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Beyond PNT

  1. Aug 23, 2006 #1
    In fact if PNT says that the series [tex] \sum_{p<x}1 \sim Li(x) [/tex]

    My question is if we can't conjecture or prove that:

    [tex] \sum_{p<x}p^{q} \sim Li(x^{q+1}) \sim \pi(x^{q+1}) [/tex] q>0

    In asymptotic notation.....
     
  2. jcsd
  3. Aug 23, 2006 #2

    matt grime

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    Of course we can conjecture it, Jose (I presume this a new account for eljose). Have *you* tried to prove it? Does it even seem reasonable? Have you run it through a computer at all? Why do you even think it might be true?
     
  4. Aug 23, 2006 #3

    shmoe

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  5. Aug 24, 2006 #4
    The main key is that according to the manual..."Mathematical Handbook of Formulas and Tables"..the integral:

    [tex] \int_{2}^{x} dt \frac{t^n }{log(t)}= A+log(log(x))+\sum_{k>0}(n+1)^{k}\frac{log^{k}}{k. k!} [/tex]

    Using the properties of the logarithms you get that the series above is just Li(x^{n+1}) , in fact using "this" conjecture and prime number theorem you get the (known) asymptotic result:

    [tex] \sum_{i=1}^{N}p_i \sim (1/2)N^2 log(N) [/tex]

    for the case n=-1, you get that the "Harmonic prime series" diverges as log(log(x)) ...although the constant i give is a bit different.
     
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