### Author Topic: Problem 4  (Read 15345 times)

#### Kun Guo

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##### Problem 4
« on: November 05, 2012, 11:41:38 PM »
For part c, sinx/x is an even function and its integral is Si(x). Then should we simply get positive infinity?

#### Victor Ivrii

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##### Re: Problem 4
« Reply #1 on: November 06, 2012, 02:41:32 AM »
you need to calculate it (NO tricks)

#### Ian Kivlichan

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##### Re: Problem 4
« Reply #2 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 $\int_{-\infty}^{\infty}{\frac{\sin(x)}{x}dx}$?

#### Victor Ivrii

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##### Re: Problem 4
« Reply #3 on: November 07, 2012, 04:45:49 PM »
Is there a way to go from $\int_{-\infty}^{\infty}{\frac{\sin^2(x)}{x^2}dx}$ to $\int_{-\infty}^{\infty}{\frac{\sin(x)}{x}dx}$?

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

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##### Re: Problem 4
« Reply #4 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*}
« Last Edit: November 07, 2012, 09:32:21 PM by Victor Ivrii »

#### Aida Razi

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##### Re: Problem 4
« Reply #5 on: November 07, 2012, 09:31:18 PM »
Solution is attached!

#### Fanxun Zeng

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##### Re: Problem 4
« Reply #6 on: November 07, 2012, 09:31:49 PM »
Question 4 a solution attached

#### Calvin Arnott

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• OK
##### Re: Problem 4
« Reply #7 on: November 07, 2012, 09:34:32 PM »
Problem 4
Compute the Fourier transform:

Problem a. f(x) = \left\{\begin{aligned} &1 &&: |x| \le a\\ &0 &&: |x| > a \end{aligned} \right.

$$F\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx = \int_{-a}^{a} e^{-i k x}dx \text{ when } a > 0 \text{ and } 0 \text{ otherwise.}$$

$$= -\frac{e^{-i k x}}{i k}\Bigr|_{x=-a}^{a} = \frac{1}{i k}\left(e^{i k a} - e^{-i k a}\right) = \frac{2}{k}\sin\left(a k\right) \blacksquare$$

Problem b.  f(x) = \left\{\begin{aligned} &x &&: |x| \le 1\\ &0 &&: |x| > 1 \end{aligned} \right.

$g\left(x\right) = x f\left(x\right)$ where $f\left(x\right)$ is defined as in part a. Then we have the Fourier transform of $g\left(x\right)$, is given by $G\left(k\right) = i\frac{dF}{dk}$ where $F\left(k\right) = \frac{2}{k}\sin\left(a k\right)$ as derived in part a. Then $G\left(k\right) = i \partial_k \left( \frac{2}{k}\sin\left(a k\right)\right) = \frac{2 i}{k^2} \left(a k \cos \left(a k\right) - \sin \left(a k\right)\right)$ is our transform for $g\left(x\right) \blacksquare$

Problem c. Compute $\int_{-\infty}^{\infty} \frac{\sin\left(x\right)}{x} d x$

Let $f\left(x\right)$ be defined as in part a. with $a = 1$

f(x) = \left\{\begin{aligned} &1 &&: |x| \le 1\\ &0 &&: |x| > 1 \end{aligned} \right.

The Fourier transform of $f\left(x\right)$ is given by $F\left(k\right) = \int_{-\infty}^{\infty}f\left(x\right) e^{-i k x}dx = \frac{2 sin\left( k\right)}{k}\$. Then the inverse Fourier transform is given by: $f\left(x\right) = \int_{-\infty}^{\infty}\frac{2 \sin\left(k\right)}{k} e^{i k x} \frac{dk}{2\pi} \implies \pi f\left(x\right) = \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} e^{i k x} dk$. We use the continuity of $f\left(x\right)$ at $x = 0$ and continuity of the exponential inside of the transform to take the limit as $x \rightarrow 0$.

$$\lim_{x \to 0+} \pi f\left(x\right) \rightarrow \pi \cdot 1 = \lim_{x \to 0+} \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} e^{i k x} dk \rightarrow \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} dk$$

$$\implies \pi = \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} dk \blacksquare$$
« Last Edit: November 18, 2012, 05:41:27 PM by Calvin Arnott »

#### Fanxun Zeng

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##### Re: Problem 4
« Reply #8 on: November 07, 2012, 09:38:02 PM »
Bonus Mark: for a and b, we need to consider special case if the denominator w is 0, in which case  the answer is 2a in a as I attached at 9:31pm, more complete than Aida's original solution posted at 9:30pm. (I think it is better to reply than edit to see the original)

#### Fanxun Zeng

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##### Re: Problem 4
« Reply #9 on: November 07, 2012, 09:44:12 PM »
Hi, I just see Calvin's PDF solution for 4 a and 4 b, as stated above, we need to add the second case that if the denominator w is 0, or denominator k=0, so in addition to that case's answer in part a is 2a, that case's answer in part b is 0. More complete for Bonus Mark!

#### Victor Ivrii

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##### Re: Problem 4
« Reply #10 on: November 07, 2012, 09:50:09 PM »
$k=0$ is trivial (and could be obtained by a limit transition).

#### Fanxun Zeng

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##### Re: Problem 4
« Reply #11 on: November 07, 2012, 10:05:28 PM »
Thanks professor. As I originally suggested, we do need to add k=0 to get complete answer, otherwise mark should be deducted for anyone not considering that, as it's not completely correct, as mathematics is a serious academic field and a denominator cannot be zero! Bonus mark: I have spent half an hour to edited Calvin's incomplete solution as PDF attached for both a and b to get corrected a b c, especially showing how to mathematically get F(k) in case of k=0. Thanks for karma!

#### Victor Ivrii

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##### Re: Problem 4
« Reply #12 on: November 07, 2012, 10:14:17 PM »
Fanxun, thank you very much for the lecture what mathematics is   if you wanted more rigour you should take APM351Y.

I am not sure that everyone appreciates your suggestion about mark reduction and I am looking forward for a little flame war and myself issuing warnings and bans (so, my suggestion to everyone--don't start).
« Last Edit: November 08, 2012, 02:10:48 AM by Victor Ivrii »

#### Fanxun Zeng

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##### Re: Problem 4
« Reply #13 on: November 07, 2012, 10:44:17 PM »
Thanks professor. For parts a and b denominator issues maybe 3 TAs can vote for either deduction or bonus for who did right, we students don't discuss here. Correction: as PDF can't be displayed and need to be downloaded, for convenience, in fact what I posted at 10:05pm above is not in PDF but in JPG. Mine is complete a b c solutions I edited from Calvin's incomplete one.