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Let $\pi(x)$ be the number of primes $p$ not exceeding $x, \theta(x) = \sum_{p\leq x} \log p$ and $Li(x)$ be the logarithmic integral.

Is it true that

$$\pi(x)-Li(x) = \theta(x) - x + O(x^{1/2}\log^{2}x) ?$$Or is this equivalent to the Riemann Hypothesis ?

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  • $\begingroup$ Partial summation $\endgroup$
    – reuns
    Commented Aug 18, 2019 at 14:12
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    $\begingroup$ RH certainly implies this, since both differences individually are $\ll\sqrt x\log^2x$ under RH. But $\pi(x)-\mathop{\rm Li}(x)$ is going to act like $1/\log x$ times $\theta(x)-x$; so morally speaking, the difference between the two differences will be as big as $\theta(x)-x$ itself, which is definitely not $\ll\sqrt x\log^2x$ if RH is false. So I think it is equivalent to RH. But I know this isn't a proof. $\endgroup$
    – Greg Martin
    Commented Aug 18, 2019 at 16:52

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