Problem with defining $\int_0^\infty\delta(x)dx$ within distribution theory by considering nascent dirac deltas

Question

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 $1/2$?

Answer 1

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.

Answer 2

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:

$$\int_0^\infty \delta_\epsilon(r)\space dr = 1$$