Convergence of $\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)$ as $\varepsilon \to 0$

Question

Is it true that $\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)$ converges (in some sense) to $\delta_0$, Dirac delta distribution at point $0$, as $\varepsilon \to 0$ ? Thanks.

Answer 1

So the $i$ is intentional. Then let's look how that function acts as a distribution, hence let $\varphi$ an arbitrary test function:

$$\begin{align} \frac{1}{\sqrt{2\pi i\varepsilon}}\int_{-\infty}^\infty \exp \left(i\frac{x^2}{2\varepsilon}\right)\varphi(x)\,dx &= \frac{1}{\sqrt{2\pi i\varepsilon}}\int_{-\infty}^\infty \left(\cos \frac{x^2}{2\varepsilon} + i\sin \frac{x^2}{2\varepsilon}\right)\varphi(x)\,dx\\ &= \frac{1}{\sqrt{\pi i}}\int_{-\infty}^\infty \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\ &= \frac{1}{\sqrt{\pi i}}\int_{-\infty}^0 \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\ &\qquad + \frac{1}{\sqrt{\pi i}}\int_{0}^\infty \left(\cos (y^2)+ i \sin (y^2)\right)\varphi(\sqrt{2\varepsilon}\cdot y)\,dy\\ &= \frac{1}{\sqrt{\pi i}} \int_0^\infty \left(\cos (y^2) + i\sin (y^2)\right)\left(\varphi(\sqrt{2\varepsilon}\cdot y) + \varphi(-\sqrt{2\varepsilon}\cdot y) \right)\,dy. \end{align}$$

Now for $\varepsilon = 0$, we have the Fresnel integral

$$\frac{2\varphi(0)}{\sqrt{\pi i}}\int_0^\infty \cos (y^2) + i \sin (y^2)\,dy = \frac{2\varphi(0)(1+i)}{\sqrt{\pi i}}\sqrt{\frac{\pi}{8}} = \varphi(0),$$

and it remains to argue that taking the limit is legitimate. I'll leave that to handwaving, it can be done.

Answer 2

If the $i$ wasn't there, then yes, it would converge to the Dirac delta distribution (I am going to pretend that it isn't there).

First of all, you should verify that you actually have a distribution (for $\varepsilon > 0$). I.e. that is is always positive, integrable and that the integral over the entire real line is $1$. The last part is verified with substitution and the fact that $\int_{-\infty}^\infty e^{-t^2}dt = \sqrt\pi$.

Once you have that, you can start worrying about what the distribution converges to. For any fixed $x \neq 0$, your distribution looks like $$ \frac{a}{\sqrt\varepsilon}\cdot b^{1/\varepsilon} $$ where $b < 1$. As $\varepsilon \to 0$, it is well known (and a good exercise to show if you don't know) that this does indeed converge towards $0$.

For $x = 0$, your distribution looks like $$ \frac{a}{\sqrt\varepsilon} $$ which diverges to $\infty$ when $\varepsilon \to 0$.

So we have shown that the limit distribution as $\varepsilon \to 0$ has integral $1$ (as it should, being a distribution), and that at any point except $x = 0$, it is exactly $0$. Thus it must be the Dirac delta distribution.

Answer 3

With the $i$ in there ... for $x$ real and $\varepsilon > 0$, $$ \left|\frac{1}{\sqrt{2\pi i\varepsilon}}\exp\left( -\frac{x^2}{2i\varepsilon}\right)\right| = \frac{1}{\sqrt{2\pi\varepsilon}} $$ is constant (as a function of $x$), so I don't see how it could approximate a delta function.