I am totally new to Fourier transforms, so I would be very thankful for a feedback on my calculation. I want to calculate the Fourier transform of $ f=\chi_{[-1,1]} $, that is, $f(x)=1$ for $-1\le x\le 1$ and $0$ otherwise.
My answer: \begin{align*} \widehat{f}(\omega)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}f(x)e^{-i\omega x}dx\\ &=\frac{1}{\sqrt{2\pi}}\int_{-1}^{1}e^{-i\omega x}dx\\ &=\frac{1}{\sqrt{2\pi}}\left[\frac{1}{-i\omega}e^{-i\omega x}\right]_{-1}^1\\ &=\frac{1}{\sqrt{2\pi}}\frac{1}{-i\omega}\left(e^{-i\omega}-e^{i\omega}\right)\\ &=\frac{1}{\sqrt{2\pi}}\frac{1}{-i\omega}(-2i\sin \omega)\\ &=\sqrt{\frac{2}{\pi}}\frac{1}{\omega} \sin \omega \end{align*} where I have used the relation $$ \sin \omega = \frac{1}{2i} \left( e^{i\omega} - e^{-i\omega}\right). $$ Is this correct?