[ F(\omega) = \int_-\infty^\infty f(t) e^-i\omega t dt ]
This gives ( 1/(i\omega) ), but this is not the whole story. Something is missing: the step function has a nonzero average value (1/2 over all time, if we consider symmetric limits), which implies a DC component. It turns out that the Fourier transform of the unit step function is: fourier transform step function
Now, take the limit as ( \alpha \to 0^+ ): [ F(\omega) = \int_-\infty^\infty f(t) e^-i\omega t dt
[ \boxed\mathcalFu(t) = \pi \delta(\omega) + \frac1i\omega ] if we consider symmetric limits)