- $ \hspace{1.5cm} \sum_{p\leq x}\frac{1}{p} = \log(\log x) + a+ O(\frac{1}{\log x})$
$ \hspace{1.5cm}$with $ a=\int_{2}^{\infty}R(u)\frac{du}{u(\log u)^2} +1- \log(\log 2) $
- $ \hspace{1.5cm} e^{-1/p}(1-1/p)^{-1}= 1+O(1/p^2)$
Below is the expression that needs to be proved: $$ \prod_{p\leq x}(1-1/p)^{-1}= b\log x + O(1)$$ with $b=e^{a}\prod_{p} e^{-1/p}(1-1/p)^{-1}$.
I have tried proving that but not able to. This is how proceeded:
$$ \prod_{p\leq x}(1-1/p)^{-1}= \prod_{p\leq x}e^{1/p}e^{-1/p}(1-1/p)^{-1} $$
$$ = e^{\sum_{p\leq x}1/p}\prod_{p\leq x}e^{-1/p}(1-1/p)^{-1}\\ = \log x*e^{a}*e^{O(1/\log x)}\prod_{p\leq x}e^{-1/p}(1-1/p)^{-1}\\ = \frac{e^{O(1/\log x)}(\log x*e^{a}\prod_{p}e^{-1/p}(1-1/p)^{-1})}{\prod_{p> x}e^{1/p}e^{-1/p}(1-1/p)^{-1}}$$
I do not know how to proceed further.