# Riemann hypothesis and the logarithmic integral

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.

• Come on $\text{li}(x)= \text{li}(2)+\text{Li}(x)$ so what is the point of your question ? – reuns Dec 8 '17 at 0:23
• 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. – Carlos Dec 8 '17 at 1:14
• 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. – reuns Dec 8 '17 at 1:20
• 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. – Carlos Dec 8 '17 at 1:26
• 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 $x > 1$ – reuns Dec 8 '17 at 1:32