# Weighted Prime Powers Truncation of the Asymptotic Expansion for the
Logarithmic Integral: Properties and Applications.

Shaun R. Deaton<EMAIL_ADDRESS>
(Date: 01/19/2021)

###### Abstract.

Given the asymptotic expansion for the logarithmic integral
$\int_{0}^{n}\frac{dt}{\ln(t)}$, obtained from repeated integration by parts
until the expansion terms reach a minimum; approaching zero. Which determines
a cut-off for the number of terms in the expansion and this truncation is a
function of $n$. By dropping the minimization constraint and introducing a new
variable $x$ for the number of expansion terms, consider the question: Where
to truncate the asymptotic expansion for the logarithmic integral to equal the
prime count function $\pi(n)$?. Although constructing this new truncation
function requires using the non-trivial zeros $\rho$ of the Riemann zeta
function. There exists a closed form approximation that does not utilize the
zeros at all. From this, a new bound is obtained on the summation
$\sum_{\rho}li(n^{\rho})$. Which is then compared to an equivalent form of the
Schoenfeld bound derived for this same summation. Resulting in a proof that
the Riemann Hypothesis is true.

Keywords. Logarithmic Integral, Lambert W Function, Riemann Zeta Function,
Riemann Hypothesis, Schoenfeld Bound.

###### 2020 Mathematics Subject Classification:

Primary 11M26 & Secondary 11A41

##  1\. Introduction

After a chance, re-introduction to the asymptotic expansion for the
logarithmic integral. A question arose: Where would the expansion have to be
truncated so that it equals the prime count function $\pi(n)$? Answering this
question, and following through on exploring others it will raise, shall be
the focus of all that follows.

Beginning with a definition for the logarithmic integral, with a case
including offset integration limits to account for the singularity, needed for
numeric calculations:

######  Definition 1.1.

$li(n):=\int_{0}^{n}\frac{dt}{\ln(t)}=li(2)+\int_{2}^{n}\frac{dt}{\ln(t)}\approx
1.04516+\int_{2}^{n}\frac{dt}{\ln(t)}$

The dependent variable $n$ can be real-valued, the notation adopted here
refers back to $\pi(n)$, where the input is generally taken as integer-valued.
Being a non-elementary integral, one method of solution is to use repeated
integration by parts to generate an asymptotic expansion. Given as follows:

######  Definition 1.2.

$\int_{0}^{n}\frac{dt}{\ln(t)}=\frac{n}{\ln(n)}\sum_{k=0}^{\lfloor
x\rfloor-1}\frac{k!}{\ln^{k}(n)}+x!\int_{0}^{n}\frac{dt}{\ln^{x+1}(t)}$

Since this expansion diverges with increasing $x$, to best approximate
$li(n)$, it’s truncated at the value of $x$ that minimizes the integral term.
Which will generally not be integer valued, requiring the floor function for
the summation to make sense; in addition to using $x!:=\Gamma(x+1)$. Although
the method of constructing the expansion forces $x$ to be integral valued,
minimization means it must be viewed as continuous when appearing in the
integral term. A notion that is then extended to the summation. When $x$ isn’t
integer valued, some authors have added a linear interpolating or fractional
term for real $x$ to reduce the truncation error.

Stieltjes in [Ref17] calculated the minimizing truncation, the first two terms
are $\tau=\tau(n)\approx\ln(n)-1/3$; all other terms are reciprocal powers of
$\ln(n)$. Stieltjes’s paper, originally in French, was reviewed in van Boven
et. al. [Ref3], where a fractional term is included, ”derived from a heuristic
reasoning about the remainder” of the expansion.

Another instance of using a fractional term occurs in the appendix of Stoll &
Demichael [Ref16], along with additional correction terms and error bounds.
For reference that information, preserving their notation, is reproduced here;
where they use $x$ as a general input variable, same as $n$ is used just
above. Also, notationally, $\log(x):=\ln(x)$ and
$L_{1}:=\log(x),L_{2}:=\log(\log(x)),\cdots,L_{j}:=\log(\log(\cdots\log(x)))$.

This summation begins to diverge at $k=\lfloor L_{1}\rfloor$ so we stop at
$k=\lfloor L_{1}-1\rfloor$. The error is almost logarithmically linear between
integer changes in $L_{1}$, but the envelope of the error is not symmetric:

$li(x)\cong\frac{x}{L_{1}}\sum_{k=0}^{\lfloor
L_{1}-1\rfloor}\dfrac{k!}{L_{1}^{k}}\enspace\Big{\\{}+1.2/\sqrt{L_{1}},-0.6/\sqrt{L_{1}}$

For significantly higher accuracy, $li(x)$ can be expressed as a finite sum
with smaller, symmetric errors as

$\begin{split}li(x)\cong&\frac{x}{L_{1}}\Big{(}\sum_{k=0}^{\lfloor
L_{1}-1\rfloor}\dfrac{k!}{L_{1}^{k}}+\frac{(L_{1}-\lfloor L_{1}\rfloor)\lfloor
L_{1}\rfloor!\enspace}{L_{1}^{\lfloor L_{1}\rfloor}}\Big{)}\\\
&-\Big{(}\frac{300L_{1}^{3/2}}{359L_{1}^{2}+L_{2}L_{1}-3L_{3}+3}\Big{)}-\Big{(}\frac{\sin(2\pi
L_{1})}{7/3L_{1}^{2}61(L_{1}-2)}\Big{)}.\end{split}$

The remaining error is estimated to be
$<\frac{23L_{2}^{2}}{(L_{1}^{2}+48L_{1})^{2}}$ for $x>113$.

Although there are small differences between all the expansions for $li(n)$
given so far; e.g. location of truncation point and when the floor is applied
to the upper summation limit. These are minor enough that one can still infer
properties relating them. The property needed here is the error bound applying
to an expansion with only the fractional term $\frac{(L_{1}-\lfloor
L_{1}\rfloor)\lfloor L_{1}\rfloor!\enspace}{L_{1}^{\lfloor L_{1}\rfloor}}$
added. Expecting that within the quoted range of error bounds from
$\frac{1}{\sqrt{L_{1}}}$ to
$\frac{23L_{2}^{2}}{(L_{1}^{2}+48L_{1})^{2}}\sim\frac{1}{L_{1}^{4}}$, is where
the error for including only this fractional term exists. Using this to guide
a computational investigation into the approximation error to $li(n)$ incurred
by using and expansion of the form:

######  Definition 1.3.

$li(n;x):=\frac{n}{\ln(n)}\sum_{k=0}^{\lfloor
x\rfloor-1}\frac{k!}{\ln^{k}(n)}+\frac{(x-\lfloor x\rfloor)\lfloor
x\rfloor!\enspace n}{\ln^{\lfloor x\rfloor+1}(n)},\forall n>1$, and $x\geq 1$.

The $x$ in definition (1.3) can take values independent of $n$. But to
approximate $li(n)$ by using the Stieltjes truncation gives $li(n;\tau)$, that
now fixes $x$ as a function of $n$. Computation using the pracma package in R
to compute $li(n)$, shows that $li(n;\tau)$ has an error bound of at least,
$|li(n)-li(n;\tau)|<\frac{0.5}{\ln^{2}(n)},\forall n\geq e^{4/3}$. Where the
restriction on $n$ is to keep the truncation $\tau$ from becoming negative.
From here on, this error will be assumed implicitly anytime $li(n)$ is
replaced by the asymptotic expansion with a linear interpolation term.

Definition (1.3), must be modified to restore the connection with the integral
term in definition (1.2). As it’s the summation’s source, determining both
form and number of terms by imposing a minimization constraint. It essentially
builds the expansion, except for the fractional term, and is then dropped;
being effectively zero at it’s minimum. To sufficiently generalize this
situation to be of use later, the main definition going forward is:

######  Definition 1.4.

$li(n;x):=\frac{n}{\ln(n)}\sum_{k=0}^{\lfloor
x\rfloor-1}\frac{k!}{\ln^{k}(n)}+\frac{(x-\lfloor x\rfloor)\lfloor
x\rfloor!\enspace n}{\ln^{\lfloor x\rfloor+1}(n)},\enspace\forall n>1$ &
$x\geq 1$, s.t. for some fixed $V\geq 0$ have that $x$ is defined by
$V:=x!\int_{0}^{n}\frac{dt}{\ln^{x+1}(t)}$.

For a fixed $n$, choosing $x$ to minimize $V$ gives the $li(n;\tau)$ case,
making $x$ solely a function of $n$. Definition (1.4) works by using $n$ and
$V$ to determine $x$, where the source integral term is not part of the
expansion as it is in definition (1.2). Now, even though $V\not\approx 0$,
it’s not added to the expansion, it only determines the number of terms. Using
this more general view of the asymptotic expansion for the logarithmic
integral. Treat the interpolation formula of definition (1.4) as defining a
continuous ‘asymptotic expansion variable’ $x$, with a constraint condition.
Ultimately $V$ will be a function of $n$, so it’s not included in the notation
$li(n;x)$.

The fractional term allows for the asymptotic expansion function $li(n;x)$ to
be jointly continuous. Technically, this is only the logarithmic integral for
one specific truncation point $x=\tau$, so that referring to it generally as
such is an abuse of the name. Done, as always, with the intention to maintain
simplicity.

The layout of this paper is as follows. Section 2 will define, prove the
existence of, and examine the exact prime truncation. Section 3 will do the
same for the average prime truncation. Section 4 will derive the asymptotic
form approximating an average of the prime truncation function, and then a
specific form for large values of n. Finally, section 5 closes with an
application of these new truncation functions to bounds on the summation over
the zeta functions non-trivial zeros, and then compare them to Schoenfeld’s
bound. Finally showing that the Riemann Hypothesis is true.

##  2\. Exact Prime Truncation

### 2.1.

Using definition (1.4) to re-express the question posed at the beginning of
the introduction: Where to truncate $li(n;x)$ to equal $\pi(n)$? Is expressed
in the following definition;

######  Definition 2.1.

Exact Prime Truncation Function: Let $x_{\pi}(n)$ be such that,
$li(n;x_{\pi}(n))=\pi(n).$

Note that no value for $V$ has been given, leaving the constraint condition of
definition (2.1) unfulfilled. Part of the investigation here is to find a
suitable $V$ to constrain the integral term to give $x_{\pi}(n)$. The idea
used to find such a value is to compute $x_{\pi}(n)$ directly, and then
determine the appropriate $V$. But, this will not be solved till the next
section, and apply to a variant of $x_{\pi}(n)$. From that point it will be
shown how to construct a $V$ for definition (2.1) to also fulfill the
requirements of definition (1.4). The task at hand is to determine what
constraint on the source integral’s value achieves the desired truncation of
the $li(n;x)$ function.

