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\begin{center}
\vskip 1cm{\LARGE\bf On a Family of Functions Defined\\
\vskip .1in
Over Sums of Primes}
\vskip 1cm
\large
Christian Axler\\
Department of Mathematics\\
Heinrich-Heine-University\\
40225 D\"usseldorf\\
Germany\\
\url{christian.axler@hhu.de}\\
\end{center}

\vskip .2 in

\begin{abstract}
Let $r$ and $m$ be real numbers so that the sum $S_{r,m}(x) = \sum_{p \leq x} p^r\log^mp$ diverges as $x \to \infty$. Here $p$ runs over all primes not 
exceeding $x$. In this paper, we give an asymptotic formula for each $S_{r,m}(x)$ as $x \to \infty$. The case where $x$ is the $n$th prime number is of 
particular interest. Here we use a method developed by Salvy to give an asymptotic formula for $S_{r,m}(p_n)$ as $n \to \infty$, which generalizes,
for instance, 
the previously known one for $S_{1,0}(p_n)$, the sum of the first $n$ prime numbers.
\end{abstract}

\section{Introduction}

Let $m$ and $r$ be real numbers and let $S_{r,m}: [0, \infty) \to \R$ be defined by
\begin{displaymath}
S_{r,m}(x) = \sum_{p \leq x} p^r\log^m p,
\end{displaymath}
where $p$ runs over all primes not exceeding $x$. The sum $S_{r,m}(x)$ diverges as $x \to \infty$ if and only if
\begin{displaymath}
\text{(i)} \ r &gt; -1 \text{ and } m \in \R \q\q \text{ or } \q\q \text{(ii)} \ r = -1 \text{ and } m \geq 0.
\end{displaymath}
The aim of this paper is to find the asymptotic behaviour of the sum
$S_{r,m}(x)$ in the case where $m$ and $r$ satisfy conditions
(i) or (ii). First,
we study the case (i). Let $\pi(x)$ denote the number of primes not
exceeding $x$. A well-known result (see \cite{vallee1899}) concerning
this function is the \textit{prime number theorem}, which states that
\begin{equation}\label{1.1}
\pi(x) = \text{li}(x) + O(x e^{-a\sqrt{\log x}})
\end{equation}
as $x \to \infty$, where $a$ is a positive absolute constant,
and the \emph{logarithmic integral} $\text{li}(x)$ is defined
for every real $x \geq 0$ as follows:
\begin{equation}\label{1.2}
\text{li}(x) = \int_0^x \frac{dt}{\log t} = \lim_{\varepsilon \to 0+} \left \{ \int_{0}^{1-\varepsilon}{\frac{dt}{\log t}} + 
\int_{1+\varepsilon}^{x}{\frac{dt}{\log t}} \right \}.
\end{equation}
Denoting the sum of the first prime numbers not exceeding $x$ by $S(x)$, Szalay \cite[Lemma 1]{szalay} used \eqref{1.1} and Stieltjes integration to find
\begin{equation}\label{1.3}
S(x) = \text{li}(x^2) + O ( x^2 e^{-a\sqrt{\log x}})
\end{equation}
as $x \to \infty$. The case $x=p_n$, where $p_n$ denotes the $n$th prime number, is of particular interest. Here, $S(x) = \sum_{k \leq n}p_k$ is equal to the 
sum of the first $n$ prime numbers. Massias and Robin \cite[p.\:217]{mr} found that
\begin{equation}\label{1.4}
\sum_{k=1}^n p_k = \tl((\tl^{-1}(n))^2) + O(n^2 e^{-c\sqrt{\log n}})
\end{equation}
as $n \to \infty$, where $c$ is a positive absolute constant and $\tl^{-1}(x)$ is the inverse function of $\tl(x)$. Then they \cite[p.\:217]{mr} used 
\eqref{1.4} and a result of Robin \cite{robin1988} to derive the asymptotic expansion
\begin{equation}\label{1.5}
\sum_{k=1}^n p_k = \frac{n^2}{2} \left( \log n + \sum_{i=0}^N \frac{A_{i+1}(\log \log n)}{\log^i n} \right) + O_N \left( \frac{n^2(\log \log 
n)^{N+1}}{\log^{N+1} n} \right)
\end{equation}
as $n \to \infty$, where $N$ is a nonnegative integer and the polynomials $A_k$ satisfy the formulas $A_0(x) = 1$ and
\begin{equation}\label{1.6}
A_{i+1}' = A_i' - (i-1)A_i.
\end{equation}
It follows that $\deg(A_0) = 0$, $\deg(A_1)=1$, and $\deg(A_i) = i-1$ for every integer $i \geq 2$. Unfortunately, the recursive formula \eqref{1.6} for the 
derivatives does not yield a description of the polynomials $A_i$, since the constant coefficient of the polynomials $A_i$ remains undetermined by this 
equation. This problem was fixed in \cite[Theorem 1.4]{axler2018} by applying a method developed by Salvy \cite[Theorem 2]{salvy}. We use the same method to 
give the following result concerning the sum $S_{r,m}(p_n)$ in the case where $r &gt; -1$ and $m \in \R$. Here, we use the notation
\begin{displaymath}
{\delta \choose 0} = 1 \q \text{ and } \q {\delta \choose k} = \frac{\delta(\delta-1) \cdots (\delta-k+1)}{k!}
\end{displaymath}
for a real number $\delta$ and a positive integer $k$.

