- #1
quasar987
Science Advisor
Homework Helper
Gold Member
- 4,793
- 22
I got this set of functions defined on [-pi, pi]\{0} by
[tex]f_n(x)=\frac{sin((n+1/2)x)}{sin(x/2)}[/tex]
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 [itex]4\pi/(2(n-m)+1)[/itex] while that of sin(x/2) is 4pi. So they are both of period 4pi.
[tex]f_n(x)=\frac{sin((n+1/2)x)}{sin(x/2)}[/tex]
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 [itex]4\pi/(2(n-m)+1)[/itex] while that of sin(x/2) is 4pi. So they are both of period 4pi.