A nice little Calculus challenge

[Lenny]

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

[Rule]

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 .

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

[Lenny]

In 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.

[Rule]

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.

[Lenny]

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 

And thanks for playing

[Rule]

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 

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 .