How to evaluate the Fourier Series of $\frac{\sin(\alpha\sin(\omega t))}{\alpha\sin(\omega t)}\exp(-j\beta\sin(\omega t))$?

Denote $f(t) = x(t)+iy(t)$. Then we have that

$$x(t)\pm y(t) = \frac{\sin(\alpha\sin(\omega t))\cos(\beta\sin(\omega t)) \mp \cos(\alpha\sin(\omega t))\sin(\beta\sin(\omega t))}{\alpha \sin(\omega t)} = \frac{\sin((\alpha\mp \beta)\sin(\omega t))}{\alpha \sin(\omega t)}$$

$$= \frac{1}{\alpha}\sum_{n=0}^\infty\frac{(-1)^n(\alpha\mp\beta)^{2n+1}}{(2n+1)!}\left(\frac{e^{i\omega t}-e^{-i\omega t}}{2i}\right)^{2n} = \frac{\alpha\mp\beta}{\alpha}\sum_{n=0}^\infty\sum_{k=0}^{2n}\frac{\left(\frac{\alpha\mp\beta}{2}\right)^{2n}(-1)^k}{(2n+1)!}{2n \choose k}e^{i2(n-k)\omega t}$$

Immediately the last representation implies that $\hat{f}_l = 0$ for $l$ odd. So now consider $l=2m$. The only term to survive the Fourier series integral is the term $k=n-m$

$$\hat{x}_{2m}\pm\hat{y}_{2m} = \frac{\alpha\mp\beta}{\alpha}\sum_{n=m}^\infty\frac{(-1)^{n-m}}{(2n+1)!}{2n \choose n-m}\left(\frac{\alpha\mp\beta}{2}\right)^{2n}$$

Adding and subtracting these series gives the coefficients for $f$ by linearity.