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Motivating by this post Is the function $g(n) = \displaystyle\sum_{k=1}^{\pi(n) - 1}\left(\frac{1}{n-p_{k}}\right)$ bounded?, we get by partial summation $$\sum_{p\le x-1}\frac{1}{x-p}=\pi\left(x-1\right)-\int_{2}^{x-1}\frac{\pi\left(u\right)}{\left(x-u\right)^{2}}du.$$ I'm trying to find a tight-as-possible upper bound (building on the cited post, it is probably unbounded, so tight lower bound, e.g. $\Omega(1)$, is maybe out of reach without assuming second Hardy-Littlewood conjecture) using the asymptotic $$\pi\left(u\right)=\mathrm{li}\left(u\right)+O\left(u\exp\left(-c\sqrt{\log u}\right)\right)$$ for some positive constant $c>0$. Unfortunately, I couldn't proceed much further because of the complexity of calculating the integral. Any suggestion on how to continue will be very much appreciated!

Omer Simhi
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  • I believe we can integrate by parts to turn the integral into $$\frac{\pi(u)}{x-u}\Bigg|_2^{x-1} - \displaystyle\int_2^{x-1} \frac{(\pi(u))'}{x-u},du$$ ...unless my math is wrong which it very well might be. – Mathemagician314 Sep 26 '24 at 12:45
  • @Mathemagician314 The fundamental problem is that $\pi(u)$ is not even continuous, let alone differentiable. You can do partial summation with $\mathrm{li}(u)$ instead, but I'm afraid that we will run into the same issues with the error term. – Omer Simhi Sep 26 '24 at 13:10
  • Oh, I see, sorry about that. Hmm... you're probably right about $\text{li}(u)$, but it might be worth a try... – Mathemagician314 Sep 26 '24 at 13:36
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    Use Fubini's theorem to rewrite $\int_{2}^{x - 1} \frac{\operatorname{Li}(u)}{(x - u)^2} , du$. – Peter Humphries Sep 26 '24 at 17:03
  • @PeterHumphries If my calculation is valid, this integral is $\mathrm{Li}(x-1) + O(1)$. We still have problems with the error term, right? – Omer Simhi Sep 26 '24 at 23:30

1 Answers1

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Let $R(x)=\pi(x)-\operatorname{Li}(x)$. Then in the Riemann-Stieltjes sense, there is

$$ \mathrm d\pi(x)={\mathrm dx\over\log x}+\mathrm dR(x), $$

so

$$ S(x)=\sum_{p\le x-1}{1\over x-p}=\int_2^{x-1}{1\over x-u}{\mathrm du\over\log u}+\int_{2^-}^{x-1}{\mathrm dR(u)\over x-u}. $$

For the first term, set $\delta=(\log x)^{-0.99}$, so

$$ \int_2^{x-1}{1\over x-u}{\mathrm du\over\log u}=O\left(x^{1-\delta}\over x-x^{1-\delta}\right)+\int_{x^{1-\delta}}^{x-1}{\mathrm du\over(x-u)\log u}. $$

The O-term simplifies into $O(e^{-(\log x)^{0.01}})$. In the remaining integral, there is

$$ \log x\ge\log u\ge(1-\delta)\log x, $$

so ${\log u\over\log x}=1+O((\log x)^{-0.99})$. Plugging this into the integral gives

$$ \int_2^{x-1}{1\over x-u}{\mathrm du\over\log u}=1+O\left\{1\over(\log x)^{0.99}\right\} $$

For the second term, notice that $R(2^-)=0$, so

$$ \int_{2^-}^{x-1}{\mathrm dR(u)\over x-u}=R(x-1)-\int_2^{x-1}{R(u)\over(x-u)^2}\mathrm du. $$

By a theorem of Littlewood, $R(x)$ is changing signs infinitely often as $x\to+\infty$, so it is expected that $S(x)$ is oscillating around $1$.

To obtain more specific estimates, one needs to substitute in the explicit formula of $R(x)$ involving a sum over the nontrivial zeros of $\zeta(s)$.

TravorLZH
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  • Thank you for the answer! Although it doesn't provide more insight than what I already have, I'll accept it since this discussion in the forum seems to be coming to a conclusion. – Omer Simhi Sep 28 '24 at 14:53