Prime-Counting Function in terms of Eulerian Logarithmic Integral/Riemann Hypothesis Holds

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Theorem

Let $\map \pi x$ denote the prime-counting function of a number $x$.

Let $\map \Li x$ denote the Eulerian logarithmic integral of $x$:

$\map \Li x := \ds \int_2^x \dfrac {\d t} {\ln t}$


If the Riemann Hypothesis holds, then:

$\map \pi x = \map \Li x + \map \OO {\sqrt x \ln x}$

where $\OO$ is the big-$\OO$ notation.


Proof


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Let $x$ denote a real number.

Let $n$ denote a natural number.

Let $p$ denote a prime number.

Let $\map \li x$ denote the logarithmic integral.


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Definition 1

Let $\sequence {a_n}$ be a sequence of complex numbers.

Define the partial sum $\ds \map A x = \sum_{0 \mathop \le n \mathop \le x} a_n$.

Let $f$ be a real valued and continuously differentiable function on $\closedint y x$.

Then, Abel's summation is defined by:

$\ds \sum_{y \mathop \le n \mathop \le x} a_n \map f n = \map A x \map f x - \map A y \map f y - \int_y^x \map A u \map {f'} u \rd u$

Definition 2

Let $m$ and $n$ be natural numbers and let $f$ be a real valued and continuously differentiable function on $ [m,n] $.

Then, the Euler-Maclaurin summation formula is defined by:

$\ds \sum_{i \mathop = m}^n \map f i = \int_m^n \map f x \rd x + \frac {\map f n + \map f m} 2 + \frac {\map {f'} n - \map {f'} m} {12} + R_2$

where:

$\ds \size {R_2} \le \frac 1 {12} \int_m^n \size {\map {f^2} x} \rd x$

Since the remainder term $R_2$ only has a negligible value for these purposes, this inequality is sufficient for this proof.

Definition 3

Recall the definition of the Meissel-Mertens constant:

\(\ds M\) \(=\) \(\ds \map {\lim_{n \mathop \to \infty} } {\sum_{\substack {p \mathop \le n \\ \text {$p$ prime} } } \dfrac 1 p - \ln \ln n}\)
\(\ds \) \(\approx\) \(\ds 0.2614972127 \ldots\)


Definition 4

Let $\map E x$ be the error term defined by:

$\ds \map E x = \sum_{p \mathop \le x} \frac 1 p - \ln \ln x - M$


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It is proved that $\map E x$ changes sign infinitely often.

For $x > 1$

$\size {\map E x} < \dfrac 1 {\ln^2 x}$


Lemma 1

For $n \ge 2$:

$(1): \quad \ds \sum_{p \mathop \le n} \frac 1 p = \frac {\map \pi n} n + \int_2^n \frac {\map \pi x} {x^2} \rd x$

Proof

Let $a_n= \cases {1 & : $n = p$ \cr 0 & : $n \neq p$}$

Let $\map A x = \map \pi x$.

Let $\map f x = \dfrac 1 x$ and $ f'(x) = - \dfrac 1 {x^2}$.

Using Definition 1:

\(\ds \ds \sum_{p \mathop \le n} \frac 1 p\) \(=\) \(\ds \frac {\map \pi n} n + \int_2^n \frac {\map \pi x} {x^2} \rd x\) $y = 2$ since $p_1 = 2$

$\Box$


Lemma 2

For $n \ge 2$:

$(2): \quad \ds \sum_{k \mathop = 2}^n \sum_{p \mathop \le k} \frac 1 p = \paren {n + 1} \sum_{p \mathop \le n} \frac 1 p - \map \pi n$


Proof

Let $p_i$ denote the $i$th prime.

