# An analytic formula for the sum of the logs of primes.

I just read in Martin Klazar's Intoduction to Number Theory (page 53), that $\sum_{p\leq x} \log p - \log (p-1) = \log\log x + \gamma + O(1/\log x)$.

Where $\gamma$ is the Euler-Mascheroni constant, $p$ is a prime,

and the error term $O(1/\log x)$ was evaluated from the integral $\int_x^{\infty} \frac{z(t)}{t\log^{2}t} \mathrm {d}t$ and $z(t) = O(1)$.

My question is on this error term: is it nonnegative such that we have

$\sum_{p\leq x} (\log p - \log (p-1)) > \log\log x + \gamma$ for sufficiently large $x$ ?

Or is there any other means by which we can deduce (or refute) this inequality ?

• Is the index $p$ here referring to prime numbers only? – Clement C. Jan 30 '16 at 4:01
• Yes, let me specify that in the edit. – User1 Jan 30 '16 at 10:26
• I only find $\sum_{p\le x}\log(1-1/p)$ at Martin Klazar's text here on page $53$. But then the sum is not only over $\log(p)$. Did you mean $(\log(p)-\log(p-1))$ in brackets ? Also, we would have $c_2-c_1$ as constants. – Dietrich Burde Jan 30 '16 at 12:34
• @DietrichBurde, yes, in brackets. The constant $c_{2} - c_{1}$ was evaluated as $\gamma$. – User1 Jan 30 '16 at 13:33