# An inequality involving Chebyshev's theta and the Meissel-Mertens constant

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

• It would be nice if you could provide some motivation for this question. – Sylvain JULIEN Mar 19 '19 at 21:01
• 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. – Greg Martin Mar 19 '19 at 23:12
• @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. – Greg Martin Mar 19 '19 at 23:13
• @Greg Do you mean $\sum_{p\leq x}1/p$ ? – EGME Mar 19 '19 at 23:17
• $\frac{\theta(x)}{x\log x}\big|_{2-}^\infty = 0$. – Greg Martin Mar 20 '19 at 7:13