Dirac Delta definition in non-standard analysis?

Question

What is the definition of Dirac Delta in non-standard analysis?

I would define it either as a standard distribution with $\sigma=\epsilon$ or maximum equal to $\omega$. Which is the correct answer?

Answer 1

I found the answer.

$$\delta(x)=\frac{\omega e^{-(\omega x)^2}}{\sqrt{\pi }}$$

And its integral, Heaviside Theta is

$$\theta(x)=\frac12\operatorname{erf}(\omega x)+1/2$$

As such,

$$\delta(0)=\frac{\omega}{\sqrt{\pi}}$$

Answer 2

Just like in standard analysis where there are lots of ways to represent the Dirac delta distribution as a limit of a sequence, in nonstandard analysis there are lots of ways to represent the Dirac delta as a nonstandard function.

The simplest is probably to pick an infinite $H$ and set

$$ d(x) = \begin{cases} 0 & |x| > \frac{1}{2H} \\ H & |x| \leq \frac{1}{2H} \end{cases} $$

Convolving $d$ with any standard continuous function gives

$$ \begin{align}\int_{-\infty}^{\infty} d(x) f(x) \, dx &= H \int_{-\frac{1}{2H}}^{\frac{1}{2H}} f(x) \, dx \\&= H f(\epsilon) \int_{-\frac{1}{2H}}^{\frac{1}{2H}} 1 \, dx \\&= H f(\epsilon) \frac{1}{H} \\&= f(\epsilon) \\&\approx f(0) \end{align}$$

where I've used the mean value theorem to go from line one to line two. (one could instead use $|f(x) - f(0)| < e$ for $|x| < \frac{1}{2H}$ if they liked)