Intuitive reason for why the Gaussian integral converges to the square root of pi?

This is a very famous problem, which is commonly taught when students begin learning about multivariable integration in polar coordinates. However, it has always bothered me that we recieved such an answer. There has always been a voice nagging at me, saying "this is just a clever approximation based upon our axioms."

Pi and Euler's constant are both very special transcendental numbers, with a very special relationship. However, the antecedent of pi comes from a physical feature which connects all curved metrics/spaces/objects. I do believe that non-algebraic numbers are proof that our theory of "natural numbers" is quite flawed [my opinion is also related to Godel's second incompleteness theorem], and that pi is a natural phenomenon which we were able to observe [quite incompletely] with the method that we chose [natural numbers].

Is there an intuitive reason why the integral converges to such a value? We were only able to analyze it like this due to a very clever trick [which I am in no way doubting the aunthenticity of]. I do not want to come off as childish, however, I would be much appreciative if an experienced mathematician (which are ubiquitous on this site) were to explain it to me.

Answer

Let $ V_n(h) $ is the $ n$ -dimensional volume of a hyperspherical cap (of unit radius) of height $ 1+ h (-1 \le h \le 1) $ in $ n$ -dimensional Euclidean space, $ F()$ - the integral of the standard normal distribution.

It’s known
( Wikipedia, Hyperspherical cap) that
if $ n \to \infty $ then $ V_n(h)/V_n(1) \to F(h \sqrt n ). $

Here I’ll try to show some details and interesting (I think) connections between $ e, \pi , V_n(h) $ and $ F(). $

$ V_n(h) = \int _{-1}^h \int _{-z_{2}}^{z_{2}} ... \int _{-z_{n}}^{z_{n}} dt_{n}... dt_{2} dt_{1}= C_{n-1} \int _{-1}^h (1-t^2)^{(n-1)/2} dt, $

where $ C_{n-1}= V_{n-1}(1) $ - the volume of the unit $(n-1)$-dimensional ball,

$ z_i= (1- \sum_{j=1}^{i-1} t_j^2)^{1/2}$.

Now we can write

$ P_n(x)=_{def} \frac {V_n(x/n)} { V_n(1)} $ $ = \frac { C_{n-1}} { C_{n}}\int _{-1}^{x/\sqrt {n}} (1-t^2)^{(n-1)/2} dt. $

Than using the formulas:

$ C_{2k}=\ 2^k /k!, C_{2k+1}= 2^{k+1} \pi ^k /(2k+1)!!, $
$ (2k)!!=2^k/k!, (2k+1)!!=n!/(2^k k!) $

and Stirling's approximation for factorials, it’s not very difficult to get:

$ C_{n-1}/ C_{n}= \sqrt{ \frac {n-1} {2 \pi}}$ (here I put $ n=2k $).

Thus for $ n \to \infty $:

$ P_n(x) \to \sqrt{ \frac {n-1} {2 \pi}} \int_{-1}^{x/\sqrt {n}} (1-t^2)^{(n-1)/2} dt, $

$ \frac {dP_n(x)}{dx} \to (2 \pi)^{-1/2} (1-x^2/n)^{(n-1)/2} \to (2 \pi)^{-1/2} \exp (- x^2 / 2) $

and $ P_{\infty}(x) =F(x) $.