The Heaviside function is an asymmetric step function that may be defined either as a [[Distribution]] or a piece-wise constant function.  <font size="2"> The Heaviside function can be thought of as a constant piece-wise function with a non-differentiable jump, (a singular point at $x=0$), as shown here, or a [[Distribution]] when used in an integral.</font> # As a distribution Given the [[Dirac delta function]], $\delta(x),$ $\int_{-\infty}^{\infty} dx f(x) \delta(x) = f(0),$ However, consider the case where we take some finite upper limit of the integral instead of $\infty$: $\int_{-\infty}^{a} dx f(x) \delta(x) = \begin{cases} 0, & a < 0\\ \frac{1}{2}f(0), & a = 0\\ f(0), & a > 0 \end{cases}$ The reasons for the 1st and 3rd cases are straightforward given how the $\delta$ function acts. The middle case yields $\frac{1}{2}f(0)$ simply because $\delta(x)$ is symmetrical and normalized to $1$ while $x=0$ is its center, therefore cutting the integral in half. Equivalently, $\int_{-\infty}^{a} dx f(x) \delta(x) = \Theta(x)f(0)$ # As a piece-wise function We may also define the Heaviside function as a _constant piecewise function_ as follows: $\Theta(x) = \begin{cases} 0, & x < 0\\ \frac{1}{2}, & x = 0\\ 1, & x > 0 \end{cases}$ # Derivative of the Heaviside function Notice that $\int_{-\infty}^{a} dx' \delta(x') = \Theta(x)$ which alone implies $\Theta'(x)=\delta(x)$ However, this derivative must be understood as being the limit of a defined $\Theta_n$ function since the Heaviside function isn't truly differentiable. The derivative is [Singularity](Singularity.md) at $x=0$. # Integrating the Heaviside function Finally, if we treat $\Theta(x)$ [as a distribution](Heaviside%20function.md#As%20a%20distribution), it is defined via, $\int_{-\infty}^{\infty}\Theta(x)f'(x) = -f(0)$ where $f$ is a smooth function. #MathematicalFoundations/Analysis/GeneralizedFunctions