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Math Help - How can i explain???

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

    I have shown that:

    When  \rho = \beta + i \gamma

    Then

    \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:

    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  \rho = \beta + i \gamma

    Then

    \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:

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

    Let  t=x^\rho . Now integrate by parts:  \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  \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  \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@; June 7th 2010 at 08:12 AM.
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