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

== 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 ==
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.
=== 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:
=== Definition 4 ===
Let $\map E x$ be the error term defined by:
:$\ds \map E x = \sum_{p \mathop \le x} \frac 1 p - \ln \ln x - M$
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: 
 
=== 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.
=== 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 ====
=== 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 ====
=== 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:
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:
:$C_1 = 2 M + \dfrac {3 \ln \ln 2} 2 - \map \li 2 - \dfrac 1 {12} \paren {\dfrac 1 {n \ln n} - \dfrac 1 {2 \ln 2} } - R_2$
=== 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:
Combining these results with $(7)$ we obtain $(6)$, where:
:$\ds C_2 = M + \dfrac {\map \pi {2657} } {2657} + \ln \ln 2657 - \sum_{p \mathop \le 2657} \frac 1 p - \frac {\map \li {2657} } {2657}$
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:
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:
:$C_3 = \dfrac 2 {\ln 2} - \map \li 2 + \dfrac 1 2 \paren {\dfrac 1 {\map {\ln^2} {n - 1} } + \dfrac 1 {\ln^2 2} } + \dfrac 1 {12} \paren {-\dfrac 2 {\paren {n - 1} \map {\ln^3} {n - 1} } + \dfrac 2 {2 \ln^3 2} } + R_2$
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 $n \ge 2658$.
=== 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 $n \ge 6859948$.
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 $n > 2657$.
== Sources ==
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Category:Analytic Number Theory