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?
 A: 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)
A: 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,
$$\delta(0)=\frac{\omega}{\sqrt{\pi}}$$
