Note: $j=i=\sqrt{-1}$
What is the Fourier transform of $h[n]=\frac{2\sin(0.5\pi)}{\pi}$?
$$h=\frac{2}{\pi n}\times \frac{e^{j0.5\pi n}-e^{-j0.5\pi n}}{2j}$$
I know by memorization of the conversion formula that the Fourier transform of $e^{j\omega_o n}$ is $\sum\limits_{k=-\infty}^\infty 2\pi\delta(w-w_o+2\pi k)$ but I don't don't what to do with "n" in the denominator.