MATH529: L20 Integration in $ℂ$

Overview

• Review of Riemann integral

• Contour integrals in $ℂ$

• Solved examples

Review of Riemann integral

• Consider function $f:[a,b]\to ℝ$

• Partition on interval $[a,b]$ is $𝒫=\left\{x_i|i=0,\dots ,n.\right\}$, $a=x_0<x_1<\cdots <x_n=b$

• Norm of partition $||𝒫||={max}_{0⩽i⩽n-1}|x_{i+1}-x_i|$

• Tagged partition = partition and set $𝒯=\left\{t_i|i=1,\dots ,.n\right\}$, $t_i\in [x_{i-1},x_i]$

• Riemann sum $S_{𝒫,𝒯}={\sum }_{i=1}^{n}f\left(t_i\right)(x_i-x_{i-1})$

• If ${lim}_{||𝒫||\to 0}{S}_{𝒫,𝒯}$ exists, $f$ is said to be Riemann-integrable and

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={lim}_{||𝒫||\to 0}\sum _{i=1}^{n}f\left(t_i\right)(x_i-x_{i-1})}$

Review of path integral - parametric form

• Consider a curve $C$ in ${ℝ}^2$, i.e., a pair of univariate functions $x,y:[a,b]\to ℝ$

  ${\displaystyle C:x\left(s\right),y\left(s\right),}$

  $s$ is the curvilinear parameter.

• Partition the interval by $𝒫=\left\{s_i|i=0,\dots ,n.\right\}$, $a=s_0<s_1<\cdots <s_n=b$, with $||𝒫||={max}_{0⩽i⩽n-1}|s_{i+1}-s_i|$

• Form tagged partition from $𝒫$ and $𝒯=\left\{t_i|i=1,\dots ,.n\right\}$, $t_i\in [s_{i-1},s_i]$

• Consider $F:{ℝ}^2\to ℝ$ a real, continuous, bivariate function

• Form the sum $S_{𝒫,𝒯}={\sum }_{i=1}^{n}F(x\left(t_i\right),y\left(t_i\right))(s_i-s_{i-1})$

• If ${lim}_{||𝒫||\to 0}{S}_{𝒫,𝒯}$ exists, $F$ is said to be path-integrable, and

  ${\displaystyle \underset{C}{\int }F(x,y)ds=\underset{a}{\overset{b}{\int }}F(x\left(s\right),y\left(s\right))ds={lim}_{||𝒫||\to 0}\sum _{i=1}^{n}F(x\left(t_i\right),y\left(t_i\right))(s_i-s_{i-1})}$

Review of path integral - explicit form

• An integral on the curve $C:x\left(s\right),y\left(s\right),x,y:[a,b]\to ℝ$, can also be stated by explicit dependence on $(x,y)\in {ℝ}^2$

  ${\displaystyle I=\underset{C}{\int }P(x,y)dx+\underset{C}{\int }Q(x,y)dy=\underset{C}{\int }Pdx+Qdy}$

• The curve $C$ is smooth if $x,y\in {𝒞}^1(a,b)$ in which case $dx=x^{\text{'}}ds,dy=y^{\text{'}}ds$

  ${\displaystyle I=\underset{a}{\overset{b}{\int }}[P(x\left(s\right),y\left(s\right))x^{\text{'}}\left(s\right)+Q(x\left(s\right),y\left(s\right))y^{\text{'}}\left(s\right)]ds}$

• Recall that the vector with components $(x^{\text{'}}\left(s\right),y^{\text{'}}\left(s\right))$ is tangent to $C$

$ℂ$ integral

• Analogous to path integral in ${ℝ}^2$, let $z\left(s\right)=x\left(s\right)+iy\left(s\right)$, $z:[a,b]\to ℂ$ define a path or contour in $ℂ$.

• Partition $C$ by $𝒫=\{z_0=z\left(s_0\right),z_1=z\left(s_1\right),\dots ,z_n=z\left(s_n\right)\}$, $a=s_0<\dots <s_n=b$, with $||𝒫||={max}_{1⩽k⩽n}|z_k-z_{k-1}|$, with ${\zeta }_k=z\left(t_k\right)$, $t_k\in [s_{k-1},s_k]$.

