Heaviside step function

In mathematics, physics, and engineering the Heaviside step function is the following function,
 * $$ H(x) = \begin{cases} 1 &\quad\hbox{if}\quad x > 0\\ 0 &\quad\hbox{if}\quad x < 0\\ \end{cases} $$

The function is undetermined for x = 0, sometimes one defines $$H(0) = 1/2\,$$.

From the definition it follows immediately that
 * $$ H(x-a) = \begin{cases} 1 &\quad\hbox{if}\quad x > a\\ 0 &\quad\hbox{if}\quad x < a\\ \end{cases} \qquad\quad\hbox{and}\quad H(-x) =  \begin{cases} 0 &\quad\hbox{if}\quad x > 0\\ 1 &\quad\hbox{if}\quad x < 0\\ \end{cases} $$

The function is named after the English mathematician Oliver Heaviside.

Derivative
Note that a block ("boxcar") function B&Delta; of width &Delta; and height 1/&Delta; can be given in terms of step functions (for positive &Delta;), namely
 * $$ B_\Delta(x) = \frac{ H(x+\Delta/2) - H(x-\Delta/2)}{\Delta}= \begin{cases} \frac{0 - 0}{\Delta} = 0 & \quad\hbox{if}\quad x < -\Delta/2 \\ \frac{1 - 0}{\Delta} = \frac{1}{\Delta} & \quad\hbox{if}\quad -\Delta/2 < x < \Delta/2 \\ \frac{1 - 1}{\Delta} = 0 & \quad\hbox{if}\quad x > \Delta/2 \\ \end{cases} $$

Knowing this, the derivative of H follows easily
 * $$ \frac{dH(x)}{dx} = \lim_{\Delta\rightarrow 0} \frac{H(x+\Delta/2) -H(x-\Delta/2)}{\Delta} = \lim_{\Delta\rightarrow 0} B_\Delta(x) =\delta(x), $$

where &delta;(x) is the Dirac delta function, which may be defined as the block function in the limit of zero width, see the article on the Dirac delta function.

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,
 * $$ \int_{-\infty}^{\infty} H(x) \frac{df(x)}{dx} \mathrm{d}x = - \int_{-\infty}^{\infty} \frac{dH(x)}{dx} f(x) \mathrm{d}x = - \int_{-\infty}^{\infty}  \delta(x) f(x) \mathrm{d}x = -f(0). $$

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.

H(x) as limit of arctan
In the figure it is shown that
 * $$ \frac{1}{\pi} \lim_{\epsilon \rightarrow 0^+} \arctan(\epsilon/x) = H(-x). $$

Note here that the function arctan returns angles on the full interval 0 to 2&pi;. In particular, if a point x + iy in the complex plane has x < 0 and y approaches zero from above, then the function arctan returns a value approaching &pi;. Most computer languages use a two parameter function for this form of the inverse tangent.

Fourier transform

 * $$ \mathcal{F}(H) \equiv \sqrt{\frac{1}{2\pi}} \int_{-\infty}^{\infty} e^{-iux}H(x) dx = \sqrt{\frac{1}{2\pi}} \left( \pi \delta(u) - i \mathrm{P}\Big(\frac{1}{u}\Big)\right), $$

where &delta;(u) is the Dirac delta function and P stands for the Cauchy principal value.

Proof
Write
 * $$ \begin{align} \mathcal{F}(H) &= \sqrt{\frac{1}{2\pi}} \int_{0}^{\infty} e^{-iux} dx = \lim_{\epsilon \rightarrow 0^+}  \sqrt{\frac{1}{2\pi}} \int_{0}^{\infty} e^{-iux-\epsilon x} dx  =   \\ &= \lim_{\epsilon \rightarrow 0^+}  \sqrt{\frac{1}{2\pi}} \left[ \frac{e^{-iux}e^{-\epsilon x}}{-iu-\epsilon} \right]^{\infty}_0 =\sqrt{\frac{1}{2\pi}} \lim_{\epsilon \rightarrow 0^+}  \frac{1}{iu+\epsilon}, \end{align} $$

where we used
 * $$ \lim_{x \rightarrow \infty} e^{-iux}\,e^{-\epsilon x} = 0,\quad \epsilon > 0. $$

Now use the following relation,
 * $$ \lim_{\epsilon \rightarrow 0^+} \frac{-i}{u-i\epsilon} = -i \mathrm{P}\Big(\frac{1}{u}\Big) + \pi \delta(u) $$

and the result is proved.

In order to prove the last relation we write a complex number in polar form
 * $$ x+iy = (x^2+y^2)^{1/2}\; e^{i\alpha}\quad \hbox{with}\quad \alpha = \arctan(y/x). $$

Take the natural logarithm and the limit for y &rarr; 0
 * $$ \lim_{y \rightarrow 0^+} \ln(x+iy) = \lim_{y \rightarrow 0^+} \frac{1}{2} \ln(x^2+y^2) + i\lim_{y \rightarrow 0^+} \arctan(y/x) = \ln|x| + i \pi H(-x). $$.

Differentiation of the last expression gives gives
 * $$ \lim_{y \rightarrow 0^+} \frac{1}{x+iy} = \frac{1}{x} -i\pi \delta(x). $$

In fact, the functions in this expression are distributions (generalized functions) and are to be used in an integrand multiplied by a well-behaved function f(x). Since 1/x is singular for x = 0, we must take the Cauchy principal value of the integral, i.e., exclude the singularity from the integral; we make the replacement
 * $$ \int_{-\infty}^{\infty} \frac{f(x)}{x} \mathrm{d}x \Rightarrow \mathrm{P}\int_{-\infty}^{\infty} \frac{f(x)}{x} \mathrm{d}x, $$

and the result follows.