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I think, here, I found $$ P_\color{red}x(\color{blue}s)=\sum_{p<\color{red}x} \frac{1}{p^{\color{blue}s}} =\sum_{\color{green}n=1}^{\infty}\frac{ \mu (\color{green}n)}{\color{green}n} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac z{\color{green}n}-\color{blue}s}) \right]^{\color{red}x}_2 \tag{7}, $$ 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 $$ P(\color{blue}{s})=\sum_{\color{green}n>0}\frac {\mu(\color{green}n)}{\color{green}n}{\log\zeta(\color{green}n\color{blue}s)} $$ (if you forgot, see here/WIKI or here/MathWorld).

So my question is

If $\lim_{\color{red}x\to \infty} P_{\color{red}x}(\color{blue}s)=P(\color{blue}s) $ then is it true that $$ \log\zeta(\color{green}n\color{blue}s)=\lim_{\color{red}x\to \infty} \sum_{z\in\{1,\rho\}}(-1)^{1-\delta_{1z}} \left[ {\rm li}(t^{\frac1n(z-\color{green}n\color{blue}s)}) \right]^{\color{red}x}_2 ? $$

What I got so far is:

  • Thanks to robjohn it possible to see that both coincide at least some special values:
    If $\color{green}n\color{blue}s=1$ or $\color{green}n\color{blue}s=\rho_k$, one addend in the sum diverges like $\lim_{\color{red}x\to\infty} \log\left(\frac{\log(\color{red}x)}{\log(2)}\right)=\infty$. So we get $$ \begin{eqnarray} \color{green}n\color{blue}s=1: & \log(\zeta(1)) &=& +\infty &=& \phantom{-}\left[ {\rm li}(t^{0}) \right]^{\color{red}x}_2 +\sum_{z\in\{\rho\}} \cdots \\ \color{green}n\color{blue}s=\rho_k: & \log(\zeta(\rho)) &=& -\infty &=& -\left[ {\rm li}(t^{0}) \right]^{\color{red}x}_2 +\sum_{z\in\{1,\rho\} / \rho_k} \cdots \\ \end{eqnarray} $$

  • I looked at the series expansion at $\color{blue}s=0$ for ${\rm li}(\color{red}x^{\frac1{\color{green}n}(z-\color{green}i\color{green}n\color{blue}s)})=$ ${\rm Ei}\biggr((\frac z{\color{green}n}-i\color{blue}s)\ln(\color{red}x)\biggr)$ and $\log\zeta(i\color{green}n\color{blue}s)$ (Forget about the $i$, if you like. It was just a try). Assuming I'm not wrong, you'll get the following when you compare the linear terms $$ \log(2\pi) \overset{?}{=} \lim_{\color{red}x\to \infty}\sum_{z\in\{1,\rho\}}(-1)^{\delta_{1z}} \left[ \frac{ t^{\frac{z}{\color{green}n}}}{z}\right]_2^{\color{red}x} , $$ which looks a little irritating, since the RHS has to be independent of $\color{red}x$ and $\color{green}n$. Does this show that it's wrong at all?

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

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Cross-posted at MO... –  draks ... Jan 17 '13 at 20:49
    
Im pretty sure an inverse laplace transform is used with that integral to get an explict formula for the reimann prime counting function, though I don't know much about complex analysis. –  Ethan Jan 24 '13 at 0:13
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