122B Homework 8 - Evaluating Improper Integrals.tex

\documentclass[11pt]{article}
\usepackage[margin=1.25in]{geometry}

\usepackage{graphicx,tikz}
\usepackage{amsmath,amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{boondox-cal}
\title{ }

\newtheorem*{thm}{Theorem}

\begin{document}
	%\maketitle
	%\date
	\begin{center}	% centers
		\Large{Homework 8}	% Large makes the font larger, put title inside { }
	\end{center}
	\begin{center}
		Vincent La \\
		Math 122B \\
		September 5, 2017
	\end{center}
	
\begin{enumerate}
	\item Evaluate the improper integral \[\int_0^\infty \frac{dx}{(x^2 + 1)^2} \]
	\paragraph{Solution}
	We'll solve this by evaluating it over the complex axis of the real plane. First, notice that this is equivalent to the integral 
	\[
	\int^R_{-R} \frac{dx}{(x^2 + 1)^2} + \int_{C_R} \frac{dz}{(z^2 + 1)^2} = 2\pi i Res f(z)
	\]
	
	Now, notice that our function has isolated singularities whenever $(z^2 + 1)^2 = 0 \implies z^2 = -1$ implying that $z = \pm i$. Therefore, we'll want to integrate over a semicircle of radius $R > 1$.
	
	\bigskip
	
	Moreover, whenever $z \in C_R$, this implies that $|z| = R$. Therefore, 
	\[(z^2 + 1)^2 \geq ||z|^2 + 1|^2 = |R^2 + 1|^2 \]
	
	This then implies
	\[\int_{C_R} \frac{dz}{(z^2 + 1)^2} \leq \frac{1}{|R^2 + 1|^2}  \cdot 
      \int_{C_R} dz = \frac{\pi i}{(R^2 + 1)^2}\]
      
    Since the right hand side goes to zero as $R \rightarrow \infty$, the entire complex integral disappears. Finally, we just need to calculate the residue of $f(z)$ in $C_R$ at $z = i$. Recall that the residue of quotients, and we get
    \[\begin{aligned}
	\frac{p(z_0)}{q'(z_0)}
	&= \frac{1}{[(z^2 + 1)^2]'} \\
	&= \frac{1}{2(z^2 + 1) \cdot 2z} \\
	&= 4z^3 + 4z
    \end{aligned}\]
    
    Because f is even,
    
    \[\begin{aligned}
    \int^\infty_{-\infty} \frac{dx}{(x^2 + 1)^2}
    &= 2\pi i Res f(i) \\
    &= 2\pi i(\frac{1}{4i^3 + 4})  \\
    &= 2\pi i \cdot \frac{1}{4 - 4i}
    \end{aligned}\]
	
	\item Evaluate the improper integral \[\int_0^\infty \frac{dx}{(x^2 + 1)(x^2 + 4)} \]
	
	\paragraph{Solution} We'll take roughly the same steps as above.
	
	
	 First, notice that this is equivalent to the integral 
	 \[
	 \int^R_{-R} \frac{dx}{(x^2 + 1)(x^2 + 4)} +
	 \int_{C_R} \frac{dz}{(z^2 + 1)(z^2 + 4)} = 2\pi i Res f(z)
	 \]
	
	Using a similar argument to problem 1, we can show that the complex integral disappears. If I weren't so tired, I could probably come up with a better argument that this flimsy sentence.
	
	\bigskip Now, notice that the function
	\[\frac{1}{(z^2 + 1)(z^2 + 4)}\] has isolated singularities whenever
	 $(z^2 + 1)(z^2 + 4) = 0$ or equivalently, whenever $z^4 + 5z^2 + 4 = 0$.
	 
	Applying the quadratic formula, this implies
	\[\begin{aligned}
	z^2
	&= \frac{-5 + \sqrt{5^2 - 4\cdot 1 \cdot 4}}{2} \\
	&= \frac{-5 + \sqrt{25 - 16}}{2} \\
	&= \frac{-5 + 3}{2} \\
	&= -4
	\end{aligned}\]
	
	Therefore, we have singularities whenever $z = 2i$. Finally, we just have to compute the residues and I can finally be done with this problem. But before doing that, notice that
	\[\begin{aligned}
	\frac{d}{dz} (z^2 + 1)(z^2 + 4) &=
	2z \cdot (z^2 + 4) - 2z \cdot (z^2 + 1) \\
	8z - 2z = 6z
	\end{aligned} \]
	
	Furthermore, recall that the residue of a quotient implies that our residue is simply
    $\frac{1}{6z}$ evaluated at $z = 2i$. Therefore, our original integral is equal to $2\pi i \cdot \frac{\pi}{12i} = \frac{1}{6i}$.
	
	\item Find the Cauchy principal value of
	\[\int_{-\infty}^{\infty} \frac{x dx}{(x^2 + 1)(x^2 + 2x + 2)} \]
	
	\paragraph{Solution}
	
	First, let us identify the singularities of 
	\[\frac{z}{(z^2 + 1)(z^2 + 2z + 2) = 0} \]
	
	This function is singular whenever either $(z^2 + 1) = 0$ or $z^2 + 2z + 2 = 0$. For the first equation, this implies there are isolated singularities at $z = \pm i$. For the second, applying the quadratic equation and we get
	\[\begin{aligned}
	z
	&= \frac{-2 \pm \sqrt{2^2 - 4(1)(2)}}{2} \\
	&= \frac{-2 \pm \sqrt{4 - 8}}{2} \\
	&= \frac{-2 \pm 2i}{2}
	&= -1 \pm 1i
	\end{aligned}\]
	
	Because the principal value is simply
	\[P.V. \int^\infty_{-\infty} f(x) dx = 2\pi i Res f(x) \]

	 and the residue of the quotient is simply the following quantity evaluated at
	 $z = -1 + i$
	\[\begin{aligned}
	\frac{z}{(z^2 + 1)(z^2 + 2z + 2)} 
	&= \frac{z}{[(z^2 + 1)(z^2 + 2z + 2)]']} \\
	&= \frac{z}{2z(z^2 + 2z + 2) - (2z + 2)(z^2 + 1)} \\
	&= 2z \\
	\end{aligned}\]
	
	The principal value is $2\pi i \cdot 2 \cdot (-1 + i) = 4(-1 + i)$.
\end{enumerate}
\end{document}