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Euler product and Goldbach conjecture

  1. Apr 26, 2006 #1
    In an anlaogy with the Euler product of the Riemann function we make:

    [tex] \prod_{p}(1+e^{-sp})=f(s) [/tex] of course we have that:

    [tex] f(p1+p2+p3)=f(p1)f(p2)f(p3) [/tex] f(x)=exp(-ax) if Goldbach Conjecture is true then p1+p2= even and p5+p6+p8=Odd for integer n>5? then this product should be equal to:

    [tex] f(s)=\sum_{n=0}^{\infty}a(n)e^{-sn} [/tex] where the a(n) is the function that tells in how many ways and odd or even number can be descomposed as a sum of 2 or 3 primes, we now define:

    [tex] A(x)=\sum_{n=0}^{x}a(n) [/tex] if a(n)=1 for every n then using Perron formula we get that A(x)=[x] (floor function) so we would have that:

    [tex] f(s)=(1-e^{-s}) [/tex] then if correct take numerical values to proof that the product and f(s) are equal.
  2. jcsd
  3. May 2, 2006 #2
    sorry there,s a mistake if we set:

    [tex] \prod_{p} (1-e^{-sp}=f(s)= \sum_{n=0}^{\infty} a(n)e^{-sn}= \int_{0}^{\infty}A(x)e^{-sx} [/tex]

    And using the same trick Riemann did we have that:

    [tex] f(s)= \sum_{n=1}^{\infty} \frac{g(x^{1/n})}{n} [/tex]

    where [tex] g(x)=\pi (lnx) [/tex] so from this we could conclude that:

    [tex] f(x)= \frac{1}{2 \pi i}\int_{C} ds \frac{ ln \zeta(s) }{s}ln^{s} (t) [/tex]

    Where C is the real line Re(s)>c for a real c constant
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