integrating normal density

**garymarkhov** · March 8th 2010, 01:06 PM

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 $x$ ?

**Focus** · March 8th 2010, 02:33 PM

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

**garymarkhov** · March 8th 2010, 02:56 PM

Originally Posted by Focus

Try differentiating $-\frac{1}{x}e^{-\frac{1}{2}x^2}$ . Remember to use the product/quotient rule

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

**Focus** · March 8th 2010, 04:45 PM

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).

**mr fantastic** · March 8th 2010, 04:46 PM

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.

**garymarkhov** · March 8th 2010, 04:59 PM

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).

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