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If you downvote please leave some constructive feedback.

I would like to compare and visualize/gain insight about the zeros of two functions, $F(s)$ and $G(s).$ $\pi(m)$ is the prime counting function.

$$F(s)=\sum_{m=2}^\infty \pi(m)^{-s}=\sum_{n=1}^\infty n^{-s} (p_{n+1}-p_n)= \sum_{n=1}^\infty p_{n+1} (n^{-s}-(n+1)^{-s})\\ =\sum_{n=1}^\infty n \ln n (1+o(1)) (s n^{-s-1}+O(s(s+1)n^{-s-2}).$$

Therefore it converges and is analytic for $\Re(s) > 1$ and as $s \to 1,$ $F(s) \sim -s\zeta'(s) \sim \frac{1}{(s-1)^2}.$

For the analytic continuation under the RH, $n=\pi(p_n) = Li(p_n)+O(p_n^{1/2+\epsilon}),$ thus

$p_n = Li^{-1}(n+O(n^{1/2+\epsilon}))=Li^{-1}(n)+O(n^{1/2+\epsilon}),$

and,

$$F(s)-s\sum_{n=1}^\infty n^{-s-1} Li^{-1}(n)$$ is analytic for $\Re(s) > 1/2.$

So now the question is, what is the asymptotic expansion of $Li^{-1}?$

Is the remainder $O(x^a)$ with the remainder $a<1?$

After analytic continuation where does $F(s)=0?$

Where does $G(s)=0?$

$$G(s)=\sum_{n=1}^\infty p_nn^{-s}= \frac{2}{1^s}+\frac{3}{2^s}+\frac{5}{3^s}+\frac{7}{4^s}+\frac{11}{5^s}+... ,$$ where $p_n$ is the nth prime.

Plots would be preferable.

Is there literature on $G(s)?$

Thanks.

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First of all $$\mathrm{li}(z)=\mathrm{Ei}(\ln(z))$$ and the problem would be to find the inverse of the exponential integral function.

For that, you could find a very interesting approach in the paper entitled

"On the function inverse to the exponential integral function"

published by P. Pecina in

Astronomical Institutes of Czechoslovakia, Bulletin (ISSN 0004-6248), vol. 37, Jan. 1986, pp 8-12

Have a look here and you can access the pdf of the full paper which even contains the Fortran code for high accuracy.

On the other hand, if I properly remember,

$$Li^{-1}(n) \sim n \left(\log (n)+\log (\log (n))-1+\frac{\log (\log (n))-2}{\log (n)}+\cdots\right)$$

Edit

After @reuns's comment, I asked a former colleague of mine; he gave me the next term inside brackets $$-\frac{\log ^2(\log (n))-6 \log (\log (n))+11}{2 \log ^2(n)}$$ This was published by Cesaro (have a look here) in 1894.

So, in a more compact form

$$Li^{-1}(n) \sim n \left(L_1+L_2-1+\frac{L_2-2}{L_1}-\frac{L_2^2-6 L_2+11}{2 L_1^2}+\cdots\right)$$ where $L_1=\log(n)$ and $L_2=\log(L_1)$. The part in brackets "looks" like the expansion of Lambert function for infinitely large values of the argument.

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  • $\begingroup$ thanks, anything on the zeros? $\endgroup$ – Ultradark Apr 28 at 11:41
  • $\begingroup$ @Ultradark It doesn't make any sense to ask about the zeros. It is not for nothing that locating the zeros of $\zeta(s)$ is so hard. @ ClaudeLeibovici What are the next terms, what is the growth of the coefficients, does it mean there is any chance its Mellin transform has an analytic continuation to $\Re(s) \ge 2$ $\endgroup$ – reuns Apr 28 at 22:50
  • $\begingroup$ @reuns. See my edit. In the linked paper, there is much more about this inverse. Cheers. $\endgroup$ – Claude Leibovici Apr 29 at 6:22
  • $\begingroup$ So is there any chance its Mellin transform has an analytic continuation to $\Re(s) \ge 2$? $\endgroup$ – Ultradark May 2 at 0:14
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If you control the error terms then $Li^{-1}(n) \sim p_n$. Hence all you need is the asymptotic expansion of the $n$-th prime. This was first proved by Marin Cipolla way back in 1902. See Theorem 2.1 here.

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