Derivation of a generalization of Mertens' Third Theorem.

Question

One of Mertens' Theorems states $$\prod_{p\le x}(1-\frac{1}{p})\sim \frac{e^{-\gamma}}{\ln(x)}.$$ I have seen a generalized version that states $$\prod_{m<p\le x}(1-\frac{m}{p})\sim \frac{c(m)}{(\ln(x))^m},$$ where $m$ is a positive integer and $c(m)$ is a real number that depends on $m$. How can I derive this result? Is there a way to find what the values of $c(m)$ are? Thank you very much.

Answer 1

In this answer, the asymptotic was given in detail for $m=2$. In general, by noting that $$\mathfrak{S}=\prod_{p>m}\left(1-\frac{m}{p}\right)\left(1-\frac{1}{p}\right)^{-m} $$ converges, we may rewrite $$\prod_{m<p\leq x}\left(1-\frac{m}{p}\right)=\prod_{m<p\leq x}\left(1-\frac{1}{p}\right)^{m}\prod_{m<p\leq x}\left(1-\frac{m}{p}\right)\left(1-\frac{1}{p}\right)^{-m}.$$ The error from ending at $x$ in the product for $\mathfrak{S}$ will be very little, and so $$\prod_{m<p\leq x}\left(1-\frac{m}{p}\right)\sim\mathfrak{S}\prod_{p\leq m}\left(1-\frac{1}{p}\right)^{-m}\prod_{p\leq x}\left(1-\frac{1}{p}\right)^{m}.$$

Hence $$\prod_{m<p\leq x}\left(1-\frac{m}{p}\right)\sim\frac{c_{m}}{\log^{m}x}$$ where $$c_m=\mathfrak{S}e^{-m\gamma}\prod_{p\leq m}\left(1-\frac{1}{p}\right)^{-m}.$$

Answer 2

For a fixed $b$ and $m$ in $\mathbb{N}$, we have as $x \to + \infty$ : $${\small \prod_{\substack{b \leq p \leq x \\ \text{p prime}}} \left({\normalsize 1-\frac{m}{p}}\right)} \sim \left( \prod_{\substack{b \leq p \\ \text{p prime}}}\frac{1-\frac{m}{p}}{(1-\frac1p)^{m}} \right) \, {\small \prod_{\substack{p < b \\ \text{p prime}}} \left({\normalsize 1-\frac{1}{p}}\right)}^{-m} \, \dfrac{e^{-m \gamma}}{\log(x)^m}$$ You can see the proof in section 3 in my article here.