Fourier transform of simple function I'm stuck on this problem
Calculate the Fourier transform of
$$
f(t) = \sin t \, , \, |t|<\pi \\
f(t) = 0 \, , \, |t|≥ \pi
$$
So the start is really simple
$$
\hat{f}(\omega) = \int_{-\pi}^{\pi}\sin(t)e^{-i \omega t}dt = ... = \frac{1}{2i}\left[\frac{e^{-(\omega-1)it}}{-i(\omega -1)} \right]_{-\pi}^{\pi}-\frac{1}{2i}\left[\frac{e^{-(\omega+1)it}}{-i(\omega +1)} \right]_{-\pi}^{\pi}
$$
and now I am asked to simplify this so i get $\hat{f}(\omega)= 2i\frac{\sin \pi \omega}{\omega^2-1}$, I have been trying for hours and I simply can't make it! Is there anyone here that knows the trick? Maybe it have been answered before somewhere?

You should use this identity after you plug your bounds in the antiderivative:
$$\sin \theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$$
With this, you get
$$\frac{\sin(\pi(w-1))}{i(w-1)} -\frac{\sin(\pi(w+1))}{i(w+1)}$$
Observe
$$\frac{\sin(\pi(w-1))}{i(w-1)} -\frac{\sin(\pi(w+1))}{i(w+1)} = \frac{-\sin(\pi w)}{i(w-1)} -\frac{-\sin(\pi w)}{i(w+1)} = \frac{-\sin(\pi w)}{i(w-1)} +\frac{\sin(\pi w)}{i(w+1)}.$$
Combine denominators and you'll get exactly what you want:
$$\frac{-(w+1)\sin(\pi w) +(w-1)\sin(\pi w)}{i(w+1)(w-1)} = \frac{-2\sin(\pi w)}{i(w^2-1)} = \frac{2i\sin(\pi w)}{(w^2-1)}$$
