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Hey all!

i want to find the integral pr = Integral(limits from a constant>0 to +infinite, and the function inside is the PDF of Gauss distribution)..

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

in this link u can see the PDF function..

Does anyone knows how to do this?

thank you very much in advance.

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This is not an elementary function. But it can be done in terms of a special function known as the error function $$ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} dt $$

edit Oct 28

I mean this. If $$ f \bigl(x,\mu,\sigma^{2}\bigr) = \frac{\operatorname{e} ^{\frac{-(-x + \mu)^{2}}{2 \sigma^{2}}}}{\sigma \sqrt{2 \pi}} $$ then evaluate the quantity in your question in terms of erf as follows: $$ \int_{c}^{\infty} f \bigl(x,\mu,\sigma^{2}\bigr) d x = \frac{1-\mathrm{erf} \biggl(\frac{(c - \mu)}{\sqrt{2}\; \sigma}\biggr)}{2} $$

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    $\begingroup$ Thanks Gerald for your answer, but this erf(x) corresponds to what? you mean that i have to find the CDF? I want the Integral{PDF, from known const to +inf} the CDF i think finds the Integral{PDF, from -inf to zero}, right? $\endgroup$
    – Nakamura
    Oct 18, 2012 at 8:06

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