Using the programming language R, and prime counting function from the
RcppAlgos package, the question posed above was computationally examined.
Solving the equation in definition (2.1) for $x$, gives the exact prime
truncation point $x_{\pi}(n)$, to obtain the prime count from the asymptotic
expansion function of the logarithmic integral. The result has strong local
variations in value, that can be viewed as an average variation about a
steadily increasing function; shown in figure (1). Most variation about this
average is in fact determined by the zeta-zeros of Riemann’s exact formula, as
it is the same information contained within $\pi(n)$. Something to be verified
later on; in addition to clarifying what is meant by average in this case.
Now, presenting graphs of the truncation value $x_{\pi}=x_{\pi}(n)$ that
satisfies definition (2.1).

(a) Prime Truncation $x_{\pi}$ of $li(n),\ \forall n\in[2,100]$.

(b) Prime Truncation $x_{\pi}$ of $li(n),\ \forall n\in[2,1000]$.

Figure 1. Prime Truncation $x_{\pi}$ of $li(n)$. Figure 2. Prime Truncation
$x_{\pi}$ of $li(n)$ such that $10^{12}\leq n\leq 10^{13}$.

The overall shape, seen in figures (1) & (2) seems to possibly be derivable
analytically, in part since it is fit well by the initial estimates
$[\ln(n)-1/3]^{1/2}$ and $1+\log_{10}(n)/2$. Looking at figure
(LABEL:PRIMETRUNC2TO100), can see that these functions miss the prime
truncation function. Yet rather quickly assume something similar to an average
for $n>100$, as seen in figure (LABEL:PRIMETRUNC2TO1000). The vertical line
occurs at $n=11$, while the horizontal line is at a truncation value of $1$;
signifying that the asymptotic expansion is undefined below this line, as
cannot have less than one term, but is valid $\forall n\geq 11$. Altering the
asymptotic expansion’s definition to allow the higher order terms to be
subtractive to the initial term when $2\leq n<11$ provides a nice fit, since
that is what’s shown in figure (LABEL:PRIMETRUNC2TO100); a bit more detail is
given in second remark of Theorem 2.4. More investigation shows that
$[\ln(n)-1/3]^{1/2}$ begins to fall short at about n $=10^{6}$, only becoming
a worse fit for larger $n$; as seen in figure (2). After more examination,
found $1+\log_{10}(n)/2$ to be a better fit up to about $n=10^{12}$; though it
seems to surpass the mark as $n\rightarrow\infty$. Later on, will show why
this function is so close to the average, and why it appears to exceed it.

Another feature to note is that the magnitude of the variation about what
would be the average of the prime truncation is rather small to start, around
$1/2$ and seems to decrease for $n>10^{12}$ to about $1/4$. The variations
themselves are a direct consequence of the jumps and plateaus in the $\pi(n)$
function. Additionally, on a computational note, figure (1) displays all
integer values in the specified ranges, while figure (2) only samples $301$
values.

Even though can compute such a solution for the equation in definition (2.1)
up to $n\leq 10^{16}$, a proof is need $\for alln\geq 2$. Utilizing an
analytical approach, such as that used by Axler [Ref1] and Dusart [Ref6] to
prove various bounding inequalities for $\pi(n)$ shows the prime truncation
solution always exists. In fact, these inequalities use specific truncations
or alterations of the logarithmic integral’s asymptotic expansion. Which
requires a specified range of applicability for the inequality, tying the
truncation variable $x$ of the expansion to the input variable $n$. While
generalizing this connection for such inequalities, and referencing Dusart’s
work in [Ref7] on similar inequalities. Axler in [Ref1] packaged this idea
into a functional form $g_{1}(x)$, taking integral values. E.g. Axler
referenced Dusart obtaining the results, $g_{1}(2)=599$ & $g_{1}(3)=88,783$.
Meaning that a 2-term expansion bounds $\pi(599)$, while a 3-term bounds
$\pi(88,783)$.

Some notational differences to note from the referenced works. Use of the
variables $n$ and $x$ is reversed, and the truncation $x_{\pi}$ in definition
(2.1) is related to the inverse of $g_{1}(x)$, denoted here simply as $g(n)$;
re-defined by the following:

######  Definition 2.2.

Let $g(n)$ be such that,
$\pi(n)>li(n;g(n))=\frac{n}{\ln(n)}\sum_{k=0}^{g(n)-1}\dfrac{k!}{\ln^{k}(n)}.$

Although, a few values for $g(n)$ where given in relation to the inequalities
discussed, no general form was shown or conjectured. Compared to the
referenced examples, have $g(599)=2$ & $g(88,783)=3$; the actual computed
values using the prime truncation are $x_{\pi}(599)=2.15$ &
$x_{\pi}(88,783)=3.0$. Bringing Axler’s idea full circle; the same argument
implies there exists a $G(n)$, essentially $1+g(n)$ for next tightest bound,
that reverses the inequality in definition (2.2):

######  Definition 2.3.

Let $G(n)$ be such that,
$\pi(n)<li(n;G(n))=\frac{n}{\ln(n)}\sum_{k=0}^{G(n)-1}\dfrac{k!}{\ln^{k}(n)}.$

Taken together, there must be an integral valued $g_{\pi}(n)$ such that:

(2.1) $li(n;g(n))\leq\pi(n)\approx li(n;g_{\pi}(n))\leq li(n;G(n)).$

Allowing $g_{\pi}(n)$ to be real-valued is the prime truncation $x_{\pi}(n)$
of the logarithmic integral; and the above equivalence can then be set to an
equality. Note that since the exact solution will pass through integral values
the inequalities have been change to include an equality condition. The linear
interpolation term must also be added; since it is zero at integral values the
original definitions are recovered. After searching all available references,
Axler [Ref1] was the most explicit mention that is the closest to this idea.
Proving the existence and uniqueness of such a function will use the basic
method underlying the inequalities above.

######  Theorem 2.4.

(Prime Truncation) There exists, $\forall n\geq 11$, a unique continuous real-
valued function $x_{\pi}(n)=g_{\pi}(n)$ such that;

(2.2) $\pi(n)=li(n;g_{\pi}(n))$

###### Proof.

By the prime number theorem $\pi(n)=li(n)+error(n)$; where the error is
generally negative, except for Skewes regions where it must be positive.
Normally, have no way to adjust $li(n)$, but introducing $li(n;x)$ allows for
a fix. For some value of $x$ have that:

(2.3) $\pi(n)=li(n)+error(n)=li(n;x)$

By definition (2.1) $x=x_{\pi}$, which must be shown to exist and be unique.
Where the idea is that by controlling the truncation point for the asymptotic
expansion of $li(n)$ the target function $\pi(n)$ can be hit.

Consider the asymptotic expansion of $li(n)$ as defining a multiplicative
correction factor $C(n,x)$ for $\frac{n}{\ln(n)}$. A view that comes from the
asymptotic expansion function:

(2.4)
$li(n;x)=\frac{n}{\ln(n)}\sum_{k=0}^{x-1}\dfrac{k!}{\ln^{k}(n)}=\frac{n}{\ln(n)}(1+\frac{1}{\ln(n)}+\frac{2}{\ln^{2}(n)}+...)=\frac{n}{\ln(n)}C(n;x)$

(2.5) $C(n;x)=\sum_{k=0}^{x-1}\dfrac{k!}{\ln^{k}(n)}$

The summation is Stieltjes form, used definition (1.4), denoted here by
removing the floor on the upper summation limit; while also suppressing the
fractional term, assuming it’s absorbed into the $C(n;x)$ function. Allowing
for continuous adjustment of $\frac{n}{\ln(n)}$ by $C(n;x)$ through the
asymptotic truncation variable $x$.

Have that $\frac{n}{\ln(n)}\leq\pi(n),\forall n\geq 11$, along with the fact
that the correction factor $C(n,x)$ is a continuous monotonically increasing
function $\geq 1$; since it’s a sum of positive terms. $C(n;x)$ also
approaches infinity as $x$ does, because the asymptotic expansion is
divergent. Giving a continuous range; $1\leq C(n,x)\leq\infty$. As
$\frac{n}{\ln(n)}$ is a continuous monotonically increasing function, then so
is $\frac{n}{\ln(n)}C(n;x)$.

For most $2\leq n<11$, it would require $C(n,x)<1$, which cannot happen with
existing definition. When $n\approx 7$ again have $\frac{n}{\ln(n)}<\pi(n)$.
Implying the need for a more complicated $C(n,x)$ when $2\leq n<11$; see the
second remark following this proof for more elaboration.

Then $\forall n\geq 11$, since $\frac{n}{\ln(n)}\leq\pi(n)$, the correction
$C(n;x)$ provides a continuous monotonic increase for $\frac{n}{\ln(n)}$
allowing $li(n;x)$ to meet then surpass $\pi(n)$; as the truncation variable
$x$ increases. When $x=\tau$, the Stieltjes truncation, essentially have an
equivalent form of the prime number theorem; $\pi(n)=li(n)+error(n)\approx
li(n;\tau)+error(n)$. Since generally $error(n)<0$, the usual truncation
$x=\tau$ must be cut short before $li(n;x)$ exceeds $\pi(n)$.

Since $\pi(n)$ is a piece-wise continuous, monotonically increasing step
function. There must be a crossing point between $li(n;x)$ and $\pi(n)$ for
some value of $x$, call it $x_{\pi}$; and by montonicity of both functions it
must also be unique in the interval $[2,n]$. Embodied by the equality
$C(n,x_{\pi})=\pi(n)\frac{\ln(n)}{n}$.

Therefore for all $n\geq 11$, a unique $x_{\pi}(n)=g_{\pi}(n)$ exists having
the property that; $\pi(n)=li(n;g_{\pi}(n))$.

∎

######  Remark 2.5.

For Skewes regions where $li(n)>\pi(n)$; the theorem still holds, since have
infinitely large correction factors $C(n,x)$ to choose from, these finite
spikes can always be hit. Requiring a truncation past the usual Stieltjes
point $\tau$, and into divergent territory of the asymptotic expansion.

######  Remark 2.6.

The Stieltjes Interpolation can be altered to allow for $0<C(n,x)<1$, when
$2\leq n<11$; which actually works well in practice, again as it’s how the
prime truncation $x_{\pi}$ is obtained for the figure (LABEL:PRIMETRUNC2TO100)
plot in this range. I.e. subtracting terms of the form
$\frac{k!}{\ln^{k}(n)},\ \forall k\geq 1$, from the initial $k=0$ term of $1$;
representing a decrease in $\frac{n}{\ln(n)}$.

