A work in progress.
The Continuous Time Fourier Transform (CTFT)
CTFT:
$ X(\omega) = \int_{-\infty}^{\infty} \! x(t)e^{-j \omega t} dt $
Inverse CTFT:
$ x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \! X(\omega)e^{j \omega t} dw $
Example:
Let x(t) = rect(t)
$ X(\omega) = \int_{-\infty}^{\infty} \! x(t)e^{-j \omega t} dt = \int_{-\infty}^{\infty} \! rect(t)e^{-j \omega t} dt = \int_{-\frac{1}{2}}^{\frac{1}{2}} \! rect(t)e^{-j \omega t} dt = \int_{-\frac{1}{2}}^{\frac{1}{2}} \! e^{-j \omega t} dt $
...because rect(t) has an area of 1 over the limits $ [-\frac{1}{2}, \frac{1}{2}] $. So,
$ \int_{-\frac{1}{2}}^{\frac{1}{2}} \! e^{-j \omega t} dt = \frac{1}{-j\omega}(e^{-j\frac{\omega}{2}} - e^{j\frac{\omega}{2}}) = \frac{1}{j\omega}(e^{j\frac{\omega}{2}} - e^{-j\frac{\omega}{2}}) = \frac{2}{\omega}(\frac{e^{j\frac{\omega}{2}} - e^{-j\frac{\omega}{2}}}{2j}) = \frac{2}{\omega}sin(\frac{\omega}{2}) $
Therefore,
$ X(\omega) = \frac{2}{\omega}sin(\frac{\omega}{2}) \ \ or \ \ \frac{1}{\omega/2}sin(\frac{\omega}{2}) = sinc(\frac{\omega}{2\pi}) $
