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Let $B$ be the Meissel-Mertens constant, and $\theta(x)$ be Chebyshev's first function. Would there exist an $x_0$ such that for all $x>x_0$ the following inequalities are true (clearly, if the first one is true, so is the second one)? Can $x_0=3$? if not, 121?

1) $\int_2^x \frac{\theta(y)(1+\log y)}{y^2\log^2 y}dy-\log\log x>B$

2) $\int_2^x \frac{\theta(y)(1+\log y)}{y^2\log^2 y}dy-\log\log x+\frac{1}{\log x}>B$

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    $\begingroup$ It would be nice if you could provide some motivation for this question. $\endgroup$ Mar 19, 2019 at 21:01
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    $\begingroup$ The integral that appears twice is simply $\sum_{p\le x} 1/p$. I don't know why the OP thought the integral version would be better received. $\endgroup$ Mar 19, 2019 at 23:12
  • $\begingroup$ @EGME: You should look at, for example, Lamzouri's paper "A bias in Mertens' product formula". It doesn't answer the exact question you ask, but the methods therein probably show that (1) has arbitrary large counterexamples, and would allow you to determine the truth of (2) as well. $\endgroup$ Mar 19, 2019 at 23:13
  • $\begingroup$ @Greg Do you mean $\sum_{p\leq x}1/p$ ? $\endgroup$
    – EGME
    Mar 19, 2019 at 23:17
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    $\begingroup$ $\frac{\theta(x)}{x\log x}\big|_{2-}^\infty = 0$. $\endgroup$ Mar 20, 2019 at 7:13

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