• The path integral of $f:ℂ\to ℂ$ is

  ${\displaystyle \underset{C}{\overset{}{\int }}f\left(z\right)dz={lim}_{||𝒫||\to 0}\sum _{k=1}^{n}f\left({\zeta }_k\right)(z_k-z_{k-1})}$

• With $f\left(z\right)=u(x,y)+iv(x,y)$, $z=x+iy$, $dz=dx+idy$

  ${\displaystyle \underset{C}{\overset{}{\int }}f\left(z\right)dz=\underset{C}{\overset{}{\int }}udx-vdy+i\underset{C}{\overset{}{\int }}vdx+udy=\underset{a}{\overset{b}{\int }}f\left(z\left(s\right)\right)z^{\text{'}}\left(s\right)ds}$

Examples

1. $I={\int }_C\bar{z}dz$, $C:x\left(t\right)=3t$, $y\left(t\right)=t^2$, $0⩽t⩽2\pi$, $f\left(z\right)=$$\bar{z}=x-iy=u+iv$

   ${\displaystyle C-R:u_x=1\ne -1=v_y}$ ${\displaystyle I=\underset{C}{\int }\bar{z}{z}^{\text{'}}\left(t\right)dt=\underset{0}{\overset{2\pi }{\int }}(x\left(t\right)-iy\left(t\right))(x^{\text{'}}\left(t\right)+i{y}^{\text{'}}\left(t\right))dt\Rightarrow }$ ${\displaystyle I=\underset{0}{\overset{2\pi }{\int }}(x\left(t\right)x^{\text{'}}\left(t\right)+y\left(t\right)y^{\text{'}}\left(t\right)+i[x\left(t\right)y^{\text{'}}\left(t\right)-x^{\text{'}}\left(t\right)y\left(t\right)])dt}$ ${\displaystyle I=\underset{0}{\overset{2\pi }{\int }}[9t+2t^2]dt+i\underset{0}{\overset{2\pi }{\int }}[6t^2-3t^2]dt=\frac{9}{2}4{\pi }^2+\frac{2}{3}8{\pi }^3+i8{\pi }^4}$

Examples

1. $I={\oint }_C\frac{1}{z}dz$, $C:x\left(t\right)=\cos \left(t\right),y\left(t\right)=\sin \left(t\right)$, $0⩽t⩽2\pi$

   ${\displaystyle z\left(t\right)=e^{it},\frac{1}{z}=e^{-it},z^{\text{'}}\left(t\right)=i{e}^{it}}$ ${\displaystyle I=\underset{0}{\overset{2\pi }{\int }}{e}^{-it}i{e}^{it}dt=2\pi i}$

2. $I={\oint }_{C}zdz=0$, $C:x\left(t\right)=\cos \left(t\right),y\left(t\right)=\sin \left(t\right)$, $0⩽t⩽2\pi$

   ${\displaystyle z\left(t\right)=e^{it},z^{\text{'}}\left(t\right)=i{e}^{it}}$ ${\displaystyle I=\underset{0}{\overset{2\pi }{\int }}{e}^{it}i{e}^{it}dt=i\underset{0}{\overset{2\pi }{\int }}{e}^{2it}dt=\frac{i}{2i}{\left[e^{2it}\right]}_{t=0}^{t=2\pi }=0}$

Properties of integration in $ℂ$

• ${\int }_{C}af\left(z\right)dz=a{\int }_{C}f\left(z\right)dz$, $a\in ℂ$ a constant

• ${\int }_C[f\left(z\right)+g\left(z\right)]dz={\int }_{C}f\left(z\right)dz+{\int }_{C}g\left(z\right)dz$

• ${\int }_{C}f\left(z\right)dz=-{\int }_{-C}f\left(z\right)dz$, $-C$ denotes contour in reverse direction

• If $f$ continuous on smooth curve $C$ and $\left|f\left(z\right)\right|⩽M$ for $\forall z\in C$ then

  ${\displaystyle |\underset{C}{\int }f\left(z\right)dz|⩽ML}$

  with $L$ the curve length

Example: bound on complex integral

• Example

  ${\displaystyle I=\underset{\left|z\right|=4}{\oint }\frac{e^z}{z+1}dz}$ ${\displaystyle |z_1+z_2|⩽\left|z_1\right|+\left|z_2\right|}$ ${\displaystyle \left|z_1\right|⩽|z_1+z_2+(-z_2)|⩽|z_1+z_2|+\left|z_2\right|\Rightarrow \left|z_1\right|-\left|z_2\right|⩽|z_1+z_2|}$ ${\displaystyle \left|\frac{e^z}{z+1}\right|=\frac{\left|e^z\right|}{|z+1|}⩽\frac{\left|e^z\right|}{\left|z\right|-1}=\frac{\left|e^x{e}^{iy}\right|}{\left|z\right|-1}=\frac{\left|e^x\right|}{\left|z\right|-1}⩽\frac{e^4}{3}\Rightarrow I⩽\frac{e^4}{3}2\pi 4=\frac{8\pi e^4}{3}}$