Need help solving CDF (Cumulative distribution function)

[invictor]

Hello, I am new. I been looking on the net for a guide how to solve the CDF by hand, i know the answer and I am about to crack this baby but I got stuck...

Im trying to calculate Cumulative distribution function by hand:
$\int^{1}_{-1}\frac{1}{2\pi} e^{\frac{-z^{2}}{2}} dz$ or wolfram alpha: integrate 1/sqrt(2*pi) * e^(-z^2 /2) dz from -1 to 1

Anyway, this is the tricky part, how do this? (I left out the lefthand part above part for easier readability):

$\int e^{\frac{-z^{2}}{2}} dz =$

$u = \frac{-z^{2}}{2}$

$du = -z dz$

$\frac{du}{-z} = dz$

$\int e^{u} \frac{du}{-z} =$

then? How do i need to do?. Can any friendly soul here show me step by step how to solve this?

best regrads
invictor

 

[CompuChip]

You can define
$$\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt$$
and express the result in terms of that, getting $\operatorname{erf}(1/\sqrt{2})$ as the answer.

You cannot give a (more) exact answer than that (otherwise we wouldn't be needing erf in the first place).

 

[kdbnlin78]

Ok so now you have the CDF in terms of the error function erf(/frac{1}{/sqrt{2}}) what is the associating PDF?

 

[mathman]

The density function for the error function is the normal density function, which is the integrand you started with.

 

[invictor]

Thanks i got it now :)