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Math Help - Convergence test

  1. #1
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    Convergence test

    Hi,

    I have to test the integral from 0 to pi of sin(x)/sqrt(pi-x) for convergence using the direct comparison test or the limit comparison test.

    I know I can't integrate so I have to find a function that is either always greater than sin(x)/sqrt(pi-x) or do the limit comparison test on it.

    I'm not quite sure how to approach this one.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Is it compulsory to use those test?. Take into account that:

    \displaystyle\lim_{x \to \pi^-}{\dfrac{\sin x}{\sqrt{\pi -x}}}=\ldots=0

    so, we can expand by continuity the function taking f(\pi)=0 hence, the integral is convergent.

    Alternatively:

    0\leq \dfrac{\sin x}{\sqrt{\pi -x}}\leq \dfrac{1}{\sqrt{\pi -x}}

    etc.
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  3. #3
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    Well, I think as long as we can prove it he won't care, but that's a guess.

    Also, I didn't tackle the problem as a whole...sin(x)<1, always, and I forgot that.

    Thanks!
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Wolvenmoon View Post
    Well, I think as long as we can prove it he won't care, but that's a guess.
    If so, the former is more elegant.
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