Integration involving two DiracDelta with variable limit

Question

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?

Answer 1

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 :

Answer 2

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.

Answer 3

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 !