Why is the Gaussian function easier to integrate using polar coordinates?

[g.lemaitre]

Homework Statement

Homework Equations

The Attempt at a Solution

Looking at equations 17 and 18, I don't see how that follows. If you substitute infinity for x you're going to get infinity divided by some real number which is infinity

 

[chiro]

Hey g.lemaitre.

The first result is not intuitive and it is based on what is calle the error function or erf(x). The function does converge because e^(-x) when x gets really large goes quickly to 0.

http://en.wikipedia.org/wiki/Error_function

For the second one, you need to use a substitution of u = x^2. If you have done a year of calculus, this should be straight-forward.

 

[gabbagabbahey]

Looking at equations 17 and 18, I don't see how that follows. If you substitute infinity for x you're going to get infinity divided by some real number which is infinity

e^{- \infty}=0

Hey g.lemaitre.

The first result is not intuitive and it is based on what is called the error function or erf(x). The function does converge because e^(-x) when x gets really large goes quickly to 0.

There is no need for the error function when evaluating the first integral, just use the fact that \int_{ - \infty }^{ \infty } f(x)dx = \int_{ - \infty }^{ \infty } f(y)dy to calculate the square of the integral, by switching to polar coordinates.

 

[chiro]

Do you mean ydx and xdy instead of f(x)dx f(y)dy? Sorry to nitpick but changing x to y is change a dummy variable change rather than a variable description change.

 

[gabbagabbahey]

Do you mean ydx and xdy instead of f(x)dx f(y)dy? Sorry to nitpick but changing x to y is change a dummy variable change rather than a variable description change.

The whole point is to exploit a change of the "dummy" integration variable, to transform the problem from one of finding a one-dimensional integral, to one of finding a two-dimensional integral, as the latter turns out to be easier:

\left( \int_{-\infty}^{\infty} f(x)dx \right)^2 = \left( \int_{-\infty}^{\infty} f(x)dx \right)\left( \int_{-\infty}^{\infty} f(y)dy \right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)f(y)dxdy

To see why it's easier, just switch to polar coordinates.

 

[chiro]

The whole point is to exploit a change of the "dummy" integration variable, to transform the problem from one of finding a one-dimensional integral, to one of finding a two-dimensional integral, as the latter turns out to be easier:

\left( \int_{-\infty}^{\infty} f(x)dx \right)^2 = \left( \int_{-\infty}^{\infty} f(x)dx \right)\left( \int_{-\infty}^{\infty} f(y)dy \right) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)f(y)dxdy

To see why it's easier, just switch to polar coordinates.

That makes it a lot clearer. Thanks.