# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment 6 => Topic started by: Kun Guo on November 05, 2012, 11:41:38 PM

Title: Problem 4
Post by: Kun Guo 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?
Title: Re: Problem 4
Post by: Victor Ivrii on November 06, 2012, 02:41:32 AM
you need to calculate it (NO tricks)
Title: Re: Problem 4
Post by: 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 $\int_{-\infty}^{\infty}{\frac{\sin(x)}{x}dx}$?
Title: Re: Problem 4
Post by: Victor Ivrii 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
Title: Re: Problem 4
Post by: Zarak Mahmud 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*}
Title: Re: Problem 4
Post by: Aida Razi on November 07, 2012, 09:31:18 PM
Solution is attached!
Title: Re: Problem 4
Post by: Fanxun Zeng on November 07, 2012, 09:31:49 PM
Question 4 a solution attached
Title: Re: Problem 4
Post by: Calvin Arnott 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 $x \rightarrow 0$.

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

$$\implies \pi = \int_{-\infty}^{\infty}\frac{\sin\left(k\right)}{k} dk \blacksquare$$
Title: Re: Problem 4
Post by: Fanxun Zeng 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)
Title: Re: Problem 4
Post by: Fanxun Zeng 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!
Title: Re: Problem 4
Post by: Victor Ivrii on November 07, 2012, 09:50:09 PM
$k=0$ is trivial (and could be obtained by a limit transition).
Title: Re: Problem 4
Post by: Fanxun Zeng 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!
Title: Re: Problem 4
Post by: Victor Ivrii on November 07, 2012, 10:14:17 PM
Fanxun, thank you very much for the lecture what mathematics is :D  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).
Title: Re: Problem 4
Post by: Fanxun Zeng 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.
Title: Re: Problem 4
Post by: Victor Ivrii on November 08, 2012, 02:17:21 AM
No votes. This course is really not very rigorous but APM351 is much less rigorous than any comparable math course and the reason is simple: the really rigorous and in the same time meaningful PDE course requires serious Real Analysis (Complex Analysis does not hurt either) which Math Specialist students have in the 4th year (so Graduate PDE class is rigorous). You attempts to bring more rigour here are well intended but not productive (sure, we should maintain some level of rigour but it is not very high).
Title: Re: Problem 4
Post by: Victor Ivrii on November 08, 2012, 04:32:56 PM
Fanxun, I decided that you deserve a karma, for raising a good question. In Lecture 15 (http://www.math.toronto.edu/courses/apm346h1/20129/L15.html) we proved Theorem 2 (http://www.math.toronto.edu/courses/apm346h1/20129/L15.html#thm-15.2):

Theorem 2. Let $f$ be a piecewise continuously differentiable function. Then the Fourier series
\begin{equation}
\frac{a_0}{2}+\sum_{n=1}^\infty \Bigl(a_n\cos \bigl(\frac{\pi n x}{l}\bigr) + a_n\cos \bigl(\frac{\pi n x}{l}\bigr)\Bigr)
\label{eq-1}
\end{equation}
converges to

(b) $\frac{1}{2}\bigl(f(x+0)+f(x-0)\bigr)$  if $x$ is internal point and $f$ is discontinuous at $x$.

Exactly the same statement holds for Integral Fourier
\begin{equation}
\int_0^\infty \Bigl(A(\omega) \cos (\omega x) + B(\omega)\sin (\omega x)\Bigr)\,d\omega
\label{eq-2}
\end{equation}
where $A(\omega)$ and $B(\omega)$ are $\cos$- and $\sin$-Fourier transforms.

None of them however holds for Fourier series or Fourier Integral in the complex form:
\begin{gather}
\sum_{n=-\infty}^\infty c\_n e^{i\frac{\pi n x}{l}},\label{eq-3}\\
\int_{-\infty}^\infty C(\omega)e^{i\omega x}\,d\omega.\label{eq-4}
\end{gather}

Why and what remedy do we have? If we consider definition of the partial sum of (\ref{eq-1}) and then rewrite in the complex form  and similar deal with (\ref{eq-4}) we see that in fact we should look at
\begin{gather}
\lim_{N\to \infty} \sum_{n=-N}^N c_n e^{i\frac{\pi n x}{l}},\label{eq-5}\\
\lim_{N\to \infty} \int_{-N}^N C(\omega)e^{i\omega x}\,d\omega\label{eq-6}.
\end{gather}

Meanwhile convergence in (\ref{eq-3}) and (\ref{eq-4}) means more than this:
\begin{gather}
\lim_{M,N\to \infty} \sum_{n=-M}^N c_n e^{i\frac{\pi n x}{l}},\label{eq-7}\\
\lim_{M,N\to \infty} \int_{-M}^N C(\omega)e^{i\omega x}\,d\omega\label{eq-8}
\end{gather}
where $M,N$ tend to $\infty$ independently. So the remedy is simple: understand convergence as in (\ref{eq-5}), (\ref{eq-6}) rather than as in (\ref{eq-7}), (\ref{eq-8}) . For integrals such limit is called essential value of integral and is denoted by
\begin{equation*}
\operatorname{ess}-\int_{-\infty}^\infty G(\omega)\,d\omega
\end{equation*}
or
\begin{equation*}
\operatorname{vrai}-\int_{-\infty}^\infty G(\omega)\,d\omega
\end{equation*}

BTW similarly is defined \begin{equation*}
\operatorname{ess}-\int_{a}^b G(\omega)\,d\omega:= \lim_{\epsilon\to +0} \Bigl(\int_a^{c-\epsilon}G(\omega)\,d\omega+
\int_{c+\epsilon}^bG(\omega)\,d\omega\Bigr)
\end{equation*}
if there is a singularity at $c\in (a,b)$.

This is more general than improper integrals studied in the end of Calculus I. Those who took Complex Variables encountered such notion.