######  Remark 2.7.

The two notations $x_{\pi}$ and $g_{\pi}$ occur because $x$ was used to
represent a specific truncation point for $li(n;x)$, before the work in Axler
[Ref1] and Dusart [Ref7] was known. Where the $g$ notation from Axler and
Dusart references a function that controls the number of terms in the
asymptotic expansion for $li(n)$, that when allowed to take real values,
coincides with $x_{\pi}$. From this point the $g$ notation will be used.

Although the prime truncation exists and is unique, solving for it requires
knowing $\pi(n)$, or equivalently Riemann’s explicit formula and all the zeros
of the zeta function; regardless of where they may lie. Making this function
of interest, but not necessarily useful. Instead the remainder of this paper
will focus on finding a closed form solution for it’s average behavior. Along
with some applications.

##  3\. Sum of Weighted Prime Powers & Its Relation to $g_{\pi}(n)$

### 3.1.

Riemann [Ref13] constructed an explicit formula for the prime count function
$\pi(n)$, that was later proven by Mangoldt [Ref11], and others. Since then, a
number of authors have studied and re-derived the formula; reproduced here
from Dittrich [Ref5]. Having a slight symbol change of $Li$ to $li$ in the RHS
summation; the source material defines $Li$ from $0$ to $n$; which makes
better notational sense compared with $li$; i.e. an uppercase letter is larger
than lowercase.

(3.1)
$\sum_{r=1}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}=\int_{0}^{n}\frac{dt}{\ln(t)}-\sum_{\rho}^{zeta-
zeros}li(n^{\rho})+\int_{n}^{\infty}\frac{dt}{(t^{2}-1)t\ln(t)}-\ln(2)$

Where the summation over non-trivial zeros of the zeta function must be
addressed. Questions concerning the Riemann Hypothesis (RH) now enter,
regarding the zeros positioning in the complex plane. To be clear, the RH is
not assumed. Everything that follows relies only on the validity of equation
(3.1) and the fact that the non-trivial zeros are complex valued. Often,
M$\ddot{o}$bius inversion is performed on equation (3.1) to obtain the final
form used in practice, Kotnik [Ref10]. In this case the inversion is not
needed, instead will work from equation (3.1). Also, note that the sum of
weighted prime counts evaluated at the roots of $n$ will be referred to as the
sum of weighted prime powers. Highlighting the terms source, rather than the
term itself.

### 3.2.

For now, drop all but the logarithmic integral from the RHS of equation (3.1),
and work with the equivalence:

(3.2)
$\sum_{r=1}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}\approx\int_{0}^{n}\frac{dt}{\ln(t)}$

Then separate out the first term of the summation on LHS of equation (3.2),
while replacing RHS with its asymptotic expansion from definition (1.4), again
suppressing the floor on the upper summation limit, denoting that fractional
term is part of the sum. With the offset source integral added back to
illustrate its connection to the sum of prime power counts.

(3.3)
$\pi(n)+\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}\approx\frac{n}{\ln(n)}\sum_{k=0}^{x-1}\dfrac{k!}{\ln^{k}(n)}+x!\int_{2}^{n}\frac{dt}{\ln^{x+1}(t)}+li(2)$

Essentially, by equating the first terms on each side and then the second
terms. Have that the weighted sum of prime powers determines the source
integral term, thus giving a truncation value when solving for $x$. Providing
an idea on what the missing constraint in definition (2.1) should be. Now, by
construction, the remaining expansion in this case is equivalent to $\pi(n)$.
Requiring the next step of solving for $x$ in:

(3.4)
$\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}=x!\int_{2}^{n}\frac{dt}{\ln^{x+1}(t)}+li(2)$

This method does not seek to minimize the integral term, but set it to a
specific value, correcting for the over-count error added to $\pi(n)$.

The LHS of equation (3.4) is the sum weighted prime power counts, and from
this point on will refer to only powers $>1$. While the RHS of equation (3.4)
will require multiple levels of approximation; e.g. moving to the asymptotic
form of the integral and then approximating the inverse factorial. First, a
computational investigation into the exact solution to equation (3.4) will be
presented before delving into the approximating solution.

Using the prime count function in R to calculate the LHS of equation (3.4),
and numeric integration for the RHS, then searching for $x$ values satisfying
the equality. Produces the following series of graphs shown in figures (3),
(4), & (5). These solutions are denoted by $\bar{g}_{\pi}(n)$, which will be
referred to as the average prime truncation.

Figure 3. Prime $g_{\pi}(n)$ & Average Prime $\bar{g}_{\pi}(n)$ Truncation
Functions; $2\leq n\leq 100$. Figure 4. Prime $g_{\pi}(n)$ & Average Prime
$\bar{g}_{\pi}(n)$ Truncation Functions; $2\leq n\leq 10^{4}$. Figure 5. Prime
$g_{\pi}(n)$ & Average Prime $\bar{g}_{\pi}(n)$ Truncation Functions;
$10^{12}\leq n\leq 10^{13}$.

Notice how the solution follows closely what one would think is an averaging
value of the prime truncation function. Here the basic notion of average for a
continuous function is extended to a windowed or running average; defined by
$average(f(x)):=\frac{1}{b_{x}-a_{x}}\int_{b_{x}}^{a_{x}}f(t)dt$, where the
integration limits move with $x$, creating a window $[a_{x},b_{x}]$. In this
case, e.g. a possible window would be $[x-1/2,x+1/2]$.

It does start a bit off, likely for two main reasons; its origin is from an
equivalence after dropping terms from Riemann’s exact formula equation (3.1),
and it relies on prime powers that take time to accumulate. Some computational
errors do exist, mostly from numeric integration; e.g. “probably divergent”
warnings and jumps in the minimum error. Mostly they are accurate, since extra
code was added to check for the proper values; though it did increase run-
time. Figure (3) plots all integral values in the given range, while figure
(4) plots $1000$ values, and figure (5) plots $301$ values; and is just a
close-up of the plot in figure (2).

This computationally verifies that the weighted prime powers determine an
approximate average of the prime truncation function. Doing so by not
minimizing the integral error term, as with the usual approach when computing
$li(n)$, instead setting it equal to the error added to $\pi(n)$ by the over-
count introduced from the weighted prime powers. Providing a value for the
constraint $V$, that is determined by $n$.

Next, proving this computational verification by looking to Riemann’s equation
(3.1). For clarity, will generally refer to $\bar{g}_{\pi}(n)$ as ’the
average’ when really it’s just ’an average’, one of possibly many
approximating $average(g_{\pi}(n)))$. Only set apart by being a consequence of
the relation above involving the sum over weighted prime powers. Again, the
averaging is meant in the context of smoothing out the jagged variation in the
prime truncation function $g_{\pi}(n)$.

######  Definition 3.1.

(Average Prime Truncation Function) Let $x=\bar{g}_{\pi}(n)$ solve equation
(3.4). I.e. $\bar{g}_{\pi}(n)$ is the truncation determined by the weighted
sum of prime power counts as the constraint $V$ on the integral term in
definition (1.4). So that;

(3.5)
$\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}=\bar{g}_{\pi}(n)!\int_{2}^{n}\frac{dt}{\ln^{\bar{g}_{\pi}(n)+1}(t)}+li(2)$

Next, to prove that $\bar{g}_{\pi}(n)$ behaves as it was computationally shown
to in the above figures. While also relating back to $g_{\pi}(n)$, and
explaining its variations seen in the figures up to this point. Moreover, it
better clarifies the averaging behavior of the function.

######  Theorem 3.2.

(Average Prime Truncation) The function $\bar{g}_{\pi}(n)$ provides an initial
approximation to the average behavior of ${g}_{\pi}(n)$, requiring no
information from the non-trivial zeta zeros.

###### Proof.

Consider that any truncation can be viewed as splitting $li(n)$ into two
functions at the truncation point $x$, a head and a tail such that;
$li(n)\approx li(n;x)+[li(n;\tau)-li(n;x)]$. Using the truncation implied by
equation (3.5), and letting $x=\bar{g}_{\pi}(n)$, then substituting into
Riemmann’s equation (3.1).

(3.6)
$\begin{split}\pi(n)+\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}&\approx
li(n;\bar{g}_{\pi}(n))+[li(n;\tau)-li(n;\bar{g}_{\pi}(n))]\\\
&-\sum_{\rho}^{zeta-
zeros}li(n^{\rho})+\int_{n}^{\infty}\frac{dt}{(t^{2}-1)t\ln(t)}-\ln(2)\end{split}$

Since, by construction, the prime powers sum on the LHS is equal to the tail
function in square brackets on the RHS, they cancel. This tail is the ”error”
determined by the source integral, which was set to the sum of weighted prime
power counts. Leaving,

(3.7) $\pi(n)-li(n;\bar{g}_{\pi}(n))\approx-\sum_{\rho}^{zeta-
zeros}li(n^{\rho})+\int_{n}^{\infty}\frac{dt}{(t^{2}-1)t\ln(t)}-\ln(2)$

As the constant and integral terms on the RHS provide diminishing and little
influence over the zeros summation, the difference on the LHS must be
predominantly controlled by the summation over non-trivial zeta function
zeros. Whose oscillatory behavior is well known, shown specifically by Stoll &
Demichael [Ref16] to be expressible as a summation in terms of cosines; or
sines. When $n$ is considered to be a continuous variable it means that
occasionally the LHS difference must be zero, giving an exact solution; the
prime truncation from Theorem 2.4. Even though the integral and constant terms
will give a small shift to where this zero occurs, it still will occur. In
general the RHS summation term’s oscillation about $0$ both bounds and
ultimately controls, the LHS difference.

By using the first few zeta-zeros can move some summation terms $li(n^{\rho})$
to modify the constraint
$V=\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}$. Now as more
zeros are used, have a sequence of constraints and their related truncations,
that move from $\bar{g}_{\pi}$ to $g_{\pi}$. So, $\bar{g}_{\pi}$ can be
described as a baseline, or initial term, subject to corrections induced by
$li(n^{\rho})$.

Therefore $\bar{g}_{\pi}(n)$ must be an approximation to the average behavior
of $g_{\pi}(n)$, determined by equation (3.5); requiring no non-trivial zeta
zeros.

∎

######  Remark 3.3.

The oscillation of the zeta zeros summation about zero requires that the
average prime truncation $\bar{g}_{\pi}(n)$ periodically satisfy the
definition of the prime truncation $g_{\pi}(n)$, and differ from it by an
amount determined by the zeta zeros summation. This implies that given
$g_{\pi}$ and then removing the oscillations, have $\bar{g}_{\pi}$ left over.
From this; and with the aid of figures (3), (4), & (5); it can been seen that
the use of ”average”, as described, is justified. Though possibly baseline or
initial term would be better.

