0
$\begingroup$
Integrate[DiracDelta'[y - z] DiracDelta[x - z], {z, -Infinity, Infinity}, 
 Assumptions -> x \[Element] Reals && x \[Element] Reals]
$\endgroup$
6
  • $\begingroup$ is this what you want to input? Integrate[ DiracDelta[y - z] DiracDelta[x - z], {z, -Infinity, Infinity}] $\endgroup$
    – chris
    Oct 13, 2012 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, 2012 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, 2012 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, 2012 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, 2012 at 2:54

2 Answers 2

4
$\begingroup$

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.

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5
  • $\begingroup$ 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, 2012 at 19:40
  • $\begingroup$ @chris Please see edit for additional support to the result I obtained. $\endgroup$ Oct 13, 2012 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, 2012 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. $\endgroup$ Oct 14, 2012 at 8:53
  • $\begingroup$ I am not sure changing the sign inside the DiracDelta is sufficient: DiracDelta[-x]== DiracDelta[x] $\endgroup$
    – chris
    Oct 14, 2012 at 9:45
1
$\begingroup$

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

$\endgroup$

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