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Prime number theorem

  1. Nov 9, 2008 #1
    We have that

    $P(x) = \sum_{k=1}^{\infty} \frac 1k \pi(x^{1/k})$
    and
    $Li(x) = \int_2^n \frac {dt}{\log t}$

    And the prime number theorem is:

    $$\pi(n) \sim \frac{n}{\log n }$$
    I want to show that $$P(x) \sim Li(x)$$ is equivalent to prime number theorem.

    Can some body please help me with this.
     
  2. jcsd
  3. Nov 9, 2008 #2

    CRGreathouse

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    \sim is transitive, so it suffices to show that
    [tex]\operatorname{Li}(x)\sim x/\log(x)[/tex]
     
  4. Nov 9, 2008 #3
    I have shown this but still I need to show that P(x) \sim Li(x).
     
  5. Nov 9, 2008 #4

    CRGreathouse

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    Then you're done:
    1. [tex]\pi(x)\sim x/\log(x)[/tex] (Prime Number Theorem)
    2. [tex]\operatorname{Li}(x)\sim x/\log(x)[/tex] (you said you proved it already)
    3. [tex]\pi(x)\sim\operatorname{Li}(x)[/tex] (by transitivity of ~)
     
  6. Nov 10, 2008 #5
    I wish I was done but I am not how can I prove that P(x) \sim Li(x).

    P is different from /pi. Give any suggestion how can I prove this.
     
  7. Nov 10, 2008 #6

    CRGreathouse

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    Ah, sorry, I forgot your notation. But P(x) is just pi(x) plus some insignificant terms. It suffices to show that P(x) = pi(x) + O(sqrt(x)).
     
  8. Nov 10, 2008 #7
    yes, but how.
     
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