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

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