What is inverse fourier transform of $\displaystyle\,\,{\sin\left(3\omega\right) \over \omega}$ ?.
It says it's $\displaystyle{1 \over 2}\left[\mathrm{u}\left(t + 3\right) - \mathrm{u}\left(t - 3\right)\right] = {1 \over 2}\quad\mbox{for}\quad\left\vert t\right\vert < 3\ \mbox{and}\ 0\ \mbox{for}\ \left\vert t\right\vert > 3$
I understand that applying the inverse fransform yields the periodic square wave but I'm not sure where $\displaystyle 1/2$ in the equation came from and why the answer is equal to $\displaystyle 1/2$ instead of $\displaystyle 1$.
Here I'm using the Fourier transform $$ \mathcal{F}\{f(t)\} = \int_{-\infty}^{\infty} f(t) \, e^{-i\omega t} dt . $$
You should easily be able to confirm that $$ \mathcal{F}\{\chi_{[-T,T]}(t)\} = 2\frac{\sin\omega T}{\omega} $$ so the inverse transform you want to find is $$ \mathcal{F}^{-1}\{\frac{\sin 3\omega}{\omega}\} = \frac12 \mathcal{F}^{-1}\{2\frac{\sin 3\omega}{\omega}\} = \frac12 \chi_{[-3,3]}(x) . $$