\begin{theorem} \label{thm101}
Let $r$ and $m$ be real numbers with $r &gt; -1$ and let $N$ be a nonnegative integer. As $n \to \infty$, we have
\begin{equation}\label{1.7}
\sum_{k=1}^n p_k^r \log^m p_k = \frac{n^{r+1}\log^{r+m}n}{r+1} \left( \sum_{i=0}^N \frac{A_{r,m,i}( \log \log n)}{\log^in} + O_{r,m,N} \left( \frac{(\log \log 
n)^{N+1}}{\log^{N+1}n} \right) \right),
\end{equation}
where the polynomials $A_{r,m,i} \in \R[x]$ are defined by
\begin{equation}\label{1.8}
A_{r,m,0} = 1, \q A_{r,m,i+1}' = A_{r,m,i}' + (m + r - i)A_{r,m,i}.
\end{equation}
The polynomials $A_{r,m,i}$ can be computed explicitly. In particular,
\begin{displaymath}
A_{r,m,1}(x) = (r+m)x - \frac{m-1}{r+1} - r - 1
\end{displaymath}
and
\begin{displaymath}
A_{r,m,2}(x) = {r+m \choose 2} x^2 + \frac{(-r^3 - mr^2 + (-2m+3)r - m^2 + 2m)x}{r+1} + \lambda_{r,m},
\end{displaymath}
where
\begin{displaymath}
\lambda_{r,m} = \frac{(m-1)(m-2)}{(r+1)^2} + \frac{r(r - 3)}{2} - 2.
\end{displaymath}
\end{theorem}

\begin{remark}
The polynomials $A_i$ given in \eqref{1.6} and $A_{r,m,i}$ are connected by the formula $A_i = A_{1,0,i}$.
\end{remark}

\begin{remark}
Since $S_{1,0}(p_n) = S(p_n)$, Theorem \ref{thm101} yields a generalization of \eqref{1.5}.
\end{remark}

\begin{remark}
For some alternative asymptotic formulae for $S(p_n)$, see \cite[Theorems 1 and 2]{axler2015}.
\end{remark}

In the second part of this paper, we study the case (ii); i.e. $r=-1$ and $m \geq 0$. If $m = 0$, we see that $S_{r,m}(x)$ is equal to the sum of the 
reciprocals of all prime numbers not exceeding $x$. Mertens \cite[p.\:52]{mertens1874} proved that $\log \log x$ is the right order of magnitude for this sum
by showing 
\begin{displaymath}
\sum_{p \leq x} \frac{1}{p} = \log \log x + B + O \left( \frac{1}{\log x} \right)
\end{displaymath}
as $x \to \infty$. Here $B$ denotes the Mertens' constant and is defined by
\begin{displaymath}
B = \gamma + \sum_{p} \left( \log \left( 1 - \frac{1}{p} \right) + \frac{1}{p} \right) = 0.261 \ldots,
\end{displaymath}
where $\gamma = 0.577\ldots$ denotes the Euler--Mascheroni constant. So it suffices to consider the case where $r = -1$ and $m&gt;0$. Here, we find the following 
result.

\begin{theorem} \label{thm107}
Let $m$ be a positive real number and let $N$ be a nonnegative integer. As $n \to \infty$, we have
\begin{equation}\label{1.9}
\sum_{k=1}^n \frac{\log^mp_k}{p_k} = \frac{\log^mn}{m} \left( \sum_{i=0}^N \frac{B_{m,i}( \log \log n)}{\log^in} + O_{N,m} \left( \frac{(\log \log 
n)^{N+1}}{\log^{N+1}n} \right) \right),
\end{equation}
where the polynomials $B_{m,i} \in \R[x]$ are defined by
\begin{displaymath}
B_{m,0} = 1, \q B_{m,i+1}' = B_{m,i}' + (m - i)B_{m,i}.
\end{displaymath}
The polynomials $B_{m,i}$ can be computed explicitly. For example, we have
\begin{displaymath}
B_{m,1}(x) = m x \q\q \text{ and } \q\q B_{m,2}(x) = {m \choose 2}x^2 + m x - m.
\end{displaymath}
\end{theorem}

