## Tuesday, June 30, 2009

### Derivative and Integral of the Heaviside Step Function

The Setup
(a) Large horizontal scale

(b) ``Zoomed in''
Figure 1: The Heaviside step function. Note how it doesn't matter how close we get to
$x=0$ the function looks exactly the same.

The Heaviside step function $H(x)$, sometimes called the Heaviside theta function, appears in many places in physics, see [1] for a brief discussion. Simply put, it is a function whose value is zero for $x<0$ and one for $x>0$. Explicitly,
$H(x)=\begin{cases}0& x<0,\\1& x>0\end{cases}.$
(1)
We won't worry about precisely what its value is at zero for now, since it won't effect our discussion, see [2] for a lengthier discussion. Fig. 1 plots $H(x)$. The key point is that crossing zero flips the function from 0 to 1.

Derivative -- The Dirac Delta Function
(a) Dirac delta function

(b) Ramp function
Figure 2: The derivative (a), and the integral (b) of the Heaviside step function.
Say we wanted to take the derivative of $H$. Recall that a derivative is the slope of the curve at at point. One way of formulating this is
$\frac{dH}{dx}=\lim_{\Delta x \rightarrow 0} \frac{\Delta H}{\Delta x}.$
(2)
Now, for any points $x<0$ or $x>0$, graphically, the derivative is very clear: $H$ is a flat line in those regions, and the slope of a flat line is zero. In terms of (2), $H$ does not change, so $\Delta H=0$ and $dH/dx=0$. But if we pick two points, equally spaced on opposite sides of $x=0$, say $x_-=-a/2$ and $x_+=a/2$, then $\Delta H=1$ and $\Delta x=a$. It doesn't matter how small we make $a$, $\Delta H$ stays the same. Thus, the fraction in (2) is
$\frac{dH}{dx}=\lim_{a \rightarrow 0} \frac{1}{a}$
$=\infty.$
(3)
Graphically, again, this is very clear: $H$ jumps from 0 to 1 at zero, so it's slope is essentially vertical, i.e. infinite. So basically, we have
$\delta(x)\equiv\frac{dH}{dx}=\begin{cases}0& x<0\\\infty& x=0\\0& x>0\end{cases}.$
(4)
This function is, loosely speaking, a ``Dirac Delta'' function, usually written as $\delta(x)$, which has seemingly endless uses in physics.

We'll note a few properties of the delta function that we can derive from (4). First, integrating it from $-\infty$ to any $x_-<0$:
$\int_{-\infty}^{x_-}\delta(x)dx=\int_{-\infty}^{x_-}\left(\frac{dH}{dx}\right)dx$
$=H(x_-)-H(-\infty)$
$=0$
(5)
since $H(x_-)=H(-\infty)=0$. On the other hand, integrating the delta function to any point greater than $x=0$:
$\int_{-\infty}^{x_+}\delta(x)dx=\int_{-\infty}^{x_+}\left(\frac{dH}{dx}\right)dx$
$=H(x_+)-H(-\infty)$
$=1$
(6)
since $H(x_+)=1$.

At this point, I should point out that although the delta function blows up to infinity at $x=0$, it still has a finite integral. An easy way of seeing how this is possible is shown in Fig. 2(a). If the width of the box is $1/a$ and the height is $a$, the area of the box (i.e. its integral) is $1$, no matter how large $a$ is. By letting $a$ go to infinity we have a box with infinite height, yet, when integrated, has finite area.

Integral -- The Ramp Function
Now that we know about the derivative, it's time to evaluate the integral. I have two methods of doing this. The most straightforward way, which I first saw from Prof. T.H. Boyer, is to integrate $H$ piece by piece. The integral of a function is the area under the curve,1 and when $x<0$ there is no area, so the integral from $-\infty$ to any point less than zero is zero. On the right side, the integral to a point $x$ is the area of a rectangle of height 1 and length $x$, see Fig. 1(a). So, we have
$\int_{-\infty}^x H\,dx=\begin{cases}0& x<0,\\x& x>0\end{cases}.$
(7)
We'll call this function a ``ramp function,'' $R(x)$. We can actually make use of the definition of $H$ and simplify the notation:
$R(x)\equiv\int H\,dx=xH(x)$
(8)
since $0\times x=0$ and $1\times x=x$. See Fig. 2(b) for a graph -- and the reason for calling this a ``ramp'' function.

But I have another way of doing this which makes use of a trick that's often used by physicists: We can always add zero for free, since ${anything}+0={anything}$. Often we do this by adding and subtracting the same thing,
$A=(A+B)-B,$
(9)
for example. But we can use the delta function (4) to add zero in the form
$0=x\,\delta(x).$
(10)
Since $\delta(x)$ is zero for $x\neq 0$, the $x$ part doesn't do anything in those regions and this expression is zero. And, although $\delta(x)=\infty$ at $x=0$, $x=0$ at $x=0$, so the expression is still zero.

So we'll add this on to $H$:
$H=H+0$
$=H+x\,\delta(x)$
$=H+x\frac{dH}{dx} \quad \quad \quad \text{by (4)}$
$=\frac{dx}{dx}H+x\frac{dH}{dx}$
$=\frac{d}{dx}\left[xH(x)\right],$
(11)
where the last step follows from the ``product rule'' for differentiation. At this point, to take the integral of a full differential is trivial, and we get (8).

References
[1] M. Springer. Sunday function [online]. February 2009. Available from: http://scienceblogs.com/builtonfacts/2009/02/sunday_function_22.php [cited 30 June 2009].
[2] E.W. Weisstein. Heaviside step function [online]. Available from: http://mathworld.wolfram.com/HeavisideStepFunction.html [cited 30 June 2009].

1 To be completely precise, it's the (signed) area between the curve and the line $x=0$.