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Thread: Gamma function

  1. #1
    Super Member Random Variable's Avatar
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    Gamma function

    Can you say that $\displaystyle \displaystyle \Gamma \Big(-\frac{1}{4} \Big) = \int_{0}^{\infty} x^{-5/4} e^{-x} \ dx $ even though the integral diverges and by definition $\displaystyle \displaystyle \Gamma \Big(-\frac{1}{4} \Big) \ne \infty $ ?

    Specifically what I want to know is does $\displaystyle \displaystyle \Gamma \Big(\frac{5}{4} \Big) \int_{0}^{\infty} x^{-5/4} e^{-x} \ dx $ diverge, or does it equal $\displaystyle \displaystyle \Gamma \Big(\frac{5}{4} \Big) \Gamma \Big(\frac{-1}{4} \Big) = \frac{\pi}{sin \frac{5 \pi}{4}} = - \sqrt{2} \pi $ ?
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  2. #2
    MHF Contributor chisigma's Avatar
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    The definition...

    $\displaystyle \displaystyle \Gamma(z)= \int_{0}^{\infty} t^{z-1}\ e^{-t}\ dt $ (1)

    ... is valid only for $\displaystyle \mathcal{R} (z) >0$. For the value of z with non positive real part You have to compute the so called 'analytic extension' of $\displaystyle \Gamma (*)$, that pratically means to use the 'Euler's formula'...

    $\displaystyle \displaystyle \Gamma(z)\ \Gamma(1-z)= \frac{\pi}{\sin \pi\ z}$ (2)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    Super Member Random Variable's Avatar
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    So then $\displaystyle \displaystyle \Gamma \Big(\frac{5}{4} \Big) \int_{0}^{\infty} x^{-5/4} e^{-x} \ dx $ diverges?
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  4. #4
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    Quote Originally Posted by Random Variable View Post
    $\displaystyle \displaystyle \Gamma \Big(\frac{5}{4} \Big) \Gamma \Big(\frac{-1}{4} \Big) = \frac{\pi}{sin \frac{5 \pi}{4}} = - \sqrt{2} \pi $ ?
    $\displaystyle \displaystyle\Gamma\left(-\frac{1}{2}\right)=-\sqrt{2}\pi$

    I know

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

    I am not sure if that implies

    $\displaystyle \displaystyle\Gamma\left(-\frac{1}{4}\right)=-4\Gamma\left(\frac{1}{4}\right)\mbox{?}$
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Random Variable View Post
    So then $\displaystyle \displaystyle \Gamma \Big(\frac{5}{4} \Big) \int_{0}^{\infty} x^{-5/4} e^{-x} \ dx $ diverges?
    ... the integral $\displaystyle \displaystyle \int_{0}^{\infty} t^{-\frac{5}{4}}\ e^{-t}\ dt $ diverges...

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    $\displaystyle \chi$ $\displaystyle \sigma$
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  6. #6
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    The gamma integral doesn't converge $\displaystyle \forall x<0$.
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