I have the following Gaussian function $c(x,x^{\prime})=\frac{1}{a\sqrt{2\pi}}e^{-\frac{(x-x^{\prime})^{2}}{2a^{2}}}$, for $a>0$, which can be thought of as a nascent delta function, in such a way that
$$ \lim_{a\rightarrow 0}c(x,x^{\prime}) = \lim_{a\rightarrow 0}\frac{1}{a\sqrt{2\pi}}e^{-\frac{(x-x^{\prime})^{2}}{2a^{2}}} = \delta(x-x^{\prime}) $$
Suppose I perform a change in variable, such that $\xi = x - x^{\prime}$, it is easy to integrate and check that
$$ \int_{-\infty}^{\infty}\frac{1}{a\sqrt{2\pi}}e^{-\frac{\xi^{2}}{2a^{2}}}d\xi =1 $$
Hence $c(\xi)$ is indeed, a suitable representation of the Dirac delta function. Suppose I now take the double derivative of $c(x,x^{\prime})$, such that
$$ \frac{\partial^{2}}{\partial x\partial x^{\prime}}c(x,x^{\prime})= \frac{c(x,x^{\prime})}{a^{4}}\left(a^{2} - (x-x^{\prime})^{2}\right) = D(x,x^{\prime}) $$
I wish to represent $D(x,x^{\prime})$ as a Dirac delta function as it does behave like a Dirac delta. For instance, consider the case where $x=x^{\prime}$, then
$$ \lim_{a\rightarrow 0}D(x,x) = \lim_{a\rightarrow 0}\frac{1}{a^{3}\sqrt{2\pi}} = +\infty $$
whereas when $x\neq x^{\prime}$
$$ \lim_{a\rightarrow 0}D(x,x) = 0 $$
(this can be checked with L'Hopital's rule). Which suggests that $D(x,x^{\prime})\sim\delta(x-x^{\prime}) \longrightarrow D(\xi)\sim\delta(\xi)$. However, the integral of $D(\xi)$ is such that
$$ \int_{-\infty}^{\infty}D(\xi)d\xi = \int_{-\infty}^{\infty}\frac{c(\xi)}{a^{4}}\left(a^{2} - \xi^{2}\right) d\xi = 0 $$ Hence $D(\xi)$ does not normalize to 1. How can $D(\xi)$ possess the property of Dirac delta yet the integral doesn't converge to 1? What is my mistake here? Is there any way I can normalize $D(\xi)$ so that I can represent it as a Dirac delta function $(D(\xi)\sim\delta(\xi))$?