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.

## Hint

• We can express the Heaviside function as: $$\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega=\theta(t)$$

## Solution

In order to prove the relation above we perform the inverse Fourier transform:

$\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}2\pi\bar{\theta}(\omega)e^{-i\omega t}d\omega=\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\int\limits _{-\infty}^{\infty}\theta(t')e^{i\omega t'}dt'\right)e^{-i\omega t}d\omega=\theta(t).$

Therefore, on the one hand we must show that:

$\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega=\theta(t),$

where $$\lim_{\varepsilon\rightarrow0}\frac{i}{\omega+i\varepsilon}$$ is an expression for $$2\pi\bar{\theta}(\omega)$$ and on the other hand we also need to prove:

$\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\frac{i}{\omega}+\pi\delta(\omega)\right)e^{-i\omega t}d\omega=\theta(t).$

## 1. Proving $$\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega=\theta(t)$$:

As we see the integrand has a pole at $$\omega=-i\varepsilon$$, which is in the negative complex plane. Solving the integral we need to consider two cases, where the contour is a closed semicircle in the complex plane:

• Integral in the upper complexe plane $$(t<0)$$:
Here we obtain upon splitting the integral into its components:
\begin{eqnarray*}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega&=&\mathcal{P}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega+\int_{C_{2}}\frac{i}{\omega''+i\varepsilon}e^{\omega''t}d\omega\\&=&\mathcal{P}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{i\omega|t|}d\omega+\int_{C_{2}}\frac{i}{\omega''+i\varepsilon}e^{-\omega''|t|}d\omega=0,\end{eqnarray*}
because there are no poles in the upper complex plane and the integral along $$C_{2}$$ vanishes for $$\omega^{\prime\prime}\longrightarrow\infty$$.
• Integral in the lower complexe plane $$(t>0)$$:
Using residue theorem the solution is:
$\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\left(\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega\right)=-\lim_{\varepsilon\rightarrow0}\frac{1}{2\pi}\left(2\pi i\,\textrm{Res}\left(\frac{i}{\omega+i\varepsilon}e^{-i\omega t};-i\varepsilon\right)\right)=1.$
Therefore we get:
$\Theta\left(t\right)=\lim_{\varepsilon\rightarrow0}\int\limits _{-\infty}^{\infty}\frac{i}{\omega+i\varepsilon}e^{-i\omega t}d\omega=\begin{cases} 1 & t>0\\ 0 & else \end{cases}$

## 2. Proving $$\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\frac{i}{\omega}+\pi\delta(\omega)\right)e^{-i\omega t}d\omega=\theta(t)$$:

The first step is easy:

$\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\frac{i}{\omega}+\pi\delta(\omega)\right)e^{-i\omega t}d\omega=\frac{1}{2}-\frac{1}{2\pi i}\int\limits _{-\infty}^{\infty}\frac{1}{\omega}e^{-i\omega t}d\omega.$

Now we will rearrange the second integral, obtaining:

$\int\limits _{-\infty}^{\infty}\frac{1}{\omega}e^{-i\omega t}d\omega=\int\limits _{-\infty}^{0}\frac{1}{\omega}e^{-i\omega t}d\omega+\int\limits _{0}^{\infty}\frac{1}{\omega}e^{-i\omega t}d\omega.$

In order to solve this integral we will use the trick, that we already used here, deriving the KKR, i.e. we will consider a contour integral in the complex plane with a small semicircle around $$\omega=0$$, where we obtain:

$0=\oint\frac{e^{-i\omega t}}{\omega}d\omega=\int\limits _{-\infty}^{-0}\frac{e^{-i\omega t}}{\omega}d\omega+\int\limits _{+0}^{\infty}\frac{e^{-i\omega t}}{\omega}d\omega+\int\limits _{\pi}^{0}\frac{e^{-i\omega t}}{\omega}d\omega+\int_{C_{2}}\frac{e^{-\omega''|t|}}{\omega''}d\omega''$

Since we just look at the case $$t<0$$ the $$C_{2}$$-integral vanishes, as shown above. Using the parametrization $$\omega=\rho e^{i\phi}\quad\rightarrow\quad d\omega=i\rho e^{i\phi}d\phi$$ we obtain:

$\int\limits _{\pi}^{0}\frac{e^{-i\omega t}}{\omega}d\omega=\int\limits _{\pi}^{0}\frac{i\rho e^{i\phi}}{\rho e^{i\phi}}e^{-i\omega t}d\phi=-i\pi.$

Also we get depending on the signum of $$t$$, i.e. direction of the integration:

$\int\limits _{-\infty}^{-0}\frac{e^{-i\omega t}}{\omega}d\omega+\int\limits _{+0}^{\infty}\frac{e^{-i\omega t}}{\omega}d\omega=-i\pi\cdot\textrm{sign}(t).$

$\frac{1}{2\pi}\int\limits _{-\infty}^{\infty}\left(\frac{i}{\omega}+\pi\delta(\omega)\right)e^{-i\omega t}d\omega=\frac{1}{2}\left[1+\textrm{sign}(t)\right]=\theta(t).$