Symbolic integral with distributions

Question

Why is

Integrate[ HeavisideTheta[1 - x^2] DiracDelta[1 - x^2] , {x, 0, 1} ]
(*0*)

It should be HeavisideTheta[0]?

Answer 1

I think what you want is

Integrate[HeavisideTheta[1 - x^2] DiracDelta[1 - x^2], {x, 0, 1 + $MachineEpsilon}]

This gives

HeavisideTheta[0]/2

The $MachineEpsilon (or any number > 0) is necessary because you need to integrate across the singularity of the Dirac Delta (i.e. - in a range including at least a little bit above and below it) to capture its behavior as a distribution.

The factor of 1/2 is due to a delta function having a nontrivial differentiable function (like your 1-x^2) as its argument. Wikipedia and other sources have more complete explanations.

Answer 2

I do not think the question is asking for workarounds, is it? it is asking why is the answer given is not HeavisideTheta[0]

Since I have to make this as an answer, ok, then my answer is this: It should have, and this is a bug.

from http://mathworld.wolfram.com/DeltaFunction.html

$$ \int_{-\infty}^{\infty}f\left( t\right) \delta\left( t-a\right) dt=f\left( a\right) $$

So when $\delta(t-a)$ is under the integral it acts to pull out the value of the function at the point where $t=a$, i.e. where $\delta(0)$

In the example given $\int_{0}^{1}\theta\left( 1-x^{2}\right) \delta\left( 1-x^{2}\right) dx$, when $x=1$ then $\delta\left( 0\right) $, hence the result should be $\theta\left(1-1^{2}\right) =\theta\left( 0\right) $.

But $\theta\left( x\right) $ is not defined at $x=0$

http://reference.wolfram.com/mathematica/ref/HeavisideTheta.html

hence Mathematica should have returned HeavisideTheta[0]