Equation (3.5) provides a path towards solving for the average behavior of the
prime truncation function. Showing that the truncation point determined by
using the proper constraint, $V$ as the sum of weighted prime powers,
corresponds to ‘an averaging’ of the prime truncation function. Additionally,
the proof shows, in a reverse of the direction used above, that a bound on the
LHS difference would also apply to the summation over the zeta-zeros. After a
final closed form asymptotic solution for $\bar{g}_{\pi}(n)$ is obtained, will
return to this line of inquiry.

As the proof and remark show, using
$V=\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}+\sum_{\rho}^{zeta-
zeros}li(n^{\rho})+\int_{n}^{\infty}\frac{dt}{(t^{2}-1)t\ln(t)}-\ln(2)$ to
determine a truncation point for $li(n;x)$, gives $x=g_{\pi}(n)$. Fulfilling
the constraint lacking in definition (2.1). Since the zeros of the zeta
function are not needed to calculate the constraint to obtain
$\bar{g}_{\pi}(n)$, this average will be the focus.

### 3.3.

Briefly looking a little more into how well $li(n;\bar{g}_{\pi}(n)$
approximates $\pi(n)$. Which is largely because $li(n;\bar{g}_{\pi}(n)$
closely approximates Riemann’s correction $R(n)$ for $li(n)$, defined next.
Normally, the M$\ddot{o}$bius inversion of Riemann’s explicit equation (3.1),
in part, produces a prime power correction function:

######  Definition 3.4.

(3.8)
$R(n)=\sum_{r=1}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\mu(r)}{r}li(n^{1/r})$

Whose extra terms work to correct the over count in $li(n)$ from the weighted
prime powers. The same goal as $li(n;\bar{g}_{\pi}(n))$. Except $R(n)$
requires more computation on at least two levels; have many more terms to
compute with all the instances of $li(n^{1/r})$, and the M$\ddot{o}$bius
function’s cost.

A further consequence from the proof of Theorem 3.2, is that the function
$li(n;\bar{g}_{\pi}(n))$ behaves a lot like $R(n)$. Below, in figures (6),
(7), & (8), are a few graphs to show their relative behavior. Using the exact
count $\pi(10^{27})$ from the On-Line Encyclopedia of Integer Sequences OEIS:
A006880, to compare with the truncation $li(10^{27};\bar{g}_{\pi}(10^{27}))$,
giving a relative error of $3.81\cdot 10^{-13}$ %. Compared to $li(10^{27})$
itself which has relative error of $1.64$ % to $\pi(10^{27})$, it’s a much
better estimate with less calculation. While $R(10^{27})$ has relative error
$1.05\cdot 10^{-13}$ %, nearly four times better than the truncated
$li(10^{27})$ function, $li(10^{27};\bar{g}_{\pi}(10^{27}))$. This only adds
to the evidence that they are approximately the same function.

As expected, $li(n;\bar{g}_{\pi}(n))$, starts a little off, before it better
approximates $\pi(n)$; following from the same behavior of $\bar{g}_{\pi}(n)$.
When comparing $R(n)$ and the truncated function $li(n;\bar{g}_{\pi}(n))$,
they will generally alternate on which is a better fit to $\pi(n)$. Looking at
figure (7), shows $li(n;\bar{g}_{\pi}(n))$ a better fit. While figure (8),
shows that $R(n)$ is better; though not as good as $li(n;\bar{g}_{\pi}(n))$
was in the previous range, $R(n)$ can still achieve the same level of accuracy
depending on chosen range examined. Overall will state, without proof, that
$R(n)$ is the better approximation to $\pi(n)$; precisely how much so is the
question.

Figure 6. $\pi(n)$ vs. $R(n)$ vs. $li(n;\bar{g}_{\pi})\equiv
li(n;x\\_Ave);\forall n\in[2,150]$ Figure 7. $\pi(n)$ vs. $R(n)$ vs.
$li(n;\bar{g}_{\pi})\equiv li(n;x\\_Ave);\forall n\in[10^{10},10^{10}+1000]$
Figure 8. $\pi(n)$ vs. $R(n)$ vs. $li(n;\bar{g}_{\pi})\equiv
li(n;x\\_Ave);\forall n\in[10^{11},10^{11}+1000]$

##  4\. Asymptotic Solution(s) To
$\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}=\bar{g}_{\pi}(n)!\int_{0}^{n}\frac{dt}{\ln^{\bar{g}_{\pi}(n)+1}(t)}$

### 4.1.

The difficulties that arise when solving equation (3.4) are the following: 1)
The RHS is a non-elementary anti-derivative. 2) Inverting the factorial
function. 3) On LHS, require a means to generally calculate $\pi(n)$. 4)
Obtaining a non-asymptotic solution for $x$ solely in terms of $n$; valid for
all $n\geq 2$.

Focusing on the third issue for the moment. Up to this point have relied on
sheer computational capacity to obtain $\pi(n)$ for up to $n=2^{53}<10^{16}$;
currently not much better can be done, at least in a reasonable amount of time
for the basic desktop PC tasked here. A Dell Optiplex 7040, with an I7-6700
cpu @3.4GHz and 16GB DDR4 RAM. Using a version the prime number theorem is the
only recourse; falling again to the question of what degree of approximation.
For the $\frac{n}{ln(n)}$ route, can bound via the asymptotic expansion of
$li(n)$; as described above. Instead, using Riemann’s equation (3.1), or its
M$\ddot{o}$bius inversion is a possibility, but would still have to mine
infinitely many, or a great many, zeros of the zeta function.

A reasonable solution is to use the first few terms of $R(n)$. Normally, only
the first two terms are used, since higher order terms are dominated by these
two as well as the zeta-zero corrections. Still, to have good behavior for
small $n$ as well, will use first five terms; i.e. up to first positive
correction, occurring for $r=6$ in equation (3.8). Now, on to the other
issues.

First, the anti-derivative can be replaced by its asymptotic approximation.
Which is generally less than the integral for small $n$. With an error on the
order of $O((x+1)!\ n/\ln^{x+2}(n))$; where $x$ is dependent on $n$.

Because the overall goal is to obtain an asymptotic form for the average prime
truncation function, this approximation is valid. Although, its worthwhile to
consider introducing a temporary multiplicative correction factor to remove
any approximation. Defined by the ratio; where the tilde on $\tilde{\sigma}$
signifies its correcting for an integral:

(4.1)
$\tilde{\sigma}(n,x)=\frac{li(2)+\int_{2}^{n}\frac{dt}{\ln^{x+1}(t)}}{\frac{n}{\ln^{x+1}(n)}}$

Implying a sigmoid-type correction , approaching $1$ as $n\rightarrow\infty$.
Where sigmoid-type is meant to describe a generally monotonic function having
two horizontal asymptotes. In this case, since concerned with increasing $n$,
only require it to have one asymptote. This correction applies to small $n$,
as it will essentially be set to $1$ for large $n$.

Noting that must input both an $n$ and its corresponding $x$ value to
accurately check this function. Having to calculate the truncation value $x$,
for each $n$, increases the complexity of the investigation into the precise
asymptotic behavior of the integral in equation (3.5). Which is then further
complicated by being tied to the prime count, causing this integral and its
asymptotic form to fluctuate similarly as the other functions seen so far.

Approximating this ratio by a sigmoidal function will allow its product with
the asymptotic form to fit the integral over a larger range; i.e. small $n$.
With the correction dying off as the asymptotic equivalence is reached. The
idea, only to undo some error from forced simplification of the integral for a
more precise relationship.

Such a correction would be nice to have on its own, but introduced at this
stage will over-complicate the solution. As it interferes with the method used
to deal with both instances of the variable $x$, in the factorial and the
exponent in the denominator

###  4.2. For General and Large $\mathbf{n}$

Next will need $W_{-1}(t)$ the lower branch of the Lambert $W$ function. Where
$W(t)$ is the compositional inverse for the function $te^{t}$; making $W(t)$
something reminiscent of a logarithm. It takes two values on the negative
real-axis, so is not actually a function, but by splitting into two branches,
usually denoted $W_{0}$ and $W_{-1}$, single-valued functions are defined. The
lower branch, $W_{-1}$, with domain $[-1/e,0)$; and tends back towards the
y-axis from $t=-1/e$, taking only negative values.

Finally, an answer to the question posed at the introduction.

######  Theorem 4.1.

(Asymptotic Form of the Average Prime Truncation) Let $D_{n}$ be the density
obtained from dividing the sum of weighted prime powers by $n$, then;

(4.2)
$\bar{g}_{\pi}(n)\sim\frac{\ln(D_{n})}{W_{-1}\Big{[}\frac{\ln(D_{n})}{e\ln(n)}\Big{]}}\enspace\forall
n\geq 9$

###### Proof.

Starting with equation (3.5), then moving to the asymptotic form of the
integral $x!\int_{0}^{n}\frac{dt}{\ln^{x+1}(t)}\sim\frac{x!\ n}{\ln^{x+1}(n)}$
and dividing through by $n$; to obtain a density. Defined by the following;

(4.3)
$D_{n}:=\frac{1}{n}\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}$

Need to solve for $x$ in;

(4.4) $D_{n}=\frac{x!}{\ln^{x+1}(n)}$

Inverting the factorial function is only part of what is needed. The
denominator’s exponent $(x+1)$ must also be dealt with. From Borwein et al
[Ref2], a collection of results on the gamma function, showed that in Mumma
[Ref12]:

(4.5) $\lim_{x\rightarrow\infty}\frac{x!^{1/x}}{x}=\frac{1}{e}$

Additionally, Borwein et al [Ref2] showed that the gamma function can be
inverted via the Lambert $W$ function. Interesting, though not used here to
directly solve for the factorial, the $W$ function still appears as a means of
solution. Next, take the $x^{th}$ root of equation (4.4) then apply relation
(4.5):

(4.6)
$D_{n}^{1/x}=\Big{[}\frac{x!}{\ln^{x+1}(n)}\Big{]}^{1/x}\approx\frac{x}{e\ln^{1+1/x}(n)}$

Since $x$ increases with $n$, for large enough $n$, can assume $1/x\rightarrow
0$; at least much smaller than $1$. Gives a simplified form of equation (4.6)
as:

(4.7) $D_{n}\approx\Big{[}\frac{x}{e\ln(n)}\Big{]}^{x}$

Using a particular case of a formula in Edwards [Ref8], after a substitution
and rearrangement, provides the general solution, using the Lambert $W$
function’s lower branch, is given by:

(4.8) $x=\frac{\ln(D_{n})}{W_{-1}\Big{[}\frac{\ln(D_{n})}{e\ln(n)}\Big{]}}$

The details of the derivation are given in this paper’s appendix A1.
Therefore, have that equation (4.8) gives an asymptotic solution to equation
(3.5) for the average prime truncation function. Lastly, computation shows
that for $4\leq n\leq 8$, the $W_{-1}$ function’s input causes it to be
undefined for this range; otherwise it’s well-defined $\forall n\geq 9$.

