# Asymptotic expansion of $Li^{-1}$ and zeros of $F(s)$ and $G(s)$

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.

## 2 Answers

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.

• thanks, anything on the zeros? – Ultradark Apr 28 at 11:41
• @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 $\Re(s) \ge 2$ – reuns Apr 28 at 22:50
• @reuns. See my edit. In the linked paper, there is much more about this inverse. Cheers. – Claude Leibovici Apr 29 at 6:22
• So is there any chance its Mellin transform has an analytic continuation to $\Re(s) \ge 2$? – Ultradark May 2 at 0:14

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.