[ \lim_\alpha \to 0^+ \frac1\alpha + i\omega = \frac1i\omega ]
[ \mathcalFu(t) = \frac12 \cdot 2\pi\delta(\omega) + \frac12 \cdot \frac2i\omega = \pi\delta(\omega) + \frac1i\omega ] fourier transform step function
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: [ \lim_\alpha \to 0^+ \frac1\alpha + i\omega =
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: This integral does not converge in the usual
[ F(\omega) = \int_-\infty^\infty f(t) e^-i\omega t dt ]
[ u(t) = \begincases 0, & t < 0 \ 1, & t > 0 \endcases ]