7:48 PM
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Integrating DiracDelta in 3D in Cartesian coordinates works just fine i.e. gives vecf[{x, y, z}] vecr = {x, y, z}; vecrp = {xp, yp, zp}; Assuming[(vecr | vecrp | vecf[_]) ∈ Vectors[3, Reals], Simplify[Integrate[DiracDelta[-xp + x] DiracDelta[-yp + y] DiracDelta[-zp + z]*vecf[vecrp], ...

If you look in Wiki en.wikipedia.org/wiki/Distribution_(mathematics) , you will know that the integral under consideration has no sense. Upgrade your math.
The quote "Adding and multiplying distributions Distributions may be multiplied by real numbers and added together, so they form a real vector space. Distributions may also be multiplied by infinitely differentiable functions, but it is not possible to define a product of general distributions that extends the usual pointwise product of functions and has the same algebraic properties". Don't hesitate to ask for further explanation in need.
This is not my point. Up to Wiki, the integrand (as the product of three distributions) is not defined at all. Don't hesitate to ask for further explanation in need.
Sorry, I cannot discuss with you anymore.

@user64494 Multiplying the distributions DiracDelta[-xp + x] DiracDelta[-yp + y] DiracDelta[-zp + z] makes sense indeed for the integration variables are with respect to different coordinate directions. The resulting distribution is $\delta_{x_p,y_p,z_p}$ in $\mathbb{R}^3$. One cannot define a product of distributions in general but for two given ones, one can try to decide wether that is possible or not. There is actually a whole theory about that (e.g. see works by Hörmander). I would appreciate if you would hold back yourself a bit in the future and to try to be a bit more polite.
@user64494 And Integrate[DiracDelta[x],{x,-Infinity,Infinity}] is Mathematica's (and not only Mathematica's) way to write the pairing $\langle \delta_0, 1\rangle$.

@Henrik Schumacher: Can you base your statement by arguments and exact references? How about Integrate[DiracDelta[x],{x,-Infinity,Infinity}] which is simply ignorant? Thank you anyway.
Somebody edited my above comment which was as follows "Sorry, I have nothing to discuss with you in such manner. BTW, the notation Integrate[DiracDelta[x],{x,-Infinity,Infinity}] is simply ignorant: eg see en.wikipedia.org/wiki/Dirac_delta_function " . See dropbox.com/s/9id0ocydludljdy/screen%2026.09.18.docx?dl=0 and dropbox.com/s/uzyh131wkagvh86/screen%202%2026.09.18.docx?dl=‌​0 This is not a good practice.
@Henrik Shumacher: BTW, we read in Mathematica help "DiracDelta[x] returns 0 for all real numeric x other than 0". Nothing about any pairing.

@user64494 This "somebody" must have moderator priviledges. If I were you, I would start now to be a bit more careful. For the references: See here and here. Again, in particular the works by Hörmander are relevant.

I read in the first reference "pairs of distributions with disjoint singular support" Do DiracDelta[x] and DiracDelta[y] have such singular supports?

7:49 PM
I just tried to outline that there might be ways to make sense of that. Nothing more.
And no: They don't have singular support: The point {0,0} lies in their intersection if you treat x and y as integration variables. But that is only point (a) in that reference.
Point (b) is much more relevant.

"But that is only point (a) in that reference". It's enough to fail the construction.

No it does not. There are several cases in which a product of two distributions can be defined. Case (a) just does not apply.
And of course there are theoretical results that state that a uniform way to define "the" product of two arbitrary distributions cannot be defined consistently (consistently in the sense that some of the properties of products between functions have to be lost on the way).
But Mathematica's way of thinking about DiracDelta[-xp + x] DiracDelta[-yp + y] DiracDelta[-zp + z] is much simpler: It interprets Integrate[
DiracDelta[-xp + x] DiracDelta[-yp + y] DiracDelta[-zp + z] phi[x, y,
z], {x, -Infinity, Infinity}, {y, -Infinity,
Infinity}, {z, -Infinity, Infinity}] as Integrate[
DiracDelta[-zp + z] Integrate[
DiracDelta[-yp + y] Integrate[
DiracDelta[-xp + x] phi[x, y, z], {x, -Infinity,
Infinity}], {y, -Infinity, Infinity}], {z, -Infinity, Infinity}] which makes perfect sense.

8:08 PM
"There are several cases in which a product of two distributions can be defined". The question is still open: how to define the productDiracDelta[x]*DiracDelta[y]? I repeat that Integrate[Delta[x],{x,-Infinity,Infinity}] has no sense (see Wiki).

As I already said: Integrate[Delta[x],{x,-Infinity,Infinity}]  is Mathematica's way of applying $\delta_0$ to the constant function 1 - or to any other function that is constant 1 in a neighborhood of 0.
$\delta_0$ is not only interpretable as linear form on the space of (compactly) supported test functions. It is also a point measure.
So it can be applied to all continuous functions on $\mathbb{R}$.

We read in Mathematica help "DiracDelta[x] returns 0 for all real numeric x other than 0". Nothing about any pairing. Yuor above comment is wishful thinking.

That's another topic. Not the one with which we started.

And how about the product of the deltas? You get rid of answering.

Remember: You said, a multiplication between distributions was not possible. And I tried to give examples where is is possible.
And as I also wrote already, Mathematica interprets DiracDelta[x]*DiracDelta[y] as DiracDelta[0,0] or $\delta_{(0,0)}$ in $\mathbb{R}^2$.

8:19 PM
The question deals with the product of deltas.

Or more correct: Mathematica interprets DiracDelta[x]*DiracDelta[y] as DiracDelta[0,0] or $\delta_{(0,0)}$ in $\mathbb{R}^2$ whenever it occurs in an integral with x and y as integration variables. Otherwise, it does not have any meaning.

"And as I also wrote already, Mathematica interprets DiracDelta[x]*DiracDelta[y] as DiracDelta[0,0] or $\delta_{(0,0)}$ in $\mathbb{R}^2$". One more wishful thinking since this is not documented.

Actually, when within an Integrate expression with the integration variables x and y, Mathematica interprets DiracDelta[x] as the the measure $mu$ defined by $\mu(A) = \int_A \mathrm{d} \mathscr{H}^1$ where $\mathscr{H}^1$ is the one-dimensional Hausdorff measure.
Maybe not documented, but easy to check: Integrate[(DiracDelta[x] DiracDelta[y] -
DiracDelta[x] DiracDelta[y]) \[Phi][x, y], {x, -Infinity,
Infinity}, {y, -Infinity, Infinity}]

Wishful thinking again: this is not documented. What I read is "DiracDelta[x] returns 0 for all real numeric x other than 0".

And that is what it does. I don't know why people implemented it like that. But what I try to say is that DiracDelta does not have a real meaning when not wrapped by an Integrate. And when it is wrapped by an Integrate, it behaves different.
Moreover, "DiracDelta[x] returns 0 for all real numeric x other than 0" is completely unrelated to the question of products.

8:26 PM
No arguments.

Okay, I finish here. You seem to have a bad day. Let's stop here.
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Integrate[(DiracDelta[x, y] - DiracDelta[x] DiracDelta[y]) \[Phi][x,
y], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] irs running on my comp without any output for ages.

Are you running a fresh kernel of Mathematica 11.3?
It takes 20 milliseconds on my machine.

You are right : Integrate[(DiracDelta[x, y] - DiracDelta[x] DiracDelta[y]) \[Phi][x,
y], {x, -Infinity, Infinity}, {y, -Infinity, Infinity}] performs 0 for any function \[Phi][x,
y]. This is simply wrong result.

i think that user64494 is just trolling
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