L lamhy8 New member Joined Aug 8, 2006 Messages 2 Aug 8, 2006 #1 Pls help me to solve this integration- Q(z)= Integrate from z to infinity [ e^(-0.5x^2)] it's related to error function complementary. Really need it urgently ..Thanksssss alot. [/img]
Pls help me to solve this integration- Q(z)= Integrate from z to infinity [ e^(-0.5x^2)] it's related to error function complementary. Really need it urgently ..Thanksssss alot. [/img]
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Aug 8, 2006 #2 This can not be done by elementary means. I can tell you from 0 to infinity it equals \(\displaystyle \frac{\sqrt{2{\pi}}}{2}\) Try a Google. You can probably find a site which steps through it, but I must warn you, it'll be rather complicated.
This can not be done by elementary means. I can tell you from 0 to infinity it equals \(\displaystyle \frac{\sqrt{2{\pi}}}{2}\) Try a Google. You can probably find a site which steps through it, but I must warn you, it'll be rather complicated.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Aug 9, 2006 #3 Here's a popular technique for integrating from -infinity to infinity. \(\displaystyle \L\\\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}}dx\) Square it: \(\displaystyle \L\\\left(\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}}dx\right)^{2}\) =\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}}dx\int_{-\infty}^{\infty}e^{\frac{-y^{2}}{2}}dy\) =\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dydx\) Convert to polar: =\(\displaystyle \L\\\int_{0}^{\infty}\int_{0}^{2{\pi}}e^{\frac{-r^{2}}{2}} r d{\theta}dr\) =\(\displaystyle \L\\\int_{0}^{2{\pi}}d{\theta}\int_{0}^{\infty}e^{\frac{-r^{2}}{2}} r dr\) Let \(\displaystyle u=\frac{r^{2}}{2}\;\ du=rdr\) \(\displaystyle \L\\2{\pi}\int_{0}^{\infty}e^{-u} du\) =\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}} dx=\sqrt{2{\pi}}\) Therefore, \(\displaystyle \L\\\int_{0}^{\infty}e^{\frac{-x^{2}}{2}} dx=\frac{\sqrt{2{\pi}}}{2}\)
Here's a popular technique for integrating from -infinity to infinity. \(\displaystyle \L\\\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}}dx\) Square it: \(\displaystyle \L\\\left(\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}}dx\right)^{2}\) =\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}}dx\int_{-\infty}^{\infty}e^{\frac{-y^{2}}{2}}dy\) =\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}}e^{\frac{-y^{2}}{2}}dydx\) Convert to polar: =\(\displaystyle \L\\\int_{0}^{\infty}\int_{0}^{2{\pi}}e^{\frac{-r^{2}}{2}} r d{\theta}dr\) =\(\displaystyle \L\\\int_{0}^{2{\pi}}d{\theta}\int_{0}^{\infty}e^{\frac{-r^{2}}{2}} r dr\) Let \(\displaystyle u=\frac{r^{2}}{2}\;\ du=rdr\) \(\displaystyle \L\\2{\pi}\int_{0}^{\infty}e^{-u} du\) =\(\displaystyle \L\\\int_{-\infty}^{\infty}e^{\frac{-x^{2}}{2}} dx=\sqrt{2{\pi}}\) Therefore, \(\displaystyle \L\\\int_{0}^{\infty}e^{\frac{-x^{2}}{2}} dx=\frac{\sqrt{2{\pi}}}{2}\)
