Let $\delta(t)$ be defined as the limit of a Gaussian pdf with 'zero variance'. What is then the result of $$I=\int_0^0 \delta(t)dt\quad?$$
on the one hand, "$\delta(0)=\infty$", but the length of the interval is zero. I think that since $\delta$ is defined by a limit, whereas the interval is zero (without a limit), then the integral will be zero, but I'm not sure.
Thanks.
On
There's a subtle difference between "zero variance" and "limit when variance goes to zero". Try computing the limit
$$\lim_{\sigma\to 0}\int_{-\omega}^{\omega}\frac{1}{\sigma\sqrt{2\pi}}e^{t^2/2\sigma^2} dt$$
The answer is
$$\lim_{\sigma\to 0^+} erf\bigg(\frac{\omega}{\sigma\sqrt{2}} \bigg) = 1 ^*$$
which is what you expect when you integrate the Dirac delta function $\delta(t-t_0)$ over the interval $-\omega<t_0<\omega$:
$$\int_{-\omega}^{\omega}\delta(t-t_0)dt = 1 $$
$^*$ Notice that I put $\lim_{\sigma\to 0^+}$, because the variance is a positive quantity and thus the limit is taken from the right. The limit from the left of this error function would be -1, but again, the variance is positive and this would make no sense.
Without the limit, it's undefined, because it's an improper integral which can be assigned a value only as a limit of well-defined integrals.