Revision as of 08:10, 23 December 2008 by imported>Paul Wormer
In mathematics, physics, and engineering the Heaviside step function is the following function,

The function is named after the English mathematician Oliver Heaviside.
Derivative
Note that a block function BΔ of width Δ and height 1/Δ can be given in terms of step functions (for positive Δ), namely

Knowing this, the derivative of H follows easily

where δ(x) is the Dirac delta function, which may be defined as the block function in the limit of zero width, see this article.
The step function is a generalized function (a distribution).
When H(x) is multiplied under the integral by the derivative of an arbitrary differentiable function f(x) that vanishes for plus/minus infinity, the result of the integral is minus the function value for x = 0,

Here the "turnover rule" for d/dx is used, which may be proved by integration by parts and which holds when f(x) vanishes at the integration limits.