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I think, here, I found $$ P_x(s)=\sum_{p < x} \frac{1}{p^s} =\sum_{n=1}^{\infty}\frac{ \mu (n)}{n} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac zn-s}) \right]^{x}_2 \tag{7} $$

where $\rho$ are all the zeros (trivial and non-trivial) of $\zeta$ function. See the linked question for more detail, corrections are welcome. Further we know, that $$ P(s)=\sum_{n> 0}\frac {\mu(n)}n{\log\zeta(ns)} . $$

So my question is

If $\lim_{x\to \infty} P_x(s)=P(s) $ then $$ \log\zeta(ns)=\lim_{x\to \infty} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac1n( z-ns)}) \right]^{x}_2 ? $$

What I got so far is:

  • Could $ \displaystyle \log \zeta(s) = s \int_0^\infty \frac{\pi(x)}{x(x^s-1)}\,dx $ be useful somehow?

  • Thanks to robjohn it possible to see that both coincide at least some special values:
    If $ns=1$ or $ns=\rho$, one addend in the sum diverges like $\lim_{x\to\infty} \log\left(\frac{\log(x)}{\log(2)}\right)=\infty$. So we get $$ \begin{eqnarray} ns=1: & \log(\zeta(1)) &=& +\infty\\ ns=\rho: & \log(\zeta(\rho)) &=& -\infty \end{eqnarray} $$

  • I looked at the series expansion at $s=0$ for ${\rm li}(x^{\frac1n(z-ns)})=$ ${\rm Ei}((\frac zn-s)\ln(x))$ and $\log\zeta(ns)$. Assuming I'm not wrong, you'll get the following when you compare the linear terms $$ \color{grey}{ins}\log(2\pi) \overset{?}{=} \color{grey}{ins} \lim_{x\to \infty}\sum_{z\in\{1,\rho\}}(-1)^{\delta_{1z}} \left[ \frac{ t^{\frac{z}n}}{z}\right]_2^x , $$ which looks a little irritating, since the RHS has to be independent of $x$ an $n$. Does this show that it's wrong at all?

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Cross-posted at math.stackexchange.com/q/280984/19341 –  draks ... Jan 17 '13 at 20:50
I think by $z\in 1,\rho$ you mean to sum over the (one) pole and all the zeros of $\zeta(s)$. This might be clearer if you separated out the contribution of the pole. –  Stopple Jan 17 '13 at 21:43
Yes but I wanted a compact, notation. –  draks ... Jan 17 '13 at 22:01

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