1. ## integrating normal density

Can someone tell me why it's not possible to integrate the following?

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

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

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

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

What's wrong with $\displaystyle -\frac{1}{x}e^{-\frac{1}{2}x^2}$ evaluated from $\displaystyle -\infty$ to $\displaystyle x$ ?
Try differentiating $\displaystyle -\frac{1}{x}e^{-\frac{1}{2}x^2}$. Remember to use the product/quotient rule

3. Originally Posted by Focus
Try differentiating $\displaystyle -\frac{1}{x}e^{-\frac{1}{2}x^2}$. Remember to use the product/quotient rule
Hmm, I see your point. So there's no way to take the integral?

4. 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).

5. 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.

6. 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).