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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.

Anixx
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user56829
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    All integrals of the form $\mathcal G_a(n)=\displaystyle\int_0^\infty a^{-\color{red}{x^n}}dx$ are intimately connected to geometric shapes of the form $x^n+y^n=r^n$, called superellipses. More to the point, $\mathcal G_a\bigg(\dfrac1n\bigg)=\dfrac{n!}{\ln^na}$ , for $n>0$ and $a>1$. Why are factorials connected to sums of powers ? Because the power of a sum is expressed in terms of binomial coefficients, $(x+y)^n=\displaystyle\sum_{k=0}^n{n\choose k}~x^k~y^{n-k},~$ where $\displaystyle{n\choose k}=\frac{n!}{k!~(n-k)!}$ – Lucian Aug 21 '14 at 02:49
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    In particular, $\mathcal G_{\large e}(2)=\frac12!$ is connected to the circle, defined by the algebraic equation $x^2+y^2=r^2$, whose characteristic constant is $\pi$. – Lucian Aug 21 '14 at 02:50
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    Intuitive argument: the bivariate normal pdf is of the form $e^{-c(x^2 + y^2)}$, which is circularly symmetric, so I would be somewhat surprised if $\pi$ did not show up when you integrate it over $\mathbb{R}^2$. Then, since $e^{-c(x^2 + y^2)} = e^{-cx^2}e^{-cy^2}$, the double integral equals the square of the integral over $\mathbb{R}$ of $e^{-cx^2}$, we would expect $\sqrt{\pi}$ to show up in the latter. –  Aug 21 '14 at 23:04
  • brilliant guys, this era is quite easy to gain information in. 20 years ago, you couldn't write a book report if you were unable find the right text. – user56829 Aug 22 '14 at 00:41
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    Now, proffessional mathematicians are answering my questions for free. thank you very muc. – user56829 Aug 22 '14 at 00:41
  • Bungo's answer is expanded in Spivak's excellent slim book Calculus on Manifolds. See Chris Leary's answer at

    http://math.stackexchange.com/questions/96840/does-the-improper-integral-int-0-infty-e-x2dx-converge/96854#96854

     – Colin McLarty Aug 22 '14 at 00:47
  • I see now, according to Lorde Kelvin I was no mathematician. The geometry is very self evident to me now, and I have been given more than good reason to believe in this notion now. Unfortunately, in undergraduate, my proffesors just said "this is the equation," and skipped any geometric intuition (I work in biological sciences now, but I wish I had taken more in depth math courses when I was able to). – user56829 Aug 22 '14 at 00:59

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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) $.

Yog Urt
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