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Gaussian Integral

  • Thread starter Adoniram
  • Start date
  • #1
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Homework Statement


I'm encountering these integrals a lot lately, and I can solve them because I know the "trick" but I'd like to know actually how the cartesian to polar conversion works:
##\int_{-\infty}^{\infty}e^{-x^2}dx##

Homework Equations


##\int_{-\infty}^{\infty} e^{-x^2} = I##
##I^2=\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy##
##=\int_{-\infty}^{\infty} e^{-r^2}r dr dθ = π##

The Attempt at a Solution


So, if I look at ##r=\sqrt{x^2+y^2}##, it's easy to see that ##dr=(1/r)(x dx+y dy)##
Which leads me to believe that ##dθ=(x dx+y dy)## ... ???

If ##θ=ArcTan(y/x)##, then how does ##dθ=(x dx+y dy)##?

Say ##y=Tan(θ)x##, then taking ##dy## and simplifying, I can get:
##x dy - y dx=(x^2+y^2) dθ##

So unless ##x/(x^2+y^2) = y## and ##y/(x^2+y^2) = -x##, I am at a loss... :(
 

Answers and Replies

  • #3
Ray Vickson
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Homework Statement


I'm encountering these integrals a lot lately, and I can solve them because I know the "trick" but I'd like to know actually how the cartesian to polar conversion works:
##\int_{-\infty}^{\infty}e^{-x^2}dx##

Homework Equations


##\int_{-\infty}^{\infty} e^{-x^2} = I##
##I^2=\int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy##
##=\int_{-\infty}^{\infty} e^{-r^2}r dr dθ = π##

The Attempt at a Solution


So, if I look at ##r=\sqrt{x^2+y^2}##, it's easy to see that ##dr=(1/r)(x dx+y dy)##
Which leads me to believe that ##dθ=(x dx+y dy)## ... ???

If ##θ=ArcTan(y/x)##, then how does ##dθ=(x dx+y dy)##?

Say ##y=Tan(θ)x##, then taking ##dy## and simplifying, I can get:
##x dy - y dx=(x^2+y^2) dθ##

So unless ##x/(x^2+y^2) = y## and ##y/(x^2+y^2) = -x##, I am at a loss... :(
We can write
[tex] I^2 = \int_{-\infty}^{\infty} e^{-x^2}dx \int_{-\infty}^{\infty} e^{-y^2}dy = \int_{R^2} e^{-x^2-y^2} \, dA, [/tex]
where ##dA = dx dy## is the two-dimensional "area element". When we switch to polar coordinates, the area element becomes ##dA = r \, dr \, d\theta##. Basically, we have ## I^2 \approx \sum_i \Delta A_i e^{-r^2_i} ##, where the ##\Delta A_1, \Delta A_2, \ldots ## are finite but small areas of little regions and the ##r_i## are the ##\sqrt{x_i^2 + y_i^2}## values at some point ##(x_i,y_i)## inside the regions ##i = 1,2, \ldots##. We can either take the little regions to be rectangular, with horizontal and vertical sides ##\Delta x_i## and ##\Delta y_i##, or else we can take them to be slices of radial wedges with sides along radial lines at angles ##\theta_i## and ##\theta_i + \Delta \theta_i## and inner-outer radii at ##r_i## and ##r_i + \Delta r_i##. In rectangular coordinates the area element is ##\Delta A_i = \Delta x_i \, \Delta y_i##, while in polar coordinates it is ##\Delta A_i = r_i \, \Delta r_i \, \Delta \theta_i##.

All this was supposed to have been covered thoroughly in Calculus II, where things such as Jacobians and the like were introduced and motivated.
 
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