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*"The individual has always had to struggle to keep from being overwhelmed by the tribe. If you try it, you will be lonely often, and sometimes frightened. But no price is too high to pay for the privilege of owning yourself."*

-Nietzsche

Whence one might presume that the integral, from the negative of infinity towards the infinite, of the simple function exp(-x^2) might equate to the root of pi? I shall set forth to prove to you this assertion, that you should be struck by its audacity as much as I on this Sunday's eve.

We begin by considering the square, such that the integral under examination would be the product of an integral of exp(-x^2) and of an integral of exp(-y^2), for {x:-r,r} and {y:-r,r}.

As we should desire r to tend toward an infinite quantity, it necessarily must be that we have, indeed, a product of two limits. As one must recall from elementary calculus,

Lim(r->oo) int[-r,r]exp(-x^2)dx * Lim(r->oo) int[-r,r]e^(-y^2)dy

is equivalent to

Lim(r->oo) { int[-r,r]exp(-x^2) int[-r,r]exp(-y^2)dy }

One may then be struck with an occurrence: a theorem of Fubini may be applied to conjoin the integration. The result follows as

Lim[r->oo] int[(-r <= x <= r), (-r <= y <= r)]exp(-(x^2+y^2))dxdy

This is, of course, the same as int[R^2]exp(-(x^2+y^2))dxdy; we are therefore not constrained by a square modicum of integration, and indeed we should choose a circular region to evaluate this integral. Thus, we have the result

Lim(q->oo) int[x^2+y^2<=q^2]exp(-(x^2+y^2))dxdy

When considering circular regions it is most convenient to use a polar co-ordinate system, thus we apply the angular transformation to result in

Lim(q->oo) int[0,q]int[0,2pi]q*exp(-q^2)dtdq

Which can be evaluated by elementary calculus techniques to yield Lim(q->oo) (2pi)*(1 - exp(-q^2))/2 = pi.

We recall that we have calculated the square of the integral of interest. Thus, it follows by taking the principal root that the infinite-integral of exp(-x^2) is the root of pi.

tl; dr

[Edited on 02.12.2012 11:49 PM PST]