# 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.