Why is the Dirac delta “function” often presented as the limit of a Gaussian with a fraction in the exponent?

I often see the Dirac delta function defined as $$\delta(x)=\lim_{a \to 0}\frac{1}{a \sqrt{\pi}}e^{-x^2/a^2}$$ Yet this is clearly equivalent to $$\delta(x)=\lim_{b \to \infty}\frac{b}{\sqrt{\pi}}e^{-b^2x^2}$$ and you might say that it's a bit neater, getting rid of the division in the exponent.

Why is the first definition so commonly used?

• When go to actually evaluate the second limit (for $x\neq0$ of course) you find that it is an indeterminate form of the kind $\infty\cdot0$, whereas the first limit is of the kind $\frac{0}{0}$. In order to apply L'Hopital's Rule to evaluate the second limit, we would first have transform it back to the first expression anyway. As such, I would regard the first limit representation as the "neater" of the two, despite the use of division in the expression. – David H Apr 21 '15 at 5:55
• This is a question of dimensions. You can only plug to the exp a dimensionless quantity. So on original definition, $a$ and $x$ are of the same dimension, say meters. And they have real physical meanings in most applications. On the other hand, the result is a density which is natural to measure in 1/meter(s) when $x$ is in meters. But your $b$ has no simple physical meaning. – Alexandre Eremenko Apr 21 '15 at 21:22
• How is this a historical question? – hjhjhj57 Apr 23 '15 at 3:08

Paul Dirac was a physicist and I think it's in that regard that a division in the exponent seems much more interesting and common than a product, because it is easier to assign a physical interpretation: $a$ can have the same dimension as $x$ which allows you to give $a$ a significance - for example the width of a potential well which area is constant.
Perhaps people are following the Wikipedia, but this form is by no means exclusive, here the one you prefer is used instead. If there is a physically canonical one it is $\delta(x)=\lim_{\sigma \to 0}\frac{1}{\sigma\sqrt{2\pi}}e^{-x^2/2\sigma^2}$, which contains the Gaussian distribution density with the standard deviation $\sigma$, see here, or a version with the Planck constant in it. Other characteristics of the distribution, like entropy for example, are traditionally expressed in terms of $\sigma$ as well, making this form most convenient for look ups.
In quantum mechanics the Gaussian distribution represents a localized "wave packet" with "characteristic width" $\sigma$, so the limit is a transition to the complete localization of the packet in the "position space". Gaussian wave packets are often used as "typical" in heuristic quantum mechanical calculations, and the ground state of the quantum oscillator happens to be of this form. The Heisenberg uncertainty principle is also usually expressed in terms of standard deviations, $\sigma_x\sigma_p\geq1/2$ (setting the Planck constant to $1$), where $\sigma_x$, $\sigma_p$ are the standard deviations of a wave function and its Fourier transform, which represents the wave function in the "momentum space".
So the $\sigma$ version is most natural from Dirac's point of view, my guess is that rescaling to $a=\sigma\sqrt{2}$ is done by industrious mathematicians, who are less interested in physical meanings. But reversing to $1/a$ is perhaps a step too far even for many of them.
Writing the density as $$x\mapsto \frac 1 {a\sqrt{2\pi}} e^{-x^2/(2a^2)}$$ is a natural way to do it since that makes $a$ the standard deviation of the distribution. If it is written as $$x\mapsto \frac 1 {a\sqrt{\pi}} e^{-x^2/a^2}$$ then $a$ is not the standard deviation but $a$ is still a scale parameter.