While trying to understand a proof in a paper, I came upon the following a calculation needing the following identity:
$$\lim\_{t\to 0} \int\_{-\infty}^\infty \left(e^{-\log(4\pi i t)/2} e^{ik^2/4t} -\delta(k)\right)f(k)\,dk=0.$$
for $f\in\mathcal{S}(\mathbb{R})$ and $t>0$. Of course, this means that the exponential kernel converges to the Dirac measure in the limit $t \to 0$ in the sense of distributions. I'm able to show this is the case if $t$ approaches $0$ along the negative imaginary axis in the complex plane, since then, we can apply the dominated convergence theorem after the substitution $k\mapsto k\sqrt{t}$ and exchange limit and integral. But in this case, this doesn't help us because $e^{-ik^2/4}$ lies on the unit circle for all $k\in\mathbb{R}$, and thus its modulus isn't integrable. We can also try shifting the contour of integration by $\pi/4$ radians and attempt to apply Cauchy's formula to take the contour back to the real axis, but I run into issues where $f$, extended to the complex plane, may not be bounded on the rays/curves we are interested in.


Perhaps this equality does not hold the way I've asked it. What is a way to derive such an equality and the context in which it is true?



How to prove that this one-parameter family of distributions converges to the Dirac measure?
Allow me to replace $k$ by $x$. The kernel
$$G(x,t)=(4\pi it)^{-1/2}e^{ix^2/4t}$$
is the Green function of the Schrödinger equation, which can be written in the integral form
$$G(x,t)=\frac{1}{2\pi}\int\_{-\infty}^\infty e^{-ikx}e^{-ik^2 t}dk.$$
In the limit $t\rightarrow 0$ we then have an integral representation of the delta function,
$$2\pi\delta(x)=\int\_{-\infty}^\infty e^{-ikx}dk.$$
Both these integral representations need to be understood in the distributional sense (they are formally divergent for real $x$ and $t$; they need to be multiplied by a test function and integrated for a finite answer).


