Antiderivative of Heaviside step function with absolute-value-argument

[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

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

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

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

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[vela]

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.

 

[Orodruin]

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