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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
[itex]e^{- \infty}=0[/itex]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
There is no need for the error function when evaluating the first integral, just use the fact that [itex]\int_{ - \infty }^{ \infty } f(x)dx = \int_{ - \infty }^{ \infty } f(y)dy [/itex] to calculate the square of the integral, by switching to polar coordinates.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.
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: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.
That makes it a lot clearer. Thanks.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:
[tex] \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[/tex]
To see why it's easier, just switch to polar coordinates.