The Heaviside step function is very important in physics. It often models a sudden switch-on phenomenon and is therefore present in a lot of integrals. For example, the derivation of the Kramers-Kronig Relations can be significantly simplified once we know the Fourier-Transform \(\bar{\theta}(\omega)\) of the Heaviside function \(\theta(t)\). Although the function appears to be quite simple, the calculation of its Fourier transform can be quite challenging. Let's find out!

Problem Statement

Show that the Fourier transform is given by the following expression:

\[2\pi\bar{\theta}(\omega)=\int\limits _{-\infty}^{\infty}\theta(t)e^{i\omega t}dt=P\frac{i}{\omega}+\pi\delta(\omega),\]

where \(\delta(\omega)\) is the Dirac delta distribution and the functional \(P\) refers to the Cauchy Principal value.

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