Limit & Sum Problem (Parseval's Theorem)

[daon]

Its been while since I asked a question... So here goes nothing!

This is from "Baby Rudin." Pg. 199, #12, (d).

Using:

\(\displaystyle \lim _{\delta \rightarrow 0} \,\, \,\, \,\, \sum_{n \ge 1} \frac{sin^2(n\delta)}{n^2\delta} = \frac{\pi}{2}\)

How to show:

\(\displaystyle \int_0^{\infty} (\frac{\sin{x}}{x})^2dx = \frac{\pi}{2}\)

Parts (a),(b),(c) have been proven. They are quoted here:

Part a: Compute the Fourier coefficients of f.
\(\displaystyle f(x)=0\) when \(\displaystyle \delta < |x| \le \pi\) and \(\displaystyle f(x)=1\) when \(\displaystyle |x| \le \delta\).
\(\displaystyle f(x+2\pi)=f(x) \,\, \forall x\), \(\displaystyle 0 < \delta < \pi\)

Part b: Show
\(\displaystyle \sum _{n\ge1} \frac{\sin{n\delta}}{n} = \frac{\pi-\delta}{2}\)

Part c: Show, using Parseval's Thm:
\(\displaystyle \sum _{n\ge1} \frac{(\sin{n\delta})^2}{n^2\delta} = \frac{\pi-\delta}{2}\)

Thanks!

 

[pka]

To tell my age, my copy of “Baby Rudin” does not even discuss that problem.
Moreover, I am not really sure of what you as asking.
At first I though that applying an approximating sum to that improper integral.
That is just a guess.

Here is an interesting collection of problems that go with Rudin.
But I did not find reference to this problem in it.
http://www.math.berkeley.edu/~gbergman/ug.hndts

I wish I could be more helpful on this.

 

[daon]

This is the 3rd edition copy that I'm using. The international edition is available online somewhere for free, but I forget the address.

Anyway... I was thinking I might be able to use upper and lower sums to "trap" the integral in \(\displaystyle [\frac{\pi}{2} - \delta, \frac{\pi}{2} + \delta]\). The only problem is that it is improper, so I cannot assume there is a specific partition.

edit: Here is a copy: PMA 3rd, but I can't open it at this computer to verify its language.