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Let us execute in version 12.0

Integrate[DiracDelta[x]*DiracDelta'[x], x]

DiracDelta[x]^2/2

and this lasts at least since version 7. However, the documentation to DiracDelta says "Products of distributions with coinciding singular support cannot be defined " and Encyclopedia of Mathematics seconds it. Therefore, neither DiracDelta[x]*DiracDelta'[x] nor DiracDelta[x]^2 make sense. The question arises: how to interpret the above result?

user64494
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  • One possibility is that Integrate simply does the integration treating DiracDelta as an unknown function with no special properties. (You get the same result if you replace DiracDelta with e.g. f) - I doubt that every transformation rule for Integrate checks for distributions, it's more likely that there are simply some special cases for DiracDelta. – Lukas Lang Oct 17 '19 at 22:35
  • I think this is more a math question though of course it becomes a Mathematica question if Mathematica is giving an incorrect result. First note that the linked reference actually states that noncoinciding supports is a sufficient condition, not that it is necessary. After playing around using two approximations to the Dirac delta, one for the function itself and another for the derivative, and integrating by parts against a test function, the claimed result seems plausible. I've been wrong about this kind of thing before though. – Daniel Lichtblau Oct 17 '19 at 22:38
  • There is a difference between the antiderivative (that you compute and that is evaluated formally) and the integral. Only the latter makes sense mathematically. Here the result is as expected Integrate::idiv: Integral of DiracDelta[x] (DiracDelta^\[Prime])[x] does not converge on {-\[Infinity],\[Infinity]}. – yarchik Oct 18 '19 at 06:28
  • @Daniel Lichtblau : Approximations converge to the delta distribution in the weak topology. Its definition includes the forall quantifier so the way suggested by you is difficulty realized in Wolfram Language. – user64494 Oct 18 '19 at 13:33
  • @yarchik : The command Integrate[HeavisideTheta[x]*DiracDelta[x], x] returns the input. – user64494 Oct 18 '19 at 13:35
  • Indeed Integrate[HeavisideTheta[x] DiracDelta[x], x] and Integrate[DiracDelta[x] DiracDelta'[x], x] yield inconsistent results. But what the problem with that? Either are mathematically invalid. Only the definite integral makes sense. You seems to be interested in products of generalized functions. I find this post very elucidating: https://mathoverflow.net/questions/48067/is-square-of-delta-function-defined-somewhere Have a look at the answer of Denis Serre. I don't know of an example where this tool (the Colombeau algebra) solved an open problem This pretty much says it all. – yarchik Oct 18 '19 at 14:59
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    Yeah, it is not an ideal result, and unevaluated would perhaps be preferred. The current behavior is a manifestation of garbage in, garbage out so I would not call it a high priority item. Maybe a better way to state it is "If this has a result, here is what it is" without caring about first addressing the existence quastion. – Daniel Lichtblau Oct 20 '19 at 15:07
  • Look in that. – user64494 Mar 18 '21 at 12:21

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