Disambiguation
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 coordinate 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 infiniteintegral 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]

Prometheus
2/15/2012 5:13:02 AM Permalink[quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] prometheus25 Never got too into those fields. Closest I got was control systems, but I didn't like my prof much: too easy so I just ended up doing the bare minimum.[/quote]Control theory was one of the easiest senior level courses I've taken. Applying it is another matter altogether; analog control systems are a huge pain in the ass to design and implement. We decided to go with this digital microcontroller for our electric drives project this term, and it's so much easier to write some software and let an IC do all the work than it is to figure out the exact values for a bunch of R's and C's and then realize you can't buy the values you need (lol).[/quote] I can understand that! We had to design amplifiers for a class project. It's such a pain getting those [i]just[/i] right to meet the requirements, rather than just uploading to Matlab and filtering through multiplication in the frequency domain. But that's terribly off topic. 
Disambiguation
2/15/2012 4:01:07 AM Permalink[quote][b]Posted by:[/b] prometheus25 Never got too into those fields. Closest I got was control systems, but I didn't like my prof much: too easy so I just ended up doing the bare minimum.[/quote]Control theory was one of the easiest senior level courses I've taken. Applying it is another matter altogether; analog control systems are a huge pain in the ass to design and implement. We decided to go with this digital microcontroller for our electric drives project this term, and it's so much easier to write some software and let an IC do all the work than it is to figure out the exact values for a bunch of R's and C's and then realize you can't buy the values you need (lol). [Edited on 02.14.2012 8:04 PM PST] 
Prometheus
2/15/2012 3:53:28 AM Permalink[quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] prometheus25 I am [i]rather[/i] enjoying this. It's been a while since I've been pulled from signal processing and transforms into pure math.[/quote]Ahh, fun. I focused on electric motors and power electronics.[/quote] Never got too into those fields. Closest I got was control systems, but I didn't like my prof much: too easy so I just ended up doing the bare minimum. 

Disambiguation
2/15/2012 3:38:58 AM Permalink[quote][b]Posted by:[/b] prometheus25 I am [i]rather[/i] enjoying this. It's been a while since I've been pulled from signal processing and transforms into pure math.[/quote]Ahh, fun. I focused on electric motors and power electronics. [quote]And I would like to know the last time a thread on integration made it to 100 posts.[/quote]I had a thread on [url=http://www.bungie.net/Forums/posts.aspx?postID=54009081]vector integration[/url] get to 100, but it was more for humor (as you can see from the title). 



Prometheus
2/15/2012 3:18:29 AM PermalinkI am [i]rather[/i] enjoying this. It's been a while since I've been pulled from signal processing and transforms into pure math. And I would like to know the last time a thread on integration made it to 100 posts. [Edited on 02.14.2012 7:19 PM PST] 


Disambiguation
2/15/2012 3:05:29 AM Permalink[quote][b]Posted by:[/b] prometheus25 I was merely implying that I am not pulling these figures out of my ass[/quote]And I respect you for it, but you're still wrong. 
Prometheus
2/15/2012 3:03:46 AM Permalink[quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] prometheus25 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.[/quote]I am a 4th year BSEE and I have a BA in mathematics. Your credentials are not the same as mine, and none of this is at all relevant to the question at hand. Also, see my edit.[/quote] I was merely implying that I am not pulling these figures out of my ass 
Disambiguation
2/15/2012 3:02:36 AM Permalink[quote][b]Posted by:[/b] prometheus25 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.[/quote]I am a 4th year BSEE and I have a BA in mathematics. Your credentials are not the same as mine, and none of this is at all relevant to the question at hand. Also, see my edit. [Edited on 02.14.2012 7:03 PM PST] 
Prometheus
2/15/2012 3:00:45 AM Permalink[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^%28r^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 twobladed 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] 
Disambiguation
2/15/2012 2:55:42 AM Permalink[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^%28r^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. By symmetry, your integral should be equivalent to [url=http://www.wolframalpha.com/input/?i=2+*+%28integral+%28integral+re%5E%28r%5E2%29+theta+%3D+0+to+pi%29+r+%3D+0+to+infinity%29]this one[/url], which it isn't. [Edited on 02.14.2012 6:59 PM PST] 
Prometheus
2/15/2012 2:54:06 AM Permalink[quote][b]Posted by:[/b] Disambiguation [quote][b]Posted by:[/b] prometheus25 I have some issues with this proof, even though it mathematically is sound. They can be seen in the document linked in [url=http://www.bungie.net/Forums/posts.aspx?postID=70094942&postRepeater1p=4#70136471]this post[/url].[/quote][url=http://imgur.com/RmHbS]I don't see the problem.[/url][/quote] 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^%28r^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. 

