You are correct. Since g(x) is neither even nor odd, you have to split your integrals over the appropriate intervals:
g is periodic of period 2piand, for x 2 (−pi, pi] are
given by the formulae:
g(x) = 0, -pi < x </= 0
sin x 0 < x < pi
I really dont understand how to go about this, since it seems neither to be an even or odd function. Do I need to integrate separately over -pi to 0 and then 0 to pi?
Could someone please check my working so far..
I get a[n] = 1/2pi[((-1)^n+1)/(n+1) + ((-1)^(n+1))/(1-n))]
a = 1/2
Is this correct? What about when n = 1? I did as above but this meant I could not eliminate anything for being an odd/even function..