3

Yesterday I came up with an asymptotic expansion for the partial sums of the prime zeta function $$\mathcal P(x)=\sum_{p\le x}\frac1{p^s},\quad p\in\Bbb P$$ with the extra constraint that $s\in\Bbb Z^+\setminus\{1\}$. This was done by first considering the sum $$A(x)=\sum_{n\le x}\frac{a(n)}n\log n=\log x+\mathcal O(1)$$ where $a(n)=1$ if and only if $n\in\Bbb P$, and using Abel's summation formula to give $$B(x)=\sum_{n\le x}\frac{a(n)}{n^s}\log n=x^{1-s}A(x)+(s-1)\int_1^x\frac{A(t)}{t^s}\,dt=\mathcal O(1-x^{1-s}).$$ Abel's summation formula was used once more to give $$\mathcal P(x)=\frac{B(x)}{\log x}+\int_2^x\frac{B(t)}{t\log^2t}\,dt=\mathcal O\left(\int_2^x\frac{1-t^{1-s}}{t\log^2t}\,dt\right)$$ as an asymptotic expansion (this can also be written in terms of the exponential integral function). However, this is not too meaningful, as the RHS consists of $\mathcal O$ terms only.

Are there better asymptotic expansions of $\mathcal P(x)$ (in literature or otherwise) that include non-$\cal O$ terms as well?

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005

1 Answers1

1

You really need to understand that $\sum_{p\le x} p^{-s}$ is just an obfuscated version of $$\sum_{n\le x} \Lambda(n) n^{-s}= \frac1{2i\pi} \int_{2-i\infty}^{2+i\infty} \frac{-\zeta'(z+s)}{z\zeta(z+s)}x^{z}dz$$ $$=Res(\frac{-\zeta'(z+s)}{z\zeta(z+s)}x^{z},1-s)+Res(\frac{-\zeta'(z+s)}{z\zeta(z+s)}x^{-z},0)-\frac1{2i\pi} \int_{\sigma_0-\Re(s)+\epsilon-i\infty}^{\sigma_0-\Re(s)+\epsilon+i\infty} \frac{-\zeta'(z+s)}{z\zeta(z+s)}x^{z}dz$$ $$ = \frac{x^{1-s}}{1-s}-\frac{\zeta'(z)}{\zeta(z)}+O(x^{\sigma_0-\Re(s)+\epsilon})$$

where $\sigma_0=\sup_\rho \Re(\rho)$ is $1/2$ iff the RH is true.

Under the PNT we only know that $\sigma_0\le 1$ and we need to replace $\int_{\sigma_0-\Re(s)+\epsilon-i\infty}^{\sigma_0-\Re(s)+\epsilon+i\infty}$ by an integral over the boundary of the zero-free region obtaining $\sum_{n\le x} \Lambda(n) n^{-s}=\frac{x^{1-s}}{1-s}-\frac{\zeta'(z)}{\zeta(z)}+ O(x^{1-\Re(s)}e^{-\log^{1/10}(x)})$.

reuns
  • 77,999