[ \lim_\epsilon \to 0^+ \frac1\epsilon + i\omega = \frac1i\omega + \pi \delta(\omega) \quad \text(in the sense of distributions) ]
[ \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 ] fourier transform of heaviside step function
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^+): [ \lim_\epsilon \to 0^+ \frac1\epsilon + i\omega =
[ H(t) = \begincases 1, & t > 0 \ \frac12, & t = 0 \ 0, & t < 0 \endcases ] dt = \int_0^\infty e^-(\epsilon + i\omega)t
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