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This actually is undefined..

@@rauldurand The Dirac Delta Function is always a pointed subject!

The delta function is not a function between real numbers, but a construct only to be used as a limit expression of the functional defined by various models for an approximation of this delta function with the pertinent function being integrated alongside with it.

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nice! Thank you very much 👍

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gracias!

Thank you!

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It is a generalization of a function, in this case a functional. The mapping that relates variables for a functional is not necessarily the same as a traditional function (although your comment hints that you understand it isn't a standard function), so it's not fair to call his definition "wrong" if you're not specifying this difference of mapping between a standard function and another definition that generalizes functions to functionals. You're arguing against convention rather than utility at that point. Neither of you are wrong. You can certainly argue that a function is just a limiting case of a functional over a given vector space, but when most people talk about functionals they are referring to a mapping of a vector space V over a field F onto a field F (a function with a function as the input). A traditional function is the limiting case where the input itself is constant.

thanks

No, you don't, because the integral sign is actually an abuse of notation. What you do is not an integral operation but the application of a functional to a test function.

This is not only true for the delta function! Any function f(x) shifts to the positive x-direction, if you subtract x0 from the argument: f(x-x0). And if you add x0 to the argument, f(x+x0), then the function shifts to the negative x-direction. Plot the quadratic functions f(x) = x^2 and f(x-x0) = (x-x0)^2 then you will see that f(x-x0) = (x-x0)^2 is shifted to the positive direction. Feel free to join the polls I'm doing in the community tab. There you can vote for the next video topic: kzfaq.infocommunity If you regularly visit my channel and like my videos, I would be happy about 1 x coffee ☕ per month: kzfaq.infojoin

The way I personally like to think of the delta function is δ(x) = 1/dx if x = 0, and δ(x) = 0 otherwise. That way, if you do the integral ∫ f(x)δ(x − a)dx over ℝ, you first integrate over ℝ\{a}, and since for all x in ℝ different than a, δ(x − a) = 0, you get 0 for that whole portion of the integral. What's left to integrate over is the singleton {a}. Summing over a single element is like getting rid of the integral sign and simply plugging-in x = a. So in the end we have f(a)δ(0)dx = f(a)dx/dx = f(a).

consider the functions g, h defined by : g(x) = f(x)δ(x) h(x) = f(a)δ(x) for any a fixed and f a random function if those are equal then their integral from -∞ to +∞ must be too so ∫f(x)δ(x)dx = f(a)∫δ(x)dx so f(0) = f(a) which is absurd as we chose f any function unless a=0. We can do the same to disprove f(x)δ(x) = δ(x) By integrating we get f(0) = 1 for any function f

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