Why is there a 2 in the pdf for the normal distribution? Why does the variance in the Normal density have a $2$ in it? I can make sense of the rest of the function, but I do not understand what the $2$ adds to the equation or why is it there. Wouldn't the version on the left give a proper density, with about the same properties as the normal density on the right?
$$ \underbrace{\int_{-\infty}^{\infty} \frac{1}{\sqrt{\pi\sigma^2}}\exp{\frac{-(x-\mu)^2}{\sigma^2}}}_{\text{(without 2)}}\qquad\qquad\qquad \underbrace{\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi\sigma^2}}\exp{\frac{-(x-\mu)^2}{2\sigma^2}}}_{\text{(standard version)}}$$
 A: You can find a good derivation of the univariate normal function using calculus in this document $[1]$. (I did try to reproduce the essential excerpt of the derivation here, but its still too lengthy, so you're better off having a look at the linked pdf file itself.)

Keep a lookout for the part, in the linked document, where there is an integral of x, which we know is:
  $\int x dx = \frac{x^{2}}{\mathbf{2}} + c$.
  That is where the 2 'enters the picture' in the final form of the univariate normal distribution.

One fairly good reason not to substitute for the standard deviation with $\sigma\prime = \sqrt{2}\sigma$, is that the points of inflection of normal distribution are exactly 1 standard deviation ($\sigma$) from mean ($\mu$) on both sides. (Point of inflection is where the second order derivative is zero). I'd rather have the inflection point at $\sigma$ and keep the 2 as 'variance coefficient'.
--
The mention of 'coefficient of parameter' got me confused and curious. Have a look at coefficient and parameter.

$[1]$ www.planetmathematics.com
