[ \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 ]
[ \hatH(\omega) = \int_-\infty^\infty H(t) , e^-i\omega t , dt = \int_0^\infty e^-i\omega t , dt ] fourier transform of heaviside step function
(At (t=0), the value is often taken as (1/2) for symmetry in Fourier analysis, but it’s a set of measure zero, so it doesn’t affect the transform in the (L^2) sense.) The Fourier transform (using the unitary, angular frequency convention) is: dt = \int_0^\infty e^-(\epsilon + i\omega)t