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Thread: How can i explain???

  1. #1
    Junior Member
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    How can i explain???

    I have shown that:

    When $\displaystyle \rho = \beta + i \gamma$

    Then

    $\displaystyle \frac{x^{\rho}}{\rho} +\frac{x^{\overline \rho}}{\overline\rho} = \frac{2x^{\beta}}{|\rho|} cos(\gamma In(x) - \theta)$

    However I don't how I can explain why:

    $\displaystyle Li(x^{\rho}) \sim \frac{x^{\rho}}{\rho Inx}$
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  2. #2
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by signature View Post
    I have shown that:

    When $\displaystyle \rho = \beta + i \gamma$

    Then

    $\displaystyle \frac{x^{\rho}}{\rho} +\frac{x^{\overline \rho}}{\overline\rho} = \frac{2x^{\beta}}{|\rho|} cos(\gamma In(x) - \theta)$

    However I don't how I can explain why:

    $\displaystyle Li(x^{\rho}) \sim \frac{x^{\rho}}{\rho Inx}$
    First note that $\displaystyle \frac{x^\rho}{\rho\log(x)}=\frac{x^\rho}{\log(x^\r ho)} $.

    Let $\displaystyle t=x^\rho $. Now integrate by parts: $\displaystyle \text{li}(t)=\int_2^t\frac{1}{\log(t)}dt=\frac{t}{ \log(t)}-\frac{2}{\log(2)}+\int_2^t\frac{1}{\log^2(t)}dt $.

    What's left to show is $\displaystyle \lim_{t\to\infty} \frac{\int_2^t\frac{1}{\log^2(t)}dt}{\frac{t}{\log (t)}}=0 $. I'll let you do it... apply L'hopital's rule to get it.

    Thus $\displaystyle \lim_{t\to\infty} \frac{\text{li}(t)}{\frac{t}{\log(t)}}=1 \implies \text{li}(t)\sim\frac{t}{\log(t)} $
    Last edited by chiph588@; Jun 7th 2010 at 07:12 AM.
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