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Math Help - similar to gamma function

  1. #1
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    similar to gamma function

    ive got this question in for tomorrow and i have no idea whatsoever how to attempt it

    evaluate the following intergral. your final answer should not contain the gamma function.

    maths 6.JPG


    thank you
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  2. #2
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    Since you can't use gamma, you could try rewriting and making a substitution.

    (1-x^{3})^{\frac{5}{2}}\cdot{x^{\frac{7}{2}}}

    =(1-x^{3})^{2}(1-x^{3})^{\frac{1}{2}}x^{\frac{7}{2}}

    Let u=\sqrt{1-x^{3}}, \;\ du=\frac{-3x^{2}}{2\sqrt{1-x^{3}}}

    Making the subs gives:

    \frac{-2}{3}\int{u^{6}\sqrt{1-u^{2}}}du

    Now, let u=sin{\theta}, \;\ du=cos{\theta}d{\theta}

    Make the subs and you get:

    \frac{-2}{3}\int{sin^{6}{\theta}cos^{2}{\theta}}d{\theta}

    =\frac{-2}{3}\int[sin^{6}{\theta}-sin^{8}{\theta}]d{\theta}

    Now, you can use the reduction formulas:

    \int{sin^{n}{\theta}}d{\theta}=\frac{sin^{n-1}{\theta}cos{\theta}}{n}+\frac{n-1}{n}\int{sin^{n-2}}d{\theta}


    Just a thought. It ain't pretty no matter how you look at it.
    Last edited by galactus; July 9th 2007 at 05:24 PM.
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  3. #3
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    This can be achieved with the Beta Integral.

    \int_0^1 (1-x^3)^{5/2} x^{7/2} dx
    We can write this as,
    \frac{1}{3} \int_0^1 (1-x^3)^{5/2} \cdot x^{3/2} \cdot (3x^2) \ dx

    Let t=x^3 \implies t'=3x^2 thus,

    \frac{1}{3}\int_0^1 (1-t)^{5/2} \cdot t^{1/2} dt

    Use the Beta Integral and get,
    \bold{B}\left(\frac{7}{2} , \frac{3}{2} \right)

    Converte to Gamma Function,

    \frac{\Gamma \left( \frac{7}{2} \right) \cdot \Gamma \left( \frac{3}{2} \right) }{\Gamma (5)}

    But,
    \Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}

    Thus, using reduction identity for Gamma functions,
    \Gamma \left( \frac{7}{2} \right) = \frac{15\sqrt{\pi}}{8}

    And,
    \Gamma \left( \frac{3}{2} \right) = \frac{\sqrt{\pi}}{2}

    And,
    \Gamma (5) = 4!=24

    Thus,
    \mbox{ answer } = \frac{15\pi}{384}
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  4. #4
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    Galactus' approach is not completely way off. He arrives at the integral,
    \int_0^{\pi/2} \sin ^{\alpha} x \cos^{\beta} x \ dx

    Which is discussed here.
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