APM346-2012 > Home Assignment 6

Problem 4

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Kun Guo:
For part c, sinx/x is an even function and its integral is Si(x). Then should we simply get positive infinity?

Victor Ivrii:
you need to calculate it (NO tricks)

Ian Kivlichan:
Is there a way to go from $\int_{-\infty}^{\infty}{\frac{\sin^2(x)}{x^2}dx}$ to $\int_{-\infty}^{\infty}{\frac{\sin(x)}{x}dx}$?

Victor Ivrii:

--- Quote from: Ian Kivlichan on November 07, 2012, 04:32:43 PM ---Is there a way to go from $\int_{-\infty}^{\infty}{\frac{\sin^2(x)}{x^2}dx}$ to $\int_{-\infty}^{\infty}{\frac{\sin(x)}{x}dx}$?

--- End quote ---

Sure:
$$\int_0^{\infty} \frac{\sin^2(x)}{x^2}dx = -\int_0^{\infty} \sin^2(x) dx^{-1} $$
and integrate by parts.

This is not part of HW

Zarak Mahmud:
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*}

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