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I read on a physics textbook by Griffiths that the Dirac delta function is defined by $\delta(x)=0$ if $x\not=0$ and $\int_{-\infty}^{\infty}\delta(x)dx=1$. As a distribution, the Dirac delta function is defined by $\delta(f):=f(0)$, where $f:\Bbb R\to\Bbb R$ is infinitely differentiable and has compact support. However, I feel like the latter definition does not capture the idea of the former definition for several reasons:

$1.$ A relation is given in the textbook as $$\delta(kx)=\frac{1}{|k|}\delta(x)$$ for nonzero constant $k$. This concept is only captured by using change of variable, but not by defining Dirac delta explicitly as a distribution.

$2.$ The author gives several problems on evaluating integrals involving Dirac delta, with an example (after changing some numbers, I don't know if there are copyright issues) being $\int_{-8}^{8}x\delta(x-5)dx$. It is obvious that $x$ is not a test function. But several expressions similar to this are written in the book anyway.

$3.$ The author gives several problems on evaluating integral involving Dirac delta function over a compact set (like a region bounded by a cube or sphere), just like the above example. We can use case-defined functions or step functions to simulate integration over a compact set, but the result is that we get a function that is not a test function.

Could it be that definition of Dirac delta function as a distribution didn't really capture physicists' idea? Or is the definition already good enough for all practical applications in physics and other fields (that is, good enough for solving real problems in their fields rather than calculating an integral that comes out of nowhere)?

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both definitions lead to exactly the same properties. but you must recognise how are they related.

$$ \delta[f]=f(0) $$

$$ \int_{\mathbf{R}}\delta(x)f(x)\,\textrm{dx}=f(0)\int_{\mathbf{R}}\delta(x)\,\textrm{dx}=f(0) $$ So the distribution and the delta function are related this way: $$ \delta[f]=\int_{\mathbf{R}}\delta(x)f(x)\,\textrm{dx} $$ and since they are equivalent in both settings you can derive the same properties

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  • $\begingroup$ You didn't read all the details I have given... $\endgroup$ – edm Aug 21 '16 at 14:19
  • $\begingroup$ i've read them and maybe i didn't understand where your problem lies, physicists prefer the integral representation since it directly leads to useful computations, mathematicians prefer the distribution definition since it's more rigourous. $\endgroup$ – Francesco Alem. Aug 21 '16 at 14:24
  • $\begingroup$ The problem is that I found something that seems to be used by physicists, and those may not be accurately represented in terms of distributions. $\endgroup$ – edm Aug 21 '16 at 14:27
  • $\begingroup$ Look here and you'll see that it can be done. en.wikipedia.org/wiki/Distribution_(mathematics) $\endgroup$ – Francesco Alem. Aug 21 '16 at 14:29
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I think you are completely correct that the more definitional/formal, super-careful version of Dirac's $\delta$ (and other related notions) does not immediately capture the sense or use of it by physicists, although it is a good approximation. Further, with some sophistication, rigorous mathematical rationalizations for some seemingly glib physicists' tricks can be found, although often these rigorizations require a level of sophistication that's hard to anticipate.

For example, $\int_{-8}^8 x\,\delta(x-3)\;dx$ is not immediately legitimate, since the function $f(x)$ that is $x$ on $[-8,8]$ and $0$ outside is not a test function, nor is it a Schwartz function. There are at least two ways to make this "ok". First, we can use the notion of support of a distribution, and the support of $\delta$ is $\{0\}$. Then (making a choice) we can modify $f(x)$ by multiplying by a smooth cut-off function that is identically $1$ near $0$... thus creating a test function out of $f$. This is a bit of a kludge, because we must show that the outcome is independent of the smooth cut-off... and, anyway, far more importantly, how would we know that this process is compatible with all our other expectations? :)

A second, slightly fancier, approach is to observe that $f(x)$ and $\delta$ have disjoint _singular_support_, so can be multiplied, as distributions. They have compact support, so can be applied to the smooth function $1$. (One should prove that the dual of the space of all smooth functions on $\mathbb R$, suitably topologized, is compactly-supported distributions.)

And there are more rationalizations available... From a strict mathematical viewpoint, one would want to know that they are all inevitably the same. Well, one approach is to once-and-for-all determine a very large space of functions on which $\delta$ is indeed a continuous functional, and prove some uniqueness results... For example, spaces of functions which are continuous on a neighborhood of $0$ will suffice... but without more information this is too self-referential. Further, without more information, the seminorm $\nu(f)=|f(0)|$ is not separating on many interesting spaces of functions... so there is considerable latitude in choosing a topology on a more-global space of functions... to make $\delta$ legit.

Partly because various uniqueness theorems are provable, much of the time the naive form of a computation does yield the correct outcome... whether or not we know why. A miraculous idiot-proof-ness of this bit of mathematics. But, yes, it is still possible to create/find seeming paradoxes if one presses too hard from a naive viewpoint (in effect disconnecting from physical intuition, for example).

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