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Math Help - Trouble with Gamma function

  1. #1
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    Trouble with Gamma function

    Solve for \Gamma(-3/2).

    \Gamma(-3/2) = \frac{1}{2} \cdot \frac{3}{2} \Gamma(1/2)
    = \frac{4}{3} \int_{0}^\infty t^{1/2} e^{-t} dt <-- With t = u^2,
    = \frac{8}{3} \int_{0}^\infty e^{-u^2} du

    My book says that I need to square the left side and turn the right side into a double integral of e^{-(u^2 + v^2)} and then convert u into r cos theta, v into r sin theta, and go from there. But that is where I get lost. Obviously that makes u^2 + v^2 = r^2, but what the heck do I do with (a) du and dv, and (b) the limits of integration?
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  2. #2
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    Re: Trouble with Gamma function

    Are you familiar with polar coordinates? You express the function value in terms of r and theta. The transformation is x = rcos(theta), y = rsin(theta), so the Jacobian will be r. But even if you don't know this stuff, the integral should still be solvable by other means.

    Gaussian integral - Wikipedia, the free encyclopedia

    This will help.
    Last edited by SworD; September 10th 2012 at 02:41 PM.
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  3. #3
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    Re: Trouble with Gamma function

    Oh whoa. That's basically the normal curve. So really, I could just stick with u and integrate, getting \frac {\sqrt{\pi}}{2}? And then multiply times 8/3 to get the final answer?
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  4. #4
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    Re: Trouble with Gamma function

    Yes, that would be precisely correct. But I can't tell you whether or not the teacher wants more work. It would be the correct answer though.
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