final updates

pages/Math416/Math416_L25.md

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 QED
 
-## Application ot valuating definite integrals
+## Application to evaluating definite integrals
 
 Idea:
 

pages/Math416/Math416_L26.md

@@ -1 +1,148 @@
+# Math416 Lecture 26
 
+## Continue on Application to evaluating definite integrals
+
+Note: Contour can never go through a singularity.
+
+Recall the semi annulus contour.
+
+Know that $\int_\gamma f(z)dz=0$.
+
+So $\int_A+\int_B+\int_C+\int_D=0$.
+
+From last lecture, we know that $\int_D=0$ and $\int_A+\int_C=2i\int_0^\infty \frac{\sin x}{x}dx$.
+
+### Integrating over $B$
+
+Do $B$, we have $\gamma(t)=\epsilon e^{it}$ for $t\in[0,\pi]$.
+
+$\int_B=-\int_0^\pi f(\epsilon e^{it})\epsilon i e^{it}dt$.
+
+$f(z)=\frac{e^{iz}}{z}=\frac{1}{z}(1+iz-\frac{z^2}{2!}+\cdots)$.
+
+So $z f(z)=1+O(\epsilon)$ and $f(z)=\frac{1}{z}+O(\frac{\epsilon}{z})$.
+
+$$
+\begin{aligned}
+\int_B&=-\int_0^\pi (\frac{1}{\epsilon}e^{it}+O(1))\epsilon i e^{it}dt\\
+&=-i\int_0^\pi 1dt+O(\epsilon)\\
+&=-i\pi+O(\epsilon)
+\end{aligned}
+$$
+
+### Integrating over $D$
+
+#### Method 1: Using estimate
+
+$z=Re^{it}$ for $t\in[0,\pi]$.
+
+$f(z)=\frac{e^{iz}}{z}=\frac{e^{iRe^{it}}}{Re^{it}}$.
+
+$Re^{it}=R(\cos t+i\sin t)$, $iRe^{it}=-R(\sin t-i\cos t)$.
+
+$e^{iRe^{it}}=e^{-R\sin t}e^{iR\cos t}$.
+
+$\max|f(z)|=\max\frac{|e^{iR\cos t}|}{|R e^{it}|}=\frac{1}{R}$.
+
+This only bounds the function $|\int_D|\leq \pi R\frac{1}{R}=\pi$.
+
+This is not a good estimate.
+
+#### Method 2: Hard core integration
+
+$\gamma(t)=Re^{it}$ for $t\in[0,\pi]$.
+
+$$
+\begin{aligned}
+\int_D&=\int_0^\pi \frac{e^{iRe^{it}}}{R e^{it}}iR e^{it}dt\\
+&=i\int_0^\pi e^{iR\cos t}e^{-R\sin t}dt\\
+\end{aligned}
+$$
+
+Notice that we can use $\frac{2}{\pi}t$ to replace $\sin t$.
+
+$$
+\begin{aligned}
+\left|\int_D\right|&\leq\int_0^\pi e^{-R\sin t}dt\\
+&=2\int_0^{\pi/2} e^{-R\sin t}dt\\
+&\leq 2\int_0^{\pi/2} e^{-2Rt/\pi}dt\\
+&=-\frac{2\pi}{R}(e^{-\frac{R\pi}{2}t})|_0^{\pi/2}\\
+&\leq\frac{\pi}{R}
+\end{aligned}
+$$
+
+As $R\to\infty$, $\left|\int_D\right|\to 0$.
+
+So $\int_D=0$.
+
+So we have $\int_A+\int_C=2i\int_0^\infty \frac{\sin x}{x}dx=i\pi$.
+
+So $\int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}$.
+
+## Application to evaluate $\int_{-\infty}^\infty \frac{\cos x}{1+x^4}dx$
+
+$f(z)=\frac{e^{iz}}{1+z^4}=\frac{\cos z+i\sin z}{1+z^4}$.
+
+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
+
+$$
+\begin{aligned}
+\operatorname{Res}_{z=e^{i\pi/4}}(f)+\operatorname{Res}_{z=e^{i3\pi/4}}(f)&=\frac{e^{ie^{i\pi/4}}}{4e^{i3\pi/4}}+\frac{e^{ie^{i3\pi/4}}}{4e^{i\pi/4}}\\
+&=\frac{e^{\frac{i}{\sqrt{2}}} e^{-\frac{1}{\sqrt{2}}}}{4[-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}]}+\frac{e^{-\frac{i}{\sqrt{2}}}-e^{-\frac{1}{\sqrt{2}}}}{4[\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}]}\\
+&=\frac{\pi\sqrt{2}}{2}e^{-\frac{1}{\sqrt{2}}}(\cos\frac{1}{\sqrt{2}}+\sin\frac{1}{\sqrt{2}})
+\end{aligned}
+$$
+
+SKIP
+
+Review on next lecture.