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Theorem (2016)

If we use the value $H_0=30\,610\,046\,000$ obtained by D. Platt in , these bounds are thus valid for $x\le 1.8\cdot 10^{21}$. In , the following bounds are also obtained.Assume the Riemann Hypothesis has been checked up to height $H_0$. Then when $x$ satisfies $\sqrt{x/\log x}\le H_0/4.92$, we have

- $|\psi(x)-x|\le \frac{\sqrt{x}}{8\pi}\log^2x$ when $x > 59$,
- $|\theta(x)-x|\le \frac{\sqrt{x}}{8\pi}\log^2x$ when $x > 599$,
- $|\pi(x)-\text{li}(x)|\le \frac{\sqrt{x}}{8\pi}\log x$ when $x > 2657$.

Theorem (2018)

The subject really started with the four papers , , and . We recall the usual notation: $\pi(x)$ is the number of primes up to $x$ (so that $\pi(3)=2$), the function $\psi(x)$ is the summatory function of the van Mangold function $\Lambda$, i.e. $\psi(x)=\sum_{n\le x}\Lambda(n)$, while we also define $\vartheta(x)=\sum_{p\le x}\log p$. Here are some elegant bounds that one can find in these papers.We have

- $|\psi(x)-x|\le 0.94\sqrt{x}$ when $11 < x\le 10^{19}$,
- $0<\text{li}(x)-\pi(x)\le\frac{\sqrt{x}}{\log x}\Bigl(1.95+\frac{3.9}{\log x}+\frac{19.5}{\log^2x}\Bigr)$ when $2\le x\le 10^{19}$.

Theorem (1962)

There are many other results in these papers. In , on can find among other things the inequality $$ \text{When $x\ge17$, we have } \pi(x) > \frac{x}{\log x -1}. $$ And also

- For $x > 0$, we have $\psi(x)\le 1.03883 x$ and the maximum of $\psi(x)/x$ is attained at $x=113$.
- When $x\ge17$, we have $\pi(x) > x/\log x$.
- When $x > 1$, we have $\displaystyle \sum_{p\le x}1/p > \log\log x$.
- When $x > 1$, we have $\displaystyle \sum_{p\le x}(\log p)/p < \log x$.

Theorem (1999)

This is improved in , and in particular, it is shown that the 515 above can be replaced by 20.83 and also that $$ \text{When $x\ge 89\,967\,803$, we have } \vartheta(x)=x+O^*\Bigl(\frac{x}{\log^3 x}\Bigr). $$ Bounds of the shape $|\psi(x)-x|\le \epsilon x$ have started appearing in . The latest paper is with its corrigendum , where the explicit density estimate from is put to contribution, even for moderate values of the variable. In particular $$ \text{When $x\ge 485\,165\,196$, we have } \psi(x)=x+O^*(0.00053699\,x). $$ In , one find the following estimate

- When $x\ge e^{22}$, we have $\displaystyle\psi(x)=x+O^*\Bigl(0.006409\frac{x}{\log x}\Bigr)$.
- When $x\ge 10\,544\,111$, we have $\displaystyle\vartheta(x)=x+O^*\Bigl(0.006788\frac{x}{\log x}\Bigr)$.
- When $x\ge 3\,594\,641$, we have $\displaystyle\vartheta(x)=x+O^*\Bigl(0.2\frac{x}{\log^2 x}\Bigr)$.
- When $x > 1$, we have $\displaystyle\vartheta(x)=x+O^*\Bigl(515\frac{x}{\log^3 x}\Bigr)$.

Theorem (2021)

Refined bounds for $\pi(x)$ are to be found in and in . By comparing in an efficient manner with $\psi(x)-x$, , obtained the next two results. [There was an error in this paper which is corrected below].When $x\ge e^{2000}$, we have $\biggl|\frac{\psi(x)-x}{x}\biggr|\le 235\,(\log x)^{0.52}\exp-\sqrt{\frac{\log x}{5.573412}}\;.$

Theorem (2013)

For $x > 1$, we have $\sum_{n\le x}\Lambda(n)/n=\log x-\gamma+O^*(1.833/\log^2x)$. When $x\ge 1\,520\,000$, we can replace the error term by $O^*(0.0067/\log x)$. When $x\ge 468\,000$, we can replace the error term by $O^*(0.01/\log x)$. Furthermore, when $1\le x\le 10^{10}$, this error term can be replaced by $O^*(1.31/\sqrt{x})$.

Theorem (2013)

developed the method to incorporate more functions (and corrected the initial work), extending results of . Here are some of his results.For $x\ge 8950$, we have $$ \sum_{n\le x}\Lambda(n)/n=\log x-\gamma +\frac{\psi(x)-x}{x}+O^*\Bigl(\frac{1}{2\sqrt{x}}+1.75\cdot 10^{-12}\Bigl) $$.

Theorem (2017)

For $x\ge 2$, we have $$ \sum_{p\le x}\frac1p=\log\log x+B+O^*\Bigl(\frac{4}{\log^3x}\Bigr). $$ When $x\ge 1000$, one can replace the 4 in the error term by 2.3, and when $x\ge24284$, by 1. The constant $B$ is the expected one.

