How can the integral of a normal distribution be solved using substitution?

[wombat4000]

Homework Statement

I'm having difficulty integrating something,
click http://en.wikipedia.org/wiki/Normal_distribution
and under Cumulative distribution function, there is an integral - how do you get to the next line?

Homework Equations

The Attempt at a Solution

i have tried the substitution of t=u-\mu which gives
\int_{-\infty}^{x}e^{-t^{2}/2}dt

but can't integrate this.

 

[tiny-tim]

see error functions

i have tried the substitution of t=u-\mu which gives
\int_{-\infty}^{x}e^{-t^{2}/2}dt
but can't integrate this.

Hi wombat! Welcome to PF

All you need is in http://en.wikipedia.org/wiki/Error_function

Good luck!

[size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​

 

[wombat4000]

Still don't really understand what's what - this to clarify this is what i was originally referring to

and this is what i got from the error function page

but i need to integrate from -\infty to x.

 

[wombat4000]

Thanks for your help by the way!

 

[tiny-tim]

An even function!

Ah! You didn't take account of the fact that the integrand is the same for -t as for t (an "even function"), so you can mutiply the limits by -1:

\int_{-\infty}^{x}e^{-t^{2}/2}dt​

is the same as:

\int_{-x}^{\infty}e^{-t^{2}/2}dt\qquad,​

which is (a multiple of) erfc(-x).

Then the top of the wikipedia page gives you the answer!

[size=-2](if you're happy, don't forget to mark thread "solved"!)[/size]​

 

[wombat4000]

i still don't get it.

 

[wombat4000]

i know that

and

but i don't end up with

 

[wombat4000]

should i being using this?

\int_{-x}^{\infty}e^{-t^{2}/2}dt\qquad = \int_{-x}^{\infty}{e^{-t^{2}}e^{t^{2}/2}}dt\qquad

 

[wombat4000]

what should \phi equal?

?

 

[nowoman]

I think in this case, x should be greater than zero.