# Math416 Lecture 27

## Continue on Application to evaluate $\int_{-\infty}^\infty \frac{\cos x}{1+x^4}dx$

Consider the function$f(z)=\frac{e^{iz}}{1+z^4}=\frac{\cos z+i\sin z}{1+z^4}$.

Our desired integral can be evaluated by $\int_{-R}^R f(z)dz$

To evaluate the singularity, $z^4=-1$ has four roots by the De Moivre's theorem.

$z^4=-1=e^{i\pi+2k\pi i}$ for $k=0,1,2,3$.

So $z=e^{i\theta}$ for $\theta=\frac{\pi}{4}+\frac{k\pi}{2}$ for $k=0,1,2,3$.

So the singularities are $z=e^{i\pi/4},e^{i3\pi/4},e^{i5\pi/4},e^{i7\pi/4}$.

Only $z=e^{i\pi/4},e^{i3\pi/4}$ are in the upper half plane.

So we can use the semi-circle contour to evaluate the integral. Name the path as $\gamma$.

$\int_\gamma f(z)dz=2\pi i\left[\operatorname{Res}_{z=e^{i\pi/4}}(f)+\operatorname{Res}_{z=e^{i3\pi/4}}(f)\right]$.

The two poles are simple poles.

$\operatorname{Res}_{z_0}(f)=\lim_{z\to z_0}(z-z_0)f(z)$.

So 

$$
\begin{aligned}
\operatorname{Res}_{z=e^{i\pi/4}}(f)&=\lim_{z\to e^{i\pi/4}}(z-e^{i\pi/4})\frac{e^{iz}}{1+z^4}\\
&=\frac{(z-e^{i\pi/4})e^{iz}}{(z-e^{i\pi/4})(z-e^{i3\pi/4})(z-e^{i5\pi/4})(z-e^{i7\pi/4})}\\
&=\frac{e^{ie^{i\pi/4}}}{(e^{i\pi/4}-e^{i3\pi/4})(e^{i\pi/4}-e^{i5\pi/4})(e^{i\pi/4}-e^{i7\pi/4})}
\end{aligned}
$$

A short cut goes as follows:

We know $p(z)=1+z^4$ has four roots $z_1,z_2,z_3,z_4$.

$$
\lim_{z\to z_0}\frac{(z-z_0)}{p(z)}=\frac{1}{p'(z_0)}
$$

So

$$
\operatorname{Res}_{z=e^{i\pi/4}}(f)=\frac{e^{ie^{i\pi/4}}}{4e^{i3\pi/4}}
$$

Similarly,

$$
\operatorname{Res}_{z=e^{i3\pi/4}}(f)=\frac{e^{ie^{i3\pi/4}}}{4e^{i\pi/4}}
$$

So the sum of the residues is

$$
\begin{aligned}
\operatorname{Res}_{z=e^{i\pi/4}}(f)+\operatorname{Res}_{z=e^{i3\pi/4}}(f)&=\frac{e^{ie^{i\pi/4}}}{4e^{i3\pi/4}}+\frac{e^{ie^{i3\pi/4}}}{4e^{i\pi/4}}\\
&=\frac{e^{\frac{i}{\sqrt{2}}} e^{-\frac{1}{\sqrt{2}}}}{4[-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}]}+\frac{e^{-\frac{i}{\sqrt{2}}}-e^{-\frac{1}{\sqrt{2}}}}{4[\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}]}\\
&=\frac{\pi\sqrt{2}}{2}e^{-\frac{1}{\sqrt{2}}}(\cos\frac{1}{\sqrt{2}}+\sin\frac{1}{\sqrt{2}})
\end{aligned}
$$

For the semicircle part, we can bound our estimate by

$$
\left|\int_{C_R}f(z)dz\right|\leq\pi R\max_{z\in C_R}|f(z)|\leq \pi \frac{1}{R^4}\to 0
$$

as $R\to\infty$.

So

$$
\int_{-\infty}^\infty\frac{\cos x}{1+x^4}dx=\frac{\pi\sqrt{2}}{2}e^{-\frac{1}{\sqrt{2}}}(\cos\frac{1}{\sqrt{2}}+\sin\frac{1}{\sqrt{2}})
$$

## Big idea of this course

$f$ is holomorphic $\iff$ $f$ has complex derivative.

