You are not defining the integral limits correctly.
For example:
syms a b k L w0 x
krnl = symfun(a*cos(-j*k*w0*x) + j*b*sin(-j*k*w0*x), x) % Transform kernel
F1 = int(-1*krnl, x, -L, 0); % -L -> 0
F2 = int(+1*krnl, x, 0, L); % 0 -> +L
FT = F1 + F2; % Add integrals
FT = subs(FT, L, 1) % L = 1
FT = simplify(collect(FT))
produces (in R2014a):
FT =
(4*b*sinh((k*w0)/2)^2)/(k*w0)
MATLAB (correctly) substituted sinh for the sin of an imaginary argument, so this is equivalent to:
FT =
(4*b*sin((j*k*w0)/2)^2)/(k*w0)
The individual ‘b’ coefficients are functions of k. The ‘a’ coefficients are identically zero.
The frequency argument, w0, is the fundamental frequency of the sampled signal. I refer you to the Wikipedia article on Fourier series for the details.
