3.9 KiB
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.