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I want to write a code to evaluate

$$ \int_0^{t-1}\frac{\delta(z-1)}{z}\frac{\delta(t-z-1)}{t-z}dz$$

which should gives the answer $$\frac{\delta(t-2)}{t-1}.$$

I tried to write the code

Integrate[DiracDelta[z - 1]/z * DiracDelta[t - z - 1]/(t - z), {z, 0, t - 1}]

but it didn't give the answer.

How should I write the code to give the expected answer?

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  • $\begingroup$ Why should (DiracDelta[-2 + t] HeavisideTheta[-2 + t])/(-1 + t) be true? Can you kindly give a reference? $\endgroup$
    – user64494
    Sep 9 at 14:42
  • $\begingroup$ Are you asking why mathematically should we get that answer? $\endgroup$
    – Mohamed Mostafa
    Sep 9 at 17:09
  • $\begingroup$ I asked for a reference for it. What is the reason you think that the right answer should be $\frac{\delta(t-2)}{t-1}$? Hope I am clear. $\endgroup$
    – user64494
    Sep 9 at 17:27
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    $\begingroup$ Just calculate it by hand ✋ $\endgroup$
    – Mohamed Mostafa
    Sep 9 at 17:41
  • $\begingroup$ This assumes t>2? $\endgroup$
    – Daniel Lichtblau
    Sep 9 at 19:12

3 Answers 3

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Workaround:

FourierTransform[Integrate[InverseFourierTransform[
DiracDelta[z - 1]/z*DiracDelta[t - q z - 1]/(t - z), q, w], {z, 0,
t - 1}, Assumptions -> t \[Element] Reals], w, q] /. q -> 1

(*(DiracDelta[-2 + t] HeavisideTheta[-2 + t])/(-1 + t)*)

Solution with Maple 2023.1 :

enter image description here

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  • $\begingroup$ Fourier transforms of generalized functions are hard matter. It's unclear to me how to justify FourierTransform[Integrate[InverseFourierTransform.... Also InverseFourierTransform[ DiracDelta[z - 1]/z*DiracDelta[t - q z - 1]/(t - z), q, w] is running without any response for a long time. All that is built on the sand. Last, but not least: why is (DiracDelta[-2 + t] HeavisideTheta[-2 + t])/(-1 + t) true? $\endgroup$
    – user64494
    Sep 9 at 14:25
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    $\begingroup$ @user64494 .My sand is GOLD. $\endgroup$
    – Mariusz Iwaniuk
    Sep 9 at 14:30
  • $\begingroup$ MariuszIwaniuk (@ does not work.): I prefer arguments over emotional words. $\endgroup$
    – user64494
    Sep 9 at 14:43
  • $\begingroup$ InverseFourierTransform[ DiracDelta[z - 1]/z*DiracDelta[t - q z - 1]/(t - z), q, w] results in (E^(-I (-1 + t) w) DiracDelta[-1 + z])/(Sqrt[2 \[Pi]] (-1 + t)). Can somebody give a reference to this result? It would be kind of that person. $\endgroup$
    – user64494
    Sep 9 at 15:41
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    $\begingroup$ It's undefined. The integral cannot have a member of the singular set as an endpoint of the integration. $\endgroup$
    – Daniel Lichtblau
    Sep 10 at 21:21
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Approximate the DiracDelta by the normal distribution

   Integrate[DiracDelta[z-1 ]/z 1/Sqrt[2 \[Pi] s] E^(-(t-z-1)^2/(2s^2)),  
     {z,0,t-1}]

$$\fbox{$\frac{\theta (t-2) e^{-\frac{(t-2)^2}{2 s^2}}}{\sqrt{2 \pi } \sqrt{s}}\text{ if }t\in \mathbb{R}$}$$

Now let $s\to 0$, evaluating $\theta$ at its jump. Good example to show that products of singular distributions are undefined.

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    $\begingroup$ All that is built on the sand. If you look in W. Rudin, Functional analysis, Chap. 6,7, you will see that $\frac{e^{-\frac{(t-2)^2}{2 s^2}}}{\sqrt{2 \pi } \sqrt{s}}$ converges to the $\delta$-distribution in the weak topology. The pointwise convergence makes no sense. $\endgroup$
    – user64494
    Sep 9 at 14:46
  • $\begingroup$ The product DiracDelta[-2 + t] HeavisideTheta[-2 + t] is not defined because both distributions have $ \{2\}$ as their support.See "Problem of multiplying distributions" in the Wiki article as a first reading. $\endgroup$
    – user64494
    Sep 9 at 17:48
  • $\begingroup$ Also the documentation says "Products of distributions with coinciding singular support cannot be defined" and presents as an example Integrate[HeavisideTheta[x] DiracDelta[x], {x, -1, 1}]. $\endgroup$
    – user64494
    Sep 9 at 18:33
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Mathematica needs the info t \[Element] Reals !

Using DiracDelta as a limit (see @RolandF's approach) we get

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

Integrate[DiracDelta [z - 1]/z* dirac[t - z - 1]/(t - z), {z, 0, t - 1},Assumptions ->{ Element[t, Reals] }]
(*(E^(-((-2 + t)^2/(2 eps))) HeavisideTheta[-2 + t])/(Sqrt[eps]Sqrt[2 \[Pi]] (-1 + t))*)

Using the definition of dirac the result follows to

DiracDelta[t-2] HeavisideTheta[-2 + t] / (-1 + t)

which confirms the expectation of QP and the result @MariuszIwaniuk !

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  • $\begingroup$ The product DiracDelta[t-2] HeavisideTheta[-2 + t] is not defined because both distributions have $\{2\}$ as their support (see "Problem of multiplying distributions" in the Wiki article). $\endgroup$
    – user64494
    Sep 9 at 17:41
  • $\begingroup$ Knowing that the definitions of both distributions are well defined I see no reason why the product shouldn't be defined too! $\endgroup$
    – Ulrich Neumann
    Sep 9 at 17:43
  • $\begingroup$ Did you look in the linked Wiki article before having replied? $\endgroup$
    – user64494
    Sep 9 at 17:50
  • $\begingroup$ Also the documentation says "Products of distributions with coinciding singular support cannot be defined" and presents as an example Integrate[HeavisideTheta[x] DiracDelta[x], {x, -1, 1}]. $\endgroup$
    – user64494
    Sep 9 at 18:32

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