# Contents

## Idea

Generally, a step function is a function from the real numbers to themselves which is constant everywhere except at one single point (or a finite number of points).

Specifically the function

$\Theta \colon x \mapsto \left\{ \array{ 0 &\vert& x \lt 0 \\ 1 &\vert& x \geq 0 } \right.$

is sometimes called the Heaviside step function.

This may be regarded as the generalized function-expression for the distributional density $\Theta \in \mathcal{D}'(\mathbb{R})$ which sends a bump function $b \in C^\infty_c(\mathbb{R})$ to its integral over the positive half-axis:

\begin{aligned} \int_{\mathbb{R}} b(x) \Theta(x) d x & \coloneqq \langle \Theta, b\rangle \\ & \coloneqq \int_0^\infty b(x) d x \end{aligned} \,.

As such $\Theta$ is also called the Heaviside distribution.

The distributional derivative of the Heaviside distribution is the Dirac delta distribution (prop. 1 below).

## Properties

###### Proposition

The distributional derivative of the Heaviside distribution $\Theta \in \mathcal{D}'(\mathbb{R})$ is the delta distribution $\delta \in \mathcal{D}'(\mathbb{R})$:

$\partial \Theta = \delta \,.$
###### Proof

For $b \in C^\infty_c(\mathbb{R})$ any bump function we compute:

\begin{aligned} \int \partial\Theta(x) b(x) \, d x & = - \int \Theta(x) \partial b(x)\, dx \\ & = - \int_0^\infty \partial b(x) d x \\ & = - \left( b(x)\vert_{x \to \infty} - b(0) \right) \\ & = b(0) \\ & = \int \delta(x) b(x) \, dx \,. \end{aligned}
###### Definition

(Fourier integral formula for step function)

The Heaviside distribution $\Theta \in \mathcal{D}'(\mathbb{R})$ is equivalently the following limit of Fourier integrals (see at Cauchy principal value)

\begin{aligned} \Theta(x) & = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{e^{i \omega x}}{\omega - i 0^+} \\ & \coloneqq \underset{ \epsilon \to 0^+}{\lim} \frac{1}{2 \pi i} \int_{-\infty}^\infty \frac{e^{i \omega x}}{\omega - i \epsilon} d\omega \,, \end{aligned}

where the limit is taken over sequences of positive real numbers $\epsilon \in (-\infty,0)$ tending to zero.

###### Proof

We may think of the integrand $\frac{e^{i \omega x}}{\omega - i \epsilon}$ uniquely extended to a holomorphic function on the complex plane and consider computing the given real line integral for fixed $\epsilon$ as a contour integral in the complex plane.

If $x \in (0,\infty)$ is positive, then the exponent

$i \omega x = - Im(\omega) x + i Re(\omega) x$

has negative real part for positive imaginary part of $\omega$. This means that the line integral equals the complex contour integral over a contour $C_+ \subset \mathbb{C}$ closing in the upper half plane. Since $i \epsilon$ has positive imaginary part by construction, this contour does encircle the pole of the integrand $\frac{e^{i \omega x}}{\omega - i \epsilon}$ at $\omega = i \epsilon$. Hence by the Cauchy integral formula in the case $x \gt 0$ one gets

\begin{aligned} \underset{\epsilon \to 0^+}{\lim} \frac{1}{2 \pi i} \int_{-\infty}^\infty \frac{e^{i \omega x}}{\omega - i \epsilon} d\omega & = \underset{\epsilon \to 0^+}{\lim} \frac{1}{2 \pi i} \oint_{C_+} \frac{e^{i \omega x}}{\omega - i \epsilon} d \omega \\ & = \underset{\epsilon \to 0^+}{\lim} \left(e^{i \omega x}\vert_{\omega = i \epsilon}\right) \\ & = \underset{\epsilon \to 0^+}{\lim} e^{- \epsilon x} \\ & = e^0 = 1 \end{aligned} \,.

Conversely, for $x \lt 0$ the real part of the integrand decays as the negative imaginary part increases, and hence in this case the given line integral equals the contour integral for a contour $C_- \subset \mathbb{C}$ closing in the lower half plane. Since the integrand has no pole in the lower half plane, in this case the Cauchy integral formula says that this integral is zero.

###### Remark

The Fourier form of the step function in prop. 1 gives rise to the standard expression for the advanced propagator, retarded propagator and Feynman propagator used in perturbative quantum field theory. See at Feynman propagator for more.

## References

Revised on November 20, 2017 10:12:34 by Urs Schreiber (94.220.49.59)