\section{Proof of Theorem \ref{thm101}}

In order to prove Theorem \ref{thm101}, we first note \emph{Abel's summation formula}, which can be found in \cite{ap}.

\begin{lemma}[Abel's summation formula]  \label{lem201}
Let $a : \mathds{N} \rightarrow \mathds{C}$ be a function, and let $A(x) = \sum_{n \leq x} a(n)$, where $A(x)=0$ if $x&lt;1$. If $g$ has a continuous derivative on
the interval $[y,x]$, where $0 &lt; y &lt; x$, then
\begin{displaymath}
\sum_{y&lt; n \leq x} a(n)g(n) = A(x)g(x) - A(y)g(y) - \int_y^x A(t)g'(t)\, \emph{d}t.
\end{displaymath}
\end{lemma}

\begin{proof}
See \cite[Theorem 4.2]{ap}.
\end{proof}

Using this Lemma, we get the following result.

\begin{proposition} \label{prop202}
Let $r$ and $m$ be real numbers with $r &gt; -1$ and let $N$ be a nonnegative integer. For a nonnegative integer $j$, we set
\begin{equation}\label{2.1}
\chi_{r,m,j} = \frac{(-1)^jj!}{(r+1)^j}{m-1 \choose j}.
\end{equation}
As $x \to \infty$, we have
\begin{displaymath}
\int_2^x t^r \log^{m-1} t \,\emph{d}t = \frac{x^{r+1} \log^{m-1}x}{r+1} \sum_{j=0}^N \frac{\chi_{r,m,j}}{\log^jx} + \chi_{r,m,N+1} \int_2^x t^r \log^{m-2-N} t 
\,\emph{d}t + O(1).
\end{displaymath}
\end{proposition}

\begin{proof}
Integration by parts and induction.
\end{proof}

\begin{remark}
In the case where $m$ is a positive integer and $N \geq m-1$, we get $\chi_{r,m,N+1} = 0$. Here, Proposition \ref{prop202} gives
\begin{displaymath}
\int_2^x t^r \log^{m-1} t \,\text{d}t = \frac{x^{r+1} \log^{m-1}x}{r+1} \sum_{j=0}^N \frac{\chi_{r,m,j}}{\log^jx} + O(1)
\end{displaymath}
as $x \to \infty$.
\end{remark}

Proposition \ref{prop202} implies the following asymptotic formula.

\begin{corollary} \label{kor203}
Let $r$ and $m$ be real numbers with $r &gt; -1$ and let $N$ be a nonnegative integer. As $x \to \infty$, we have
\begin{displaymath}
\int_2^x t^r \log^{m-1} t \,\emph{d}t = \frac{x^{r+1} \log^{m-1}x}{r+1} \left( \sum_{j=0}^N \frac{\chi_{r,m,j}}{\log^jx} + O_{r,m,N} \left( 
\frac{1}{\log^{N+1}x} \right) \right).
\end{displaymath}
\end{corollary}

\begin{proof}
Using Lâ€™Hospitalâ€™s rule, we see that
\begin{displaymath} 
\int_2^x t^r \log^{m-2-N} t \,\text{d}t = O_{r,m,N}\left( \frac{x^{r+1}}{\log^{N+2-m}x} \right)
\end{displaymath}
as $x \to \infty$, and it suffices to apply this equation to Proposition \ref{prop202}.
\end{proof}

Now we apply Corollary \ref{kor203} to find the following result.

