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In this MO question, it says that we have

$$ \sum_{p<n} \frac{1}{\log{p}} =\frac{n}{\log^2 n}+O\left(\frac{n\log\log n}{\log^3 n}\right).$$

where the sum is on all primes $p$, up to some max prime $n$. This is derived from the prime number theorem.

My question is, does there exist a similar result for the order of magnitude of the sum of the logarithms of primes, squared?

$$ \sum_{p<n} \frac{1}{\log^2{p}} = ?$$

Or, when raised to an arbitrary power, let's say $a$?

$$ \sum_{p<n} \frac{1}{\log^a{p}} = ?$$

I would be happy to get an analogous result for either of these.

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    $\begingroup$ The accepted answer to the linked question contains a proof that generalizes immediately to your desired sums. Have you tried working through it? $\endgroup$
    – Greg Martin
    Commented Apr 11, 2019 at 5:33
  • $\begingroup$ I get the gist of how the prime number theorem is being used, and how this leads to some of the asymptotics, but I don't understand how this gives an answer for the sum of powers of logs for the primes. Feel free to write as an answer, though, if you see how to do it. $\endgroup$
    – Mike Battaglia
    Commented Apr 11, 2019 at 5:38
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    $\begingroup$ For any regularly-varying, slowly growing/decreasing function, by the argument mentioned above, you have $\sum_{p<n}f(p)\sim \int_1^n f(x)/\log x\,dx$. $\endgroup$
    – Anthony Quas
    Commented Apr 11, 2019 at 5:59

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For $a > 0$, partial summation yields $$\sum_{p \leq x} \frac{1}{\log^a{p}} = \frac{\pi(x)}{\log^a x} + a \int_2^x \frac{\pi(t)}{t \log^{a+1} t} \,dt.$$ The first term is $\frac{\text{li(x)}}{\log^a x} + O\left(\frac{x}{\log^a x}e^{-c \sqrt{\log x}}\right)$ by the prime number theorem.

Cut the integral in half : $$\int_2^x \frac{\pi(t)}{t \log^{a+1} t} \,dt = \int_2^{\sqrt x} \frac{\pi(t)}{t \log^{a+1} t} \,dt + \int_{\sqrt x}^x \frac{\pi(t)}{t \log^{a+1} t} \,dt.$$ The first one is clearly $O(\sqrt{x})$, while the second is $$O\left( \frac{1}{\log^{a+1}(x)} \int_{\sqrt x}^x \frac{\pi(t)}{t} \,dt\right) = O\left( \frac{1}{\log^{a+1}(x)} \int_{\sqrt x}^x \frac{dt}{\log t}\right) = O\left(\frac{\text{li(x)}}{\log^{a+1}(x)}\right).$$

In the end, $$\sum_{p \leq x} \frac{1}{\log^a{p}} = \frac{\text{li(x)}}{\log^a x} + O\left(a \frac{\text{li(x)}}{\log^{a+1}(x)}\right).$$

One could probably get a better second term by using integration by parts in a more clever way.

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  • $\begingroup$ When taking the log term out of the second integral, do you incur a dependence on $a$ in the implied constant? $\endgroup$
    – Dror Speiser
    Commented Apr 11, 2019 at 10:08
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    $\begingroup$ Indeed, maybe I should add it, though uniformity in $a$ wasn't in the original question. $\endgroup$
    – A. Bailleul
    Commented Apr 11, 2019 at 13:25

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