## gaussian function

1. The problem statement, all variables and given/known data

2. Relevant equations
3. 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
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 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.

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 Quote by g.lemaitre 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
$e^{- \infty}=0$

 Quote by chiro 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.

## gaussian function

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.

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 Quote by 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.
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.

 Quote by gabbagabbahey 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.

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