Integrate[DiracDelta'[y - z] DiracDelta[x - z], {z, -Infinity, Infinity}, 
 Assumptions -> x \[Element] Reals && x \[Element] Reals]
  • $\begingroup$ is this what you want to input? Integrate[ DiracDelta[y - z] DiracDelta[x - z], {z, -Infinity, Infinity}] $\endgroup$ – chris Oct 13 '12 at 13:56
  • $\begingroup$ @belisarius: Are you sure that what you edited away isn't actually the source of his problem? (I suspect he typed something wrong, because normally I don't get superscript boxes from copying Mathematica content). $\endgroup$ – celtschk Oct 13 '12 at 14:26
  • 1
    $\begingroup$ @Sergej: Could you please say what exactly is your problem? What does Mathematica give you? An unexpected result (which?), giving back the expression unevaluated? $\endgroup$ – celtschk Oct 13 '12 at 14:32
  • $\begingroup$ @celtschk I think mathematica returns incorrectly 0 to Integrate[DiracDelta'[y - z] DiracDelta[x - z], {z, -Infinity, Infinity}] but I am guessing... $\endgroup$ – chris Oct 13 '12 at 15:02
  • $\begingroup$ By the way, Sergei, welcome to Mathematica.SE! Please consider registering your account so that any upvotes you get on this question are added to those you might get on future questions and answers. That way, over time you will be able to do more on the site (post graphics, edit things, etc). $\endgroup$ – Verbeia Oct 15 '12 at 2:54

I'll go with the same interpretation as @chris (and derivative wrt y) and suggest :

Integrate[f'[y - z] DiracDelta[x - z], {z, -Infinity, Infinity}, 
    Assumptions -> x \[Element] Reals && x \[Element] Reals] 
 /. f -> DiracDelta

(* Derivative[1][DiracDelta][-x + y] *)

Another approach is the one suggested by @chris : we can consider

$\delta(x) = \lim_{\epsilon \rightarrow 0} \frac{1}{\sqrt{2 \pi \epsilon^2}} \ \ \exp(-\frac{x^2}{2 \epsilon^2}) $

dirac[x_, \[Epsilon]_] = PDF[NormalDistribution[0, \[Epsilon]], x]

Integrate[Derivative[1, 0][dirac][y - z, \[Epsilon]] dirac[x - z, \[Epsilon]], 
  {z, -\[Infinity], \[Infinity]}, Assumptions -> {\[Epsilon] > 0}] == 
Derivative[1, 0][dirac][y - x, Sqrt[2] \[Epsilon]] // Simplify

(* True *)

In the limit $\epsilon \rightarrow 0$ this result matches the previous one.

It is mentioned in the documentation that

Products of distributions with coinciding singular support cannot be defined.

  • $\begingroup$ Do we know for a fact this is the correct answer though? It seems odd that when taking the limit of the Gaussian (as I did below), one gets a different answer. $\endgroup$ – chris Oct 13 '12 at 19:40
  • $\begingroup$ @chris Please see edit for additional support to the result I obtained. $\endgroup$ – b.gates.you.know.what Oct 13 '12 at 20:29
  • $\begingroup$ I agree with your second edit but note that it differs through a sqrt(2) and a sign from the naive conclusion one would draw from the first edit. $\endgroup$ – chris Oct 14 '12 at 8:33
  • $\begingroup$ @chris You're right, thanks for pointing this out. In the first case I had considered the derivative wrt to y, in the second wrt z - fixed now. I think the $\sqrt{2}$ factor is ok, as I'm taking the limit $\epsilon \rightarrow 0$ anyways. $\endgroup$ – b.gates.you.know.what Oct 14 '12 at 8:53
  • $\begingroup$ I am not sure changing the sign inside the DiracDelta is sufficient: DiracDelta[-x]== DiracDelta[x] $\endgroup$ – chris Oct 14 '12 at 9:45

You can define

dirac[x_] = PDF[NormalDistribution[0, 1/\[Epsilon]], x]

and check that

Integrate[dirac'[y - z] dirac[x - z], {z, -Infinity, Infinity}] + 
1/2 (D[dirac[z], z] /. z -> (x - y)/Sqrt[2]) // PowerExpand

returns 0


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.