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Math Help - Gamma(i)

  1. #1
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    Gamma(i)

    Express Gamma(i) in terms of elementary functions, if possible.

    (i is the imaginary unit and Gamma is Euler's Gamma Function)
    (Elementary functions are as defined in Wickepedia)

    Bob.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bobbyk View Post
    Express Gamma(i) in terms of elementary functions, if possible.

    (i is the imaginary unit and Gamma is Euler's Gamma Function)
    (Elementary functions are as defined in Wickepedia)

    Bob.
    do you mean: \Gamma (p) = \int_{0}^{\infty} e^{-x}x^{p-1}~dx ? that's just the definition though
    Last edited by Jhevon; September 24th 2007 at 06:32 PM.
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    Quote Originally Posted by Jhevon View Post
    do you mean: \Gamma (x) = \int_{0}^{\infty} e^{-x}x^{p-1}~dx ? that's just the definition though
    (It should be \Gamma(p)).
    Yes, but the beautiful Gamma function is defined for complex numbers with real part positive as well.

    \Gamma(i) = \int_0^{\infty} e^{-z}z^{1-i} dz

    But I am afraid I cannot do this. I do not even want to try. MathWorld has so many identities and non involving imaginary numbers.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    (It should be \Gamma(p)).
    Yes, but the beautiful Gamma function is defined for complex numbers with real part positive as well.

    \Gamma(i) = \int_0^{\infty} e^{-z}z^{1-i} dz

    But I am afraid I cannot do this. I do not even want to try. MathWorld has so many identities and non involving imaginary numbers.
    Shouldn't the contour integration of this function should be the same as in the real case? I'm not even going to try it, but that might be a possible way to evaluate it.

    -Dan
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    Gamma(i)

    Thanks, all!

    I have the numerical value of Gamma(i). Its modulus is an elementary
    function: Sqrt(pi/sinh(pi)).
    So its real and imaginary parts COULD be elementary functions too, but
    I haven't been able to find them.

    Bob.
    Last edited by bobbyk; September 24th 2007 at 01:53 PM. Reason: Add signature.
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    Quote Originally Posted by topsquark View Post
    Shouldn't the contour integration of this function should be the same as in the real case? I'm not even going to try it, but that might be a possible way to evaluate it.

    -Dan
    I am not exactly sure what you mean.
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    \Gamma (i) = \int_0^{\infty} e^{-z} z^{i-1} dz

    Now, z^{i} = e^{\log(z)i} = \cos (\log z)+i\sin (\log z)

    Thus,
    \Gamma (i) = \int_0^{\infty} z^{-1}e^{-z}\cos (\log z) dz + i\int_0^{\infty} z^{-1}e^{-z} \sin (\log z) dz

    So we end up with,
    \int_{-\infty}^{\infty} e^{-e^x} \cos x dx + i \int_{-\infty}^{\infty} e^{-e^x} \sin x dx

    What next?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I am not exactly sure what you mean.
    (sigh) I can't find a picture of the contour on the net. But the gamma function may be integrated using complex integration. I was wondering if it could be used for this problem.

    -Dan
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