1

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.

flag

closed as off topic by Noah Stein, Andreas Blass, George Lowther, Douglas Zare, Michael Renardy Oct 18 at 18:52

1 Answer

1

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: $$ \int_{c}^{\infty} f \bigl(x,\mu,\sigma^{2}\bigr) d x = \frac{1-\mathrm{erf} \biggl(\frac{(c - \mu)}{\sqrt{2}\; \sigma}\biggr)}{2} $$

link|flag
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? – Nakamura Oct 18 at 8:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.