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« **on:** November 07, 2012, 09:30:01 PM »
**Part (a):**

\begin{equation*}

\hat{f}(\omega)=

\frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx\\

= \frac{1}{2\pi}\int_{-a}^{a} f(x)e^{-i\omega x}\,dx\\

= \frac{1}{2\pi} \frac{e^{-\omega x}}{i\omega} \big|_{-a}^{a}\\

= \frac{i}{2\pi \omega}(e^{-i\omega a}-e^{i\omega a})\\

= \frac{e^{i\omega a}-e^{-i\omega a}}{2i\pi \omega}\\

=\frac{sin(\omega a)}{\pi \omega}.

\end{equation*}

**Part (b):**

Using the result from **part (a)** along with Theorem 3d:

\begin{equation*}

g = xf(x)\implies

\hat{g}(\omega) = i\hat{f}(\omega)\\

=i\frac{d}{d\omega}\big(\frac{sin(\omega a)}{\pi \omega} \big)\\

=i \frac{a\omega \cos{\omega a} - \sin{\omega a}}{\pi \omega^2}\\

=\frac{ia\cos{\omega a}}{\pi \omega} - \frac{i\sin{}\omega a}{\pi \omega^2}.

\end{equation*}

**Part (c):**

Let $$ f(x)=\left\{\begin{aligned} & 1&& |x|\le a,\\ & 0 && |x|> a;\end{aligned}\right.$$

Then using the result obtained from **part (a)**, we have a fourier transform pair:

\begin{equation*}

f(x) = \int_{-\infty}^\infty \hat{f}(\omega)e^{i\omega x}\,d\omega\\

\implies f(x) = \int_{-\infty}^\infty \frac{\sin{\omega a}}{\pi \omega}e^{i\omega x}\,d\omega\\

\end{equation*}

Switch $\omega$ with $x$.

\begin{equation*}

\implies \left\{\begin{aligned} & \pi&& |\omega|\le a,\\ & 0 && |\omega|> a;\end{aligned}\right.= \int_{-\infty}^\infty \frac{\sin{x a}}{x}e^{i\omega x}\,dx\\

\end{equation*}

Now let $a = 1$, and $\omega = 0$. For these values the function gives us $\pi$.

Thus,

\begin{equation*}

\int_{-\infty}^\infty \frac{\sin{x}}{x}\,dx = \pi.

\end{equation*}