How Can Fourier Series Aid in Solving the Sinc Function Integral?

[Incand]

Homework Statement

Compute $\int_0^\infty \frac{\sin x}{x}dx$
using that $\frac{\sin x}{x} = \frac{b_0}{2} +\sum_1^\infty b_n \cos nx \; \; , \; \; 0 < x < \pi$
with
$b_n = \frac{1}{\pi} \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}{x}dx$.

Homework Equations

Perhaps the following convergence theorem:
If $f$ is $2\pi$-periodic and piecewise smooth on $\mathbf R$ then
$\lim_{N\to \infty} S_:N^f(\theta ) = \frac{1}{2}\left[ f(\theta -)+ f(\theta +) \right]$
where $S_N^f(\theta )$ is the partial sum of the Fourier series of $f(\theta )$.

The Attempt at a Solution

If we choose $x = 0$ we have for the right side
$\frac{1}{2\pi} \int_{-\pi}^\pi \frac{\sin y}{y}dy + \frac{1}{\pi}\sum_1^\infty \int_{(n-1)\pi}^{(n+1)\pi} =\frac{1}{\pi} \int_0^\pi \frac{\sin y}{y}dy + \int_0^\infty \frac{\sin x}{x}dx + \int_\pi^\infty \frac{\sin y}{y}dy = \frac{2}{\pi}\int_0^\infty \frac{\sin y}{y}dy$
The left side should be $\frac{1}{2}$ since $\lim_{x\to \pi} \frac{\sin x}{x} = 0$ and $\lim_{x\to 0} \frac{\sin x}{x} = 1$. we get
$\int_0^\infty \frac{\sin x}{x}dx = \frac{\pi}{4}$. However the integral should be $\frac{\pi}{2}$ so i made a mistake somewhere.

 

[vela]

Homework Statement

Compute $\int_0^\infty \frac{\sin x}{x}dx$
using that $\frac{\sin x}{x} = \frac{b_0}{2} +\sum_1^\infty b_n \cos nx \; \; , \; \; 0 < x < \pi$
with
$b_n = \frac{1}{\pi} \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin x}{x}dx$.

Homework Equations

Perhaps the following convergence theorem:
If $f$ is $2\pi$-periodic and piecewise smooth on $\mathbf R$ then
$\lim_{N\to \infty} S_:N^f(\theta ) = \frac{1}{2}\left[ f(\theta -)+ f(\theta +) \right]$
where $S_N^f(\theta )$ is the partial sum of the Fourier series of $f(\theta )$.

The Attempt at a Solution

If we choose $x = 0$ we have for the right side
$$\frac{1}{2\pi} \int_{-\pi}^\pi \frac{\sin y}{y}dy + \frac{1}{\pi}\sum_1^\infty \int_{(n-1)\pi}^{(n+1)\pi} =\frac{1}{\pi} \int_0^\pi \frac{\sin y}{y}dy + \int_0^\infty \frac{\sin x}{x}dx + \int_\pi^\infty \frac{\sin y}{y}dy = \frac{2}{\pi}\int_0^\infty \frac{\sin y}{y}dy$$

What happened to $b_n$ in the summation? I have no idea what you did subsequently.

The left side should be $\frac{1}{2}$ since $\lim_{x\to \pi} \frac{\sin x}{x} = 0$ and $\lim_{x\to 0} \frac{\sin x}{x} = 1$. we get
$\int_0^\infty \frac{\sin x}{x}dx = \frac{\pi}{4}$. However the integral should be $\frac{\pi}{2}$ so i made a mistake somewhere.

$\frac{\sin x}{x}$ is continuous, so the theorem you're trying isn't really useful.

 

[Incand]

What happened to $b_n$ in the summation? I have no idea what you did subsequently..