Theorem (2017)

For $\log x\ge 4635$, we have $$ \sum_{p\le x}\frac1p=\log\log x+B+O^*\Bigl(1.1\frac{\exp(-\sqrt{0.175\log x})}{(\log x)^{3/4}}\Bigr). $$

When truncating sums over primes, Lemma 3.2 of is handy.

Theorem (2016)

The above result relies on (5.1*) of because it is easily accessible. However on using Proposition 5.1 of , one has access to $\epsilon=1/36260$.Let $f$ be a C${}^1$ non-negative, non-increasing function over $[P,\infty[$, where $P\ge 3\,600\,000$ is a real number and such that $\lim_{t\rightarrow\infty}tf(t)=0$. We have \begin{equation*} \sum_{p\ge P} f(p)\log p \le (1+\epsilon) \int_P^\infty f(t) dt + \epsilon P f(P) + P f(P) / (5 \log^2 P) \end{equation*} with $\epsilon=1/914$. When we can only ensure $P\ge2$, then a similar inequality holds, simply replacing the last $1/5$ by a 4.

Here is a result due to .

Theorem (2012)

For $x$ a positive real number. If $x \geq x_0$ then there exist $c_1$ and $c_2$ depending on $x_0$ such that $$ \frac{x^2}{2\log{x}} + \frac{c_1 x^2}{\log^2{x}} \leq \sum_{p \leq x} p \leq \frac{x^2}{2\log{x}} + \frac{c_2 x^2}{\log^2{x}}. $$ The constants $c_1$ and $c_2$ are given for various values of $x_0$ in the next table.

$x_0$ $c_1$ $c_2$ 315437 0.205448 0.330479 468577 0.211359 0.32593 486377 0.212904 0.325537 644123 0.21429 0.322609 678407 0.214931 0.322326 758231 0.215541 0.321504 758711 0.215939 0.321489 10544111 0.239818 0.29251

Further estimates can be found in (Proposition 2.7 and Corollary 2.8). In (Proposition 2.7 and Corollary 2.8) and (Proposition 2.5, Corollary 2.6, 2.7 and 2.8), we find among other results the next two.

Theorem (2019)

Denote by $p_n$ the $n$-th prime. Thus $p_1=2,\;p_2=3,\; p_4=5,\cdots$. The classical form of prime number theorem yields easily $p_n \sim n \log n.$ shows that this equivalence does not oscillate by proving that $p_n$ is greater than $n\log n$ for $n\geq 2$. The asymptotic formula for $p_n$ can be developped as shown in : $$ p_n\sim n\left(\log n+\log\log n -1+\frac{\log\log n-2}{\log n} -\frac{(\ln\ln n)^2-6\log\log n +11}{2\log^2 n}+\cdots\right). $$ This asymptotic expansion is the inverse of the logarithmic integral $\mbox{Li}(x)$ obtained by series reversion. But also proved that for every $n> 1$: $$ n (\log n + \log \log n - 10) < p_n < n (\log n + \log\log n +8). $$ He improves these results in : for every $n\geq 55$, $$ n (\log n + \log \log n - 4) < p_n < n (\log n + \log\log n +2). $$ This result was subsequently improved by Rosser and Schoenfeld in 1962 to $$ n (\log n + \log \log n - 3/2) < p_n < n (\log n + \log\log n -1/2), $$ for $n > 1$ and $n > 19$ respectively. The constants were subsequently reduced by . In particular, the lower bound $$ n (\log n + \log \log n - 1.0072629) < p_n $$ is valid for $n>1$ and the constant $1.0072629$ can be replaced by 1 for $p_k\leq 10^{11}$. Then showed that the lower bound constant equals to 1 was admissible for $p_k < e^{598}$ and $p_k > e^{1800}$. Finally, showed that $$ n(\log n - \log \log n - 1) < p_n $$ for all $n > 1$, and also that $$ p_n < n (\log n + \log\log n - 0.9484) $$ for $n > 39017$ i.e. $p_n > 467\,473$. In , the authors prove the next result.On setting $\pi_r(x)=\sum_{p\le x}p^r$, we have $$ \frac{3x^2}{20(\log x)^4} \le \pi_1(x)-\biggl( \frac{x^2}{2\log x} +\frac{x^2}{4(\log x)^2} \frac{x^2}{4(\log x)^3} \biggr) \le \frac{107x^2}{160(\log x)^4} $$ where the upper estimate is valid when $x\ge 110117910$ and the lower one when $x\ge905238547$.

We have $$ \frac{-1069x^3}{648(\log x)^4} \le \pi_2(x)-\biggl( \frac{x^3}{3\log x} +\frac{x^3}{9(\log x)^2} \frac{x^3}{27(\log x)^3} \biggr) \le \frac{11181x^3}{648(\log x)^4} $$ where the upper estimate is valid when $x\ge 60173$ and the lower one when $x\ge 1091239$.

$$ \pi_3(x)\le 0.271\frac{x^4}{\log x}\quad\text{for $x\ge 664$},$$ $$ \pi_4(x)\le 0.237\frac{x^5}{\log x}\quad\text{for $x\ge 200$},$$ $$ \pi_5(x)\le 0.226\frac{x^5}{\log x}\quad\text{for $x\ge 44$},$$ For $x\ge 1$ and $r\ge5$, we have $$ \pi_r(x)\le \frac{\log 3}{3}\bigl(1+(2/3)^r\bigr) \frac{x^{r+1}}{\log x}$$.

