# Antiderivative of Heaviside step function with absolute-value-argument

• schniefen

#### schniefen

Homework Statement
Find the antiderivative of ##\Theta (R-|x|)##, where ##\Theta## is the Heaviside step function and ##R## is a given constant.
Relevant Equations
The derivative of ##\Theta## is the Dirac delta function ##\delta## and ##\frac{x}{|x|}=\Theta(x) -\Theta(-x)##.
For ##R<0##, the antiderivative is just a constant, since then ##R-|x|## is negative for all values of ##x##, which in turn implies ##\Theta(R-|x|)## is zero for all values of ##x##. For ##R\geq 0##, and by inspection apparently, the antiderivative is

##(R+x)\Theta(R-|x|)+2R\Theta(x-R)+C.##
I'd like to confirm this is really the antiderivative by computing the derivative. I get

##-R\frac{x}{|x|}\delta(R-|x|)+\Theta(R-|x|)-x \frac{x}{|x|}\delta(R-|x|)+2R\delta(x-R).##
Using the identity ##\frac{x}{|x|}=\Theta(x) -\Theta(-x)##, one can simplify further

##-R\Theta(x)\delta(R-|x|)+R\Theta(-x)\delta(R-|x|)+\Theta(R-|x|)-x \Theta(x)\delta(R-|x|)+x \Theta(-x)\delta(R-|x|)+2R\delta(x-R).##
This should be possible to simplify even further, although I am stuck here. Any help is appreciated.

For ##R\geq 0##, and by inspection apparently, the antiderivative is

##(R+x)\Theta(R-|x|)+2R\Theta(x-R)+C.##
I'd like to confirm this is really the antiderivative by computing the derivative. I get

##-R\frac{x}{|x|}\delta(R-|x|)+\Theta(R-|x|)-x \frac{x}{|x|}\delta(R-|x|)+2R\delta(x-R).##​
I suppose you can get the calculation to work out, but I think it's easier to analyze what the function is doing at ##x=-R## and ##x=+R##. That way, you don't have to deal with the complications from the absolute value.

In the neighborhood of ##x=-R##, the antiderivative is ##(x+R)\Theta(x+R)+C##, so its derivative is ##\Theta(x+R) + (x+R)\delta(x+R)##, which simplifies to ##\Theta(x+R)##. In the neighborhood of ##x=+R##, the antiderivative is ##(x+R)\Theta(R-x)+2R\Theta(x-R)##. Its derivative is ##\Theta(R-x) - (x+R)\delta(R-x) + 2R \delta(x-R)##, which simplifies to ##\Theta(R-x)##.

So the derivative goes from 0 to 1 at ##x=-R## and from 1 to 0 at ##x=+R##, which is exactly the function you started with in this problem.

schniefen
I suggest rewriting ##\Theta(R-|x|) = \Theta(R+x) - \Theta(x-R)## and work from there.

schniefen