Now take the limit (\epsilon \to 0^+):
[ \lim_\epsilon \to 0^+ \frac1\epsilon + i\omega = \frac1i\omega + \pi \delta(\omega) \quad \text(in the sense of distributions) ] fourier transform of heaviside step function
[ \hatH(\omega) = \int_-\infty^\infty H(t) , e^-i\omega t , dt = \int_0^\infty e^-i\omega t , dt ] Now take the limit (\epsilon \to 0^+): [
This integral does not converge in the usual sense because (e^-i\omega t) does not decay at (t \to \infty). Introduce an exponential decay factor (e^-\epsilon t) with (\epsilon > 0), then let (\epsilon \to 0^+): dt = \int_0^\infty e^-i\omega t
[ \hatH_\epsilon(\omega) = \int_0^\infty e^-\epsilon t e^-i\omega t , dt = \int_0^\infty e^-(\epsilon + i\omega)t , dt = \frac1\epsilon + i\omega ]
