5
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If I calculate

Integrate[DiracDelta[x - y] DiracDelta[x - z], {x, -∞, ∞}, 
 GenerateConditions -> False]

I get this right answer, viz.

DiracDelta[-y + z]

But if I calculate a similar integral

Integrate[DiracDelta'[x - y] DiracDelta[x - z], {x, -∞, ∞},
 GenerateConditions -> False]

I get the wrong answer 0 when in fact the right answer is

DiracDelta'[z-y]
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  • 1
    $\begingroup$ what is DiracDelta'[x - y] supposed to mean? $\endgroup$
    – Nasser
    Aug 21, 2023 at 6:24
  • 1
    $\begingroup$ Derivative of the Dirac Delta. \int dx DiracDelta'[x-a] f[x] = -f'[a] if a is real is given correctly by Mathematica. $\endgroup$ Aug 21, 2023 at 6:57
  • $\begingroup$ All the above integrals are an (unsuccessful) attempt to implement the $\delta$-distribution into Mathematica (see, for example, Encyclopedia of Mathematics (The Wiki article on this topic is permanently edited. The latest change is dated 7, August 2023, at 22:09 (UTC).)). $\endgroup$
    – user64494
    Aug 21, 2023 at 7:34
  • $\begingroup$ Why does Wolfram Inc retain concepts that give manifestly wrong answers? $\endgroup$ Aug 21, 2023 at 7:44
  • $\begingroup$ The reason why Mathematica answers wrong questions is unclear. Products of distributions are undefined, in general. Either by conventional evaluation $$\int f(x) \delta (x-z) \delta '(x-y) \, dx=-\int \delta (x-y) (f(x) \delta (x-z))' \, dx=-\int f'(x) \delta (x-y) \delta (x-z) \, dx-\int f(x) \delta (x-y) \delta '(x-z) \, dx$$ or by Fourier transform $F(\delta delta') (k)= 1 \star k$$ $\endgroup$
    – Roland F
    Aug 21, 2023 at 10:19

2 Answers 2

0
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The reason why Mathematica answers wrong questions is unclear. Products of distributions are undefined, in general. Either by conventional evaluation

$$\int f(x) \delta (x-z) \delta '(x-y) \, dx=-\int \delta (x-y) (f(x) \delta (x-z))' \, dx=-\int f'(x) \delta (x-y) \delta (x-z) \, dx-\int f(x) \delta (x-y) \delta '(x-z) \, dx$$

or by Fourier transform where products go to convolutions

$$\mathit{F}(\delta * \delta') (k)= \int dk' 1(k') (k-k')$$

Perhaps its an too ambitious project, to exclude nonsensical input from the algorithm searching closed integral forms.

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2
  • $\begingroup$ Product of distributions aren't undefined and lead sometimes to distributions. $\endgroup$ Aug 21, 2023 at 15:19
  • $\begingroup$ If $y=z$, then all the products under "the integrals" are undefined since the supports are identical. $\endgroup$
    – user64494
    Aug 21, 2023 at 16:53
0
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Mathematica DiracDelta especially DiracDelta' sometimes doesn't evaluate correctly.

Often it helps to define DiracDelta as a limit, here examplary

dirac = Function[x, Exp[-(x^2/(2 eps))]/Sqrt[2 Pi eps]]  (* eps->0   *)

Mathematica evaluates

int=Integrate[dirac'[x - y] dirac[x - z], {x, -\[Infinity], \[Infinity]},Assumptions -> eps > 0 ]
(*(E^(-((y - z)^2/(4 eps))) (y - z))/(4 eps^(3/2) Sqrt[\[Pi]])*)

Result is identical to dirac[(y-z)/Sqrt[2]] (y-z)/eps and "diverges" for eps->0

But result is a distribution too, because int==-1/2 dirac'[(y - z)/Sqrt[2]]=!=dirac'[z-y] and "confirms" expectation of QP !

Hope it helps!

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4
  • $\begingroup$ How about Limit[(E^(-((y - z)^2/(4 eps))) (y - z))/(4 eps^(3/2) Sqrt[\[Pi]]), eps -> 0, Direction -> "FromAbove", Assumptions -> {y, z} \[Element] Reals] which results in 0? $\endgroup$
    – user64494
    Aug 21, 2023 at 14:28
  • $\begingroup$ @user64494 Similar Limit[DiracDelta[x], x -> 0] gives 0 too, but doesn't disprove distribution theory! $\endgroup$ Aug 21, 2023 at 15:40
  • $\begingroup$ This is not similar, that's another story. $\endgroup$
    – user64494
    Aug 21, 2023 at 16:50
  • $\begingroup$ It's the same story, both expressions are distributions! $\endgroup$ Aug 21, 2023 at 17:13

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