Exponential Integration: int[z to inf][e^(-0.5x^2)]

[lamhy8]

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]

 

[galactus]

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.

 

[galactus]

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, \(\displaystyle \L\\\int_{0}^{\infty}e^{\frac{-x^{2}}{2}} dx=\frac{\sqrt{2{\pi}}}{2}\)