<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}

NOTE REFERENCE TO QUESTION 1

{\bf Question}

Verify the following integral.

$\ds\int_0^{\infty}dx \ds\frac{\cos
mx}{x^2+1}=\ds\frac{\pi}{2}\exp(-m),\ m&gt;0$
\medskip

{\bf Answer}

Consider $J=\ds\oint_C \ds\frac{dz}{e^{imz}}{(z^2+1)}$ where $C$
is the $D$-contour of Q1.

The integrand has simple poles at $z=\pm i$, but only $z=i$ lies
in $C$.

Residue at $z=i$ is $\lim_{z \to i}
\left\{\ds\frac{(z-i)e^{imz}}{(z-i)(z+i)}\right\}=\ds\frac{e^{-m}}{2i}$

Then

$$J=2\pi i\ds\frac{e^-m}{2i}=\pi e^{-m}$$

Now $J=\ds\int_{-R PICTURE}^{+R}\ds\frac{dx
e^{imx}}{x^2+1}+\ds\int\ds\frac{dz e^{imz}}{z^2+1}=\pi e^{-m}$

or

$J=\ds\int_{-R}^{+R} \ds\frac{\cos mx}{x^2+1} \,dx +
i\ds\int_{-R}^{+R}\ds\frac{\sin mx}{x^2+1} \,dx +
\ds\int_{PICTURE}\ds\frac{dz e^{imz}}{z^2+1}=\pi e^{-m}$

Now sin is an \un{odd} function so $\ds\int_{-R}^{+R} \sin mx
\cdots =0$

Also as $R \to \infty \ds\int_{PICTURE} \to 0$ so

\begin{eqnarray*} \lim_{R \to \infty} J &amp; = &amp; \lim_{R \to \infty}
\ds\int_{-R}^{+R} \ds\frac{\cos mx}{x^2+1} dx =\pi e^{-m}\\ &amp;
\Rightarrow &amp; 2\ds\int_0^{\infty} \ds\frac{\cos mx}{x^2+1} dx =
\ds\frac{\pi e^{-m}}{2} \end{eqnarray*}


\end{document}
</pre></body></html>