Dirac Delta definition in non-standard analysis? 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?
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

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

Convolving $d$ with any standard continuous function gives

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

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

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