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Math Help - integrating normal density

  1. #1
    Member garymarkhov's Avatar
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    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 ?
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  2. #2
    Member Focus's Avatar
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    Quote Originally Posted by garymarkhov View Post
    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
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  3. #3
    Member garymarkhov's Avatar
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    Quote Originally Posted by Focus View Post
    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?
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  4. #4
    Member Focus's Avatar
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    Quote Originally Posted by garymarkhov View Post
    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).
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  5. #5
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    Quote Originally Posted by Focus View Post
    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.
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  6. #6
    Member garymarkhov's Avatar
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    Quote Originally Posted by Focus View Post
    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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