Error term in Mertens' third theorem

Question

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: $$\prod_{\substack{ p \leq x \\ \text{p prime} }} \left( 1 - \dfrac{1}{p} \right) = \dfrac{e^{-\gamma}}{\log(x)} + \mathcal{O}(f(x))$$

Answer 1

This is not an answer but it is a key idea may lead to a correct bound .

The product $\prod_{\substack{ p \leq x \\ \text{p prime} }} \left( 1 - \dfrac{1}{p} \right) = \dfrac{e^{-\gamma}+ \mathcal{O}(1)}{\log(x)} \tag 1 $ it is the exponential form of Merten's theorem(logarithmic form), we may try to approximation $\mathcal{O}(1)$ usig $\tau(n)$ the number divisor function , we may use the informal approximation $\tau(n)\sim 1$ because “behaves like $\mathcal{O}(1)$ on the average, now we have $\mathcal{O}(1)=\sum_{n\leq x}1-x$ and try $n=dm $ to obtain $\sum_{n\leq x}1=\sum_{n,dm \leq x}\tau(n)$, The approximation of $\sum_{n,dm\leq x}1$ is montioned here in detail using Dirichlet hyperbola method to obtaine :$\sum_{n,dm \leq x}\tau(n)= x\log x+(2\gamma -1)x+\mathcal{O}(\sqrt{x})$ replace $\mathcal{O}(1)$ in the RHS of formula $(01)$ by

$x\log x+(2\gamma -1)x+\mathcal{O}(\sqrt{x})-x=\mathcal{O}(1)$, We may get this asymptotics $x+(e^{-\gamma}+2\gamma -1)x/\log x+\mathcal{O}(\sqrt{x})/\log(x)$ which probably close to $\mathcal{O}(\log \log x) $ for large $x$

Answer 2

Under PNT it was shown that $f(x) = e^{-c (\ln x)^{\alpha} }$ , i think $\alpha = \frac{3}{5} -\epsilon$ works for any $\epsilon >0$ and $c(\epsilon)$ and sufficenlty large $x$ (this estimate is unconditional)

Under R.H. it was shown that $f(x) = \frac{\ln^2 x}{\sqrt{x}}$, i am not sure if the power $2$ over the logarithm can be reduced to $1$ or something in between (this estimate is conditional on the correctness of Riemann hypothesis).

Answer 3

The error term follows from the remainder of the prime number theorem as @zeraoulia rafik suggests. This answer is just a more specific explanation that connects them. For convenience, let $\operatorname{li}(x)$ denote the logarithmic integral (regardless the lower limit), and let's write $\pi(x)$ as follows:

$$ \pi(x)=\sum_{p\le x}1=\operatorname{li}(x)+xr(x)\tag1 $$

This indicates that

\begin{aligned} \sum_{p\le x}\frac1p &=\int_0^x{\mathrm d\pi(t)\over t}=\int_2^x{\mathrm dt\over t\log t}+\int_{2^-}^x{\mathrm d(tr(t))\over t} \\ &=\log\log x-\log\log2-r(2-0)+\int_2^x{r(t)\over t}\mathrm dt \end{aligned}

In fact, it follows from de la Vallée Poussin's version of PNT that $r(t)\ll_A(\log t)^{-A}$ for any $A>0$. This indicates that there is a constant $B_1$ such that

$$ \sum_{p\le x}\frac1p=\log\log x+B_1+E(x)\tag2 $$

where $E(x)$ satisfies

$$ |E(x)|=\int_x^\infty{|r(t)|\over t}\mathrm dt $$

Before we continue, let's $r_1(x)$ be a monotonic decreasing function such that $0\le r_1(x)\ll_A(\log x)^{-A}$ and $|r(x)|\le r_1(x)$. Then, we have

$$ |E(x)|\ll(\log x)^2R(x)\int_x^\infty{\mathrm dt\over t(\log t)^2}\mathrm dt\ll r_1(x)\log x $$

Plugging this back into (2), we have

$$ \sum_{p\le x}\frac1p=\log\log x+B_1+\mathcal O(r_1(x)\log x)\tag3 $$

Since it is can be shown using partial summation that

$$ \left|\sum_{p>x}\left[\frac1p+\log\left(1-\frac1p\right)\right]\right|\asymp\sum_{p>x}{1\over p^2}\asymp{1\over x\log x} \tag4 $$

and that (See Theorem 428 of Hardy & Wright's An Introduction to the Theory of Numbers)

$$ B_1=\gamma+\sum_p\left[\frac1p+\log\left(1-\frac1p\right)\right]\tag5 $$

we can conclude by exponentiating (3) plugging in (4) that

\begin{aligned} \prod_{p\le x}\left(1-\frac1p\right) &={e^{-\gamma}\over\log x}\left\{1+\mathcal O\left\{{1\over x\log x}+r_1(x)\log x\right\}\right\} \\ &={e^{-\gamma}\over\log x}+\mathcal O\left\{{1\over x(\log x)^2}+r_1(x)\right\} \end{aligned}

By Littlewood's oscillation theorem, we know that $r_1(x)$ will grow significantly faster than $1/x(\log x)^2$. As a result, any monotonically decreasing positive function that bounds $r(x)$ in (1) can serve an error bound for Mertens' third theorem.