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Math Help - Convergence of Integral

  1. #1
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    Convergence of Integral

    Show that for s\in \mathbb{C} and Re(s) > 0 the integral \int_1^\infty \lfloor x\rfloor^{-s} - x^{-s}dx converges.
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  2. #2
    Super Member girdav's Avatar
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    You can begin to show that the sequence \left\{I_n\right\} where I_n = \int_1^n\left(\lfloor x\rfloor^{-s}-x^{-s}\right) dx is convergent.
    Last edited by girdav; May 8th 2011 at 11:53 AM.
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  3. #3
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    Well that is exactly what I want to show, I need some hint or maybe some criterion to use.
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  4. #4
    Super Member girdav's Avatar
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    If we denote s =u+iv we have \lfloor x\rfloor^{-s}-x^{-s} =\int_x^{\lfloor x\rfloor}(-s)t^{-1-s}dt =<br />
\int_{\lfloor x\rfloor}^xst^{-1-s}dt hence |\lfloor x\rfloor^{-s}-x^{-s}|\leq <br />
\int_{\lfloor x\rfloor}^x|s||t^{-1-s}|dt = \int_{\lfloor x\rfloor}^x|s||t^{-1-s}|dt<br />
=\int_{\lfloor x\rfloor}^x|s||t^{-1-u-iv}|dt=\int_{\lfloor x\rfloor}^x|s|t^{-1-u}dt =|s|\frac{-1}u(x^{-u}-\lfloor x\rfloor^{-u}) hence we only have to show the result if s is a positive real number.
    It's true if s>1 because \sum_{n=1}^{+\infty}n^{-s} and \int_1^{+\infty}t^{-s}dt are convergent. If s=1 use the fact that the sequence \left\{\sum_{k=1}^n\frac 1k-\ln n\right\} is convergent. For s<1, use inequalities involving \int_1^n t^{-s}dt and \sum_{k=1}^nk^{-s}.
    Last edited by girdav; May 9th 2011 at 01:55 AM.
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  5. #5
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    Ok, I derived something similar already as well, but for 0 < s < 1 I really don't see how to show that \int_1^\infty \lfloor x\rfloor^{-s} - x^{-s}dx converges.
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  6. #6
    Super Member girdav's Avatar
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    The function t\mapsto t^{-1-u} is decreasing hence |\lfloor x\rfloor^{-s}-x^{-s}|\leq \frac{|s|}u(\lfloor x\rfloor^{-u}-x^{-u}|)=|s|\int_{\lfloor x\rfloor}^xt^{-1-u}dt\leq |s|(x-\lfloor x\rfloor ){\lfloor x\rfloor}^{-1-u}. We have
    \begin{align*}<br />
\int_1^{n+1}|\lfloor x\rfloor^{-s}-x^{-s}|dx&\leq |s|\sum_{k=1}^n\int_k^{k+1}<br />
\left(x\lfloor x\rfloor^{-1-u}-\lfloor x\rfloor^{-u}\right)dx\\<br />
&=|s|\sum_{k=1}^n\int_k^{k+1}<br />
\left(xk^{-1-u}-k^{-u}\right)dx\\<br />
&=|s|\left(\sum_{k=1}^nk^{-1-u}\left(\frac{(k+1)^2}2-\frac{k^2}2\right)-k^{-u}\right)\\<br />
&= |s|\left(\sum_{k=1}^nk^{-1-u}\left(k+\frac 12\right)-k^{-u}\right)\\<br />
&= \frac{|s|}2\sum_{k=1}^nk^{-1-u}<br />
\end{align*}
    which is convergent (we don't have to distinguish the case s>1 or not, so it's better than my previous work).
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  7. #7
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    ok, this looks good, but what are the <br/>'s and what is exactly is u?
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  8. #8
    Super Member girdav's Avatar
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    I don't know what <br/>'s are (maybe this come from latex issue). s=u+iv, I forgot it.
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