\begin{proposition} \label{prop204}
Let $r$ and $m$ be real numbers with $r &gt; -1$ and let $N$ be a nonnegative integer. Then
\begin{displaymath}
\sum_{p \leq x} p^r\log^m p = \frac{x^{r+1} \log^{m-1}x}{r+1} \left( \sum_{j=0}^N \frac{\chi_{r,m,j}}{\log^jx} + O_{r,m,N} \left( \frac{1}{\log^{N+1}x} \right) 
\right)
\end{displaymath}
as $x \to \infty$, where $\chi_{r,m,j}$ is defined by \eqref{2.1}.
\end{proposition}

\begin{proof}
For $x \geq 2$, let $R(x) = \pi(x) - \text{li}(x)$. As $x \to \infty$, the asymptotic fomula \eqref{1.1} implies
\begin{equation} \label{2.2}
R(x) = O(xe^{-a\sqrt{\log x}})
\end{equation}
for some positive absolute constant $a$. Furthermore, let $y=3/2$, $g(x) = x^r\log^mx$, and
\begin{displaymath}
a(n) =
\begin{cases}
1, &amp;\text {if $n$ is prime;} \\
0, &amp;\text {otherwise.} \nonumber
\end{cases}
\end{displaymath}
We use Lemma \ref{lem201} to get
\begin{displaymath}
\sum_{p \leq x} p^r \log^m p = (\text{li}(x) + R(x))x^r\log^mx - \int_2^x (\text{li}(t)+R(t))t^{r-1}(r\log^mt + m\log^{m-1}t) \, 
\text{d}t.
\end{displaymath}
If we combine this with \eqref{2.2} and
\begin{displaymath}
\int_2^x t^r \log^{m-1} t \,\text{d}t = \text{li}(x)x^r\log^mx - \int_2^x \text{li}(t)t^{r-1}(r\log^mt + m\log^{m-1}t) \, \text{d}t + O(1)
\end{displaymath}
as $x \to \infty$, we see that
\begin{equation}\label{2.3}
\sum_{p \leq x} p^r \log^m p = \int_2^x t^r \log^{m-1} t \,\text{d}t + O(x^{r+1}e^{-b\sqrt{\log x}}) + O \left( \int_2^x t^re^{-b\sqrt{\log t}} \, \text{d}t 
\right)
\end{equation}
as $x \to \infty$, where the real number $b$ satisfies $0 &lt; b &lt; a$. Using Lâ€™Hospitalâ€™s rule, we get
\begin{displaymath}
\int_2^x t^re^{-a\sqrt{\log t}} \, \text{d}t = O(x^{r+1}e^{-a\sqrt{\log x}})
\end{displaymath}
as $x \to \infty$. So we can rewrite \eqref{2.3} as
\begin{equation}\label{2.4}
\sum_{p \leq x} p^r \log^m p = \int_2^x t^r \log^{m-1} t \,\text{d}t + O(x^{r+1}e^{-b\sqrt{\log x}})
\end{equation}
as $x \to \infty$. Finally, we apply Corollary \ref{kor203} to complete the proof.
\end{proof}

In the following corollary, we give a generalization of \eqref{1.3}.

\begin{corollary} \label{kor205}
Let $r$ be a real number with $r &gt; -1$ and let $m$ be a nonnegative integer. As $x \to \infty$, we have
\begin{displaymath}
\sum_{p \leq x} \frac{p^r}{\log^m p} = \frac{(r+1)^m}{m!} \left( \emph{li}(x^{r+1}) - \sum_{j=0}^{m-1} \frac{j!\,x^{r+1}}{(r+1)^{j+1}\log^{j+1}x} \right) + 
O(x^{r+1}e^{-a\sqrt{\log x}}).
\end{displaymath}
\end{corollary}

\begin{proof}
Induction over $m$ and integration by parts gives
\begin{displaymath}
\int_2^x \frac{t^r}{\log^{m+1}t} \, \text{d}t = \frac{(r+1)^m}{m!} \left( \text{li}(x^{r+1}) - \sum_{j=0}^{m-1} \frac{j!\,x^{r+1}}{(r+1)^{j+1}\log^{j+1}x} 
\right) + O(1)
\end{displaymath}
as $x \to \infty$. Now it suffices to apply the equation \eqref{2.4}.
\end{proof}

To give a proof of Theorem \ref{thm101}, we also need the following result of Salvy \cite{salvy}. Here we use the notation from Robin \cite{robin1988}.

\begin{proposition}[Salvy] \label{prop206}
Let $y = y(x)$ satisfies $e^yy^{- \alpha} D(1/y) \approx x$ as $x \to \infty$, with $D(u) = \sum_{n \geq 0}d_nu^n$ a formal power series, $\alpha \neq 0$, and 
$D(0) \neq 0$. Then for any formal power series $G$ with nonzero constant term the following asymptotic expansion hold:
\begin{displaymath}
e^{\beta y}y^{\gamma} G(1/y) \approx \left( \frac{x}{d_0} \right)^{\beta} (\log x)^{\alpha \beta + \gamma} \sum_{n \geq 0} \frac{Q_n(\log \log x)}{\log^nx} 
\q\q (x \to \infty).
\end{displaymath}
Here $Q_n$ are polynomials with $Q_0 = G(0)$ and $Q_{n+1}'/\alpha = Q_n' + (\alpha \beta + \gamma - n)Q_n$.
\end{proposition}

\begin{proof}
See \cite[Theorem 2]{salvy}.
\end{proof}

Now we can give a proof of Theorem \ref{thm101}.

