# Questions on Riemann's explicit formula

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 $$|\Im(\rho)| for some $$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.
• 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. – Steven Clark May 9 '19 at 0:37