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If we consider this version of the prime-counting function $$\pi_0(x) = \frac{1}{2} \lim_{h\to 0} (\pi(x+h) + \pi(x-h))$$ (with $\pi$ being the normal prime-counting function), then we can write $\pi_0$ as $$\pi_0(x) = \sum_n\frac{1}{n}\mu(n)f(x^{1/n}),$$ where $\mu$ is the Mertens function and $$f(x) = \operatorname{li}(x) - \sum_\rho \operatorname{Ei}(\rho \cdot \log(x)) -\log(2) +\int_x^\infty\frac{dt}{t(t^2-1)\log(t)},$$ the $\rho$ being the non-trivial zeros of the Riemann zeta function.

I am wondering, if

  • it is known how inacurrate the formula for $\pi_0$ gets (with absolute constants, not O notation), if we just consider all zeros with absolute imaginary part $|\Im(\rho)|<T$ for some $T>0$.
  • if there is a similar formula, not for $\pi_0$, but for the $n$-th prime number, and, if so, how much that would be affected by considering just the aforementioned selection non-trivial zeros.
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  • $\begingroup$ I believe you meant $\mu(n)$ is the Möbius function. The Mertens function is $M(x)=\sum\limits_{n=0}^x\mu(n)$. The integral term above can also be written as $-\sum\limits_n Ei(-2\,n\,\log(x))$ which I've found much faster for evaluation purposes. $\endgroup$
    – Steven Clark
    Commented May 9, 2019 at 0:37

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