∎

The overall form of function (4.2) fits with that of the prime truncation
following a function of logarithmic type. Such as $[\ln(n)-1/3]^{1/2}$ or
$1+log_{10}(n)/2$, for example. As the W function is like a logarithm, only
growing slower, so then function (4.2) is like a logarithm divided by a weaker
logarithm. Plots showing how this function behaves are seen in figures (9) &
(10).

Figure 9. $g_{\pi}(n)$ vs. $\bar{g}_{\pi}(n)$ (Exact) vs. $\bar{g}_{\pi}(n)$
(Asymptotic); where $n\in[10^{3},10^{6}]$ Figure 10. $g_{\pi}(n)$ vs.
$\bar{g}_{\pi}(n)$ (Exact) vs. $\bar{g}_{\pi}(n)$ (Asymptotic); where
$n\in[10^{12},10^{13}]$

In each figure the prime truncation is plotted along with an exact solution to
equation (3.5) using a numerical approximation for the integral, and a
solution to the simplified form of this equation that replaces the integral
with its asymptotic form. Have already seen plots of the first two functions
over the same or similar regions. Now, the asymptotic form can be seen to
approximate the prime truncation rather well, but soon appears to grow beyond
it. There are essentially three reasons for this behavior, all stemming from
the three approximations that where made to obtain a nice functional form;
equation (4.2). First the integral was simplified, next used an $x^{th}$-root
and limit to simplify $x!$, then finally in the exponent of $\ln$, assumed
that $(1+1/x)\rightarrow 1$ for large $n$. The influence of all
simplifications diminish as $n$ increases, because $x$ increases along with
it, though much slower being essentially logarithmicly related to $n$. The
most important is the simplification of the integral, with the $\ln$ exponent
being second, and the factorial being third.

Looking back at the sigmoidal correction $\tilde{\sigma}$ in equation (4.1).
Examining this using the proper values of $n$ and its corresponding $x$, shows
that the asymptotic form is less than the integral it approximates. At least
to start with, then the ratio gradually approaches $1$. This is one reason why
the plot of the average asymptotic form is larger than the exact solution
$\bar{g}_{\pi}(n)$. Being a smaller function requires a larger $x$ value to
hit the same mark. Figure (11) & (12) below plots this sigmoid. Note that some
spikes are likely small numeric integration errors, but generally the ratio’s
nature is shown. There is a gradual downward trend, though rather erratic
because of the variation carried over from the prime truncation.

Additionally, the exponent $1+1/x$ of $\ln$ having been replaced by $1$, means
that the denominator involving this term will now be somewhat smaller that
otherwise, increasing the ratio. This is largely why equation (4.2) starts off
looking like a good fit, then veers off for larger $x$. It unintentionally
compensates for the asymptotic form of the integral being smaller than the
integral itself.

While the factorial correction is a minor influence by comparison, decreasing
smoothly, taking values in the range $[1/e,1]$; when $x=5$, corresponding to
$n\approx 10^{9}$, the ratio of limit relation (4.5) is $\approx 0.521034$,
which is not too far from $1/e\approx 0.367879$. When $x=10$ the ratio is
$\approx 0.4528729$, for $x=150$ it’s $\approx 0.3763755$. Showing a quick
convergence on the $x$-scale; though rather slow on the $n$-scale.

Figure 11. $\tilde{\sigma}(n,x_{\pi}),\ n\in[2,10^{5}]$ Figure 12.
$\tilde{\sigma}(n,x_{\pi}),\ n\in[10^{6},10^{9}]$

Moving on to investigating the large $n$ behavior of function (4.2) by using
the first order approximation $\frac{n}{\ln(n)}$ for the prime count function
to simplify the following ratio, appearing as input to the $W$ function:

(4.9)
$\begin{split}\frac{\ln(D_{n})}{\ln(n)}&=\frac{\ln\bigg{(}\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\dfrac{\pi(n^{1/r})}{r}\bigg{)}-\ln(n)}{\ln(n)}\\\
&\approx\frac{\ln\bigg{(}\dfrac{1}{\ln(n)}\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}n^{1/r}\bigg{)}}{\ln(n)}-1\end{split}$

In the limit as $n\rightarrow\infty$, using just the first term of the sum,
this ratio approaches ${-1/2}^{-}$. Looking closer at the RHS of equation
(4.9). Can essentially view the sum of successive roots of $n$ as follows:

(4.10)
$\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}n^{1/r}=n^{\beta_{n}},1/2\leq\beta_{n}\leq
1$

Since the upper limit of the sum is an increasing function of $n$, the
summation will grow in two ways with $n$. Though still not ultimately enough
to dominate its main term and grow in any significant way.

######  Theorem 4.2.

Given $\beta_{n}$ from equation (4.10), then;

(4.11) $\lim_{n\rightarrow\infty}\beta_{n}\rightarrow{1/2}^{+}$

###### Proof.

Dividing equation (4.10) through by $n^{1/2}$ and simplifying:

(4.12)
$\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}n^{1/r-1/2}=1+\sum_{r=3}^{\lfloor\log_{2}(n)\rfloor}n^{\frac{-(r-2)}{2r}}=n^{\beta_{n}-1/2}$

Where the sum of non-constant terms has the limit.

(4.13)
$\lim_{n\rightarrow\infty}\bigg{[}\sum_{r=3}^{\lfloor\log_{2}(n)\rfloor}n^{\frac{-(r-2)}{2r}}=\frac{1}{n^{1/6}}+\frac{1}{n^{1/4}}+\frac{1}{n^{3/10}}+...\bigg{]}\rightarrow
0^{+}$

Being positive implies that beta takes values in the interval $[1/2,1]$.
Leaving the following:

(4.14) $\lim_{n\rightarrow\infty}n^{\beta_{n}-1/2}\rightarrow 1^{+}$

This can only be true if the exponent of $n$ approaches $0$ from the right.
All requiring beta to approach $1/2$ from the right. Therefore,

(4.15) $\lim_{n\rightarrow\infty}\beta_{n}\rightarrow 1/2^{+}$

∎

Using this with equation (4.9), have:

(4.16)
$\lim_{n\rightarrow\infty}\bigg{[}\frac{\ln(D_{n})}{\ln(n)}\approx\frac{\ln(n^{\beta_{n}})-\ln(\ln(n))}{\ln(n)}-1\bigg{]}\rightarrow\lim_{n\rightarrow\infty}\beta_{n}-1\rightarrow-1/2^{+}$

So that inserting this result back into the Lambert $W$ gives:

(4.17)
$\lim_{n\rightarrow\infty}W_{-1}\bigg{[}\frac{\ln(D_{n})}{e\ln(n)}\bigg{]}\rightarrow
W_{-1}\bigg{(}\frac{-1}{2e}\bigg{)}\approx-2.678347$

Remarkably close to $-e$, which is a rather small, but important, change to
the denominator in equation (4.2). After using a similar simplification on the
numerator as in the denominator, seen in equation (4.16), have that up to a
first-order approximation:

(4.18)
$\bar{g}_{\pi}(n)\approx\frac{(\beta_{n}-1)\ln(n)-\ln(\ln(n))}{W_{-1}\bigg{(}\frac{\beta_{n}-1}{e}\bigg{)}}$

Combining the last few results with the simplified form in equation (4.18),
allows for the following result.

######  Theorem 4.3.

The limit of the first order approximation to the asymptotic solution for the
Average Prime Truncation Function $\bar{g}_{\pi}(n)$ has the following forms;

(4.19)
$\lim_{n\rightarrow\infty}\bar{g}_{\pi}(n)\rightarrow\tilde{g}_{\pi}(n)=0.186682\ln(n)$

(4.20)
$\lim_{n\rightarrow\infty}\bar{g}_{\pi}(n)\rightarrow\tilde{g}_{\pi}(n)=0.186682\ln(n)+0.373365\ln(\ln(n))$

###### Proof.

Proof: Using Theorem 4.2 and equation (4.17) in taking the limit of equation
(4.18) as $n\rightarrow\infty$. Have;

(4.21)
$\lim_{n\rightarrow\infty}\bar{g}_{\pi}(n)\approx\frac{-\frac{1}{2}\ln(n)-\ln(\ln(n))}{-2.678347}\rightarrow\frac{\ln(n)}{5.356694}$

Dividing through by $\ln(n)$, shows the limit approaches the constant
$1/5.356694=0.186682$; since the $0.373365\ln(\ln(n))$ term is much smaller
than $\ln(n)$ their ratio converges to $0$. Therefore, the given asymptotic
form of the average prime truncation function approaches the stated limits, up
to first order. ∎

Saying, “limit of asymptotic form”, not only means that equations (4.2),
(4.19) and (4.20) are asymptotically equivalent; i.e. the limit of their ratio
approaches $1$ as $n\rightarrow\infty$. Additionally, they are asymptotically
equivalent to the exact solution for equation (3.5) that defines the average
prime truncation. Where equation (4.2) is the more precise relation, over its
simplified version, equation (4.20). Noting that this implies some analytical
approximation is used for calculating the density $D_{n}$. Several options for
doing so, having already been discussed. The simplest of which, a first-order
approximation, was used to obtain the results just above, starting with
equation (4.9).

Theorem 4.3 shows that the average prime truncation, as $n\rightarrow\infty$,
is a little less than a fifth of the Stieltjes truncation for the logarithmic
integral. Thus, have a better estimate of the prime count by using roughly the
first one fifth of terms in the asymptotic expansion for the logarithmic
integral. An estimate that parallels Riemann’s $R(n)$; partly shown in the
proof of Theorem 3.2 and in the discussion that followed. Though now with less
computational overhead than $R(n)$.

##  5\. Connection to the Distribution of the non-Trivial Zeros for the Zeta
Function.

### 5.1.

Returning to the line of inquiry which arose in Theorem 3.2, particularly
equation (3.7), relating a summation over non-trivial zeros of the zeta
function to a difference between the prime count $\pi(n)$ and the truncated
logarithmic integral $li(n;\bar{g}_{\pi}(n))$. Any bound on this difference is
then a bound on the summation. Since the truncated version is generally closer
to the prime count than $li(n)$. It must be able to provide a tighter bound.
With a possible exception for Skewes region’s, where $li(n)<\pi(n)$.

Again, the Riemann Hypothesis (RH) is not being assumed. The aim here is to
see what bearing a tighter bound has, if any, on the non-trivial zeta-zeros
positioning. So far nothing has relied on the position of the zeta function’s
non-trivial zeros. Other than being complex valued and occurring in complex
conjugate pairs; as used in the proof of Theorem 3.2.