\begin{proof}[Proof of Theorem \ref{thm101}]
Let $N$ be a nonnegative integer and let $D_N(u) = \sum_{j=0}^N j!\,u^j$. We define the formal power series
\begin{displaymath}
D(u) = \sum_{j=0}^\infty j!\,u^j.
\end{displaymath}
Then $D(0) = 1$. First, we note that repeated integration by parts in \eqref{1.2} gives
\begin{equation}\label{2.5}
\tl(x) \approx \frac{x}{\log x} \, D \left( \frac{1}{\log x} \right)
\end{equation}
as $x \to \infty$. For $x &gt; 1$, the logarithmic integral $\tl(x)$ is increasing with $\tl((1, \infty)) = \R$. Thus, we can define the inverse function 
$\tl^{-1} : \R \to (1, \infty)$ by
\begin{equation}\label{2.6}
\tl(\tl^{-1}(x)) = x.
\end{equation}
We combine \eqref{2.5} and \eqref{2.6} to obtain
\begin{equation}\label{2.7}
e^yy^{-1} D(1/y) \approx n
\end{equation}
as $n \to \infty$, where $y = \log \tl^{-1}(n)$. Next, we define $\delta(n)$ by $p_n = \text{li}^{-1}(n) + \delta(n)$. By Massias and Robin 
\cite[p.\:217]{mr}, we have
\begin{equation}\label{2.8}
\delta(n) = O(ne^{-c\sqrt{\log n}})
\end{equation}
as $n \to \infty$, where $c$ is a positive absolute constant. Since $p_n \sim n \log n$ as $n \to \infty$ (see, for example, \cite{salvy}), we see that 
$\text{li}^{-1}(n) \sim n \log n$ as $n \to \infty$. Substituting $x=p_n$ in Proposition \ref{prop204}, we get
\begin{equation}\label{2.9}
\sum_{k=1}^n p_k^r \log^m p_k = \frac{p_n^{r+1} \log^{m-1}p_n}{r+1} \left( \sum_{j=0}^N \frac{\chi_{r,m,j}}{\log^jp_n} + O_{r,m,N} \left( 
\frac{1}{\log^{N+1}p_n} \right) \right)
\end{equation}
as $n \to \infty$. Using the mean value theorem and \eqref{2.8}, we deduce
\begin{equation}\label{2.10}
\log^{m-1}(p_n) = y^{m-1} +  O(e^{-d\sqrt{\log n}})
\end{equation}
as $n \to \infty$, where $d$ is a real number satisfying $0 &lt; d &lt; c$. Combined with \eqref{2.9}, this gives
\begin{equation}\label{2.11}
\sum_{k=1}^n p_k^r \log^m p_k = \frac{p_n^{r+1} y^{m-1}}{r+1} \left( \sum_{j=0}^N \frac{\chi_{r,m,j}}{\log^jp_n} + O_{r,m,N} \left( \frac{1}{y^{N+1}} \right) 
\right)
\end{equation}
as $n \to \infty$. By the binomial theorem, we have
\begin{displaymath}
p_n^{r+1} = (\text{li}^{-1}(n) + \delta(n))^{r+1} = \text{li}^{-1}(n)^{r+1} + O_r(n^{r+1}e^{-c\sqrt{\log n}})
\end{displaymath}
as $n \to \infty$. Applying this to \eqref{2.11}, we see that
\begin{displaymath}
\sum_{k=1}^n p_k^r \log^m p_k = \frac{e^{(r+1)y} y^{m-1}}{r+1} \left( \sum_{j=0}^N \frac{\chi_{r,m,j}}{\log^jp_n} + O_{r,m,N} \left( \frac{1}{y^{N+1}} 
\right) \right)
\end{displaymath}
as $n \to \infty$. Let $f(x) = 1/\log^kx$, where $k$ is an integer with $0 \leq k \leq m-1$. Again, by the mean value theorem there exists a real $\xi \in 
(\min\{ p_n, \text{li}^{-1}(n) \}, \max\{ p_n, \text{li}^{-1}(n) \})$ such that $f(p_n) = f(\text{li}^{-1}(n)) + \delta(n)f'(\xi)$. Since $f'(x) = 
O(1/(x\log^{k+1}x))$ as $x \to \infty$, we get $f(p_n) = f(\text{li}^{-1}(n)) + O(e^{-c\sqrt{\log n}})$ as $n \to \infty$. Hence
\begin{equation}\label{2.12}
\sum_{k=1}^n p_k^r \log^m p_k = \frac{e^{(r+1)y} y^{m-1}}{r+1} \left( G(1/y) + O_{r,m,N} \left( \frac{1}{y^{N+1}} \right) \right)
\end{equation}
as $n \to \infty$, where
\begin{displaymath}
G(u) = G_{r,m,N}(u) = \sum_{j=0}^N \chi_{r,m,j} \, u^j.
\end{displaymath}
Since \eqref{2.7} holds, we can apply Proposition \ref{prop206} with $\alpha = 1, \beta = r+1$, and $\gamma = m-1$ to see that
\begin{equation}\label{2.13}
\frac{e^{(r+1)y} y^{m-1} G(1/y)}{r+1} = \frac{n^{r+1}\log^{r+m}n}{r+1} \left( \sum_{i=0}^N \frac{A_{r,m,i}( \log_2 n)}{\log^in} + O_{r,m,N} \left( 
\frac{(\log_2 n)^{N+1}}{\log^{N+1}n} \right) \right)
\end{equation}
as $n \to \infty$, where $\log_2x = \log \log x$ and the polynomials $A_{r,m,i} \in \R[x]$ are defined by \eqref{1.8}. If we combine \eqref{2.12} and 
\eqref{2.13}, we arrive at the end of the proof of \eqref{1.7}.
Again, we apply the symbolic algebra system
{\tt Maple} to compute the polynomials $A_{r,m,1}, \ldots, 
A_{r,m,N}$ from the appendices of \cite{salvy}. It suffices to write,
in {\tt Maple}, 
\begin{displaymath}
\texttt{'sum'(p\char'136r\char'052log(p)\char'136m) = 1/(r+1)\char'052theorem2}\_\texttt{part2(1,r+1,m-1,D}\_\texttt{N,G}\_\texttt{\{r,m,N\},n,N);}
\end{displaymath}
with $N = 2$ to get the polynomials $A_{r,m,1}$ and $A_{r,m,2}$. This completes the proof.
\end{proof}

