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Does there exists a good asymptotic formula for $$A(x) := \prod_{p\leq x}(1-\frac 1p).$$

By using a heuristic argument one can guess: $$A(x) \sim \frac{1}{2\,\mathrm{ln}(x)}.$$

Here is the argument:

$$\frac{x^2}{2\, \mathrm{ln}(x) } \sim \pi(x^2)-\pi(x) = \sum_{p_1<p_2<\dots<p_k\leq x}(-1)^k\left\lfloor \frac{x^2}{p_1p_2\cdots p_k}\right\rfloor \\ \stackrel{??}{\simeq}\sum_{p_1<p_2<\dots<p_k\leq x}(-1)^k\left( \frac{x^2}{p_1p_2\cdots p_k}\right) = x^2 \prod_{p\leq x}(1-\frac{1}{p}).$$

Is this conjectural asymptotic formula true? If so, can this heuristic argument be completed to an exact proof? The later question is more important for me!

Thanks!

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  • $\begingroup$ The constant is wrong. See en.wikipedia.org/wiki/Mertens%27_theorems $\endgroup$
    – user41593
    Commented Mar 8, 2015 at 8:14
  • $\begingroup$ Nice! But I need to know how to handle the fractional parts in the first sum. $\endgroup$
    – Mostafa - Free Palestine
    Commented Mar 8, 2015 at 8:24
  • $\begingroup$ You asked "Is this formula true? If so, can this be completed to an exact proof?": the premise about the formula is false, so no point worrying of turning a heuristic for a wrong result into a rigorous proof! Anyway, what bothers me the most is that you use the Prime Number Theorem in the very first asymptotic equivalence and still you worry about not using it in the follow-up. $\endgroup$
    – user41593
    Commented Mar 8, 2015 at 8:30
  • $\begingroup$ @EmanueleTron The main reason for the question is that one can easily relate the number of twin primes less than n to a similar sum and if we have good estimates on the fractional parts, their density can be related to $\frac 12 \prod_{2< p\leq \sqrt{x}} (1-2p^{-1})$ . $\endgroup$
    – Mostafa - Free Palestine
    Commented Mar 8, 2015 at 8:36
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    $\begingroup$ I don't know what you mean by "directly". Mertens' theorem readily gives the sum of the error terms. The idea of sieve theory is to introduce weights that cut down the number of terms in the sieving process. This is also how Zhang proved his spectacular theorem that there are infinitely many prime pairs with distance at most 70 million apart. Since then, 70 million was lowered to 246. This is a pretty good approximation to the twin prime conjecture, although still very far away from it. Number theory is hard. $\endgroup$
    – GH from MO
    Commented Mar 8, 2015 at 14:48

1 Answer 1

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Mertens' Theorem says (page 65, An Introduction to Sieve Methods and Their Applications, Cojocaru and Murty):

$$ \displaystyle\prod_{p< x}\left(1-\frac 1p\right)=\frac{e^{-\gamma}}{\log x}\left(1+O\left(\frac{1}{\log x}\right)\right), $$

where $\gamma$ is the Euler-Mascheroni constant.

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