122B Homework 10 - Evaluating Improper Trig Integrals.tex

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		\Large{Homework 10}	% Large makes the font larger, put title inside { }
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		Vincent La \\
		Math 122B \\
		September 11, 2017
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\begin{enumerate}
	\item Evaluate the improper integral
	\[ \int^{\infty}_0 \frac{x^3 \sin{x}}{(x^2 + 1)(x^2 + 9)} dx \]
	
	First, we need to evaluate
	\[
	\lim_{R\rightarrow \infty} \int^R_{-R} \frac{x^3 \sin{x}}{(x^2 + 1)(x^2 + 9)} dx
	+ \int_{C_R} \frac{z^3 \exp{iz}}{(z^2 + 1)(z^2 + 9)} dz
	= 2\pi i Res f(z)
	\]
	
	Furthermore, $f(z)$ has isolated singularities whenever
	$z^2 + 1 = 0$ and $z^2 + 9 = 0$ implying that $z = \pm i$ and $z = \pm 3i$. Thus, let $R > 3$.
	
	We are primarily concerned with the residues at $z = i$ and $z = 3i$. 
	
	\paragraph{Residue at $z = i$} First, notice that we can write $f(z)$ as
	\[f(z) = \frac{z^3 \exp{iz}}{(z^2 + 9)(z + i)} \cdot \frac{1}{z - i}
	       = \frac{\phi{(z)}}{z - i}\]
	
	Because, $\phi(i) = \frac{i^3 \exp{i^2}}{(i^2 + 9)(i + i)} \neq 0$, $q(i) = 0$, and $q'(i) = 1 \neq 0$, $z = i$ is therefore a pole of order 1 at $z = i$ where
	\[Res f(i) = \phi(i) = \frac{i^3 \cdot \exp{i^2}}{8 \cdot -1}
	= \frac{i \exp{-1}}{8}
	\]
	
	\paragraph{Residue at $z = 3i$} First, notice that we can write $f(z)$ as
	\[f(z) = \frac{z^3 \exp{iz}}{(z + 3i)(z^2 + 1)} \cdot \frac{1}{z - 3i}
	= \frac{\phi{(z)}}{z - 3i}\]
	
	Because, $\phi(i) = \frac{(3i)^3 \exp{3i^2}}{ (3i + 3i) ((3i)^2 + 1)} \neq 0$, $q(i) = 0$, and $q'(i) = 1 \neq 0$, $z = 3i$ is therefore a pole of order 1 at $z = i$ where
	\[Res f(i) =
	 \phi(i) = \frac{(3i)^3 \exp{3i^2}}{ (3i + 3i) ((3i)^2 + 1)}
	 = \frac{3^3 i^3 \exp{-3}}{6i \cdot (-8)}
	 = \frac{27 \cdot -i \cdot \exp{-3}}{-48i}
	 = \frac{9 \cdot \exp{-3}}{16} \]
	
	\paragraph{Value of $\int f(z)$}
	Whenever $z \in C_R$, because $x \geq 0$ we get that
	\[\begin{aligned}
	| Im \int \frac{z^3e^{iz}}{(z^2 + 1)(z^2 + 9)} | 
	&\leq \int | \frac{z^3e^{iz}}{(z^2 + 1)(z^2 + 9)} | \\
	&\leq | \frac{\pi \cdot R^4}{(R^2 + 1)(R^2 + 9)} | \\
	&= \frac{R^4}{R^4 + 10R^2 + 9}
	\end{aligned}\]
	
	Because the quotient dominates this fraction as $R \rightarrow \infty$, we can say that $\int f(z) \rightarrow 0$ as $R \rightarrow \infty$.
	
	\paragraph{Conclusion}
	By equating imaginary parts, we get 
	\[\begin{aligned}
	\lim_{R\rightarrow \infty} \int^R_{-R} \frac{x^3 \sin{x}}{(x^2 + 1)(x^2 + 9)} dx
	&= Im(2\pi i Res f(z)) - Im (\int_{C_R} \frac{z^3 \exp{iz}}{(z^2 + 1)(z^2 + 9)} dz) \\
	&= Im(2\pi i Res f(z)) \\
	&= 2\pi i \cdot \frac{\exp{-1}}{8}
	\end{aligned} \]
	
	\item \[\int^{\infty}_0 \frac{\cos{(ax)} - \cos{(bx)}}{x^2} dx = \frac{\pi}{2}(b - a) \]
	
	\paragraph{Solution} ...
\end{enumerate}
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