Show that for $\alpha$, not an integer, the Fourier series of
$$\frac{\pi}{\sin(\alpha n)} e^{i(\pi-x)\alpha}$$
on $[0,2\pi]$
is given by $$\sum_{n = -\infty}^{\infty} \frac{e^{inx}}{\alpha +n} $$
My attempt:
I compute $$a_n = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{\pi}{\sin(\alpha n)} e^{i(\pi-x)\alpha} e^{-inx} dx$$
$$ = \frac{1}{2\pi} \frac{\pi e^{i\alpha \pi}}{\sin(\alpha n)} \int_{0}^{2\pi} e^{-ix\alpha} e^{-inx}dx =\frac{1}{2\pi} \frac{\pi e^{i\alpha \pi}}{\sin(\alpha n)} \left(-\frac{i(1-e^{-2i\pi(\alpha +n)})}{\alpha +n}\right)$$
but I should get as a result $\frac{1}{\alpha +n}$
What I am doing wrong?