Sorry it appears there's a few typos in my post. $\frac{\sin y}{y}dy$ disappeared from one integral and the theorem should say $S_N$ of course.
What happends in the summation is that
$\sum_1^\infty \int_{(n-1)\pi}^{(n+1)\pi} \frac{\sin y}{y}dy = \int_0^{2\pi}\frac{\sin y}{y}dy + \int_\pi^{3\pi} \frac{\sin y}{y}dy + \int_{2\pi}^{4\pi} \frac{\sin y}{y}dy + \int_{3\pi}^{5\pi} \frac{\sin y}{y}dy + \dots$
Now assuming we can rearrange these terms (The series being absolute convergent) we can split the sum up into two parts
$\int_0^{2\pi} \frac{\sin y}{y}dy + \int_{2\pi}^{4\pi} \frac{\sin y}{y}dy + \dots = \int_0^\infty \frac{\sin y}{y}dy$
and
$\int_{\pi}^{3\pi} \frac{\sin y}{y}dy + \int_{3\pi}^{5\pi} \frac{\sin y}{y}dy + \dots = \int_\pi^\infty \frac{\sin y}{y}dy.$
but we also have a term $\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{\sin y}{y}dy = \frac{1}{\pi} \int_0^\pi \frac{\sin y}{y}dy$
adding these together we get that the right hand side is $\frac{2}{\pi}\int_0^\infty \frac{\sin y}{y}dy$

$\frac{\sin x}{x}$ is continuous, so the theorem you're trying isn't really useful

I disagree, $\frac{\sin x}{x}$ is continuous however we don't have that. We have $\frac{\sin x}{x}$ in an interval $x \in (0,\pi)$ where the function repeat itself in $(\pi, 2\pi)$ etc. i.e. a $\pi$-periodic function that is piecewise continuous but not continuous.

 

[vela]

Oh, OK, I understand what you did.

You're using a Fourier cosine series, so you should be looking at the even extension of the function. I don't think your expression for $b_n$ is correct because what the function is doing between $(n-1)\pi$ and $(n+1)\pi$ isn't really relevant to the periodic extension of the function restricted to the interval $[0,\pi)$.

Usually, you have
$$b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x)\cos nx\,dx.$$ Did you prove earlier that this is the same as the expression you listed in the problem statement?

You may want to provide more background if I'm way off base here.

 

[Incand]

Yes the question was originally split into two parts where the first was to prove the identify.
Here's the full original question:
Show that $\frac{\sin x}{x} = \frac{b_0}{2} + \sum_1^\infty b_n\cos nx \; \; \; \; \; (0<x<\pi)$
where
$b_n = \frac{1}{\pi}\int_{(n-1)\pi}^{(n+1)\pi}\frac{\sin x}{x} dx$.
Use this result to compute
$\int_0^\infty \frac{\sin x}{x}dx$.

Proof of first part
We take the Fourier cosine series on $(0,\pi)$.
$b_n = \frac{2}{\pi} \int_0^\pi \frac{\sin x \cos nx}{x}dx = \frac{1}{\pi} \int_0^\pi \frac{\sin (n+1)x + \sin (1-n)x}{x}dx = \frac{1}{\pi} \int_0^\pi \frac{\sin (n+1)x}{x}dx -\frac{1}{\pi}\int_{-\pi}^0 \frac{\sin(n-1)x}{x}dx =\frac{1}{\pi}\int_{(n-1)\pi}^{(n+1)\pi}\frac{\sin x}{x} dx$
Where i used the trigonometric identity $\sin x \cos y = \frac{\sin (x+y) + \sin (x-y)}{2}$ (first step) and that $\frac{\sin x}{x}$ is an even function (second step). Since the information was given in the exercise I only wrote the step I had trouble with.

When writing this out I think I see where I went wrong in the second part. Since $\frac{\sin x}{x}$ is even and the Fourier cosine series is an even extension i actually have a continuous function just like you said. And then i get the right answer. Cheers! is this correct?

 

[vela]

This is a cool problem. I'll have to make note of it for possible use later. :)

Yes, I believe you have correctly identified where you went wrong initially.