Dirac delta equalities in physics Earlier I asked this question on the Math Exchange but I'm looking for a physics point of view. How do you interpret an equation like $$x^n \delta(x) = 0, \qquad n\in \mathbb{N},$$ around $x=0$? Why does it suffice to show the integral of this expression is zero around the singularity to show the equality is valid? There are many definitions to the Dirac delta "function" so I get extremely confused with this notion. In other words, do you treat equality of a Dirac delta function and some other expression as an "integral equality" because just inputting the value doesn't make sense?
 A: There may be many different definitions of the Dirac delta, but they all share the common feature that they only make sense inside an integral, and that
$$\int_{-\infty}^\infty f(x)\delta(x-a)dx = f(a)\qquad (*)$$
My preferred perspective is that the collection of symbols on the left-hand side of $(*)$ is defined to mean $f(a)$, in the sense that the integrand is not a true function so the integral is not meant to be taken literally.
Another point of view is that the Dirac delta is the limit of a sequence of increasingly tall and narrow Gaussians,
$$\delta(x) = \lim_{\epsilon\rightarrow 0^+} \frac{e^{-x^2/\epsilon}}{\sqrt{\epsilon\pi}} \qquad (**)$$

For $x\neq 0$, the above limit evaluates to zero.  For $x=0$, the limit does not exist (or is $\infty$, if you prefer).  However,
$$\lim_{\epsilon\rightarrow 0^+}\int \frac{e^{-x^2/\epsilon}}{\sqrt{\epsilon\pi}} dx = \lim_{\epsilon\rightarrow 0^+}1 = 1$$
With a bit of work, one can show that
$$\lim_{\epsilon\rightarrow 0^+} \int f(x) \frac{e^{-(x-a)^2/\epsilon}}{\sqrt{\epsilon\pi}} dx = f(a)$$
Therefore, we are to understand the expression $(**)$ as a limit which is to be taken only after an integral sign has been applied.

In both cases, the Dirac delta is an object which is only well-defined inside an integral.  As a result, $x^n\delta(x)$ is defined in the same way.
