Concerning the Gaussion integral in polar coordinates

[naggy]

I'm looking at the proof for the Gaussion integral in polar coordinates and I don´t understand why theta reaches from 0 to 2pi in the integral since you can´t get a negative value out of an exponential function (and therefor the exponential function is never in the 3rd and fourth quadrant, which spans pi to 2pi). Here's the proof

http://upload.wikimedia.org/math/8/5/b/85bb26bab98e69735c439dcfee9807d6.png part 1

http://upload.wikimedia.org/math/f/1/0/f10de92bc974482b9f714bdaa7fda10d.png part 2

 

[tiny-tim]

I don´t understand why theta reaches from 0 to 2pi in the integral since you can´t get a negative value out of an exponential function (and therefor the exponential function is never in the 3rd and fourth quadrant, which spans pi to 2pi).

Hi naggy!

(btw, if you type alt-p, it prints π)

It has nothing to do with the integrand (in this case, an exponential).

It's only conerned with the limits of integration, in changing from ∫∫dxdy to ∫∫drdtheta.

∫∫dxdy was over the whole plane.

So ∫∫drdtheta must be also. That means over all r and all theta.

r can't be negative, so theta has to go all the way round, from 0 to 2π.

 

[flebbyman]

When you change it to a double integral, you go from $$R^2$$ to $$R^3$$, and $$e^{-x^2}$$ and $$e^{-y^2}$$ are defined for all x and all y, so in effect both exponential functions are in all two dimensional quadrants.

 

[HallsofIvy]

It is the value of the exponential function that is never never negative, not the variables.

 

[tiny-tim]

… integrating over neverland …

When you change it to a double integral, you go from $$R^2$$ to $$R^3$$

naggy, he means $$R^2$$ (… you knew that, didn't you? … )

It is the value of the exponential function that is never never negative, not the variables.

Do you mean never never never never negative, or never never never never never negative?