How can I simplify the messy integral in the Fourier series for f_n(x)?

[quasar987]

I got this set of functions defined on [-pi, pi]\{0} by

$$f_n(x)=\frac{sin((n+1/2)x)}{sin(x/2)}$$

and am asked to find its Fourier expansion.

The integral a_n is pretty messy. I've tried using the identity sinAcoB = 0.5(...) but then I'm left with two integrals of the type sin([n±m+1/2]x)/sin(x/2). I've tried pluging numerical values of n and m in "The integrator" and the results are not happy-looking. Type for exemple "Sin[1.5x]/Sin[10x]" at http://integrals.wolfram.com/.

So I was wondering if there was a shortcut to this integral? Maybe linked to the integral bounds? I've found that the (smallest )period of sin([n-m+1/2]x) is $4\pi/(2(n-m)+1)$ while that of sin(x/2) is 4pi. So they are both of period 4pi.

 

[Physics Monkey]

This is one of those problems where the easiest way to find the Fourier series is not by direct integration. That being said, the integrals can be done for any n and m using trig identities. In other words, the pesky denominator can be gotten rid of by using trig identities over and over again on the numerator.

Another way to do the problem is to write everything in terms of complex exponentials. With some simplification and an appropriate Taylor expansion you can get the right answer without ever computing an integral.

A final hint: the Fourier series is very simple and finite.

Hope you can make some progress.

 

[quasar987]

I must have made a mistake because I end up with the serie being infinite and I didn't use a Taylor expansion.

What I did:

$$a_m=\frac{1}{\pi}\int_{-\pi}^{\pi}\frac{sin([n+1/2]x)}{sin(x/2)}cos(mx)dx = \frac{1}{\pi}\int_0^{\pi}\mbox{Im}\left\{\frac{e^{i([n-m+1/2]x)}+e^{i([n+m+1/2]x)}}{e^{i(x/2)}}\right\}dx$$

where I use the trig identity sinAcosB = {sin(A-B) + sin(A+B)}/2 before switching to complex exponential notation.

Now set a = m - n and b = m + n. and multiply the integrand by 1 in the form $e^{i(x/2)}e^{-i(x/2)}$, which leads to

[tex]a_m = \frac{1}{\pi}\int_{0}^{\pi}\mbox{Im}\left\{e^{iax}+e^{ibx}\right\}dx[/itex]<br /> <br /> Anything wrong with that so far?[/tex]

 

[Physics Monkey]

Here is at least part of the problem:
$$Im\left( \frac{1}{e^{ix/2}} \right) = - \sin{x/2} \neq \frac{1}{\sin{x/2}}$$