The general idea for this section is to start from a new bound related to the
difference $|\pi(n)-li(n)|$, that does not assume the RH to be true, separate
$li(n)$ into two pieces via the average prime truncation point, and then
simplify. Finally, apply this new bound to the zeros summation in equation
(3.7), and compare to another bound equivalent to the RH.

To help simplify the argument, will refer to the two parts that $li(n)$ is
split into as the truncated logarithmic integral and the remainder; same as
that used in the proof of Theorem 3.2. Denoted by:

(5.1) $li(n)\approx
li(n;\bar{g}_{\pi})+[li(n;\tau)-li(n;\bar{g}_{\pi})]=li(n;\bar{g}_{\pi})+li_{R}(n;\bar{g}_{\pi})$

Now, referring back to equation (3.7), reproduced here for convenience.
Remembering that the remainder and weighted prime powers have been canceled
out.

$\pi(n)-li(n;\bar{g}_{\pi}(n))\approx-\sum_{\rho}^{zeta-
zeros}li(n^{\rho})+\int_{n}^{\infty}\frac{dt}{(t^{2}-1)t\ln(t)}-\ln(2)$

Require a suitable bound on the LHS difference. Dusart [Ref6] proved the
following theorem, with table 1, bounding the Tchebyshev Theta function;
defined as the following sum over prime $p$:

(5.2) $\vartheta(n)=\sum_{\begin{subarray}{c}p\leq n\\\ p\
prime\end{subarray}}\ln(p)$

Theorem 5.2 [Dusart]: $\ \forall k\in\mathbb{N}_{0},\ \exists\eta_{k}$ and
$n_{k}$, given in the following table 1, such that;

(5.3) $|\vartheta-n|<\eta_{k}\frac{n}{\ln^{k}(n)},\forall n\geq n_{k}$

Table 1. Dusart Theorem 5.2 Parameter Values

$\begin{array}[b]{|c||c|c|c|c|c|c|c|}\hline\cr k&0&1&1&2&2&2&2\\\
\hline\cr\eta_{k}&1&1.2323&0.001&3.965&0.2&0.05&0.01\\\ \hline\cr
n_{k}&1&2&908,994,923&2&3,594,641&122,568,683&7,713,133,853\\\
\hline\cr\end{array}$

$\begin{array}[b]{|c||c|c|c|c|c|c|c|}\hline\cr k&3&3&3&3&4\\\
\hline\cr\eta_{k}&20.83&10&1&0.78&1300\\\ \hline\cr
n_{k}&2&32,321&89,967,803&158,822,621&2\\\ \hline\cr\end{array}$

Where for each integral $k$ there are choices for the eta coefficient
$\eta_{k}$ and the point $n_{k}$ at which the bound becomes valid. The tables
show that several values for the coefficient $\eta_{k}$ can be chosen
depending on the smallest allowed value $n_{k}$, for the input variable $n$.
Or, the other way around, can choose a minimum $n_{k}$ and then find the
associated $\eta_{k}$. Both assume a fixed $k$.

To apply this theorem, the first perspective shall be used. Along with
choosing $\eta_{k}=(k-1)!$, which tends to be rather large, there will exist
an $n_{k}$ , that should be rather small. From the tables, can see that a
larger value for the input $n_{k}$ is required to decrease the $\eta_{k}$
coefficient’s value. Can interpolate and see for example that given $k=3$,
have $\eta_{3}=2$, implying $32,321<n_{3}<89,976,803$. The exact value of
$n_{k}$ does not matter, only its existence. While the choice for the
$\eta_{k}$ coefficient is so that it matches a general term in the asymptotic
expansion for $li(n)$.

Creating a key element to proving:

######  Theorem 5.1.

$|li(n;\bar{g}_{\pi})-\pi(n)|<\dfrac{n}{\ln(n)}\dfrac{(\bar{g}_{\pi}-1)!}{\ln(n)^{\bar{g}_{\pi}-1}},\forall
n\geq n_{\bar{g}_{\pi}}$.

###### Proof.

Using the following identity between the prime counting function and
Chebyshev’s function; Axler [Ref1]:

(5.4)
$\pi(n)=\frac{\vartheta(n)}{\ln(n)}+\int_{2}^{n}\frac{\vartheta(t)}{t\ln^{2}(t)}dt$

Starting with a reverse of the difference that one usually encounters.
Generally obtained by expanding $li(n)$ to first order, so that it has the
same basic form as equation (5.4), then subtracting the prime count function.
While also accounting for the constant term from the expansion, along with the
slight offset between the definite integral in equation (5.4) and the one
defining $li(n)$.

(5.5)
$li(n)-\pi(n)=\frac{n-\vartheta(n)}{\ln(n)}-\frac{2}{\ln(2)}+li(2)+\int_{2}^{n}\frac{t-\vartheta(t)}{t\ln^{2}(t)}dt$

The two constant terms sum is approximately $-1.840226$, a small correction
for large $n$. Next, using Theorem 5.2 [Dusart] can bound the two RHS non-
constant terms. Applying the choice $\eta_{k}=(k-1)!$ for general exponents
$k$, on the logarithm, there exists an $n_{k}$ such that:

(5.6)
$\frac{|n-\vartheta(n)|}{\ln(n)}<(k-1)!\frac{n}{\ln^{k}(n)}=\frac{n}{\ln(n)}\frac{(k-1)!}{\ln^{k-1}(n)},\forall
n\geq n_{k}$

(5.7)
$\int_{2}^{n}\frac{|t-\vartheta(t)|}{t\ln^{2}(t)}dt<(k-1)!\int_{2}^{n}\frac{dt}{\ln^{k+1}(t)},\forall
n\geq n_{k}$

Although the same index was used for inequalities (5.6) and (5.7). This still
will not yield the desired form for inequality (5.7). Want this to be of the
same form as the source integral term in the expansion for $li(n;x)$. It’s
very nearly there, just off by a factor of $k$. To fix this, only need to
multiply through by $k$. Resulting in an appropriate inequality with the right
form, given as:

(5.8)
$\int_{2}^{n}\frac{|t-\vartheta(t)|}{t\ln^{2}(t)}dt<(k-1)!\int_{2}^{n}\frac{dt}{\ln^{k+1}(t)}<k!\int_{2}^{n}\frac{dt}{\ln^{k+1}(t)},\forall
n\geq n_{k}$

Now, given equation (5.5), split $li(n)$ via equation (5.1), then apply
inequalities (5.6) and (5.8); dropping the constant $-2/\ln(2)$ term, and
using the triangle inequality, finally rearranging to obtain:

(5.9)
$\begin{split}|li(n;\bar{g}_{\pi})-\pi(n)|&=\bigg{|}\frac{n-\vartheta(n)}{\ln(n)}-\frac{2}{\ln(2)}+li(2)+\int_{2}^{n}\frac{t-\vartheta(t)}{t\ln^{2}(t)}dt-
li_{R}(n;\bar{g}_{\pi})\bigg{|}\\\
&<\frac{|n-\vartheta(n)|}{\ln(n)}+\bigg{|}li(2)+\int_{2}^{n}\frac{|t-\vartheta(t)|}{t\ln^{2}(t)}dt-
li_{R}(n;\bar{g}_{\pi})\bigg{|}\\\
&<\frac{n}{\ln(n)}\frac{(k-1)!}{\ln^{k-1}(n)}+\bigg{|}li(2)+k!\int_{2}^{n}\frac{dt}{\ln^{k+1}(t)}-li_{R}(n;\bar{g}_{\pi})\bigg{|}\end{split}$

Next step requires the index $k$ to be able to take real values. Nothing truly
forces it to be integral. The only time it needs to be integral in Dusart
[Ref6] is when it’s used to index the $k^{th}$ prime, but real values still
make sense if the index is appropriately rounded. In fact the asymptotic
expansion for the $k^{th}$ prime, and others, appear in the source paper. All
view $k$ as real-valued.

Now that $k$ can take real values, set all instances of $k$ above to the
average prime truncation function $\bar{g}_{\pi}(n)$. Then by construction,
from equations (3.5) and (3.6) used in the proof of Theorem 3.2, looking at
the last line for equation (5.9) the RHS term in absolute value is zero.
Giving the final inequality:

(5.10)
$|li(n;\bar{g}_{\pi})-\pi(n)|<\dfrac{n}{\ln(n)}\dfrac{(\bar{g}_{\pi}-1)!}{\ln(n)^{\bar{g}_{\pi}-1}},\forall
n\geq n_{\bar{g}_{\pi}}$

Where $n_{\bar{g}_{\pi}}$ also depends on which form of $\bar{g}_{\pi}$ is
used. E.g. since the closed form involving the $W_{-1}$ function is valid
$\forall n\geq 9$, in this case $n_{\bar{g}_{\pi}}=9$. For the other forms of
$\bar{g}_{\pi}$ presented have $n_{\bar{g}_{\pi}}=2$. ∎

######  Remark 5.2.

Choosing $\eta_{k}=(k-1)!$ so freely is not an issue, since normally the name
of the game is to make $\eta_{k}$ as small as possible. But, in this case it’s
essentially the opposite and making $\eta_{k}$ larger doesn’t violate the
inequality. Also, other choices can work, e.g. $\eta_{k}=(k-2)!$; giving a
less tight bound; while the original $\eta_{k}=k!$ choice has some issues with
determining where the bound holds, among others.

Independent of the RH, a bound on the zeta-zeros summation can now be
expressed in the following theorem. Basically another corollary of Theorem
5.1.

######  Theorem 5.3.

(Truncation Bound)

(5.11) $\bigg{|}\sum_{\rho}^{zeta-
zeros}li(n^{\rho})\bigg{|}<\dfrac{n}{\ln(n)}\dfrac{(\bar{g}_{\pi}-1)!}{\ln^{\bar{g}_{\pi}-1}(n)}-0.553137,\
\forall n\geq n_{\bar{g}_{\pi}}$

###### Proof.

Applying Theorem 5.1 to the RHS of equation (3.7):

(5.12) $\bigg{|}-\sum_{\rho}^{zeta-
zeros}li(n^{\rho})+\int_{n}^{\infty}\frac{dt}{(t^{2}-1)t\ln(t)}-\ln(2)\bigg{|}<\dfrac{n}{\ln(n)}\dfrac{(\bar{g}_{\pi}-1)!}{\ln^{\bar{g}_{\pi}-1}(n)}$

The integral on the LHS is a decreasing function of $n$, its maximum occurring
for $n=2$ is $0.140010$. Then subtracting the constant term $\ln(2)$ gives
$0.140010-0.693147\approx-0.553137$; which will very quickly approach
$-\ln(2)$ as $n$ increases. Combine these into a single constant. Since both
remaining terms in absolute value are negative, can factor out and eliminate
$-1$. Using the triangle inequality to obtain:

(5.13) $\bigg{|}\sum_{\rho}^{zeta-
zeros}li(n^{\rho})+0.553137\bigg{|}\leq\bigg{|}\sum_{\rho}^{zeta-
zeros}li(n^{\rho})\bigg{|}+0.553137<\dfrac{n}{\ln(n)}\dfrac{(\bar{g}_{\pi}-1)!}{\ln^{\bar{g}_{\pi}-1}(n)}$

Therefore, upon rearranging the constant term, the proof is complete.

