Order of Sum of Reciprocal of Primes

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Theorem

Let $x \ge 2$ be a real number.

We have:

$\ds \sum_{p \mathop \le x} \frac 1 p = \map \ln {\ln x} + \map \OO 1$

where:

$\ds \sum_{p \mathop \le x}$ sums over the primes less than or equal to $x$
$\OO$ is big-$\OO$ notation.


Proof

We have:

\(\ds \int_p^x \frac 1 {t \ln^2 t} \rd t\) \(=\) \(\ds \int_{\ln p}^{\ln x} \frac {e^u} {e^u u^2} \rd u\) substituting $t \mapsto e^u$
\(\ds \) \(=\) \(\ds \int_{\ln p}^{\ln x} \frac 1 {u^2} \rd u\)
\(\ds \) \(=\) \(\ds \intlimits {-\frac 1 u} {\ln p} {\ln x}\) Primitive of Power, Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds \frac 1 {\ln p} - \frac 1 {\ln x}\)

So we can write:

\(\ds \sum_{p \mathop \le x} \frac 1 p\) \(=\) \(\ds \sum_{p \mathop \le x} \paren {\frac {\ln p} p \times \frac 1 {\ln p} }\) since $p \ge 2$, $\ln p \ne 0$
\(\ds \) \(=\) \(\ds \sum_{p \mathop \le x} \frac {\ln p} p \paren {\int_p^x \frac 1 {t \ln^2 t} \rd t + \frac 1 {\ln x} }\)
\(\ds \) \(=\) \(\ds \sum_{p \mathop \le x} \frac {\ln p} p \paren {\int_p^x \frac 1 {t \ln^2 t} \rd t} + \frac 1 {\ln x} \sum_{p \mathop \le x} \frac {\ln p} p\)

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


Lemma

$\ds \sum_{p \mathop \le x} \frac {\ln p} p \paren {\int_p^x \frac 1 {t \ln^2 t} \rd t} = \int_2^x \frac 1 {t \ln^2 t} \paren {\sum_{p \mathop \le t} \frac {\ln p} p} \rd t$

$\Box$


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

$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \ln x + \map R x$

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

We then have:

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

We have, by Linear Combination of Definite Integrals:

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

Evaluating the first term:

\(\ds \int_2^x \frac 1 {t \ln t} \rd t\) \(=\) \(\ds \int_{\ln 2}^{\ln x} \frac {e^u} {e^u \ln e^u} \rd u\) substituting $t \mapsto e^u$
\(\ds \) \(=\) \(\ds \intlimits {\ln u} {\ln 2} {\ln x}\) Primitive of Reciprocal, Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds \map \ln {\ln x} - \map \ln {\ln 2}\)

We now have:

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

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

As shown in Order of Sum over Primes of $\dfrac {\ln p} p$, there exists a real constant $C > 0$ such that:

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

for $t \ge 2$.

Then:

\(\ds \size {\int_2^x \frac {\map R t} {t \ln^2 t} \rd t}\) \(\le\) \(\ds C \int_2^x \frac 1 {t \ln^2 t} \rd t\) Relative Sizes of Definite Integrals
\(\ds \) \(=\) \(\ds C \paren {\frac 1 {\ln 2} - \frac 1 {\ln x} }\) using the previous integral result with $p = 2$
\(\ds \) \(\le\) \(\ds \frac C {\ln 2}\) since $\dfrac 1 {\ln x} \ge 0$ for $x > 1$

So:

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

We also have:

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

From Logarithm is Strictly Increasing, we then have:

$\ds \frac 1 {\ln x} \le \frac 1 {\ln 2}$

for $x \ge 2$, and so:

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

for $x \ge 2$.

So:

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

Clearly:

$1 - \map \ln {\ln 2} = \map \OO 1$

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

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

giving:

$\ds \sum_{p \mathop \le x} \frac {\ln p} p = \map \ln {\ln x} + \map \OO 1$

$\blacksquare$

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