Transform Of Heaviside Step Function Fix | Fourier

[ \hatH(\omega) = \int_-\infty^\infty H(t) , e^-i\omega t , dt = \int_0^\infty e^-i\omega t , dt ]

(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: fourier transform of heaviside step function

[ \lim_\epsilon \to 0^+ \frac1\epsilon + i\omega = \frac1i\omega + \pi \delta(\omega) \quad \text(in the sense of distributions) ] [ \hatH(\omega) = \int_-\infty^\infty H(t) , e^-i\omega t