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Does the Riemann hypothesis predict an upper bound for

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

where

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

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

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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