# integrating normal density

• Mar 8th 2010, 02:06 PM
garymarkhov
integrating normal density
Can someone tell me why it's not possible to integrate the following?

$\int_{-\infty}^x e^{-\frac{1}{2}x^2}dx$

What's wrong with $-\frac{1}{x}e^{-\frac{1}{2}x^2}$ evaluated from $-\infty$ to $x$ ?
• Mar 8th 2010, 03:33 PM
Focus
Quote:

Originally Posted by garymarkhov
Can someone tell me why it's not possible to integrate the following?

$\int_{-\infty}^x e^{-\frac{1}{2}x^2}dx$

What's wrong with $-\frac{1}{x}e^{-\frac{1}{2}x^2}$ evaluated from $-\infty$ to $x$ ?

Try differentiating $-\frac{1}{x}e^{-\frac{1}{2}x^2}$. Remember to use the product/quotient rule (Happy)
• Mar 8th 2010, 03:56 PM
garymarkhov
Quote:

Originally Posted by Focus
Try differentiating $-\frac{1}{x}e^{-\frac{1}{2}x^2}$. Remember to use the product/quotient rule (Happy)

Hmm, I see your point. So there's no way to take the integral?
• Mar 8th 2010, 05:45 PM
Focus
Quote:

Originally Posted by garymarkhov
Hmm, I see your point. So there's no way to take the integral?

Well I am perfectly happy as long as it's well defined but no, there is no closed form of the integral (without an integral sign).
• Mar 8th 2010, 05:46 PM
mr fantastic
Quote:

Originally Posted by Focus
Well I am perfectly happy as long as it's well defined but no, there is no closed form of the integral (without an integral sign).

The above statement can be proved, by the way.
• Mar 8th 2010, 05:59 PM
garymarkhov
Quote:

Originally Posted by Focus
Well I am perfectly happy as long as it's well defined but no, there is no closed form of the integral (without an integral sign).

Thanks. I especially appreciate how you specified the meaning of the jargon (closed form).