### Author Topic: LEC5101-Quiz6-C  (Read 1116 times)

#### RunboZhang

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##### LEC5101-Quiz6-C
« on: November 22, 2020, 10:49:48 AM »
$\textbf{Problem:} \\\\ \text{Compute the following integral}$ $$\int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^2}} \,dx$$
$\text{with } b>0.$

$\textbf{Solution:} \\\\$
$\text{Let }$ $$f(z) = \frac{e^{iz}}{(z+a)^{2}+b^{2}}$$

$\text{Then we have two poles: } z_1 = -a-ib ,\ z_2 = -a+ib$

$\text{Since we are looking for upper-half plane, then we have pole at } z=-a+ib \text{ with order 1.}$

$\text{(Denote the path as the pic attached below)}$

$\text{By Residual Theorem we have: }$
\begin{gather} \begin{aligned} \int_{\gamma}{f(z)} \,dz &= 2 \pi i \cdot Res{f; -a+ib} \\\\ &=2 \pi i \cdot \frac{e^{iz}}{z+a+ib}|_{z=-a+ib} \\\\ &= \frac{\pi e^{-b} e^{-ia}}{b} \end{aligned} \end{gather}

$\text{For path } c_2:$

\begin{gather} \begin{aligned} |\int_{c_2}{f(z)} \, dz| &= |\int_{0}^{\pi}{f(Re^{i\theta})Re^{i\theta}} \,d\theta| \\\\ &= |\int_{0}^{pi} {\frac{e^{iRe^{i\theta}}}{(Re^{i\theta}+a)^{2}+b^{2}}} \,d\theta| \\\\ &\leq \pi R \cdot |max(f(z))| \\\\ &= \pi R \cdot max|\frac{e^{iRe^{i\theta}}}{(Re^{i\theta}+a)^{2}+b^{2}}| \\\\ &= \pi R \cdot max\frac{e^{|iR(cos(\theta)+isin(\theta))}|}{|(Re^{i\theta}+a)^{2}+b^{2}|}\\\\ &\leq \pi R \cdot \frac{e^{-Rsin(\theta)}}{|(R+a)^{2} - b^{2}|} \ \ \ \ \ \text{(since triangle inequality)} \\\\ &\leq \pi R \cdot \frac{e^{0}}{(R+a)^{2}+b^{2}} \\\\ &= 0 \ \ \text{as R} \rightarrow \infty \\\\ \end{aligned} \end{gather} $$\text{(By comparing the power of R on both numerator and denominator.)}$$

$\text{Now solve for path } c_1:$

\begin{gather} \begin{aligned} \int_{c_1}{f(z)}\,dz &= \int_{-\infty}^{\infty}{\frac{e^{ix}}{(x+a)^{2}+b^{2}}}\,dx \\\\ &= \int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^{2}}}\,dx + i\int_{-\infty}^{\infty}{\frac{sin(x)}{(x+a)^{2}+b^{2}}}\,dx \\\\ \end{aligned} \end{gather}
$\text{Since the imaginary part of the integral is an odd function, thus its integral is 0.}$

$\text{Thus we have:}$
$$\int_{-\infty}^{\infty}{\frac{cos(x)}{(x+a)^{2}+b^{2}}}\,dx = \frac{\pi e^{-b} e^{-ia}}{b}$$