As it is stated, for instance, in Wikipedia, the Riemann hypothesis is equivalent to $$ |\pi(x)-{\rm li}(x)|< \frac1{8\pi}\sqrt x\log x,\qquad \mbox{for all } x\geq 2\,657, $$ but "li" denotes there the complete logarithmic integral: $$ {\rm li}(x)=\int_0^x\frac{dt}{\log t}. $$

I have checked with Mathematica that the inequality fails for $x=2\,656$. However, what happens if the offset logarithmic integral $$ {\rm Li}(x)=\int_2^x\frac{dt}{\log t} $$ is considered instead?

I have checked that, in the range $1\leq x\leq 100\,000$, the corresponding inequality holds for $x\geq 1\,447$.

Is the Riemann hypothesis equivalent to $$ |\pi(x)-{\rm Li}(x)|< \frac1{8\pi}\sqrt x\log x,\qquad \mbox{for all } x\geq 1\,447\ ? $$ All the references I have found deal with li instead of Li.

  • $\begingroup$ Come on $\text{li}(x)= \text{li}(2)+\text{Li}(x)$ so what is the point of your question ? $\endgroup$ – reuns Dec 8 '17 at 0:23
  • $\begingroup$ The point is that this constant could invalidate the specific determination of the constant $1/8\pi$ or the first value of $x$ from which the bound is true. $\endgroup$ – Carlos Dec 8 '17 at 1:14
  • $\begingroup$ So what, who cares to improve $\frac1{8\pi}\sqrt x\log x+\text{li}(2)$ to $\frac1{8\pi}\sqrt x\log x$ ?.. What is interesting is the method and the arguments (about the distribution of non-trivial zeros) letting us obtain those bounds. $\endgroup$ – reuns Dec 8 '17 at 1:20
  • $\begingroup$ I am not saying that the difference is important. I am just asking whether the theorem is stated with li for any reason or the version with Li is also valid. In the second case, the number of exceptions is a bit lower. Nothing important, of course, but I think the statement becomes a bit nicer. $\endgroup$ – Carlos Dec 8 '17 at 1:26
  • $\begingroup$ In that case replace $\frac{1}{8\pi}$ by $1$, it will be even nicer, you can even choose a constant such that it is iff for $x > 1$ $\endgroup$ – reuns Dec 8 '17 at 1:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.