# Why this relation is true when computing the Gaussian integral?

1. Jun 24, 2012

### Arian.D

$$\int_0^\infty e^{-x^2}dx \int_0^\infty e^{-y^2}dy = \int_0^\infty \int_0^\infty e^{-(x^2+y^2)} dxdy$$

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.

2. Jun 24, 2012

### Muphrid

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

Hint: $e^{-(x^2+y^2)} = e^{-x^2} e^{-y^2}$.

3. Jun 24, 2012

### 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.

4. Jun 24, 2012

### Arian.D

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 $e^{_y^2}$ into the integral with respect to x, do you? so if I have:
$$I = \int_a^b f(x)dx . \int_a^b g(y)dy$$
Could I conclude that:
$$I = \int_a^b \int_a^b f(x).g(y) dxdy$$

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

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.

5. Jun 24, 2012

### Muphrid

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.

6. Jun 24, 2012

### Arian.D

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.

7. Jun 24, 2012

### Muphrid

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

$$I = F \int_a^b g(y) \; dy = \int_a^b F g(y) \; dy = \int_a^b \left(\int_a^b f(x) \; dx \right) g(y) \; dy$$

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

8. Jun 24, 2012

### Arian.D

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?

9. Jun 24, 2012

### Muphrid

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