Limiting case of linear canonical transform From what I understand based on reading the Wikipedia article, for $g=\big(\begin{smallmatrix}a&b\\c&e\end{smallmatrix}\big)\in\mathrm{SL}_2\mathbb{C}$ the linear canonical transform $X(u)$ of a suitable $x(t)$ is an integral transform $X(u)=\int_{\mathbb{R}} K_g(u,t)x(t)\,\mathrm{d}t$ with the kernel given by
$$ K_g(u,t) = 
\begin{cases}
(ib)^{-1/2} \exp\big(\frac{\pi i}{b}(du^2-2ut+at^2)\big) & b\ne0 \\[4pt]
d^{1/2} \exp\big(\pi i cdu^2\big)\delta(t-du) & b=0 
\end{cases} $$
What I don't get, is how is the integral transform with $b=0$ a limiting case of the one with $b\ne0$?
The Wikipedia article on delta says it's a limiting case of $\delta_p(x)=\frac{1}{|p|\sqrt{\pi}} \exp\big(-(\frac{x}{p})^2\big)$, but the formula for $K_g$ (for real $g$) has purely imaginary exponent, and it's also distributionally given by $\delta(x-\alpha)=\int_{\mathbb{R}}\exp\big(2\pi ip(x-\alpha)\big)\mathrm{d}p$, but the formula for $K_g$ (in itself) doesn't have an integral (even though applying it does). Any ideas?
 A: I realize we can use $\delta_p(x)=\displaystyle \tfrac{1}{|p|}\exp\!\Big[-\left(\tfrac{x}{p}\right)^2\Big]$ by completing the square. I think this argument plays fast-and-loose with square roots of complex numbers, but surely there's something to it...
$$ \begin{array}{l}
K_g(u,t) & =\displaystyle \sqrt{\frac{1}{ib}} \exp\left[\frac{\pi i}{b}\big(at^2-2ut+du^2\big)\right] \\[5pt] 
& \displaystyle = \frac{1}{\sqrt{ib}} \exp\left[\frac{\pi ia}{b}\left(t^2-\frac{2u}{a}t+\frac{d}{a}u^2\right)\right] \\[5pt]
& \displaystyle = \frac{1}{\sqrt{ib}} \exp\left[\frac{\pi ia}{b}\left(\left(t-\frac{u}{a}\right)^2+\frac{ad-1}{a^2}u^2\right)\right] \\[5pt]
& \displaystyle = \frac{1}{\sqrt{ib}} \exp\left[\frac{\pi ia}{b}\left(\left(t-\frac{u}{a}\right)^2+\frac{bc}{a^2}u^2\right)\right] \\[5pt]
& \displaystyle = \frac{1}{\sqrt{ib}} \exp\left[\frac{\pi ia}{b}\left(t-\frac{u}{a}\right)^2+\frac{\pi ic}{a}u^2\right] \\[5pt]
& \displaystyle = \frac{1}{\sqrt{a}} \frac{\displaystyle\exp\left[-\left(t-\frac{u}{a}\right)^2/\sqrt{\frac{bi}{a\pi}}\,\right]}{\displaystyle\sqrt{\pi}\sqrt{\frac{bi}{a\pi}}} \exp\left[\frac{\pi ic}{a}u^2\right] \\[5pt]
& = \displaystyle \frac{1}{\sqrt{a}}\,\delta_p\!\left(t-\frac{u}{a}\right)\exp\left(\frac{\pi ic}{a}u^2\right)
\end{array} $$
where $p=\displaystyle\sqrt{\frac{bi}{a\pi}}$. As $b\to0$, since $g\in\mathrm{SL}_2\mathbb{C}$, we can say $1/a\to d$, and $K_g$ becomes
$$ K_g(u,t)=\sqrt{d}~\delta(t-du)\, \exp\!\big(\pi i cd u^2\big). $$