Theorem (2019)

Under the Riemann Hypothesis we have $p_{n+1}-p_n\le\frac{22}{25}\sqrt{p_n}\log p_n$.

In , we find (Theorem 1.6 and 1.7) the next result.

Theorem (2019)

We have $$ \sum_{i\le k}p_i\ge \frac{k^2}4 +\frac{k^2}{4\log k} -\frac{k^2(\log\log k-2.9)}{4(\log k)^2}\quad(\text{for $k\ge 6309751$}), $$ as well as $$ \sum_{i\le k}p_i\le \frac{k^2}4 +\frac{k^2}{4\log k} -\frac{k^2(\log\log k-4.42)}{4(\log k)^2}\quad(\text{for $k\ge 256376$}), $$

In , we find in passing (Lemma 4.8) the next result.

Theorem (2020)

Collecting references: , , , , Lemma 10 of , section 4 of . A consequence of Theorem 1.1 and 1.2 of states thatWe have $$ \sum_{i\le k}\log p_i\le k\bigl(\log k+\log\log -3/4\Bigr)\quad(\text{for $k\ge 4$}), $$ as well as $$ \sum_{i\le k}\log\log p_i\ge k\biggl(\log\log k+\frac{\log\log -5/4}{\log k}\biggr) \quad(\text{for $k\ge319$}). $$

Theorem (2018)

Similarly, as a consequence of Theorem 1.3 of states thatWe have $\displaystyle \max_{3\le q\le 10^4}\max_{ x\ge 8\cdot 10^9}\max_{\substack{1\le a\le q,\\ (a,q)=1}} \frac{\log x}{x}\Bigl| \sum_{\substack{n\le x,\\ n\equiv a[q]}}\Lambda(n) -\frac{x}{\varphi(q)}\Bigr|\le 1/840. $

When $q\in(10^4, 10^5]$, we may replace $1/840$ by $1/160$ and when $q\ge 10^5$, we may replace $1/840$ by $\exp(0.03\sqrt{q}\log^3q)$.

Furthermore, we may replace $\sum_{\substack{n\le x,\\ n\equiv a[q]}}\Lambda(n)$ by $\sum_{\substack{p\le x,\\ p\equiv a[q]}}\log p$ with no changes in the constants.

Theorem (2018)

We have $\displaystyle \max_{3\le q\le 10^4}\max_{ x\ge 8\cdot 10^9}\max_{\substack{1\le a\le q,\\ (a,q)=1}} \frac{\log^2 x}{x}\Bigl| \sum_{\substack{p\le x,\\ p\equiv a[q]}}1 -\frac{\textrm{Li}(x)}{\varphi(q)}\Bigr|\le 1/840. $

When $q\in(10^4, 10^5]$, we may replace $1/840$ by $1/160$ and when $q\ge 10^5$, we may replace $1/840$ by $\exp(0.03\sqrt{q}\log^3q)$.

Concerning estimates with a logarithmic density, in and in , estimates for the functions $\displaystyle\sum_{\substack{n\le x,\\ n\equiv a[q]}}\Lambda(n)/n$ are considered. Extending computations from the former, the latter paper Theorem 8.1 reads as follows.

Theorem (2016)

The constant $C(q,a)$ is the one expected, i.e. such that $\sum_{\substack{n\le x,\\ n\equiv a[q]}}\frac{\Lambda(n)}{n} -\frac{\log x}{\varphi(q)}-C(q,a)$ goes to zero when $x$ goes to infinity.We have $\displaystyle \max_{q\le 1000}\max_{q\le x\le 10^5}\max_{\substack{1\le a\le q,\\ (a,q)=1}} \sqrt{x}\Bigl| \sum_{\substack{n\le x,\\ n\equiv a[q]}}\frac{\Lambda(n)}{n} -\frac{\log x}{\varphi(q)}-C(q,a)\Bigr|\in(0.8533,0.8534) $ and the maximum is attained with $q=17$, $x=606$ and $a=2$.

When $q$ belongs to "Rumely's list", i.e. in one of the following set:

- $\{k\le 72\}$
- $\{k\le 112, \text{$k$ non premier}\}$
- $\begin{aligned}\{116, 117, &120, 121, 124, 125, 128, 132, 140, 143, 144, 156, 163, \\ &169, 180, 216, 243, 256, 360, 420, 432\}\end{aligned}$

Theorem (2002)

More precise bounds are given. , ,When $q$ belongs to Rumely's list and $a$ is prime to $q$, we have $\displaystyle \sum_{\substack{n\le x,\\ n\equiv a[q]}}\frac{\Lambda(n)}{n} =\frac{\log x}{\varphi(q)}+C(q,a)+O^*(1) $ as soon as $x\ge1$.

Last updated on June 3rd, 2022, by Olivier Ramaré