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]
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[quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] prometheus25 Read over the document I linked. Orthogonality of x and y is an issue to me. Also, the polar integration can be done equivalently by [url=http://www.wolframalpha.com/input/?i=integral+%28integral+re^%28-r^2%29+theta+%3D+0+to+pi%29+r+%3D+-infinity+to+infinity]int(-a,a)int(0,pi)e^(-x^2)rdthetadr[/url], which gives you an answer of 0.[/quote]Orthogonality of x and y has nothing to do with the polar transformation; and you're still erroneously integrating r from -inf to inf, which doesn't even make sense. r is a nonnegative quantity.[/quote] You can absolutely have a negative radius. It's not mathematically incorrect. In that case, I am integrating a positive infinite radius in one direction by 180 degrees (pi radians) while simultaneousness integrating a negative radius in the exact opposite direction 180 degrees (pi radians). It's the equivalent of a two-bladed propeller (with blades of infinite length, of course) spinning 180 degrees, but each one, being mounted exactly opposite each other, cover half the area of a full circle. Mathematically this is equal to a single "blade" being spun a full 360 degrees. Edit: Further more, orthogonality has [i]everything[/i] to do with polar integration. The reason that you can substitute x^2 + y^2 for r^2 is because they are right angles to each other, allowing for the Pythagorean Theorem to be employed to find the length of the hypotenuse, which is the radius. It also means that they are out of phase. I assure you this is mathematically correct. I am a 4th year Electrical Engineer, too, ya know. That puts my credentials exactly where you are. [Edited on 02.14.2012 7:05 PM PST]