$f$ is holomorphic $\iff$ $f$ satisfies Cauchy-Riemann equations $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$

$f$ is holomorphic $\iff$ $f$ is analytic (is locally given by power series). The power series is integrable/differentiable term by term in the radius of convergence.

### Laurent series

Similar to power series both with annulus of convergence.

$f(z)=\sum_{n=-\infty}^\infty a_n(z-z_0)^n$ for $z\in A(z_0,r,R)$.

Identity theorem: If $f$ is holomorphic on a domain $\Omega$, it is uniquely determined by its values on any sets with a limit point in $\Omega$.

### Cauchy's Theorem

$$
\int_\gamma f(z)dz=0
$$

If $f$ is holomorphic on $\Omega$ and $\gamma$ is a closed path in $\Omega$ and $\gamma\cup \operatorname{int}\gamma\subset \Omega$, then $\int_\gamma f(z)dz=0$.

### Favorite estimate

$$
\left|\int_\gamma f(z)dz\right|\leq \sup_{z\in\gamma}|f(z)|\cdot \operatorname{length}(\gamma)
$$

### Cauchy's Integral Formula

$$
f(z_0)=\frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z-z_0}dz
$$

where $z_0\in \operatorname{int}\gamma$ and $\gamma$ is a closed path.

Extension: If $f$ is holomorphic on $\Omega$ and $z_0\in \Omega$, then $f$ is infinitely differentiable and

$$
f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\gamma \frac{f(z)}{(z-z_0)^{n+1}}dz
$$

### Residue theorem

If $f$ is holomorphic on $\Omega$ except for a finite number of isolated singularities $z_1,z_2,\dots,z_p$, and $\Gamma$ is a curve inside $\Omega$ that don't pass through any of the singularities ($\Gamma\subset \Omega\setminus \{z_1,z_2,\dots,z_p\}$), then

$$
\int_\Gamma f(z)dz=2\pi i\sum_{z_i}\operatorname{ind}_{\Gamma}(z_i) \operatorname{res}_{z_i}(f)
$$

### Harmonic conjugate

Locally, always have harmonic conjugates.

Globally can do this iff domain is simply connected.

### Schwarz-pick's Lemma:

If $f$ maps $D$ to $D$ and $f(0)=0$, then $|f(z)|\leq |z|$ for all $z\in D$. and $|f'(0)|\leq 1$.

For mobius map, $f:D\to D$ holds, $\varphi(f(z),f(w))=\varphi(z,w)$ for all $z,w\in D$.

$$
\varphi(z,w)=\frac{z-w}{1-\overline{w}z}
$$

### Convergence

#### Types of convergence

**Converge pointwise** (Not very strong):

$\forall x\in X, \lim_{n\to\infty}f_n(x)=f(x)$. 

Or, $\forall x\in X, \forall \epsilon>0, \exists N>0, \forall n\geq N \implies |f_n(x)-f(x)|<\epsilon$.

**Converge uniformly** (Much better):

$\forall \epsilon>0, \exists N>0, \forall n\geq N \implies \forall x\in X, |f_n(x)-f(x)|<\epsilon$.

**Converge locally uniformly** (Strong):

$\forall x\in X$, $\exists$ open $x\in U$, such that $f_n\to f$ uniformly on $U$.

**Converge uniformly on compact subsets** (Good enough for local properties):

$\forall$ compact $K\subset X$, $f_n\to f$ uniformly on $K$.

#### Weierstrass' Theorem

If $f_n\in O(\Omega)$ and $f_n\to f$ locally uniformly, then $f\in O(\Omega)$.

#### Cauchy-Hadamard's Theorem

For a power series, $\sum_{n=0}^\infty a_n(z-z_0)^n$, the radius of convergence is

$$
R=\frac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}
$$

On $B(z_0,R)$, the series converges locally uniformly and absolutely.

## Argument and Logarithm

$\arg z$ is any $\theta$ such that $z=re^{i\theta}$.

$\operatorname{Arg} z$ is the principal value of the argument, $-\pi<\operatorname{Arg} z\leq \pi$.

$\log z$ is the principal value of the logarithm, $\log z=\ln |z|+i\arg z$.

$\operatorname{Log} z$ is the set of all logarithms of $z$, $\operatorname{Log} z=\{\log z+2k\pi i: k\in\mathbb{Z}\}$.

END