1. ## Riemann-Lebesgue

For this question I have to use the Riemann-Lebesgue Lemma to prove that...

$\displaystyle \lim_{N \rightarrow \infty} \int_{-\pi}^{\pi} \Bigg{(} \frac{1}{\sin(x/2)} - \frac{2}{x} \Bigg{)} \sin((N + 1/2)x) dx = 0$.

With the Riemann-Lebesgue Lemma lemma being...

If $\displaystyle g$ is an integrable function on $\displaystyle [-\pi, \pi]$, then $\displaystyle \lim_{|n| \rightarrow \infty} G(n) = 0.$.

Where $\displaystyle G(n) = \frac{-1}{2\pi} \int_{-\pi}^{\pi} g(x - (\pi/n))e^{-inx} dx$

Now I can do this question easily by expanding the brackets but I'm unsure as to how to apply the lemma as clearly expanding the brackets never uses that...

For this question I have to use the Riemann-Lebesgue Lemma to prove that...

$\displaystyle \lim_{N \rightarrow \infty} \int_{-\pi}^{\pi} \Bigg{(} \frac{1}{\sin(x/2)} - \frac{2}{x} \Bigg{)} \sin((N + 1/2)x) dx = 0$.

With the Riemann-Lebesgue Lemma lemma being...

If $\displaystyle g$ is an integrable function on $\displaystyle [-\pi, \pi]$, then $\displaystyle \lim_{|n| \rightarrow \infty} G(n) = 0.$.

Where $\displaystyle G(n) = \frac{-1}{2\pi} \int_{-\pi}^{\pi} g(x - (\pi/n))e^{-inx} dx$

Now I can do this question easily by expanding the brackets but I'm unsure as to how to apply the lemma as clearly expanding the brackets never uses
that...
Well for the purposes of sparing myself the burdens of Latex and dodging any transcription errors, here are the main moves -- and youre right, if one
were to distribute the brackets this puppy would become sympatico, as it were -- at least from what i can tell. Instead one has to have "mathematical
maturity" and recognize something from somewhere else, even though in the end it's just a bunch of manipulations, illucidating nothing profound, at
least from my perch. This of course assumes what im doing is right, which is a distinct risk because this problem covers areas for which i never took
courses, and the courses i did take were over 5 years ago.
In this case, the function you want to isolate is:

$\displaystyle \frac{\sin((N + 1/2)t) }{\sin\frac{t}{2}}$

Clearly the way to do that is by
(1)Express $\displaystyle \Bigg{(} \frac{1}{\sin(t/2)} - \frac{2}{t} \Bigg{)}$ as $\displaystyle \Bigg{(} \frac{t-2\sin(t/2)}{t\sin(t/2)} \Bigg{)}$
(2)Distribute only $\displaystyle \sin(t/2)$ from the denominator in (1).
Now you're staring at:
$\displaystyle \lim_{N \rightarrow \infty} \int_{-\pi}^{\pi} \Bigg{(} \frac{t-2\sin(t/2)}{t} \Bigg{)}\frac{\sin((N + 1/2)t) }{\sin\frac{t}{2}}dt$
Because you can express it like this, the key observation it appears, at least to me is that the integral can then be seen as the result of a
convolution of a function $\displaystyle g$, to be defined, with that of the so-called Dirichlet kernel.
The Dirichlet kernel can be expressed as:
$\displaystyle D_n(t) = \sum_{k=-n}^ne^{ikt}$ = $\displaystyle \frac{\sin((n + 1/2)t) }{\sin\frac{t}{2}}$ (*)
see here: [html]http://planetmath.org/encyclopedia/DirchletKernel.html[/html]

Then, where $\displaystyle *$ means convolution,
$\displaystyle (D_n(t)*g)(x) = \frac{1}{2\pi} \int_{-\pi}^{\pi} g(y)D_n(x-y)dy = \sum_{k=-n}^nG(k)e^{ikx}$ (**)

where
$\displaystyle G(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} g(x)e^{-ikx}dx$

Now $\displaystyle \frac{1}{2\pi} \int_{-\pi}^{\pi} g(x)e^{-ikx}dx$ = $\displaystyle \frac{-1}{2\pi} \int_{-\pi}^{\pi}g(x-\frac{\pi}{k})$$\displaystyle e^{-ikx}dx$ since $\displaystyle e^{i\pi}=-1$
Note now that we define $\displaystyle g(x-\frac{\pi}{k}) = \frac{(x-\frac{\pi}{k})-2\sin(\frac{x-\frac{\pi}{k}}{2})}{x-\frac{\pi}{k}}$.
This function is integrable on $\displaystyle [-\pi, \pi]$.
Now that we've idenitfied our function $\displaystyle G(.)$, going back to (**) we conclude
$\displaystyle (D_N(t)*g)(x) = G(N)\frac{\sin((N + 1/2)x) }{\sin\frac{x}{2}}$ because of (*)
The result thus follows:
$\displaystyle \lim_{N \rightarrow \infty} \int_{-\pi}^{\pi} \Bigg{(} \frac{1}{\sin(x/2)} - \frac{2}{x} \Bigg{)} \sin((N + 1/2)x) dx = 0$
Note that $\displaystyle G(N)$ can be interpreted as the Nth Fourier coefficient of $\displaystyle g$.
Again, as you mentioned this problem would have followed immediately from direct application of the lemma, which shows up in the form of sines and cosines, had we not been constrained to use the more general version of the statement of the lemma.

please let me know if the reasoning above resembles that which appears in your solution set when availible. Regards.