Explicit estimate of sum of primes

Question

Let $p$ be prime. In this post, Charles gave the following answer $$\sum_{p\leq x}p=\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{3x^2}{8\log^4 x}+O\left(\frac{x^2}{\log^5 x}\right). $$ I have two problems which I don't understand well. The first is how can we get an such estimate as above? The second is that can we get an explicit form of the estimate above, that is can we compute a concrete number $c$ such that for $x>100$, $$\left|\sum_{p\leq x}p-\left(\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{3x^2}{8\log^4 x}\right)\right|\leq c\left(\frac{x^2}{\log^5 x}\right)? $$ Also if we take fewer terms to compute $k$ such that for $x>100$, $$\left|\sum_{p\leq x}p-\left(\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}\right)\right|\leq k\left(\frac{x^2}{\log^3 x}\right)? $$ Are there references on these problems?

Answer 1

I quote from Christian Axler, New bounds for the sum of the first $n$ prime numbers, https://arxiv.org/abs/1606.06874

Let $\pi(x)$ denote the number of primes not exceeding $x$. de la Vallée-Poussin estimated the error term in the Prime Number Theorem by showing that $$\pi(x)={\rm li}(x)+O(xe^{-a\sqrt{\log x}})\tag1$$ where $a$ is a positive absolute constant and the logarithmic integral ${\rm li}(x)$ is defined for every real $x\ge0$ as $${\rm li}(x)=\int_0^x{dt\over\log t}=\lim_{\epsilon\to0+}\left\{\int_0^{1-\epsilon}{dt\over\log t}+\int_{1+\epsilon}^x{dt\over\log t}\right\}\tag2$$ Denoting the sum of the first prime numbers not exceeding $x$ by $S(x)$, Szalay [24, Lemma 1] used (1) to find $$S(x)={\rm li}(x^2)+O(x^2e^{-a\sqrt{\log x}})\tag3$$ Using (3) and integration by parts in (2), we get the asymptotic expansion [given in the first line of the question].

[The Szalay reference is M. Szalay, On the maximal order in $S_n$ and $S_n^*$, Acta Arith. 37 (1980) 321–331.]

Then Axler proves various results like

For every $x \ge 110 118 925$, we have $$S(x)<\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{5.3x^2}{8\log^4 x} $$ and

For every $x \ge 905 238 547$, we have$$S(x)>\frac{x^2}{2\log x}+\frac{x^2}{2\log^2 x}+\frac{x^2}{4\log^3 x}+\frac{1.2x^2}{8\log^4 x} $$