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,



2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – M.G.
    Nov 20, 2009 at 19:09

Also, I think Selberg's result asserts 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.

  • $\begingroup$ That is interesting. Is it known whether this null set is finite? $\endgroup$
    – M.G.
    Nov 20, 2009 at 20:49
  • $\begingroup$ 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). $\endgroup$
    – maki
    Nov 20, 2009 at 23:16
  • $\begingroup$ yes, of course! ignore my last comment. $\endgroup$
    – M.G.
    Nov 21, 2009 at 23:19

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