I have seen the prime number theorem and on the of versions I know is that $\sum_{p\leq x} \log p=O(x)$ (I am counting over primes here and in the rest of the post). Are there any similar results for say, $\displaystyle \sum_{\substack{p\in [x/2, \hspace{0.1em} x]\\p \text{ prime}}} \log p$ or or $\displaystyle \sum_{\substack{p\in [x,y]\\p \text{ prime}}} \log p$ in general? Also, are there some estimates on $\displaystyle \sum_{p\in[x,y]}\frac{\log p}{p}$ or some $[x,y]$? Perhaps, these are not very hard questions but I don't know how to get good bounds without counting up to $y$ and subtracting from it the count up to $x$. Thank you!
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4$\begingroup$ I think the usual name for this sort of thing is "(a) prime number theorem in short intervals". You should have a look at the following survey by Yildirim as a starting point: math.boun.edu.tr/instructors/yildirim/paper/paper9.pdf $\endgroup$– Anurag SahayCommented Dec 14, 2020 at 0:05
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1$\begingroup$ I take it that, even though you never say so, $p$ is meant to run over primes. $\endgroup$– Gerry MyersonCommented Dec 14, 2020 at 4:41
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$\begingroup$ @GerryMyerson, edited to reflect that. $\endgroup$– user147650Commented Dec 14, 2020 at 8:00
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$\begingroup$ @asahay, thanks for pointing me to the survey! $\endgroup$– user147650Commented Dec 14, 2020 at 8:00
1 Answer
Using the various forms of Mertens' theorems and the prime number theorem, you can easily generate bounds for your sums. That is, if you have an approximation for $\sum_{n\leq y}a(n)$ and $\sum_{n\leq x}a(n)$, then subtracting one from the other, you get an approximation for $\sum_{x<n\leq y}a(n)$. However, better results are available by more advanced techniques like zero density estimates for $\zeta(s)$ or sieve methods. For example, Huxley (1972) proved that if $c\in(7/12,1]$ is fixed and $x\to\infty$, then for any $y\in[x+x^c,2x]$ we have $$\frac{1}{y-x}\sum_{x<p\leq y}\log p\to 1.$$
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$\begingroup$ Thank you for the answer! It's very helpful. $\endgroup$– user147650Commented Dec 14, 2020 at 9:45
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$\begingroup$ @user96343: If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$ Commented Dec 14, 2020 at 10:15
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1$\begingroup$ of course! Done. $\endgroup$– user147650Commented Dec 14, 2020 at 12:30