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the prime counting function obeys the integral equation

$$ log \zeta (s) =s \int_{0}^{\infty}dt \frac{\pi (e^{t})}{e^{st}-1} $$

so using the properties of the Mellin transform we have that

$$ \pi (e^{t})= \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty}dsx^{s}\frac{G(s)}{\Gamma (s+1) \zeta (s+1)}$$

with $ G(s)= \int_{0}^{\infty}\frac{dt}{t}t^{s-1}log\zeta (t)$

but why is not this representation used to evaluate the prime counting function

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1 Answer 1

up vote 2 down vote accepted

Because for evaluating $\pi(x)$ the Gram series is better, see www.tandfonline.com/doi/.../00207160903082371. The Gram series of $\pi(x)$ is given by $$ G(x)=1+\sum_{k=1}^{\infty}\frac{(\log x)^k}{k! k\zeta(k+1)}. $$

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