int. (-infinity to +infinity) e^(-ax^2) dx

I've forgotten what

S(-infinity to +infinity) e^(-ax^2) dx is. Help please?

Square it (one with x, one with y), convert to polar co-ordinates and it will be easy.

Yell if it isn't.

Sorry, just haven't had much time to look at it currently. But thanks!

It's such a neat transformation that it seems to have stuck in my mind unusually well. If you see what I mean, the rest is basic multivariate calculus (you recall Jacobians?)

ok, you've either got it or not by now, but for everyone else:

Let A = S(x = -infinity to +infinity) e^(-ax^2) dx

A^2 = [S(x = -infinity to +infinity) e^(-ax^2) dx] * [ S(y = -infinity to +infinity) e^(-ay^2) dy]

    =  S( x = -infinity to +infinity) S(y = -infinity to +infinity) e ^(-a (x^2+ y^2) dx dy

{justified since the functions are positive everywhere}

Let x = r sin(t) and y = r cos(t)

dx dy = r dr dt

{obvious if you think of area elements. r is the determinant of the 2x2 matrix d(x,y)/d(r,t)}

So A^2 = S(r=0 to +infinity) S(t=0 to 2 pi) e^(-ar^2) r dr dt

{of course x^2+y^2=r^2}

The t integral just gives a constant 2 pi, so, integrating with respect to r:

A^2 = 2 pi [ -e^(-ar^2)/2a ] (r=0 to infinity)

     = pi/a

So the original integral A = sqrt(pi/a)

[if I made no slipups along the way]

*is lazy and starts Mathematica* :P

input: Assuming[a > 0, Integrate[Exp[-a x^2], {x, -\[Infinity], \[Infinity]}]]

output: Sqrt[\[Pi]]/Sqrt[a]

So you made no slipups along the way ;)