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Let $$A(x)=x\sum_{p \leq x} 1, B(x)=\frac{3}{5}\sum_{x<p\leq 2x}p$$ Using prime number theorem, we have $A(x)\sim\frac{x^2}{\log{x}}$, but how to obtain an estimation for $B(x)$?

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Where does $B(x)$ comes from? The $\frac 35$ is really mesmerizing. –  Patrick Da Silva Dec 11 '11 at 4:01
    
I am also puzzled about that $\frac{3}{5}$. –  Rob Dec 11 '11 at 4:08
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up vote 3 down vote accepted

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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Where do you get that asymptotic formula for the sum of primes up to x? Can I find this result in Apostol? –  Rob Dec 11 '11 at 4:41
    
You can get ot out of oeis.org/A007504 but there are probably more direct ways. Maybe a better place is mathworld.wolfram.com/PrimeSums.html –  Gerry Myerson Dec 11 '11 at 4:50
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@Rob, just try to apply summation by part and use prime number theorem, you can get the asymptote directly. –  Soarer Dec 11 '11 at 6:10
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