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Mertens' third theorem states that: $$\prod_{\substack{ p \leq x \\ \text{p prime} }} \left( 1 - \dfrac{1}{p} \right) \sim \dfrac{e^{-\gamma}}{\log(x)}$$ Question: what is the best functions (unconditionally and conditionally) satisfying: $$\prod_{\substack{ p \leq x \\ \text{p prime} }} \left( 1 - \dfrac{1}{p} \right) = \dfrac{e^{-\gamma}}{\log(x)} + \mathcal{O}(f(x))$$


The question is posted here : https://math.stackexchange.com/questions/3626427/error-term-in-mertens-third-theorem

But I can't know if the answer there is correct or not without references or an exact answer.

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    $\begingroup$ Cross-post from Math.SE. You've already received an answer there so you should inducate here what's wrong with it and why you are asking again. $\endgroup$ – Wojowu Apr 16 at 9:37
  • $\begingroup$ @Wojowu, i am not sure if the answer in MSE is correct, he didn't give references, and he use terms like "i think", "i am not sure ". $\endgroup$ – LAGRIDA Apr 16 at 9:40
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    $\begingroup$ You should mention all of that in the body of the question. $\endgroup$ – Wojowu Apr 16 at 9:47
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There's been a lot of work on unconditional results of this sort.

Rosser and Schoenfeld showed in a 1962 paper that one can take

$$\dfrac{e^{-\gamma}}{\log x} \left(1- \frac{1}{2\log^2 x} \right) < \prod_{\substack{ p \leq x \\ \text{p prime} }} \left( 1 - \dfrac{1}{p} \right) < \dfrac{e^{-\gamma}}{\log x} \left(1+ \frac{1}{2\log^2 x} \right).$$ The upper bound is valid for $x>1$ and the lower bound for $x> 285.$

This was subequently improved by Dusart in a 2016 paper that if one has $ x \geq 2278382$ then one has $$\dfrac{e^{-\gamma}}{\log x} \left(1- \frac{1}{5\log^3 x} \right) < \prod_{\substack{ p \leq x \\ \text{p prime} }} \left( 1 - \dfrac{1}{p} \right) < \dfrac{e^{-\gamma}}{\log x} \left(1+ \frac{1}{5\log^3 x} \right).$$

Tighter bounds which as far as I'm aware are best known today with reasonably tight explicit constants are in a paper of Axler http://math.colgate.edu/~integers/s52/s52.pdf although the bounds as written are a bit uglier in form. Note that there's a typo in inequality 6.3 in that paper where the parenthetical should be $$\left(1+\frac{1}{20 \log^3 x} +\frac{3}{16\log^4 x} + \frac{1.02}{(x-1)\log x}\right). $$ Axler's error term is essentially of the same order as that of Dusart but with an improved constant. Axler's paper also contains full references for Dusart as well as Rosser and Schoenfeld. All of these papers also have similar bounds on other functions related to counting primes such as Chebyshev's functions.

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  • $\begingroup$ Many thanks for your answer $\endgroup$ – LAGRIDA Apr 16 at 15:22
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Unconditionally, the best known error term (f(x) in the original question) is in

E.A. Vasil’kovskaja, Mertens’ formula for an arithmetic progression, Taškent. Gos. Univ. Nauˇcn. Trudy, VoprosyMat. 548 (1977) 14–17, 139–140

See also the introduction of my paper with Zaccagnini on j.number theory, 127 (2007), 37–46, https://doi.org/10.1016/j.jnt.2006.12.015

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  • $\begingroup$ Many thanks for your answer $\endgroup$ – LAGRIDA Apr 16 at 15:33
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    $\begingroup$ You’re welcome. Assuming the Riemann Hypothesis, the best estimate is f(x)<< 1/ sqrt(x). You can find the line of the proof on lemma 9 of the paper of mine previously cited (such a lemma is stated assuming GRH because we need it for its second part, but the estimate on this product is the one in eq. (16) there and it “just” depends on RH). $\endgroup$ – Alessandro Languasco Apr 16 at 15:48

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