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Math Help - Gamma function of -1/2 = ?

  1. #1
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    Gamma function of -1/2 = ?

    I've been struggling to solve this the whole day today . Basically, I followed the same process as demonstrated in this post, but when I got to this equation
    \Gamma(-1/2)= \int_{0 }^{ \infty  } {u }^{-2 }{e }^{ {-u }^{2 }  }du
    I cannot find anyway to solve it.
    Any help would be really appreciated. Thanks a lot.
    Last edited by mr fantastic; July 15th 2010 at 05:17 PM.
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  2. #2
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    Quote Originally Posted by ttna90 View Post
    I've been struggling to solve this the whole day today . Basically, I followed the same process as demonstrated in this post, but when I got to this equation

    I cannot find anyway to solve it.
    Any help would be really appreciated. Thanks a lot.
    \displaystyle {\int_{0}^{+\infty} u^{-2} e^{-u^2} \, du = \frac{1}{2} \int_{0}^{+\infty} t^{-3/2} e^{-t} \, dt} making an obvious substitution

     \displaystyle {= \frac{1}{2} \Gamma\left(- \frac{1}{2} \right)}.

    See Gamma function - Wikipedia, the free encyclopedia
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  3. #3
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    I would probably use the following two well-known properties of the gamma function:

    \Gamma(\frac{1}{2})=\sqrt{\pi}, and
    \Gamma(z+1)=z\Gamma(z).

    Taken together, can you tell me what \Gamma(-\frac{1}{2}) is?
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