NoteNextra-origin/content/Math416/Math416_L25.md

Math416 Lecture 25

Continue on Residue Theorem

Review the definition of simply connected domain

A domain \Omega is called simply connected if \overline{C}\setminus \Omega is connected if and only if every closed curve in \Omega is null-homotopic in \Omega.

Proof:

Last time we proved \impliedby part.

If every closed curve in \Omega is null-homotopic in \Omega, then \operatorname{ind}_\Gamma(z)=0 for all z\in\mathbb{C}\setminus\Omega for all contour in \Omega.

\implies \mathbb{C}\setminus\Omega is connected.

\impliedby part:

....

Theorem 10.4-6

The following condition are equivalent:

1. \Omega is simply connected.
2. every holomorphic function on \Omega has a primitive g, i.e. g'(z)=f(z) for all z\in \Omega.
3. every non-vanishing holomorphic function on \Omega has a holomorphic logarithm.
4. every harmonic function on \Omega has a harmonic conjugate.

Residue Theorem

Theorem 10.8 The Residue Theorem

Let \Omega be a domain, \Gamma be a contour such that \Gamma\cup \operatorname{int}(\Gamma)\subset \Omega

Let f be holomorphic on \Omega\setminus \{z_1, z_2, \cdots, z_n\} where z_1, z_2, \cdots, z_n are finitely many points in \Omega, where z_1, z_2, \cdots, z_n\notin \Gamma.

Then

\int_\Gamma f(z) dz = 2\pi i \sum_{j=1}^n\operatorname{ind}_{\Gamma}(z_j) \operatorname{res}_{z_j}(f)

Proof:

For each i\leq j\leq n, let C_j be a circle centered at z_j\in \Gamma\setminus \Omega such that \operatorname{int}(C_j)\subset \Omega, counterclockwise and pairwise disjoint.

Let \Gamma_1=\Gamma\setminus\{z_1, z_2, \cdots, z_n\}, \Gamma_1=\Gamma-\sum_{j=1}^n \operatorname{ind}_{\Gamma}(z_j)C_j (This excludes the singularities outside \Gamma)

f\in O(\Omega_1), \Gamma_1\in \Omega_1

and \operatorname{ind}_{\Gamma_1}(z)=0 for all z\in \mathbb{C}\setminus \Omega_1, either z\notin \Gamma or z\in\{z_1, z_2, \cdots, z_n\}.

\operatorname{ind}_{\Gamma_1}(z_j)=\operatorname{ind}_{\Gamma}(z_j)-1\cdot\operatorname{ind}_{C_j}(z_j)=0 for all j=1, 2, \cdots, n.

By Cauchy's theorem, \int_{\Gamma_1}f(z)dz=0.

So, since f(z)=\sum_{k=-\infty}^\infty a_k(z-z_0)^k, and \gamma(t)=z_k+Re^{it} for t\in[0, 2\pi],$\gamma'(t)=iRe^{it}$,

\begin{aligned}  
\int_\Gamma f(z)dz&=\int_{\Gamma_1}f(z)dz+\sum_{j=1}^n\int_{C_j}f(z)dz\\
&=0+\sum_{j=1}^n \operatorname{ind}_{\Gamma}(z_j) \int_{C_j}f(z)dz\\
&=0+\sum_{j=1}^n \operatorname{ind}_{\Gamma}(z_j)  \int_{0}^{2\pi}f(z_j+Re^{it})ie^{i\theta}dt\\
&=0+\sum_{j=1}^n \operatorname{ind}_{\Gamma}(z_j)  \int_{0}^{2\pi}\left(\sum_{k=-\infty}^\infty a_k (z_j-z_0)^k e^{int}\right)  iRe^{i\theta}dt\\
&=0+\sum_{j=1}^n \operatorname{ind}_{\Gamma}(z_j)  i\sum_{k=-\infty}^\infty a_k R^{k+1}\left(\int_{0}^{2\pi} e^{i(k+1)t}dt\right)\\
&=\sum_{j=1}^n 2\pi i \operatorname{ind}_{\Gamma}(z_j)  \operatorname{res}_{z_j}(f)\\
\end{aligned}

QED

Corollary 10.9 Cauchy's Integral Formula

If \Gamma is a simple contour, z_0\in \operatorname{int}(\Gamma), g\in O(\Omega), then

g(z_0)=\frac{1}{2\pi i}\int_\Gamma \frac{g(z)}{z-z_0}dz

Proof:

The right hand side is the residue of g(z)/(z-z_0) at z_0.

By the residue theorem,

Notice that g(z)=a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots, and \frac{1}{z-z_0}=a_0\sum_{k=0}^\infty (z-z_0)^k.

So a_0=g(z_0), and a_k=\frac{g^{(k)}(z_0)}{k!} for k\geq 1.

\int_\Gamma \frac{g(z)}{z-z_0}dz=2\pi i \operatorname{res}_{z_0}\left(\frac{g(z)}{z-z_0}\right)=2\pi i g(z_0)

QED

Application to evaluating definite integrals

Idea:

It is easy to evaluate intervals around closed contours.

Choose contour so one side (where you want to integrate).

Handle the other side by:

• Symmetry
• length * supremum of absolute value of integrand
• Bound function by another function whose integral goes to zero.

Example:

Evaluate \int_0^\infty \frac{\sin x}{x}dx.

On the contour \gamma(t) be the semicircle in the upper half plane removed the origin.

Then let f(z)=\frac{e^{iz}}{z}=\frac{\cos z+i\sin z}{z}, by the Cauchy's theorem,

\int_\gamma f(z)dz=0

So \frac{\sin z}{z}=0 on \gamma.

If x\in \mathbb{R}, f(x)=\frac{e^{ix}}{x}=\frac{\cos x+i\sin x}{x}.

On the real axis,

\begin{aligned}  
\int_{-R}^{-\epsilon}+\int_\epsilon^R f(x)dx&=\int_{-R}^{-\epsilon}\frac{e^{ix}}{x}dx+\int_\epsilon^R \frac{e^{ix}}{x}dx\\
&=\int_{-R}^{-\epsilon}\frac{\cos x+i\sin x}{x}dx+\int_\epsilon^R \frac{\cos x+i\sin x}{x}dx\\
&=\int_{-R}^{-\epsilon}\frac{\cos x}{x}dx+i\int_{-R}^{-\epsilon}\frac{\sin x}{x}dx+\int_\epsilon^R \frac{\cos x}{x}dx+i\int_\epsilon^R \frac{\sin x}{x}dx\\
&=2i\int_0^\infty \frac{\sin x}{x}dx
\end{aligned}

For the clockwise semi-circle around the origin,

\int_{S_\epsilon} f(z)dz=\int_{S_\epsilon}\frac{e^{iz}}{z}dz

let \gamma(t)=\epsilon e^{-it}, t\in[-\pi,0].

Then \gamma'(t)=-i\epsilon e^{-it},

CONTINUE NEXT TIME.