∎

Next, need a bound equivalent to the RH. Turning to corollary $1$ of Theorem
$10$ from Schoenfeld [Ref15], reproduced here.

Corollary 1 [Schoenfeld]: If the Riemann Hypothesis (RH) is true, then;

(5.14) $|\pi(n)-li(n)|<\frac{n^{1/2}\ln(n)}{8\pi},\forall n\geq 2,657$

Next, will use this bound to find a corresponding bound for the summation over
the non-trivial zeta-zeros. Using equation (3.1), re-arrange to get the needed
difference, then take the absolute value, then simplify using the triangle
inequality:

(5.15)
$\begin{split}|\pi(n)-li(n)|&=\bigg{|}-\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\frac{\pi(n^{1/r})}{r}-\sum_{\rho}^{zeta-
zeros}li(n^{\rho})+\int_{n}^{\infty}\frac{dt}{(t^{2}-1)t\ln(t)}-\ln(2)\bigg{|}\\\
&\approx\bigg{|}-\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\frac{\pi(n^{1/r})}{r}-\sum_{\rho}^{zeta-
zeros}li(n^{\rho})-0.553137\bigg{|}\\\ &\geq\bigg{|}\sum_{\rho}^{zeta-
zeros}li(n^{\rho})\bigg{|}-\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\frac{\pi(n^{1/r})}{r}-0.553137\end{split}$

As the negative constant term always dominates the positive integral term, can
assume essentially have just three terms, all negative, in the first line.
Seen when moving to the equivalence in the next line. Combining this with
Corollary 1 [Schoenfeld], which is really just looking at other side of the
equal sign, and though it does not explicitly appear in Schoenfeld [Ref15],
will refer to as:

Corollary 1B [Schoenfeld]: If the RH is true, then;

(5.16) $\bigg{|}\sum_{\rho}^{zeta-
zeros}li(n^{\rho})\bigg{|}<\frac{n^{1/2}\ln(n)}{8\pi}+\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\frac{\pi(n^{1/r})}{r}+0.553137,\forall
n\geq 2,657$

### 5.2.

To get a idea of the relationship between these two bounds on
$|\sum_{\rho}^{zeta-zeros}li(n^{\rho})|$, will first use computational methods
then prove some related results. Since the Truncation Bound of Theorem 5.3 can
apply to the various approximations for the average prime truncation function
$\bar{g}_{\pi}(n)$. The Truncation Bound’s behavior for two of these
approximations is compared both to each other and to the related Schoenfeld
bound. The exact form of $\bar{g}_{\pi}(n)$ computed from definition (3.1)
gives the best case achievable. While the two limits $\tilde{g}_{\pi}(n)$
obtained from the closed form solution, are each worse case by comparison.
Generally intend the more accurate limit
$\tilde{g}_{\pi}(n)=0.186682\ln(n)+0.373365\ln(\ln(n))$, otherwise will
mention when dropping the $\ln(\ln(n))$ term.

In figure (13) the solid black line is Schoenfeld’s Corrollary 1B bound on
$|\sum_{\rho}^{zeta-zeros}li(n^{\rho})|$, while the other two lines are the
Truncation Bounds with $\bar{g}_{\pi}(n)$ in dot-dash red, and
$\tilde{g}_{\pi}(n)$ in dashed blue. The plot shows crossing points where each
of the Truncation Bounds drop below the Schoenfeld bound. For the
$\bar{g}_{\pi}(n)$ Truncation Bound have an intersection at $n\approx
6,062.860$, and for the $\tilde{g}_{\pi}(n)$ bound an intersection at
$n\approx 33,519.932$. For comparison, when dropping $\ln(\ln(n))$ from the
$\tilde{g}_{\pi}(n)$ Truncation Bound, now have a crossing at $n\approx
1.458\cdot 10^{12}$. A drastic change in the intersection point, but
eventually both forms of $\tilde{g}_{\pi}(n)$ produce bounds that are less
than the target bound of Schoenfeld.

Figure 13. Corollary 1B [Schoenfeld] Vs. Theorem 5.3 w/ $\bar{g}_{\pi}(n)$ and
$\tilde{g}_{\pi}(n)$, $\forall n\in[2,5\cdot 10^{4}]$

The question is now whether or not those relationships depicted in figure (13)
extend to all $n>5*10^{4}$. Computations up to $n=10^{24}$ support the same
ordering among all the bounds. Additionally, the distance between the two
Truncation Bounds stays about the same, while their distance from the
Schoenfeld bound only increases. These calculational assertions should be
proven for the Truncation Bound using $\tilde{g}_{\pi}(n)$. Because
$\bar{g}_{\pi}(n)$ is a computed object it cannot be used in such a proof,
still it exists, and would fall under similar relations as
$\tilde{g}_{\pi}(n)$. Such as the following theorem.

######  Theorem 5.4.

Let $\tilde{g}_{\pi}(n)=0.1866823\ln(n)$, then;

(5.17) $\begin{split}|\sum_{\rho}^{zeta-
zeros}li(n^{\rho})|&<\frac{n}{\ln(n)}\frac{(\tilde{g}_{\pi}-1)!}{\ln(n)^{\tilde{g}_{\pi}-1}}-0.553137\\\
&<\frac{n^{1/2}\ln(n)}{8\pi}+\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\frac{\pi(n^{1/r})}{r}+0.553137\
\forall n\geq 1.5\cdot 10^{12}\end{split}$

###### Proof.

Let $\tilde{g}_{\pi}(n)=0.1866823\ln(n)$ using the simplest form available,
and substitute into RHS of Theorem 5.3 to then define the function;

(5.18)
$f_{1}(n)\equiv\frac{n}{\ln(n)}\frac{(0.1866823\ln(n)-1)!}{\ln(n)^{0.1866823\ln(n)-1}}-0.553137$

Also, have from Robbins [Ref14] there exists a variable $z$ such that;

(5.19) $\sqrt{2\pi}z^{z+1/2}e^{-z}e^{\frac{1}{12z+1}}\leq
z!\leq\sqrt{2\pi}z^{z+1/2}e^{-z}e^{\frac{1}{12z}},\ \forall z\in\mathbb{N}$

On RHS take $z=1$ in $e^{\frac{1}{12z}}\Rightarrow e^{\frac{1}{12}}$ is max of
this factor for all $z\geq 1$; and $\sqrt{2\pi}e^{\frac{1}{12}}\approx
2.724464$ is very close to $e$, just slightly larger; denote this new constant
by $e^{+}$. From this, will use the inequality $z!\leq e^{+}z^{z+1/2}e^{-z}$
to bound $f_{1}$ from above and simplify by letting $z=0.1866823\ln(n)-1$;

(5.20)
$\begin{split}f_{1}(n)&\leq\frac{n}{\ln(n)}\frac{e^{+}[0.1866823\ln(n)-1]^{0.1866823\ln(n)-1+1/2}e^{-0.1866823\ln(n)+1}}{\ln(n)^{0.1866823\ln(n)-1}}-0.553137\\\
&\leq\frac{n}{\ln(n)}\frac{e^{+}[0.1866823\ln(n)]^{0.1866823\ln(n)-1+1/2}e^{-0.1866823\ln(n)+1}}{\ln(n)^{0.1866823\ln(n)-1}}-0.553137\\\
&=\frac{n}{\ln(n)}\frac{e^{+}e[0.1866823]^{0.1866823\ln(n)-1/2}\ln^{1/2}(n)}{n^{0.1866823}}-0.553137\\\
&=n^{0.8133177}\frac{e^{+}e[0.1866823]^{0.1866823\ln(n)-1/2}}{\ln^{1/2}(n)}-0.553137\\\
&=n^{0.8133177}\frac{e^{+}e(0.1866823)^{-1/2}[0.1866823^{0.1866823}]^{\ln(n)}}{\ln^{1/2}(n)}-0.553137\\\
&=n^{0.8133177}\frac{17.14052\ [0.7310176]^{\ln(n)}}{\ln^{1/2}(n)}-0.553137,\
\forall n\geq 2\end{split}$

Since the given inequalities on the factorial do not hold till $z\geq 1$, and
have $z=0.1866823\ln(n)-1$, solving for $n$ implies requiring $n\geq 44,954$
(the actual value is $n\approx 44,953.71$). A closer check shows the
inequality just derived actually holds $\forall n\geq 1.15$. Next let,

(5.21)
$f_{2}(n)\equiv\frac{n^{1/2}\ln(n)}{8\pi}+\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\frac{\pi(n^{1/r})}{r}+0.553137$

All terms are positive. Then have, by dropping last two terms; that
$f_{2}(n)>\frac{n^{1/2}\ln(n)}{8\pi},\ \forall n\geq 2$.

