The Fourier transform of the step function is a classic example of how generalized functions (distributions) like the delta function allow us to include non-convergent but physically meaningful signals into the frequency domain framework.
[ F(\omega) = \int_-\infty^\infty f(t) e^-i\omega t dt ]
For ( u(t) ), this becomes ( \int_0^\infty e^-i\omega t dt ). This integral does not converge in the usual sense because ( e^-i\omega t ) does not decay at infinity. So how can we proceed? The standard trick is to treat the step function as the limit of a decaying exponential:
[ \boxed\mathcalFu(t) = \pi \delta(\omega) + \frac1i\omega ]
The Fourier transform of ( \textsgn(t) ) is ( 2/(i\omega) ) (without a delta, since its average is zero). Thus:
[ u(t) = \lim_\alpha \to 0^+ e^-\alpha t u(t), \quad \alpha > 0 ]
[ \mathcalFu(t) = \frac12 \cdot 2\pi\delta(\omega) + \frac12 \cdot \frac2i\omega = \pi\delta(\omega) + \frac1i\omega ]
[ \int_0^\infty e^-\alpha t e^-i\omega t dt = \int_0^\infty e^-(\alpha + i\omega) t dt = \frac1\alpha + i\omega ]
