# Concerning the Gaussion integral in polar coordinates

1. Mar 9, 2008

### 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

2. Mar 9, 2008

### tiny-tim

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.

3. Mar 9, 2008

### 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.

4. Mar 9, 2008

### HallsofIvy

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

5. Mar 9, 2008

### tiny-tim

… integrating over neverland …

naggy, he means $$R^2$$ (… you knew that, didn't you? … )
Do you mean never never never never negative, or never never never never never negative?

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