\begin{remark}
We define $\text{lc}(P)$ to be the leading coefficient of a polynomial $P \in \R[x]$. If $r+m \in \N$, we can use \eqref{1.8} to see that the 
polynomials $A_{r,m,0}, \ldots, A_{r,m,N} \in \R[x]$ satisfy
\begin{displaymath}
\deg(A_{r,m,i}) =
\begin{cases}
i, &amp;\text {\text{if} $i \leq r+m$;}\\
i-1, &amp;\text {\text{otherwise},} \nonumber
\end{cases}
\end{displaymath}
and
\begin{displaymath}
\text{lc}(A_{r,m,i}) =
\begin{cases}
\displaystyle {r+m \choose i}, &amp;\text {\text{if} $i \leq r+m$;}\\
\displaystyle (-1)^{i-1-r-m} {i-1 \choose r+m}^{-1}, &amp;\text {\text{otherwise}.} \nonumber
\end{cases}
\end{displaymath}
In the case where $r+m \in \R \setminus (\N \cup \{ 0 \})$, we deduce from \eqref{1.8} that
\begin{displaymath}
\deg(A_{r,m,i}) = i \q \text{and} \q \text{lc}(A_{r,m,i}) = {r + m \choose i}.
\end{displaymath}
If $r+m = 0$, the equation in \eqref{1.8} implies that $A_{r,m,1}' = 0$ and we are not able to say anything in general about the degree or the leading 
coefficient of the polynomials $A_{r,m,i}$, where $i \geq 1$.
\end{remark}

