# how to calculate this exponential function integral

how to integrate this function $$\int \frac{1}{\sqrt{2\pi}}e^{-z^{2}/2}dz$$, what would the resulting function be, i tried integrating in wolframalpha but it only gave me this solution $\frac{1}{2}\text{erf}(\frac{z}{\sqrt{2}})+c$, erf stands for error function, which i have no idea what it means, can someone show me what the resulting function is going to be, btw this function is for finding area under the standard normal curve

• There's no closed form expression for the cumulative distribution function of the Normal distribution. – hejseb Jun 12 '13 at 8:56
• then how do you evaluate it for the z score to get the probability? – notamathwiz Jun 12 '13 at 9:03
• ok let me ask this then instead, having the Z score value how can i compute the probability without referring to the standard normal table? whats the easiest way of calculating this probability? – notamathwiz Jun 12 '13 at 9:52

As Sebastian mentionned, there are no closed form expression for this cdf. Many results in Probabilities and Financial Mathematics (for instance) refer to expressions with the $erf$ function.
• The erf function and the complementary error function, erfc, are available in most standard math libraries, including C/C++'s math.h so I wouldn't implement it myself if possible. – horchler Jun 12 '13 at 14:51