If we consider the Gaussian or the Lorentzian representation of $\delta$, then we obtain $\newcommand\dif{\mathop{}\!\mathrm{d}}$ \begin{equation} \lim_{\epsilon\to 0}\int_0^\infty \delta_\epsilon(x)\dif x=1/2 \end{equation} but as far as I understand the problem is that the equation is not true for other nascent dirac deltas (i.e. sequences of functions such that the associated sequences of distributions converges to $\delta$). Can we find a nascent dirac delta such that 1) the limit does not exist and 2) the limit exists and does not equal $1/2$?
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ For starters on 2), make the function not balanced around $0$, or not even. P.S. Distribution theory won't help, as the $\delta$ distribution is not "integrable." $\endgroup$– Ted ShifrinAug 7, 2022 at 19:03
-
1$\begingroup$ @filippo Pleased to see your question. $\endgroup$– Mark ViolaAug 8, 2022 at 21:17
-
$\begingroup$ Hello Filippo ! Regarding your second question, consider the case of the Dirac delta in polar or spherical coordinates. By excluding contributions from negative $r$ (for obvious reasons), all contributions are from positive values. So the limit is equal to $1$. $\endgroup$– M. WindAug 10, 2022 at 15:03
-
1$\begingroup$ @Filippo Thank you! I've edited with the correct reference. $\endgroup$– Mark ViolaAug 10, 2022 at 18:33
-
1$\begingroup$ And (+1) for the posted question $\endgroup$– Mark ViolaAug 10, 2022 at 18:56
|
Show 1 more comment
2 Answers
$\begingroup$
$\endgroup$
9
The sequence $f_n$ of piecewise-linear tent functions of height $n$ and base width $2/n$, but centered at $1/n$ rather than $0$, approximates $\delta$, but they all have integral $\int_0^\infty$ equal to $1$. Symmetrically, centering the tents at $-1/n$ have those integrals all $0$. If we interleave the two sequences, they approach $\delta$ distributionally but their integrals oscillate between $0$ and $1$, so have no limit.
answered Aug 7, 2022 at 19:04
-
$\begingroup$ Oh hah, I was making a comment along these lines. Deleted it for redundancy. Great answer as always! $\endgroup$ Aug 7, 2022 at 19:15
-
-
$\begingroup$ (+1) Hi Paul! It's always great to see you here. This answer is useful to those who have misconceptions about the meaningless funtional $\langle \delta, H\rangle$ which some wirite with abusive notation as $\int_{0}^\infty \delta (x)\,dx$. $\endgroup$ Aug 8, 2022 at 21:16
-
$\begingroup$ By the same logic one may argue that in two dimensions $\delta(x)\delta(y)$ is meaningful, but if one changes from Cartesian coordinates to polar, there is a problem. Because the integral over $\delta(r)$ runs over the positive real axis. Which is abuse of notation. Apparently because it is possible that negative values of $r$ contribute to the integral. Even though negative values of $r$ are not defined. $\endgroup$– M. WindAug 10, 2022 at 14:36
-
$\begingroup$ @M.Wind You continue to discuss "integrals" in the context of distribution theory. Distributions are linear functionals, not integrals. See THIS, which provides a rigorous expose of the Dirac Delta in spherical coordinates in $n$-dimensions. $\endgroup$ Aug 10, 2022 at 18:34
$\begingroup$
$\endgroup$
Nascent Dirac deltas are usually chosen to be symmetric. Sometimes that is not feasible. For example when one changes from Cartesian coordinates to polar, cylindrical or spherical coordinates. Then the domain of a variable gets restricted to the positive real axis. Obviously this affects the Dirac delta function.
For example consider the case of (normalized) Gaussian nascent deltas in two dimensions. We have $\delta_{\epsilon}(x) = 1/(\epsilon\sqrt(2*\pi))exp(-x^2/(2\epsilon^2)$ and $\delta_{\epsilon}(y) = 1/(\epsilon\sqrt(2*\pi))exp(-y^2/(2\epsilon^2)$. The double integral over the product of the two deltas is equal to unity. Now introduce polar coordinates. We obtain the following nascent radial Dirac delta: $\delta_\epsilon(r) = (r/\epsilon^2)exp(-r^2/(2\epsilon^2)$, valid for $r \ge 0$. It satisfies:
$$\int_0^\infty \delta_\epsilon(r)\space dr = 1$$
