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!



Asymptotic formula for $\prod_{p\leq x} (1-p^{-1})$
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.


