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\vtop{\hsize=3.86in \noindent {\bf Problem statement} The function
$\displaystyle{{\sin x}\over x}$ can be extended to have value $0$
when $x=0$ (for example, using l'H\^opital's Rule). This function
occurs in many applications, such as signal processing. A graph is
shown to the right. The ``bumps'' when $x>0$ touch the displayed
curves $\displaystyle y=\pm {1\over x}$.  When $x$ gets large, the
area under the bumps, positive and negative, almost cancels. The
quantity $\displaystyle \int_0^\infty {{\sin x}\over x}\, dx$ is
finite. Here is one way to find the exact value of this integral.}

\medskip

\noindent a) Suppose $f(t)=\displaystyle \int_0^\infty \left({{\sin
x}\over x}\right)e^{-tx}\,dx$. Compute $f'(t)$, the derivative of $f$
with respect to $t$. The resulting integral can be evaluated using
integration by parts, and you should conclude that $\displaystyle
f'(t)=-{1\over{1+t^2}}$.

\medskip

\noindent b) Solve the differential equation $\displaystyle
f'(t)=-{1\over{1+t^2}}$. If $t\to +\infty$, the value of the integral
defining $f(t)$ approaches $0$. Then the general solution of the
differential equation which involves an arbitrary additive constant
can be used to get an exact formula for $(t)$.

\medskip

\noindent c) So $f(0)$ is $\displaystyle \int_0^\infty \left({{\sin
x}\over x}\right)e^{-{\bf 0}x}\,dx= \int_0^\infty {{\sin x}\over
x}\,dx$ which can be evaluated using b).

\vfil\eject\end

A nice discussion of the method used in both sg50 and this problem is
www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf.