$\frac{1}{\sqrt{2\pi}}\int_\frac {1}{2}^0\exp(-x^2/2)dx$

Question

How do we analytically evaluate $J=\frac{1}{\sqrt{2\pi}}\int_\frac {-1}{2}^0\exp(-x^2/2)dx$? This is what I tried: $$ J^2=\frac{1}{{2\pi}}\int_\frac {-1}{2}^0\int_\frac {-1}{2}^0\exp(-(x^2+y^2)/2)dxdy \\ =\frac{1}{{2\pi}}\int_\pi ^\frac {3\pi}{2}\int_0 ^\frac {1}{2}\exp(-r^2/2)rdrd\theta \\ =\frac{1}{4}(1-\exp(-1/8))$$ WHere on, I got $J=.17$ But, this does not tally with the value from the nornal distribution table. Am I doing it wring? Is there a better way to evaluate the integral?

Answer 1

Unfortunately, you cannot evaluate it in closed form. It is the Gaussian error function and, in general, it can only be evaluated numerically.

Answer 2

You could use $$e^x=\sum_{i=1}^{\infty}\frac{x^i}{i!}$$