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Math Help - logarithmic integral

  1. #1
    Senior Member Dinkydoe's Avatar
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    logarithmic integral

    I have trouble showing that \int_{2}^x\frac{1}{(\log(x))^k}dx=O(\frac{x}{(\log  (x))^k}) as x\to\infty

    A hint I've been given, is to split the integral in 2: \int_{2}^{\sqrt{x}}+\int_{\sqrt{x}}^x

    I don't really see the point to that: I guess that: \int_{\sqrt{x}}^x\to 0 as x\to \infty

    My guess is that I need some powerseries-expansion. Any suggestion will be greatly appreciated.
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  2. #2
    Senior Member
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    Integrating this integral by parts

    <br />
\displaystyle{\int_{2}^x\frac{1}{(\log(t))^k}dt=\f  rac{t}{(\log(t))^k}|_2^x -\int_2^x td({(\log(t)^{-k})}=}<br />

    <br />
\displaystyle{=\frac{t}{(\log(t))^k}|_2^x +k \int_2^x \frac{dt}{(\log(t))^{k+1}}}<br />
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  3. #3
    Senior Member Dinkydoe's Avatar
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    Thanks!

    I shouldve thought of that. Any idea why that hint could be useful?
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