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. [url=http://www.wolframalpha.com/input/?i=integrate+e%5E(-x%5E2)+from+x+%3D+-infinity+to+infinity]tl; dr[/url] [Edited on 02.12.2012 11:49 PM PST]