\(\ds \sum_{k \mathop = 2}^n \sum_{p \mathop \le k} \frac 1 p\) \(=\) \(\ds \sum_{p \mathop \le 2} \frac 1 p + \sum_{p \mathop \le 3} \frac 1 p + \sum_{p \mathop \le 4} \frac 1 p + \cdots + \sum_{p \mathop \le n} \frac 1 p\)
\(\ds \) \(=\) \(\ds \frac 1 {p_1} + \paren {\frac 1 {p_1} + \frac 1 {p_2} } + \paren {\frac 1 {p_1} + \frac 1 {p_2} } + \cdots + \paren {\frac 1 {p_1} + \frac 1 {p_2} + \frac 1 {p_3} + \cdots + \frac 1 {p_{\map \pi n} } }\)
\(\ds \) \(=\) \(\ds \paren {n - 1} \frac 1 {p_1} + \paren {n - 1 - \paren {p_2 - p_1} } \frac 1 {p_2} + \paren {n - 1 - \paren {p_3 - p_1} } \frac 1 {p_3} + \cdots + \paren {n - 1 - \paren {p_ {\map \pi n} - p_1} } \frac 1 {p_{\map \pi n} }\) $n - 1$ since $p_1 = 2$
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^{\map \pi n} \frac {n - 1 - \paren {p_i - p_1} } {p_i}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^{\map \pi n} \frac {n + 1 - {p_i} } {p_i}\) as $p_1 = 2$
\(\ds \) \(=\) \(\ds \paren {n + 1} \sum_{i \mathop = 1}^{\map \pi n} \frac 1 {p_i} - \sum_{i \mathop = 1}^{\map \pi n} 1\)
\(\ds \) \(=\) \(\ds \paren {n + 1} \sum_{p \mathop \le n} \frac 1 p - \map \pi n\)

$\Box$


Corollary

Equating $(1)$ and $(2)$:

$\ds \sum_{k \mathop = 2}^{n - 1} \sum_{p \mathop \le k} \frac 1 p = n \int_2^n \frac {\map \pi x} {x^2}$


Proof

\(\ds \sum_{k \mathop = 2}^n \sum_{p \mathop \le k} \frac 1 p\) \(=\) \(\ds \paren {n + 1} \paren {\frac {\map \pi n} n + \int_2^n \frac {\map \pi x} {x^2} \rd x} - \map \pi n\) substituting the right hand side of $(1)$ for $\ds \sum_{p \mathop \le n} \frac 1 p$ in $(2)$
\(\ds \) \(=\) \(\ds \map \pi n + n \int_2^n \frac {\map \pi x} {x^2} \rd x + \frac {\map \pi n} n + \int_2^n \frac {\map \pi x} {x^2} - \map \pi n\)
\(\ds \) \(=\) \(\ds n \int_2^n \frac {\map \pi x} {x^2} \rd x + \sum_{p \mathop \le n} \frac 1 p\)
\(\ds \sum_{k \mathop = 2}^{n - 1} \sum_{p \mathop \le k} \frac 1 p\) \(=\) \(\ds n \int_2^n \frac {\map \pi x} {x^2} \rd x\)

$\Box$

Lemma 3

For $ n \ge 2 $:

$(3): \quad \ds \sum_{k \mathop = 2}^n \sum_{p \mathop \le k} \frac 1 p = \sum_{k \mathop = 2}^n \ln \ln k + \paren {n - 1} M + \sum_{k \mathop = 2}^n \map E k$


Proof

\(\ds \sum_{k \mathop = 2}^n \sum_{p \mathop \le k} \frac 1 p\) \(=\) \(\ds \sum_{p \mathop \le 2} \frac 1 p + \sum_{p \mathop \le 3} \frac 1 p + \sum_{p \mathop \le 4} \frac 1 p + \cdots + \sum_{p \mathop \le n} \frac 1 p\)
\(\ds \) \(=\) \(\ds ( \ln \ln 2 + M + \map E 2 ) + ( \ln \ln 3 + M + \map E 3 ) + ( \ln \ln 4 + M + \map E 4 ) + \ldots + ( \ln \ln n + M + \map E n )\) Using Definition 4
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 2}^n \ln \ln k + \paren {n - 1} M + \sum_{k \mathop = 2}^n \map E k\) $n-1$ since $p_1 = 2$

$\Box$

Theorem 1

For $n \ge 2$:

$(4): \quad \ds \map \pi n - \map \li n = n \map E n + \frac {\ln \ln n} 2 - \sum_{k \mathop = 2}^{n - 1} \map E k + C_1$

where:

$-1.07193 \ldots \le C_1 \le -1.01182 \ldots$

Proof

We start by equating $(2)$ and $(3)$:

$(5): \quad \ds \paren {n + 1} \sum_{p \mathop \le n} \frac 1 p - \map \pi n = \sum_{k \mathop = 2}^n \ln \ln k + \paren {n - 1} M + \sum_{k \mathop = 2}^n \map E k$

