                                          
                        
                                          
                        
\documentclass[12pt]{article}
  \setlength{\oddsidemargin}{0pt}
\setlength{\topmargin}{-0.5truein}
\setlength{\textwidth}{6.5truein}
\setlength{\textheight}{9truein}
  \pagestyle{empty}
                                          
                                      
                                          
                                      
\begin{document}
 \centerline{\large Math.\ 401, Sec.\ 
501\hfill    Spring, 2006}
 \bigskip
\centerline{\bfseries\large Homework 11, 
due April 17}
\bigskip
                                          
                                      
%\newcommand{\erf}{\mathop{\rm erf}\nolimits}
                                          
                                      
\begin{enumerate}
                                          
                                      


\item  {\it [Schaum's, p.\ 94, Ex.\ 5.24]\/} \     
\begin{enumerate}
\item Find the Fourier transform of 
$f(x) = \left\{\begin{array}{ll} \displaystyle\frac1{2\epsilon} & 
\mbox{if 
$|x|<\epsilon$}, 
\\ \noalign{\smallskip}
                          0 & \mbox{if $|x|\ge\epsilon$}. 
\end{array}\right.
$
\item Find the limit of this transform as $\epsilon\to 0^+$.
Discuss the result.
\end{enumerate}

\item  {\it [Schaum's, p.\ 94--95, Ex.\ 5.26 and 5.31]\/} \
Find the Fourier sine transform and the Fourier cosine transform 
of
 $$f(x)  =\left\{\begin{array}{ll} 1 &\mbox{if $0\le x<1$}, \\
\noalign{\smallskip}                   0 & \mbox{if $x \ge 1$}. 
\end{array}\right.
$$
Use the results to show that
\begin{enumerate}
\item\ $\displaystyle \int_0^\infty \left({1-\cos x\over 
x}\right)^2
\,dx = 
\frac{\pi}2$
\item  \ $\displaystyle \int_0^\infty {\sin^2 x\over x^2}  \,dx = 
\frac{\pi}2$
\end{enumerate}





\item {\it [Schaum's, p.\ 94, Ex.\ 5.37]\/} \   
Find $y(x)$ given that
$$ \int_{-\infty}^\infty y(u)y(x-u)\,du = e^{-x^2}.$$

\item {\it [Schaum's, p.\ 94, Ex.\ 5.39]\/} \   
Prove that $f*(g*h) = (f*g)*h$.

 \item  Show that  differentiation of a Fourier transform with
respect to $\omega$ corresponds to multiplication of the original
function by $-ix$,
\begin{enumerate}
 \item starting from the equation
 $$g(x) = \frac1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{i\omega 
x}\,\hat g(\omega)\, d\omega,$$
 applied to $\hat g(\omega ) = \hat f\,'(\omega)$;
 \item alternatively, starting from the equation
 $$\hat f(\omega)  = \frac1{\sqrt{2\pi}} \int_{-\infty}^\infty 
e^{-i\omega x}\, f(x)\, dx.$$
\end{enumerate}
{\sl Note:\/} With Constanda's definition of the Fourier 
transform one would get $+ix$ instead of $-ix$.

 \item  Starting from the Fourier transform formulas in
exponential form (see previous problem, with $g=f$),
 derive the formulas for the Fourier sine transform.
 (Apply the Fourier transform formulas to the odd extension of $f$,
$f$ being an arbitrary nice function defined on $[0,\infty)$.)


\item  {\it [Schaum's, p.\ 94, Ex.\ 5.27]\/} \
\begin{enumerate}
\item Find the Fourier sine transform of $e^{-x}$ 
($x\ge0$).
\item Use (a) to show that for $m>0$,
$$\int_0^\infty {x\sin mx \over x^2 +1} \, dx = \frac{\pi}2 \, 
e^{-m}.$$
\item The formula in (b) fails when $m=0$.
Why does this not contradict the basic Fourier transform theorem?
(See ``Remarks" on p.~143 of Constanda, or ``Convergence 
theorems" 
in the section ``More on Fourier transforms" of the class notes, 
or ``Fourier's integral theorem" on p.~80 of Schaum's.)
\end{enumerate}

                                           
                         

\item {\it  [Schaum's, p.\ 94, Ex.\ 5.30]\/} \
Use Parseval's identity and your knowledge of the Fourier sine 
and cosine transforms of $e^{-x}$ ($x>0$) to evaluate

\begin{enumerate}
\item \ $\displaystyle \int_0^\infty \frac{dx}{(x^2+1)^2} $

\item \ $\displaystyle \int_0^\infty \frac{x^2\, dx}{(x^2+1)^2} $
\end{enumerate}


\end{enumerate}
\end{document}
