I can prove the following fact, for $\forall f \in S(\mathbb{R})$, where $S(\mathbb{R})$ means the Schwartz space,
${\displaystyle \lim_{\tau \rightarrow 0^+}} {\displaystyle\int_{\mathbb{R}}} \sqrt{\frac{1}{2 \pi \tau}} e^{-\frac{x^2}{2 \tau}} f(x) dx = f(0)$.
The question is, when I change the $\tau$ in the above formula to $it$ where $t \in \mathbb{R}$, does the changed formula hold?
${\displaystyle \lim_{t \rightarrow 0}} {\displaystyle \int_{\mathbb{R}}} \sqrt{\frac{1}{2 \pi i t}} e^{i\frac{x^2}{2 t}} f(x) dx = f(0)$.
The issue here is that when $t$ approaches to zero, we can not let the integral of $\sqrt{\frac{1}{2 \pi i t}} e^{i\frac{x^2}{2 t}}$ outside a fixed neighborhood of the zero point approaches to zero.
So, does the changed formula hold? If it does, how to prove this? If it doesn't, can we construct a function $f\in S(\mathbb{R})$ that breaks the equality?
For $\epsilon>0,\ t>0$ let $u_{\epsilon,t} \in L^1(\mathbb{R})$ be defined by $$ u_{\epsilon,t}(x) := \frac{1}{\sqrt{2 \pi i t}} e^{(i-\epsilon)x^2/(2t)} . $$
This function satisfies the differential equation $ u_{\epsilon,t}'(x) = \frac{(i-\epsilon)}{t} x \, u_{\epsilon,t}(x) . $ Taking the Fourier transform of both sides gives $ i\xi \, \hat{u}_{\epsilon,t}(\xi) = i\frac{(i-\epsilon)}{t} \, \hat{u}_{\epsilon,t}'(\xi) , $ i.e. $$ \hat{u}_{\epsilon,t}'(\xi) = t\frac{\xi}{i-\epsilon} \hat{u}_{\epsilon,t}(\xi) . $$
Thus, $ \hat{u}_{\epsilon,t}(\xi) = C_{\epsilon,t} e^{-t(\epsilon+i)\xi^2/(2(1+\epsilon^2))} , $ where $$ C_{\epsilon,t} = \hat{u}_{\epsilon,t}(0) = \frac{1}{\sqrt{2 \pi i t}} \int_{-\infty}^{\infty} e^{(i-\epsilon)x^2/(2t)} \, dx = \frac{1}{\sqrt{1+i\epsilon}} . $$
Letting $\epsilon\to 0$ we get $ \hat{u}_{0,t}(\xi) = e^{-it\xi^2/2} . $ Now we see that $\hat{u}_{0,t}(\xi) \to 1 = \hat{\delta}(\xi)$ as $t \to 0$. From this we conclude that $u_{0,t} \to \delta$ as $t \to 0$.