Applying Definition 4:

$\ds \sum_{p \mathop \le n} \frac 1 p = \ln \ln n + M + \map E n$

Applying Definition 2:

\(\ds \sum_{k \mathop = 2}^n \ln \ln k\) \(=\) \(\ds n \ln \ln n - \map \li n - 2 \ln \ln 2 + \map \li 2\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \dfrac 1 2 \paren {\ln \ln n + \ln \ln 2} + \dfrac 1 {12} \paren {\dfrac 1 {n \ln n} - \dfrac 1 {2 \ln 2} } + R_2\)


where:

$\ds \size {R_2} \le \frac 1 {12} \int_2^n \size {-\frac {\ln x + 1} {x^2 \ln^2 x} } \rd x = \frac 1 {12} \paren {\frac 1 {2 \ln 2} - \frac 1 {n \ln n} }$

Combining these results with $(5)$ we obtain $(4)$, where:

$C_1 = 2 M + \dfrac {3 \ln \ln 2} 2 - \map \li 2 - \dfrac 1 {12} \paren {\dfrac 1 {n \ln n} - \dfrac 1 {2 \ln 2} } - R_2$

$\Box$


Theorem 2

If $\map \pi x - \map \li x = \OO (\sqrt x \ln x)$, then for $n > 2657$:

$(6): \quad \ds \size {\sum_{k \mathop = 2}^{n - 1} \map E k - \frac {\ln \ln n} 2 - C_1 + n C_2} \le \frac n {8 \pi} \paren {\frac {2 \paren {\ln 2657 + 2} } {\sqrt {2657} } - \frac {2 \paren {\ln n + 2} } {\sqrt n} }$

where:

$C_2 = - 0.00701 \ldots$


Proof

Dividing $(4)$ by $n$ and by applying Lemma 1:

$(7): \quad \ds \sum_{p \mathop \le n} \frac 1 p - \int_2^n \frac {\map \pi x} {x^2} \rd x - \frac {\map \li n} n = \map E n + \frac {\ln \ln n} {2 n} - \frac {\ds \sum_{k \mathop = 2}^{n - 1} \map E k} n + \frac {C_1} n$

Applying Definition 4:

$\ds \sum_{p \mathop \le n} \frac 1 p = \ln \ln n + M + \map E n$

It is proved that if:

$\map \pi x - \map \li x = \map \OO {\sqrt x \ln x}$

then for all $ x \ge 2657$:

$\size {\map \pi x - \map \li x} \le \dfrac {\sqrt x \ln x} {8 \pi}$

For this reason, we modify the second term of the left-hand side of $(7)$ as follows:

$\ds \int_2^n \frac {\map \pi x} {x^2} \rd x = \int_2^{2657} \frac {\map \pi x} {x^2} \rd x + \int_{2657}^n \frac {\map \pi x} {x^2} \rd x$

Substituting $\map \li x + \map \OO {\sqrt x \ln x}$ for $\map \pi x$ in the last term:

$\ds \int_2^n \frac {\map \pi x} {x^2} d x = \int_2^{2657} \frac {\map \pi x} {x^2} \rd x + \int_{2657}^n \frac {\map \li x} {x^2} \rd x + \int_{2657}^n \frac {\map \OO {\sqrt x \ln x} } {x^2} \rd x$

Applying Lemma 1:

$\ds \int_2^{2657} \frac {\map \pi x} {x^2} \rd x = \sum_{p \mathop \le 2657} - \frac {\map \pi {2657} } {2657}$

Integrating:

$\ds \int_{2657}^n \frac {\map \li x} {x^2} \rd x = \ln \ln n - \frac {\map \li n} n - \ln \ln 2657 + \frac {\map \li {2657} } {2657}$

and:

$\ds \int_{2657}^n \frac {\map \OO {\sqrt x \ln x} } {x^2} \rd x = \map \OO {\frac {2 \paren {\ln 2657 + 2} } {\sqrt {2657} } - \frac {2 \paren {\ln n + 2} } {\sqrt n} }$

So:

\(\ds \int_2^n \frac {\map \pi x} {x^2} \rd x\) \(=\) \(\ds \sum_{p \mathop \le 2657} \frac 1 p - \frac {\map \pi {2657} } {2657} + \ln \ln n - \frac {\map \li n} n - \ln \ln 2657\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \dfrac {\map \li {2657} } {2657} + \map \OO {\dfrac {2 \paren {\ln 2657 + 2} } {\sqrt{2657} } - \dfrac {2 \paren {\ln n + 2} } {\sqrt n} }\)


Combining these results with $(7)$ we obtain $(6)$, where:

$\ds C_2 = M + \dfrac {\map \pi {2657} } {2657} + \ln \ln 2657 - \sum_{p \mathop \le 2657} \frac 1 p - \frac {\map \li {2657} } {2657}$

$\Box$


We prove the validity of $(6)$ in a sequence of the following lemmata.