Disambiguation
2/15/2012 2:44:36 AM Permalink[quote][b]Posted by:[/b] prometheus25 I have some issues with this proof, even though it mathematically is sound. They can be seen in the document linked in [url=http://www.bungie.net/Forums/posts.aspx?postID=70094942&postRepeater1p=4#70136471]this post[/url].[/quote][url=http://imgur.com/RmHbS]I don't see the problem.[/url] 
Wicked Navajo
2/15/2012 2:43:54 AM Permalink[quote][b]Posted by:[/b] Disambiguation the simple function exp(x^2)[/quote] [quote][b]Posted by:[/b] Disambiguation [b]simple[/b][/quote] [quote][b]Posted by:[/b] Disambiguation [b](x^2)[/b][/quote] YOU LIE! My brain hurts just looking at it. 

tyler kell
2/15/2012 2:42:16 AM Permalink[quote][b]Posted by:[/b] Chalupa King117 Dude, Disam..... I'm sorry, but you and I both know this is something the Flood can't even begin to comprehend.[/quote] 

Prometheus
2/15/2012 2:36:28 AM Permalink[quote][b]Posted by:[/b] Alt account1 I read the OP, I don't understand. Someone explain the fact in short? I don't need to know why, just tell me the damn fact. [/quote] Basically, what is happening is that a surface is being made. The surface is a square with infinitely long sides (think of it more philosophically than literally). The height of the surface above the ground of zero at any point (x,y) is determined by e^(x^2 + y^2). [url=http://www.wolframalpha.com/input/?i=e^%28x^2+%2B+y^2%29]Here's[/url] a small portion of this surface. What integrals do is they find the area below a curve, from point a to b. In this instance, the bounds are from negative infinity to infinity, which means you are finding the area beneath the [i]entirety[/i] of the curve. By having a double integral (that's what dx and dy imply), you are finding the volume under the infinitely large square, or the entirety of the volume. X and Y (and Z) coordinates (what is referred to as the "Cartesian coordinate system" isn't the only way to represent space. It could also be done in polar coordinates, where a position is identified by a radius (usually from the 0,0 origin) and angle (from the x axis counterclockwise). The same space of an infinitely wide and tall square to a circle of infinite radius (again, "philosophically" the same thing) spun about the 0,0 point a full 360 degrees (though we use radians. It's simpler mathematically. 360 degrees = 2pi radians) When you switch coordinates, the variables need to be adjusted. See in the OP where e^1(x^2+y^2) switches to e^r^2? Think of the Pythagorean Theorem. x^2 + y^2 = z^2. z is the equivalent of a radius (Are you familiar with the Unit Circle), so this Theorem is tasked in this scenario as x^2 + y^2 = r^2, which you see in the OP. Solving integrals is a bit complicated for those unfamiliar with it. Once a certain amount of familiarity is acquired with them, it's mostly pattern recognition. At this point you're just going to have to trust me/us on the issue. So the solution here solves out to be pi. It is then assumed that since the two integrals at the beginning are so similar, you could treat them as if they are equivalent to one of them squared. This means that The square root of one of them squared, lim(a>oo)int(a,a)e^(x^2)dx, is equal to the square root of the answer, equating to sqrt(pi). I have some issues with this proof, even though it mathematically is sound. They can be seen in the document linked in [url=http://www.bungie.net/Forums/posts.aspx?postID=70094942&postRepeater1p=4#70136471]this post[/url]. 
Alt account1
2/15/2012 1:25:21 AM PermalinkI read the OP, I don't understand. Someone explain the fact in short? I don't need to know why, just tell me the damn fact. 