Squaring an Integral: Deriving ∫e-x2 = √(π/4) | Physics Forums

[weeksy]

Homework Statement

Well iv been looking around for a similar problem and found the use of the following equation. what I would like to know is if the following equation is actually correct? I believe it may be a extension of the Fubini's theorem but can't find a reference or derivation on the net. any help would be great thanks

I=∫₀^∞ e^-x2
Show I=sqrt(pi/4) by first evaluating I^2

Homework Equations

[∫f(x) dx]^2 = ∫∫f(x)f(y) dxdy ?

link- https://www.physicsforums.com/showthread.php?t=243670

The Attempt at a Solution

I²=∫∫e^-2x2dxdy?

 

[voko]

I2=∫∫exp(-2x2)dxdy is incorrect, where is y except in its differential?

 

[clamtrox]

When you multiply two integrals together, their arguments are different (even though initially you use same letter for both): use x for one and y for the other.

 

[weeksy]

ok so we would have I^2 = ∫∫e^{-x^2.y^2} dxdy

i would love to know where this rule comes from, seems very weird that we can just introduce a function dependent only on x as f(x,y).
and how do we know which order the dxdy goes?

 

[voko]

i would love to know where this rule comes from, seems very weird that we can just introduce a function dependent only on x as f(x,y).
and how do we know which order the dxdy goes?

You have number I that happens to be an integral of some f(x). You might as well look at it as the integral of f(y). You can multiply those, and you can treat them as a double integral - in any order - because their variables are independent.

 

[clamtrox]

ok so we would have I^2 = ∫∫e^{-x^2.y^2} dxdy

i would love to know where this rule comes from, seems very weird that we can just introduce a function dependent only on x as f(x,y).
and how do we know which order the dxdy goes?

You're misunderstanding something here...

I^2 = \int e^{-x^2} dx \int e^{-y^2} dy = \int e^{-x^2 - y^2} dx dy

This is basically a special case of Fubini where f(x,y) = g(x) h(y), if that helps your thinking.