\begin{example}
Let $\log_2x = \log \log x$. We write
\begin{displaymath}
\texttt{'sum'(p\char'136r\char'052log(p)\char'136m) = 1/(r+1)\char'052theorem2}\_\texttt{part2(1,r+1,m-1,D}\_\texttt{N,G}\_\texttt{\{r,m,N\},n,N);}
\end{displaymath}
with $(r,m,N) = (1,-1,2)$, $(r,m,N) = (1,-2,2)$, and $(r,m,N) = (1,-3,2)$ respectively. As $n \to \infty$, this gives the asymptotic formulae
\begin{align*}
\sum_{k=1}^n \frac{p_k}{\log p_k} &amp; = \frac{n^2}{2} \left( 1 - \frac{1}{\log n} + \frac{\log_2n - 3/2}{\log^2 n} + O \left( \frac{(\log_2n)^2}{\log^3 n} 
\right) \right), \\
\sum_{k=1}^n \frac{p_k}{\log^2 p_k} &amp; = \frac{n^2}{2\log n} \left( 1 - \frac{\log_2n + 1/2}{\log n} + \frac{(\log_2n)^2}{\log^2 n} + O \left( 
\frac{(\log_2n)^3}{\log^3 n} \right) \right), \\
\sum_{k=1}^n \frac{p_k}{\log^3 p_k} &amp; = \frac{n^2}{2\log^2 n} \left( 1 - \frac{2\log_2n}{\log n} + \frac{3(\log_2n)^2 -2\log_2n +2}{\log^2 n} + O 
\left( \frac{(\log_2n)^3}{\log^3 n} \right) \right),
\end{align*}
respectively.
\end{example}

\begin{remark}
We have $A_{1,1,0} = 1$. The last example implies that $\text{lc}(A_{1,1,1}) = 1$. Now we can use \eqref{1.8} and induction to get $\text{lc}(A_{1,1,i}) = 1$ 
for all integers $i$ with $i \geq 0$.
\end{remark}

Chebyshev's $\vartheta$-function is defined by $\vartheta(x) = \sum_{p \leq x} \log p$, where $p$ runs over primes not exceeding $x$. Notice that the prime 
number theorem \eqref{1.1} is equivalent to
\begin{displaymath}
\vartheta(x) = x + O(x e^{-c_1\sqrt{\log x}})
\end{displaymath}
as $x \to \infty$, where $c_1$ is a positive absolute constant. Applying a well-known asymptotic expansion for the $n$th prime number (see \cite{salvy}), we 
see that
\begin{displaymath}
\vartheta(p_n) = n \left( \log n + \log_2n - 1 + \frac{\log_2n - 2}{\log n} - \frac{(\log_2n)^2 - 6 \log_2n + 11}{2 \log^2 n} + O\left( 
\frac{(\log_2n)^3}{\log^3n} \right)\right)
\end{displaymath}
as $n \to \infty$, where $\log_2x = \log \log x$, which gives the asymptotic expansion for $S_{r,m}(p_n)$ in the case $r=0$ and $m=1$. For some other cases, we 
apply again the method developed by Salvy \cite[Theorem 2]{salvy} to get the following further results.

\begin{example}
Let $\log_2x = \log \log x$. Again, it suffices to write
\begin{displaymath}
\texttt{'sum'(p\char'136r\char'052log(p)\char'136m) = 1/(r+1)\char'052theorem2}\_\texttt{part2(1,r+1,m-1,D}\_\texttt{N,G}\_\texttt{\{N,r,m\},n,N);}
\end{displaymath}
with $(r,m,N) = (0,2,2)$, $(r,m,N) = (1,1,2)$, and $(r,m,N) = (1,2,1)$, respectively. As $n \to \infty$, this gives the asymptotic formulae
\begin{align*}
\sum_{k=1}^n \log^2 p_k &amp; = n\log^2 n \left( 1 + \frac{2\log_2n - 2}{\log n} + \frac{(\log_2n)^2 - 2}{\log^2 n} + O \left( \frac{(\log_2n)^2}{\log^3 n} \right) 
\right), \\
\sum_{k=1}^n p_k\log p_k &amp; = \frac{n^2\log^2 n}{2} \left( 1 + \frac{2\log_2n - 2}{\log n} + \frac{(\log_2n)^2 - 3}{\log^2 n} + 
+ O \left( \frac{(\log_2n)^2}{\log^3 n} \right) \right), \\
\sum_{k=1}^n p_k\log^2 p_k &amp; = \frac{n^2\log^3n}{2} \left( 1 + \frac{3\log_2n - 5/2}{\log n} + O \left( \frac{(\log_2n)^2}{\log^2 n} \right) \right),
\end{align*}
respectively.
\end{example}

\section{Proof of Theorem~\ref{thm107}}

In 1857, de Polignac \cite[part\:3]{polignac} stated without proof that $\log x$ is the right asymptotic behaviour for $\sum_{p \leq x} \log p/p$ as $x \to 
\infty$, where $p$ runs over primes not exceeding $x$. A rigorous proof of this statement was given by Mertens \cite{mertens1874}. He showed that
\begin{displaymath}
\sum_{p \leq x} \frac{\log p}{p} = \log x + O(1)
\end{displaymath}
as $x \to \infty$. We find the following generalization of this asymptotic formula.