To complete proof, must show that $f_{1}<f_{2},\ \forall n\geq n_{0}$; for
some $n_{0}$. Combining the expressions for these two functions:

(5.22) $\begin{split}f_{1}(n)&\leq n^{0.8133177}\frac{17.14052\
[0.7310176]^{\ln(n)}}{\ln^{1/2}(n)}-0.553137\\\
&<\frac{n^{1/2}\ln(n)}{8\pi}<f_{2}(n)\end{split}$

Rearranging the middle terms:

(5.23)
$\begin{split}[0.7310176]^{\ln(n)}&<\frac{n^{0.5-0.8133177}\ln^{3/2}(n)}{17.14052\cdot
8\pi}+\frac{0.553137\ n^{-0.8133177}\ln^{1/2}(n)}{17.14052}\\\
&<\frac{n^{-0.3133177}\ln^{3/2}(n)}{17.14052\cdot 8\pi}\end{split}$

Dropping the last term on the first line, as it decreases much faster than the
other; e.g. $n=100$ has $100^{-0.3133177}\approx 0.2362461$ and
$100^{-0.8133177}\approx 0.02362461$; and only grows smaller, somewhat
quickly, for larger inputs. Taking the natural log of both sides gives:

(5.24)
$\ln(0.7310176)\ln(n)<-0.3133177\ln(n)+\frac{3}{2}\ln(\ln(n))-\ln(17.14052\cdot
8\pi)$

Since $\ln(0.7310176)=-0.3133177$ the two logarithms on each side nicely
cancel out, and calculating the RHS constant term, leaves:

(5.25) $0<\frac{3}{2}\ln(\ln(n))-6.065617$

Which is true $\forall n\geq e^{e^{6.065617/1.5}}\approx 5.915022\cdot
10^{24}$; and agrees with the computed intersection for the full inequality
(5.23) at $n\approx 5.9144\cdot 10^{24}$. Additional computation using
$f_{1}(n)$ and $f_{2}(n)$ puts the actual intersection at $n\approx
1.4579\cdot 10^{12}$, where the inequality has been computationally verified
up to $n=10^{25}$. Therefore, have that the above string of inequalities hold,
showing:

(5.26) $f_{1}(n)<f_{2}(n),\ \forall n\geq 1.5\cdot 10^{12}$

∎

Choosing the simplest form of $\tilde{g}_{\pi}(n)$ was to facilitate a cleaner
proof. Using $\tilde{g}_{\pi}(n)=0.186682\ln(n)+0.373365\ln(\ln(n))$ in its
place, or some suitable form of $\bar{g}_{\pi}(n)$, will produce similar
theorems for these functions. They generally only differ on where the
inequality’s validity begins. Referring back to figure (13) though not shown,
the plot for the Truncation Bound using $\tilde{g}_{\pi}(n)=0.186682\ln(n)$
would be above that of $\tilde{g}_{\pi}(n)=0.186682\ln(n)+0.373365\ln(\ln(n))$
and $\bar{g}_{\pi}(n)$. Given that these functions are just a coarsening or
refinement of each other, along with the supporting computations, the
following theorems are also proven by Theorem 5.4.

######  Theorem 5.5.

Let $\tilde{g}_{\pi}(n)=0.1866823\ln(n)+0.373365\ln(\ln(n))$, then;

(5.27) $\begin{split}|\sum_{\rho}^{zeta-
zeros}li(n^{\rho})|&<\frac{n}{\ln(n)}\frac{(\tilde{g}_{\pi}-1)!}{\ln(n)^{\tilde{g}_{\pi}-1}}-0.553137\\\
&<\frac{n^{1/2}\ln(n)}{8\pi}+\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\frac{\pi(n^{1/r})}{r}+0.553137\
\forall n\geq 33,520\end{split}$

######  Theorem 5.6.

Using the exact, computed form of $\bar{g}_{\pi}(n)$ then;

(5.28) $\begin{split}|\sum_{\rho}^{zeta-
zeros}li(n^{\rho})|&<\frac{n}{\ln(n)}\frac{(\bar{g}_{\pi}-1)!}{\ln(n)^{\bar{g}_{\pi}-1}}-0.553137\\\
&<\frac{n^{1/2}\ln(n)}{8\pi}+\sum_{r=2}^{\lfloor\log_{2}(n)\rfloor}\frac{\pi(n^{1/r})}{r}+0.553137\
\forall n\geq 6,063\end{split}$

######  Remark 5.7.

Even though the computed form of $\bar{g}_{\pi}(n)$ is used, it really shows
that more accurate approximations for this function will approach this result.
Improving their range of validity for smaller values of $n$, down to a
minimum, providing tighter and tighter bounds.

### 5.3.

Though Corollary 1 [Schoenfeld] is stated as an implication, proving the bound
true, independently of the RH, is equivalent to the RH being proven true. This
bound is then equivalent to the Riemann Hypothesis. As would be any other
sufficiently tight bound. Sufficient in the general sense defined by von
Koch’s [Ref9] well known result; represented in modern form as:

(5.29) $\lim_{n\rightarrow\infty}\pi(n)=li(n)+O(\sqrt{n}\ln(n))$

Which can be recast in terms of the summation over the non-trivial zeros; just
as Schoenfeld’s original bound was altered to obtain Corollary 1B [Schoenfeld]
refocusing on $|\sum_{\rho}^{zeta-zeros}li(n^{\rho})|$.

Since Theorem 5.3 does not assume that the RH is true, and Corollary 1B
[Schoenfeld] does assume it is true. If the Truncation Bound in Theorem 5.3 is
less than or equal to the bound in Corollary 1B [Schoenfeld], then the RH must
be true in general. Which is what Theorems 5.4, 5.5, and 5.6 all show. Finally
drawing this section to a close by combining all previous results.

######  Theorem 5.8.

The Riemann Hypothesis is True.

###### Proof.

Given that the exact and average prime truncations of the asymptotic expansion
for the logarithmic function exist; by Theorems 2.4 and 3.2. With their stated
properties, the average truncation must provide a means to analyze the
variations of the exact truncation; which are mostly due to the oscillatory
behavior induced by the non-trivial zeta-zeros. Realized by using Riemann’s
explicit equation (3.1) for the prime count function. By Theorems 4.1 and 4.3
there exists analytic solutions to the average truncation. Allowing for a new
explicit bound, Theorem 5.3, on the zeta-zero summation term in equation
(3.1), that does not assume anything about the location of those zeros. Again
other than being complex-valued and occurring in complex conjugate pairs, to
imbue the summation with real-valued sinusoidal properties.

Now, given that the Schoenfeld bound, when recast in terms of the zeta-zeros
summation $|\sum_{\rho}^{zeta-zeros}li(n^{\rho})|$ , Corollary 1B
[Schoenfeld], which does assume a condition on the location of the non-trivial
zeros; i.e. that the RH is true. This bound is equivalent to the RH being
true.

Next, considering the computational investigation shown above and Theorems
5.4, 5.5, and 5.6, have that the Theorem 5.3 Truncation Bound is less than the
Corollary 1B [Schoenfeld] bound, for the given range, depending on which form
of $\bar{g}_{\pi}(n)$ is used in the Truncation Bound. Thus, showing that a
sufficiently tight bound is able to be constructed without assuming the RH;
$\forall n\geq 6,064$ in the best case.

Since an equivalent to the Schoenfeld bound is beaten by the various
Truncation Bounds, all without assuming the Riemann Hypothesis, then the RH
must be true. Therefore, all non-trivial zeta-zeros $\rho$ are on the critical
line $\Re(\rho)=1/2$.

∎

##  6\. Conclusion

### 6.1.

As stated at the beginning of this paper, the goal was to answer a question
about the asymptotic expansion of the logarithmic integral and explore its
consequences. Leading to a verification that the Riemann Hypothesis is
correct. This paper proceeded in the same way as the events of the discovery
unfolded.

At first it seemed a truncated version of $li(n)$ would just provide a cheap
computational short cut for a better approximation to $\pi(n)$, somewhat
similar to Riemann’s $R(n)$. Which it does. Instead of calculating multiple
instances of the $li(n)$ function with different inputs and combining them via
$\mu(r)$ to compute $R(n)$; now just have to calculate about a fifth of the
expansion for $li(n)$ and no $\mu(r)$.

But, the application of a truncation function to describe the variation
induced by the zeta function’s non-trivial zeros turned out to be far more
important. Essentially the main result follows from equating the weighted
prime powers over-count to the “tail” or remainder of the logarithmic
integral’s asymptotic expansion, and canceling them out. Which is equivalent,
via Theorem 3.2, to the idea of the average prime truncation function.
Additionally, gain a much tighter bound, Theorem 5.3, on the variation induced
by the summation over the zeta-zeros than that provided by Schoenfeld’s
equivalent bound on the same term.

Now having seen something of the average prime truncation function’s nature in
relation to that of the prime truncation giving $\pi(n)$. It makes sense that
it can bound the variation induced by the zeta functions non-trivial zeros so
well. Effectively making the newly introduced asymptotic expansion variable
$x$, akin to a type of logarithmic-like transform leading to an auxiliary
function.

With a possible exception to this log-like transform. No new mathematics has
really been introduced, which does make one feel like there is something not
quite right. As with some problems, especially this one, it’s generally not
only the expectation, but seemingly a necessity that new mathematics must
arise to obtain a solution. With that in mind, the work here might best be
characterized as using old mathematics by way of a new vantage point.

###  6.2. Future Work

Finding a better form of the average $\bar{g}_{\pi}(n)$; both in how well it
fits the exact solution; i.e. no simplifications; and its ease of calculation.
While another interesting route would be to determine how the $li(n^{\rho})$
terms in the zeta-zero summation map their corrections into the asymptotic
domain of the truncation functions. Which can then be applied to
$\bar{g}_{\pi}(n)$, by improving the constraint $V$ on the integral term,
giving a better approximation to $g_{\pi}(n)$. Adding the integral and
constant terms of Riemann’s explicit formula to the sum of prime powers
constraint would be a small improvement. Further improving the competition
between $li(n;\bar{g}_{\pi}(n)$ and $R(n)$.

######  Remark 6.1.

Consider a convenient notational modification: $\pi(n)\equiv$ $\pi$ $(n)$. Can
view it as a symbolic bucket for placing primes $\leq n$.

## Acknowledgments

Thanks to P. Dusart who provided a copy of a paper I had difficulty acquiring.
Thanks to J. Wood for the RcppAlgos package & replying to my questions; also
to K. Walisch for developing the algorithm implemented in the package.

## APPENDIX

### A1

Derivation of solution for linear exponential equation (4.8) from proof of
Theorem 4.1. Equation & solution (23) reproduced, respectively, from Edwards
[Ref8]:

(A-1) $y=(ax^{b})^{cx^{d}}$

(A-2)
$x=\frac{e^{\frac{W\big{(}\frac{a^{d/b}d}{bc}\ln(y)\big{)}}{d}}}{a^{1/b}}$

Now, let $b=1$, $c=1$, and $d=1$, then have,

(A-3) $y=(ax)^{x}$

(A-4) $x=\frac{e^{W(a\ln(y))}}{a}$

If $a=1/(e\ln(n))$, then,

(A-5) $x=e\ln(n)\ e^{W\big{(}\frac{\ln(y)}{e\ln(n)}\big{)}}$

Still not the form needed, but close, since from one of the basic identities
for the $W$ function derived directly from its definition is that;

(A-6) $e^{W(z)}=\frac{z}{W(z)}$

Using this identity with equation (A-5), have that;

(A-7) $x=\frac{e\ln(n)\big{[}\frac{\ln(y)}{e\
\ln(n)}\big{]}}{W\big{(}\frac{\ln(y)}{e\ln(n)}\big{)}}=\frac{\ln(y)}{W\big{(}\frac{\ln(y)}{e\ln(n)}\big{)}}$

Giving the correct form. Finally, setting $y=D_{n}$, the density provides the
required solution.

## References