Lemma 4

For $n - 1 \ge 2$:

$(8): \quad \ds \size {\sum_{k \mathop = 2}^{n - 1} \map E k} < \map \li {n - 1} - \frac {n - 1} {\map \ln {n - 1} } + C_3$

where $3.13114 \ldots \le C_3 \le 3.92159 \ldots$


Proof

Applying Definition 4, for $n \ge 3$:

$\ds \size {\sum_{k \mathop = 2}^{n - 1} \map E k} < \sum_{k \mathop = 2}^{n - 1} \frac 1 {\ln^2 k}$

Applying Definition 2:

\(\ds \sum_{k \mathop = 2}^{n - 1} \frac 1 {\ln^2 k}\) \(=\) \(\ds \map \li {n - 1} - \frac {n - 1} {\map \ln {n - 1} } - \map \li 2 + \frac 2 {\ln 2}\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \frac 1 2 \paren {\frac 1 {\map {\ln^2} {n - 1} } + \frac 1 {\ln^2 2} } + \frac 1 {12} \paren {-\frac 2 {\paren {n - 1} \map {\ln^3} {n - 1} } + \frac 2 {2 \ln^3 2} } + R_2\)

where:

$\ds \size {R_2} \le \dfrac 1 {12} \int_2^{n - 1} \size {\frac {2 \paren {\ln x + 3} } {x^2 \ln^4 x} } \rd x = \frac 1 {12} \paren {\frac 1 {\ln^3 2} - \frac 2 {\paren {n - 1} \map {\ln^3} {n - 1} } }$

and:

$C_3 = \dfrac 2 {\ln 2} - \map \li 2 + \dfrac 1 2 \paren {\dfrac 1 {\map {\ln^2} {n - 1} } + \dfrac 1 {\ln^2 2} } + \dfrac 1 {12} \paren {-\dfrac 2 {\paren {n - 1} \map {\ln^3} {n - 1} } + \dfrac 2 {2 \ln^3 2} } + R_2$

$\Box$


In the following lemmata, we solve $(6)$ where we consider the case when the sum of error terms can be negative as well.

It is sufficient if we substitute the right hand side of $(8)$ for $\sum_{k \mathop = 2}^{n - 1} \map E k$ and replace $C_1$, $C_3$ by their bounding values, adequately for each inequality, so that the inequality certainly holds.


Lemma 5

If $\ds \sum_{k \mathop = 2}^{n - 1} \map E k > 0 $, then $(6)$ holds for $n \ge 20130$.


Proof

Let $C_1 = -1.07193$, $C_3 = 3.92159$.

Then the first inequality of $(6)$ holds for $n \ge 20130$.

Let $C_1 = -1.01182$, $C_3 = 3.13114$.

Then the second inequality of $(6)$ holds for $n \ge 2658$.

$\Box$


Lemma 6

If $\ds \sum_{k \mathop = 2}^{n - 1} \map E k \le 0$, then $(6)$ holds for $n \ge 6859948$.


Proof

Let $C_1 = -1.07193$, $C_3 = 3.13114$.

Then, the first inequality of $(6)$ holds for $n \ge 2658$.

Let $C_1 = -1.01182$, $C_3 = 3.92159$.

Then, the second inequality of $(6)$ holds for $n \ge 6859948$.

$\Box$


It can be verified numerically that if $2657 < n < 20130$, then $(6)$ holds.

Because for $n < 6859948$:

$\ds \sum_{k=2}^{n - 1} \map E k > 0$

we conclude that $(6)$ holds for $n > 2657$.

$\blacksquare$


Sources


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  • 1983: Guy Robin: Sur l'ordre maximum de la fonction somme des diviseurs (Progr. Math. Vol. 38: pp. 233 – 244)
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