\begin{proposition} \label{prop301}
Let $m$ be a positive real number. As $x \to \infty$, we have
\begin{displaymath}
\sum_{p \leq x} \frac{\log^m p}{p} = \frac{\log^mx}{m} + O(1).
\end{displaymath}
\end{proposition}

\begin{proof}
Similar to Proposition \ref{prop204}.
\end{proof}

Finally, we use Proposition \ref{prop301} to find the following proof of 
Theorem \ref{thm107}.

\begin{proof}[Proof of Theorem~\ref{thm107}]
We apply Proposition \ref{prop301} with $x = p_n$ and use \eqref{2.10} to see that
\begin{displaymath}
\sum_{k=1}^n \frac{\log^mp_k}{p_k} = \frac{y^m}{m} + O(1)
\end{displaymath}
as $n \to \infty$, where $y  = \log \text{li}^{-1}(n)$. Since \eqref{2.7} holds, we can apply Proposition \ref{prop206} with $\alpha = 1, \beta = 0$, and 
$\gamma = m$ to get the asymptotic expansion \eqref{1.9}. In order to compute the polynomials $B_{m,1}, \ldots, B_{m,N}$, we use the {\tt Maple}
code given in the 
appendices of \cite{salvy}. It suffices to write
\begin{displaymath}
\texttt{'sum'(log(p)\char'136m/p) = 1/(r+1)\char'052theorem2}\_\texttt{part2(1,0,m,D}\_\texttt{N,1}\texttt{,n,N);}.
\end{displaymath}
In particular, we get the desired polynomials $B_{m,1}$ and $B_{m,2}$.
\end{proof}

\section{Acknowledgement}
I would like to express my sincere thanks to Jean-Louis Nicolas for his support. Without his helpful comments this paper would not have appeared in this form. 
I would also like to thank R. for being a never-ending inspiration. Further, I would like to thank the anonymous referees for useful suggestions to improve the 
quality of this paper.

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\href{https://cs.uwaterloo.ca/journals/JIS/VOL18/Axler/axler6.html}{Article 15.7.6}.
\bibitem{axler2018} C. Axler, On the sum of the first $n$ prime numbers, {\it J. Th\'eor. Nombres de Bordeaux}, to appear.
\bibitem{mr} J.-P. Massias and G. Robin, Bornes effectives pour certaines fonctions concernant les nombres premiers, {\it J. Th\'eor. Nombres de Bordeaux} 
\textbf{8} (1996), 213--238.
\bibitem{mertens1874} F. Mertens, Ein Beitrag zur analytischen Zahlentheorie, {\it J. Reine Angew. Math.} {\bf 78} (1874), 42--62.
\bibitem{polignac} A. de Polignac, Recherches sur les nombres premiers, {\it Comptes Rendus Acad. Sci. Paris} {\bf 45}, 406--410, 431--434, 575--580, 882--886. 
\bibitem{robin1988} G. Robin, Permanence de relations de rÃ©currence dans certains d\'eveloppements asymptotiques, {\it Publ. Inst. Math.} \textbf{43} (57) 
(1988), 17--25. 
\bibitem{salvy} B. Salvy, Fast computation of some asymptotic functional inverses, {\it J. Symbolic Comput.} \textbf{17} (1994), 227--236.
\bibitem{szalay} M. Szalay, On the maximal order in $S_n$ and $S_n^*$, {\it Acta Arith.} \textbf{37} (1980), 321--331.
\bibitem{vallee1899} C.-J. de la Vall\'{e}e Poussin, Sur la fonction $\zeta(s)$ de Riemann et le nombre des nombres premiers inf\'{e}rieurs \`{a} une limite 
donn\'{e}e, {\it Mem. Couronn\'{e}s de l'Acad. Roy. Sci. Bruxelles} \textbf{59} (1899), 1--74.
\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11N37; Secondary 11A41.

\noindent \emph{Keywords:}
asymptotic formula, logarithmic integral, sum of primes.

\bigskip
\hrule
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\noindent (Concerned with sequences \seqnum{A000040} and \seqnum{A007504}.)

\bigskip
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\bigskip

\vspace*{+.1in}
\noindent
Received December 17 2018;
revised versions received December 19 2018; March 29 2019; July 8 2019.
Published in {\it Journal of Integer Sequences}, August 23 2019.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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