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Math Help - Integration

  1. #1
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    Integration

    Show given that

    Thanks guys
    Last edited by usagi_killer; April 12th 2009 at 09:12 PM.
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  2. #2
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    I don't see any way to get directly from one of these integrals to the other. So I think that you have to evaluate each integral separately and then compare them.

    Let I_n = \int_0^\pi\!\! e^{-2t}\sin^nt\,dt. Integrate by parts twice and you will find that if n\geqslant 2 then
    \begin{aligned}I_n &= <br />
\Bigl[-\tfrac12e^{-2t}\sin^nt\Bigr]_0^\pi + \int_0^\pi\!\!\tfrac12 e^{-2t}n\sin^{n-1}t\cos t\,dt\\ &= \tfrac n2\Bigl[-\tfrac12e^{-2t}\sin^{n-1}t\cos t\Bigr]_0^\pi + \tfrac n4\int_0^\pi\!\! e^{-2t}\bigl((n-1)\sin^{n-2}t\cos^2 t - \sin^n t\bigr)\,dt\\ &= \tfrac n4\bigl((n-1)I_{n-2} - nI_n\bigr)\end{aligned}
    (using \cos^2t = 1-\sin^2t for the last step). Therefore \Bigl(1+\frac{n^2}4\Bigr)I_n = \frac{n(n-1)}4I_{n-2}, and so I_n = \frac{n(n-1)}{n^2+4}I_{n-2}.

    A similar calculation (integrating by parts twice) shows that I_1 = \tfrac15(1+e^{-2\pi}). And a straightforward integration gives I_0 = \tfrac12(1-e^{-2\pi}).

    Therefore I_3 = \tfrac6{13}I_1 = \tfrac6{65}(1+e^{-2\pi}) and I_2 = \frac14I_0 = \tfrac18(1-e^{-2\pi}). The desired connection between I_3 and I_2 then follows.
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