Order of Sum of Reciprocal of Primes

Theorem

 * $\ds \sum_{p \le x} \frac 1 p = \map \log {\log x} + \map \OO 1$

Proof
We have:

So we can write:

We look to interchange summation and integral in our first term.

Lemma
From Order of Sum over Primes of $\dfrac {\log p} p$, there exists a real function $R : \hointr 2 \infty \to \R$ such that:


 * $\ds \sum_{p \le x} \frac {\log p} p = \log x + \map R x$

with $\map R x = \map \OO 1$.

We then have:


 * $\ds \int_2^x \frac 1 {t \log^2 t} \paren {\sum_{p \le t} \frac {\log p} p} \rd t + \frac 1 {\log x} \sum_{p \le x} \frac {\log p} p = \int_2^x \frac {\log t + \map R t} {t \log^2 t} \rd t + 1 + \frac {\map R x} {\log x}$

We have, by Linear Combination of Definite Integrals:


 * $\ds \int_2^x \frac {\log t + \map R t} {t \log^2 t} \rd t = \int_2^x \frac 1 {t \log t} \rd t + \int_2^x \frac {\map R t} {t \log^2 t} \rd t$

Evaluating the first term:

We now have:


 * $\ds \sum_{p \le x} \frac {\log p} p = \map \log {\log x} + \paren {\int_2^x \frac {\map R t} {t \log^2 t} \rd t + 1 + \frac {\map R x} {\log x} - \map \log {\log 2} }$

We aim to show that the bracketed term is $\map \OO 1$.

Since $\map R t = \map \OO 1$, there exists a real constant $C > 0$ such that:


 * $-C \le \map R t \le C$

for $t \ge 2$.

Then:

So:


 * $\ds \int_2^x \frac {\map R t} {t \log^2 t} \rd t = \map \OO 1$

We also have:


 * $\ds \size {\frac {\map R x} {\log x} } \le \frac C {\log x}$

From Logarithm is Strictly Increasing, we then have:


 * $\ds \frac 1 {\log x} \le \frac 1 {\log 2}$

for $x \ge 2$, and so:


 * $\ds \size {\frac {\map R x} {\log x} } \le \frac C {\log 2}$

for $x \ge 2$.

So:


 * $\ds \frac {\map R x} {\log x} = \map \OO 1$

Clearly:


 * $1 - \map \log {\log 2} = \map \OO 1$

So from Sum of Big-$\OO$ Estimates, we have:


 * $\ds \int_2^x \frac {\map R t} {t \log^2 t} \rd t + 1 + \frac {\map R x} {\log x} - \map \log {\log 2} = \map \OO 1$

giving:


 * $\ds \sum_{p \le x} \frac {\log p} p = \map \log {\log x} + \map \OO 1$