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^+):
[ \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
Here’s a clear, rigorous explanation of the Fourier transform of the Heaviside step function ( H(t) ), suitable for a textbook, lecture notes, or technical blog. 1. Definition of the Heaviside Step Function The Heaviside step function is defined as: This integral does not converge in the usual
[ H(t) = \begincases 1, & t > 0 \ \frac12, & t = 0 \ 0, & t < 0 \endcases ] dt = \int_0^\infty e^-i\omega t