\documentclass[12pt]{article}
\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

\begin{itemize}
\item[i)]
Evaluate $\ds\int_{-\infty}^\infty\frac{dx}{(x^2+4)^2}$ by contour
integration.


\item[ii)]
Prove that if $R$ is a positive real number then

$\ds\int_R^{R+iR}\frac{e^{iz}dz}{z}, \hspace{0.5in}
\int_{-R+iR}^{R+iR}\frac{e^{iz}dz}{z} \hspace{0.3in} {\rm and}
\hspace{0.3in} \int_{-R}^{-R+iR}\frac{e^{iz}dz}{z}$

all tend to 0 as $R\to\infty$, where in each case the integral is
taken over a straight line.  Hence prove that

$$\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}.$$

\end{itemize}


\vspace{0.25in}

{\bf Answer}

\begin{itemize}
\item[i)]
Let $\ds f(z)=\frac{1}{(z^2+4)^2}$.  This is analytic except for
poles of order 2 at $z=\pm2i$.

For all $\ds x \hspace{0.2in}
\frac{1}{(x^2+4)^2}\leq\frac{1}{x^4}$ and
$\ds\int^\infty\frac{1}{x^4}$ converges.

So by comparison $\ds\int_{-\infty}^\infty\frac{1}{(x^2+4)}dx$
converges.

We integrate $f(z)$ round the contour $\Gamma$.


DIAGRAM


For $R>2$ there is a pole of order 2 at $z=2i$ inside $\Gamma$.

res$\ds(f,2i)=\left.\frac{d}{dz}(z-2i)^2f(z)\right|_{z=2i}$

$\ds=\frac{d}{dz}\frac{1}{(z+2i)^2}=
\left.\frac{-2}{(z+2i)^3}\right|_{z=2i}=\frac{-i}{32}$

$\ds\int_{\Gamma}f(z)dz=2\pi i\frac{(-1)}{32}=\frac{\pi}{16}$

On $\ds C_2 \hspace{0.2in} |f(z)|=\frac{1}{|z^2+4|^2}\leq
\frac{1}{(|z|^2-4)^2}=\frac{1}{(R^2-4)^2}$ for $R\geq2$.

So $\ds\left|\int_{C_2}f(z)dz\right|\leq\frac{\pi R}{R^2-4}\to0$
as $R\to\infty$

$\ds\int_{\Gamma}=\int_{-R}^R+\int_{C_2}$ so letting $R\to\infty$
gives
$\ds\int_{-\infty}^\infty\frac{dx}{(x^2+4)^2}=\frac{\pi}{16}$


\item[ii)]


DIAGRAM


On $\ds C_1 \hspace{0.2in} \left|\frac{e^{iz}}{z}\right|=
\frac{|e^{i(R+it)}|}{|z|}\leq\frac{e^{-t}}{R}$

So $\ds\left|\int_{C_1}\frac{e^{iz}}{z}dz\right|\leq
\int_0^R\frac{e^{-t}}{R}dt=\frac{1-e^{-R}}{R}\to0$ as $R\to\infty$

On $\ds C_2 \hspace{0.2in} \left|\frac{e^{iz}}{z}\right|=
\frac{|e^{i(t+iR)}|}{|z|}=\frac{e^{-R}}{|z|}\leq\frac{e^{-R}}{R}$

So $\ds\left|\int_{C_2}\frac{e^{iz}}{z}dz\right|\leq
\frac{e^{-R}}{R}2R\to0$ as $R\to\infty$

On $\ds C_3 \hspace{0.2in} \left|\frac{e^{iz}}{z}\right|=
\frac{|e^{i(-R+it)}|}{|z|}\leq\frac{e^{-t}}{R}$

and again $\ds\int_{C_3}\to0$ as $R\to\infty$, as with
$\ds\int_{C_1}$

Now $\ds z\frac{e^{iz}}{z}=e^{iz}\to1$ as $z\to0$ so
$\ds\frac{e^{iz}}{z}$ has a simple pole at $z=0$ with residue 1.

So $\ds\frac{e^{iz}}{z}=\frac{1}{z}+g(z)$ where $g(z)$ is analytic
near 0.

So $\exists K, \,\, M$ such that $|g(z)|\leq M$ for $|z|\leq K$.

Thus for the small semi-circle $C$, $z=-re^{-it} \hspace{0.2in}
0\leq t\leq\pi \hspace{0.2in} r\leq K$

$\ds\int_C\frac{e^{iz}}{z}=\int_C\frac{1}{z}dz+\int_C g(z)dz$

Now $\ds\left|\int_C g(z)dz\right|\leq M\pi r\to0$ as $r\to0$

and $\ds\int_C\frac{1}{z}dz=
\int_0^\pi\frac{ire^{-it}}{-re^{-it}}=-\pi i$

Inside $\ds\Gamma, \hspace{0.2in} \frac{e^{iz}}{z}$ is analytic.
Hence $\ds\int_{\Gamma}\frac{e^{it}}{z}\to0$.

So letting $R\to\infty$, $r\to0$ gives

$\ds\int_{-\infty}^0\frac{e^{ix}}{x}dx+
\int_0^\infty\frac{e^{ix}}{x}dx-\pi i=0$

So $\ds\int_{-\infty}^\infty\frac{\sin x}{x}=\pi \hspace{0.5in}$
i.e. $\ds\int_0^\infty\frac{\sin x}{x}=\frac{\pi}{2}$

\end{itemize}

\end{document}