Concerning the Gaussion integral in polar coordinates

In summary, the proof for the Gaussian integral in polar coordinates includes an integral from 0 to 2π, even though the exponential function involved is never negative. This is because the limits of integration must cover the entire plane, including all possible values of the variables. The exponential function itself is defined for all values of the variables, but its value is never negative.
  • #1
naggy
60
0
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
 
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  • #2
naggy said:
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! :smile:

(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. :smile:
 
  • #3
When you change it to a double integral, you go from [tex]R^2[/tex] to [tex]R^3[/tex], and [tex]e^{-x^2}[/tex] and [tex]e^{-y^2}[/tex] are defined for all x and all y, so in effect both exponential functions are in all two dimensional quadrants.
 
  • #4
It is the value of the exponential function that is never never negative, not the variables.
 
  • #5
… integrating over neverland …

flebbyman said:
When you change it to a double integral, you go from [tex]R^2[/tex] to [tex]R^3[/tex]

naggy, he means [tex]R^2[/tex] (… you knew that, didn't you? … :smile:)
HallsofIvy said:
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? :confused:
 

Related to Concerning the Gaussion integral in polar coordinates

1. What is the Gaussian integral in polar coordinates?

The Gaussian integral, also known as the normal distribution or bell curve, is a mathematical function that describes the probability distribution of a continuous random variable. It is often used to model natural phenomena and is commonly represented in polar coordinates.

2. How is the Gaussian integral calculated in polar coordinates?

The Gaussian integral in polar coordinates is calculated by using the polar coordinate transformation formula, which involves converting the Cartesian coordinates (x,y) into polar coordinates (r,θ) and then using the standard Gaussian integral formula.

3. What is the significance of using polar coordinates in the Gaussian integral?

Polar coordinates offer a more intuitive and convenient way to represent the Gaussian integral, as it allows for a better understanding of how the bell curve is shaped and how it relates to different variables. It also simplifies the integration process and makes it easier to visualize in certain scenarios.

4. How is the Gaussian integral used in science?

The Gaussian integral has numerous applications in various fields of science, including physics, chemistry, and biology. It is often used to model and analyze data in experiments, to determine the probability of certain outcomes, and to make predictions about natural phenomena.

5. Are there any limitations to using the Gaussian integral in polar coordinates?

One limitation of using the Gaussian integral in polar coordinates is that it assumes a symmetric and continuous distribution, which may not always be the case in real-world scenarios. Additionally, it can be challenging to calculate the integral for complex functions or in higher dimensions.

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