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        riddles >> putnam exam (pure math) >> Change the Integral
        
(Message started by: Benoit_Mandelbrot on Feb 5^th, 2004, 9:15am)

Title: Change the Integral
Post by Benoit_Mandelbrot on Feb 5^th, 2004, 9:15am

Lets say we have f '(x)=e^{-x^2}, and we want to find f([infty])-f(-[infty]). The integral looks like this: [int]_-[infty]^[infty](e^{-x^2})dx. I'm trying to find the exact value by squaring, then going to double integral, then convert to polar, but I don't exactly know how to convert to polar double integrals. Could someone show me how this is done for any double integral as well?

n²=[int]_-[infty]^[infty][int]_-[infty]^[infty](e^{-x^2-y^2})dydx

Title: Re: Change the Integral
Post by towr on Feb 5^th, 2004, 9:43am

f' is a gaussian curve, I don't think there is a way to find f
but the integral over the whole domain is known..

Title: Re: Change the Integral
Post by towr on Feb 5^th, 2004, 10:24am

http://en.wikipedia.org/wiki/Gaussian_function seems to give the final answer, but I'm missing a step..

oh wait.. I get it now.. That extra r comes from the circles you integrate over that of course get larger (dependant on r) as you get further from the origin..

Title: Re: Change the Integral
Post by Icarus on Feb 5^th, 2004, 8:38pm

When changing between two sets of coordinates, the area (or volume or hypervolume for higher dimensions) differential is multiplied by the Jacobian of the coordinate transformation.

Here is a link to the MathWorld page on the Jacobian (http://mathworld.wolfram.com/Jacobian.html).

A more heuristic but also simpler method is commonly used by physicists: look at a small element, and estimate its area in the local coordinates. For Cartesian coordinates, the element can be a square of side lengths dx and dy. The area is exactly dxdy, and this is what you integrate over. For polar coordinates, the sidelength along the radial direction is dr. The sidelength along the angular direction for a change in angle of d[theta] is rd[theta]. So the differential of area is approximately (dr)(rd[theta]). But the whole point of diffentials is that second-order contributions go away in the limit, and the approximation becomes exact. So the differential of area for polar coordinates is rdrd[theta].

That is: [int]f(x, y) dxdy = [int]f(r,[theta])r drd[theta]. This is the whole point of the "squaring and switching to polar coordinates" approach to integrating e^{-x^2}. By doing it, you pick up the extra r outside the exponent, which allows you to integrate by parts.

Another more mathematical approach - which when developed totally leads to the Jacobian, is to calculate the differentials directly using the multi-variable chain-rule: df = ([partial]f/[partial]x) dx + ([partial]f/[partial]y) dy

x = r cos [theta], y = r sin [theta]

so

[partial]x/[partial]r = cos [theta], [partial]x/[partial][theta] = -r sin [theta]
[partial]y/[partial]r = sin [theta], [partial]y/[partial][theta] = r cos [theta]

dx = cos([theta]) dr - r sin([theta]) d[theta]
dy = sin([theta]) dr + r cos([theta]) d[theta]

Multiplying:

dxdy = cos([theta])sin([theta]) dr² - r² cos([theta])sin([theta]) d[theta]² + r(cos² [theta] + sin² [theta]) drd[theta].

Now comes the magic waving of the hands: When integrated, both the dr² and d[theta]² terms drop out. Why? A very good question, and one I couldn't answer without delving into the definitions of integration, and some nasty calculations. Since I am far too lazy to do this (besides which you can find it much better presented in textboooks), you will have to take my word for it: any differential squared can be treated as 0.

This leaves only the term r(cos² [theta] + sin² [theta]) drd[theta] = r drd[theta].

Title: Re: Change the Integral
Post by Benoit_Mandelbrot on Feb 6^th, 2004, 10:24am

I got it myself! It's [hide]pi^1/2[/hide]. This is because the radius is from 0 to [infty], and the angle is from 0 to 2[pi].