Does the Riemann hypothesis predict an upper bound for

$$\left|f(x)-\left(\operatorname{li}(x)-\frac{x}{\log x} \right)\right|,\quad x\ge 2\tag{1}$$


$$f(x)=\sum\limits_{n=2}^x \frac{\Lambda(n)}{\log^2(n)}\tag{2}$$

and $\Lambda(n)$ is the von Mangoldt function?

  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented Aug 14, 2023 at 17:06

1 Answer 1


The Riemann hypothesis is equivalent to the following statement: $$f(x)=\mathrm{li(x)}-\frac{x}{\log x}+O(\sqrt{x}),\qquad x\geq 2.$$ Note that $$\mathrm{li(x)}=\mathrm{li(2)}+\frac{x}{\log x}-\frac{2}{\log 2}+\int_2^x\frac{dt}{\log^2 t},$$ hence the claim is that the Riemann hypothesis is equivalent to $$f(x)=\int_2^x\frac{dt}{\log^2 t}+O(\sqrt{x}),\qquad x\geq 2.\tag{$\ast$}$$

1. The Riemann hypothesis implies $(\ast)$ as follows: \begin{align*} f(x)&=\int_{2-}^x\frac{d\psi(t)}{\log^2 t}\\ &=\int_2^x\frac{dt}{\log^2 t}+\int_{2-}^x\frac{d(\psi(t)-t)}{\log^2 t}\\ &=\int_2^x\frac{dt}{\log^2 t}+\left[\frac{\psi(t)-t}{\log^2 t}\right]_{2-}^x+2\int_2^x\frac{\psi(t)-t}{t\log^3 t}\,dt. \end{align*} Here $\psi(t)-t=O(\sqrt{t}\log^2 t)$ by the Riemann hypothesis, hence we conclude $(\ast)$.

2. Let us denote by $g(x)$ the integral in $(\ast)$. Then $(\ast)$ implies RH as follows: \begin{align*} \psi(x)&=\int_{2-}^x(\log^2 t)\,df(t)\\ &=\int_2^x(\log^2 t)\,dg(t)+\int_{2-}^x(\log^2 t)\,d(f(t)-g(t))\\ &=x-2+\left[(\log^2 t)\,(f(t)-g(t))\right]_{2-}^x-2\int_2^x\frac{\log t}{t}(f(t)-g(t))\,dt. \end{align*} Here $f(t)-g(t)=O(\sqrt{t})$ by $(\ast)$, hence we infer that $\psi(x)=x+O(\sqrt{x}\log^2 x)$, which is a form of the Riemann hypothesis.

  • $\begingroup$ I thought the error bound would perhaps be smaller than $O(\sqrt{x})$. For example $\left|f(x)-\left(\text{li}(x)-\frac{x}{\log x}+2.5\right)\right|<1$ for $x\in\mathbb{Z}\land 2\le x<19371$. $\endgroup$ Commented Aug 14, 2023 at 15:57
  • 3
    $\begingroup$ @StevenClark For every $x_0$ there is $x>x_0$ satisfying $\left|f(x)-\left(\mathrm{li}(x)-\frac{x}{\log x}\right)\right|>\sqrt{x}/\log^2 x$. This is because $|\psi(x)-x|=\Omega(x^{1/2}\log\log\log x)$ by Theorem 15.11 in Montgomery-Vaughan: Multiplicative number theory I. Probably one needs very large values of $x$ to obtain instances $\left|f(x)-\left(\mathrm{li}(x)-\frac{x}{\log x}\right)\right|>\sqrt{x}/\log^2 x$, and I suspect that values around $e^{1000}$ should exist in abundance. Compare with en.wikipedia.org/wiki/Skewes%27s_number $\endgroup$
    – GH from MO
    Commented Aug 14, 2023 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.