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Sums over p and improper integrals

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  • Adam
    Suppose you are trying to approximate sum over large primes p of a(p) where a(p) is some monotonically decreasing to 0 function of p. Instead of writing
    Message 1 of 1 , Aug 11, 2007
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      Suppose you are trying to approximate sum over large primes p of a(p)
      where a(p) is some monotonically decreasing to 0 function of p.
      Instead of writing sum(a(p),p>=P) ~ intgeral(a(x),x=P to infinity), you
      should rather approximate it by sum(a(p),p>=P) ~ intgeral(a(x)/log
      (x),x=P to infinity), basically because of a frequency count (gap
      size). Say p and q~p+log(p) are consecutive large primes and a(x) is
      (roughly) constant on the interval [p,q]. Then integral(a(x),p<=x<=q)
      ~ a(p)+a(p+1)+ ... +a(q) ~ a(p)*log(p) and is too big to approximate
      the summand a(p). Instead, you should use a(x)/log(x) for the
      integrand. Another way to interpret this is via the dx differential
      which should be weighted: a(p) = a(p)*1 ~ a(x) * dx/log(x).
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