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Let $\psi(x):=\sum_{n\leq x}\Lambda(n)$ denotes the 2nd Chebyshev function, where $\Lambda$ stands for the von-Mangoldt function. Are there any known (and 'nice') estimates for the change rates $\psi(x+h)-\psi(x)$ for general or special $x$ and $h$?

Thanks in advance,

efq

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2 Answers 2

up vote 5 down vote accepted

There is the asymptotic estimate $\psi(x+h) - \psi(x) \sim h$ for $x^{7/12 + \epsilon} \leq h \leq x$, valid for any $\epsilon > 0$. This is due to M. N. Huxley, and dates to 1972. I am not aware of any better range for $h$ if you want asymptotic equality. But if you are satisfied with an order of magnitude result, you can have $c_1h \leq \psi(x+h) - \psi(x) \leq c_2h$ with $c_1$ and $c_2$ positive constants and $x^{\theta} \leq h \leq x$ for some $\theta$ slightly larger than $0.5$. I can't give references offhand, but you should be able to find such papers by searching on R. C. Baker, G. Harman, J. Pintz in Mathscinet.

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Thanks a lot for you answer! I believe this order-of-magnitude-result may in fact work out for me. I´ll definitely have a look into the reference you mention. –  efq Nov 20 '09 at 19:09

Also, I think Selberg's result assert that for "almost all x" we have $\psi(x + h) - \psi(x) \sim h$ for $h \asymp (\log x)^2$ this was shown to not be true "pointwise" by Mayer.

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That is interesting. Is it known whether this null set is finite? –  efq Nov 20 '09 at 20:49
    
there is a sequence of x_k -> oo such that psi(x_k + (log x_k)^2) - psi(x_k) > (1+epsilon) (log x_k)^2 ... (I hope this answers your question). –  maki Nov 20 '09 at 23:16
    
yes, of course! ignore my last comment. –  efq Nov 21 '09 at 23:19

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