Problem with a gaussian integral.

[misogynisticfeminist]

Hey, I've been learning about gaussian integrals lately. And I'm now stuck in one part. I am now trying to derive some kind of general formula for gaussian integrals

\int x^n e^{-\alpha x^2}

for the case where n is even. So they ask me to evaluate the special case n=0 and alpha=1. So its I= \int^{\infty}_{-\infty} e^{-x^2} dx. When i square this integral, they said that its I^2= ({\int^{\infty}_{-\infty} e^{-x^2} dx})({\int^{\infty}_{-\infty} e^{-y^2} dy}) with both x and y as according to them, i have to use a different variable for the first and second integral factors.

Why is this so? I have limited calc background. So i was wondering you guys could help me out. Thanks alot...

 

[inha]

I think they're trying to get you to use the gamma function.

\Gamma(z)=2{\int^{\infty}_{0}dte^{-t^2}t^{z-1}

Those two integrands you have are even and the gamma func has useful properities for situations like this.

 

[Dr Transport]

By using different variables for the squared version, you can convert to different coordinate systems, i.e. to cylindrical coordinates. What you typed above is the way you prove that \int^{\infty}_{-\infty} e^{-x^2} dx = \sqrt{\pi}

 

[HallsofIvy]

"When i square this integral, they said that its I^2= ({\int^{\infty}_{-\infty} e^{-x^2} dx})({\int^{\infty}_{-\infty} e^{-y^2} dy})"

No, they didn't say that! You put in the "when I square this integral" yourself- they didn't square the integral.

What they did say was: If I= \int^{\infty}_{-\infty} e^{-x^2} dx, then it is also true that I= \int^{\infty}_{-\infty} e^{-y^2} dy because that is just a change in the dummy variable.

It is then true that I^2= I*I= \(\int^{\infty}_{-\infty} e^{-x^2} dx\)\(\int^{\infty}_{-\infty} e^{-y^2} dy\)- just multiplying two different ways of writing the same thing.

Of course, the really important thing is that fact that that product of integrals can be written as a double integral:
(\int^{\infty}_{-\infty} e^{-x^2} dx)(\int^{\infty}_{-\infty} e^{-y^2} dy= \int_{y= -\infty}^{\infty}\int_{x=-\infty}^{\infnty}e^{-(x^2+ y^2)}dxdy.

 

[misogynisticfeminist]

ahhhh so i see, a double integral. Thanks alot. Maybe i misinterpreted what they said, cos' they made it sound that changing the dummy variable was a must.