Let $$A(x)=x\sum_{p \leq x} 1, B(x)=\frac{3}{5}\sum_{x Using prime number theorem, we have $A(x)\sim\frac{x^2}{\log{x}}$, but how to obtain an estimation for $B(x)$?
Compare two sums using prime number theorem
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prime-numbers
analytic-number-theory
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3Where does $B(x)$ comes from? The $\frac 35$ is really mesmerizing. – 2011-12-11
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0I am also puzzled about that $\frac{3}{5}$. – 2011-12-11
1 Answers
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The sum of the primes up to $x$ is, asymptotically, $(1/2)x^2/\log x$. So asymptotically, $$B(x)=(3/5)((1/2)(2x)^2/\log(2x)-(1/2)x^2/\log x))$$ Looks like about $(9/10)x^2/\log x$.
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0Where do you get that asymptotic formula for the sum of primes up to x? Can I find this result in Apostol? – 2011-12-11
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0You can get ot out of http://oeis.org/A007504 but there are probably more direct ways. Maybe a better place is http://mathworld.wolfram.com/PrimeSums.html – 2011-12-11
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1@Rob, just try to apply summation by part and use prime number theorem, you can get the asymptote directly. – 2011-12-11