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Euclid
Oct20-07, 09:27 PM
I assume this is standard but I wouldn't know where to look. How do you approximate an integral of the form
\int_{x_1}^{x_2}\exp(-x^2)dx
where possibly one of the endpoints is infinite?

Gib Z
Oct20-07, 11:05 PM
Taking into account that the integrand is an even function, and the well known result that :\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi} It remains only to calculate a integral of the form:

\int_0^x e^{-t^2} dt.

You may calculate the taylor series and integrate term by term to get a series for that integral: \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n! (2n+1)} =\left(x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}+\frac{x^9}{216}-\ \cdots\right),

Or alternatively, the Error function ( Erf(x) ) is only a constant multiple of that integral, and its values are well tabulated on the internet. http://en.wikipedia.org/wiki/Error_function

Euclid
Oct21-07, 01:37 AM
Ah... thanks. That is very useful.