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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

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  • $\begingroup$ There's no closed form expression for the cumulative distribution function of the Normal distribution. $\endgroup$ – hejseb Jun 12 '13 at 8:56
  • $\begingroup$ then how do you evaluate it for the z score to get the probability? $\endgroup$ – notamathwiz Jun 12 '13 at 9:03
  • $\begingroup$ 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? $\endgroup$ – notamathwiz Jun 12 '13 at 9:52
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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.

Should you be interested in numerically approximating it, you may apply the following classical methods : - trapezoidal rule - quadrature formulas (Gauss-Lobatto among others)

Unfortunately, they will give you a poor approximation (despite the integrand's regularity) in comparison with the classical tailored approximation mentioned in : https://stackoverflow.com/questions/457408/is-there-an-easily-available-implementation-of-erf-for-python

Watch out for the division by '0.' exception though.

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  • $\begingroup$ 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. $\endgroup$ – horchler Jun 12 '13 at 14:51
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Showing that this is, indeed, a density function (integrates to 1 over the whole real axis) is a standard example in multivariate calculus: instead of calculating it, we calculates its square, then changes to polar coordinates, which results in an integral we do can solve: \begin{eqnarray} \left( \int_{-\infty}^{\infty} e^{-z^2/2} \; dz \right)^2 \\ &=& \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2/2} e^{-y^2/2}\; dx dy \\ &=& \int_0^{2\pi} \int_0^\infty e^{-r^2/2} r \; dr d\theta \\ &=& 2\pi \int_0^{\infty} r e^{-r^2/2} \; dr \\ &=& 2\pi \cdot 1 = 2\pi \end{eqnarray}

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