I need to brush up on a few things and thought it would be an interesting challenge for some of the Math geeks here :)
This might look familiar to some of you.
|Infinity (-x^2)
| e^ dx
|0
Read as: integral with respect to x from 0 to infinity of e to the negative x squared.
(Hint: this integral converges to a value)
Go yoni :)
In other words, this is a homework problem of yours?
Consider
Integral[e^(-(x^2+y^2))dxdy] 0 < x < infinity, 0 < y < infinity
= Integral[e^(-r^2)*r dr dphi] 0 < r < infinity, 0< phi < pi/2
u = -r^2 du = -2r dr
= Integral[-1/2*e^(u) du dphi]
= Integral [ -1/2*e^(-r^2) [infinity, 0] dphi]
= Integral [ 1/2 dphi ]
= pi/4
In my opinion the next steps are rigorous, although they might not be how a pedantic mathematician would approach the problem :P.
e^-[x^2+y^2] = e^(-x^2)*e^(-y^2) .Since the bounds for x and y
are identical, the result for my double integral should be like integrating e^(-x^2) and multiplying it by the integral of e^(-x^2) again.
e.g. integrate e^(-x^2) * e^(-y^2) holding y constant, get result, then integrate
result(x) * e^(-y^2), holding the result(x) constant.
So, the answer should be Sqrt[pi/4] = Sqrt[pi]/2
QuoteIn other words, this is a homework problem of yours?
Nope...
Why would anyone assign someone a well known math proof as homework (IIRC, Gaussian distribution)?
Nonetheless, I always thought it was an interesting approach to a nonelementary integral.
But just to clarify, what Rule basically did was square the non-elementary integral (to make it elementary) and then converted it to polar coordinates. Then after finding this value, he square rooted back to get to the original answer.
If it's so well known, why would you want us to think of a solution here: "I need to brush up on a few things"?
Also, some thanks for my trouble would be nice. I think of the last 6 questions I've answered, only one person has been grateful for a solution.
Well I don't store every proof I know in my ol' noggin'. I came across this when I was reviewing. And sharing is caring ;D
And thanks for playing :)
Quote from: Lenny on April 11, 2006, 09:26 PM
Well I don't store every proof I know in my ol' noggin'. I came across this when I was reviewing. And sharing is caring ;D
And thanks for playing :)
Well, it was a fun problem. I like questions like that which don't require a ton of
busywork, and welcome more :).