Why this relation is true when computing the Gaussian integral?

[Arian.D]

[tex] \int_0^\infty e^{-x^2}dx \int_0^\infty e^{-y^2}dy = \int_0^\infty \int_0^\infty e^{-(x^2+y^2)} dxdy[/tex]

Under what conditions we could do the same for other functions? I don't get how Poisson (or Euler, or Gauss, whoever that did this for the first time) realized that this is true. It looks quite ingenious to me and I wonder if there's a theorem or something that I could apply to similar cases and this case as well.

 

[Muphrid]

You can move any multiplicative factor that doesn't depend on the integration variable into or out of the relevant integral.

Hint: [itex]e^{-(x^2+y^2)} = e^{-x^2} e^{-y^2}[/itex].

 

[homeomorphic]

That's not ingenious. That's just knowing calculus well. What's somewhat ingenious is the calculation as a whole. That step isn't really the trick. The trick is doing that change of variables at the beginning.

 

[Arian.D]

You can move any multiplicative factor that doesn't depend on the integration variable into or out of the relevant integral.

Hint: [itex]e^{-(x^2+y^2)} = e^{-x^2} e^{-y^2}[/itex].

I don't understand, would you care to explain what you mean by this?
I'm sure you don't mean that we could move [itex]e^{_y^2}[/itex] into the integral with respect to x, do you? so if I have:
[tex] I = \int_a^b f(x)dx . \int_a^b g(y)dy [/tex]
Could I conclude that:
[tex] I = \int_a^b \int_a^b f(x).g(y) dxdy [/tex]

Why so? This doesn't look obvious to me at all :-(

That's not ingenious. That's just knowing calculus well. What's somewhat ingenious is the calculation as a whole. That step isn't really the trick. The trick is doing that change of variables at the beginning.

Yea, I meant the whole calculation process of the Gaussian integral is ingenious, not particularly this step. I liked the proof so much, I also found another proof due to Laplace on wikipedia that was beautiful as well.

 

[Muphrid]

I don't understand, would you care to explain what you mean by this?
I'm sure you don't mean that we could move [itex]e^{_y^2}[/itex] into the integral with respect to x, do you? so if I have:
[tex] I = \int_a^b f(x)dx . \int_a^b g(y)dy [/tex]
Could I conclude that:
[tex] I = \int_a^b \int_a^b f(x).g(y) dxdy [/tex]

Why so? This doesn't look obvious to me at all :-(

As far as the x integral is concerned, g(y) is a constant factor that you can move into it or pull out of it as easily as you could with a true constant like, say, the number 5.

 

[Arian.D]

As far as the x integral is concerned, g(y) is a constant factor that you can move into it or pull out of it as easily as you could with a true constant like, say, the number 5.

yes, but you can't move dx into the second integral for sure. The problem is not how g(y) goes in, the problem is how the double integral appears. That's where I found it not obvious enough to be taken for granted intuitively. I'd be thankful if you care to explain it in details or show me a proof of this.

 

[Muphrid]

Let [itex]\int_a^b f(x) \; dx = F[/itex], which is just a number with no y-dependence.

[tex]I = F \int_a^b g(y) \; dy = \int_a^b F g(y) \; dy = \int_a^b \left(\int_a^b f(x) \; dx \right) g(y) \; dy[/tex]

From there, g(y) can be moved into the x-integral exactly as I described.

 

[Arian.D]

Let [itex]\int_a^b f(x) \; dx = F[/itex], which is just a number with no y-dependence.

[tex]I = F \int_a^b g(y) \; dy = \int_a^b F g(y) \; dy = \int_a^b \left(\int_a^b f(x) \; dx \right) g(y) \; dy[/tex]

From there, g(y) can be moved into the x-integral exactly as I described.

Nice. Thanks. Sounds convincing.
But we can do this if the integrals are indefinite, right? because the anti-derivative of f(x) is still a function of x and is independent of y?

Also, if the integrals are improper, then we need f or g (or maybe both) to have some nice properties, because then we'll have to move a limit into an integral which isn't always true. Right? or we could easily generalize the same argument to improper integrals as well?

 

[Muphrid]

As long as neither y nor x appear in the limits of the integrals, this approach ought to be valid.