p}\mathrm{u}\mathrm{r}(\gamma )$ with finitely many disks that are a subset of $G$ and apply the above construction to each sub-path. This implies the claim in the general case.

See also

• rectifiable curve

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