Asymptotic series for the inverse Riemann ζ-function

Question

Consider the positive branch of the Riemann $\zeta$-function: $\zeta:(1,\infty)\to(1,\infty)$ and its inverse $\zeta^{\small(-1)}:(1,\infty)\to(1,\infty)$ satisfying $\zeta\left(\zeta^{\small(-1)}(x)\right) = x$ for $x>1$.

Here are a few initial terms of an asymptotic expansion of $\zeta^{\small(-1)}$ in powers of $x$ (except for the initial term that is logarithmic): $$\begin{align}\zeta^{\scriptscriptstyle(-1)}(1+x)=-\log_2x+{\small\frac1{\ln2}}\Big(&x^{\log_2\left(\frac32\right)}+x+{\small\log_2\left(\tfrac{\sqrt2}3\right)}\cdot x^{2\log _2\left(\frac32\right)}\\+\,&x^{\log _2\left(\frac52\right)}-{\small\log_2 3}\cdot x^{\log_23}\Big)+\mathcal O\Big(x^{3\log_2\left(\frac32\right)}\Big)\color{gray}{,\,x\to0^+,}\end{align} $$ or, equivalently, $$\begin{align}\zeta^{\scriptscriptstyle(-1)}\left(1+2^{-z}\right)=z+{\small\frac1{\ln2}}\Big(&\left(\tfrac32\right)^{-z}+{\small2^{-z}}+{\small\log_2\left(\tfrac{\sqrt2}3\right)}\cdot \left(\tfrac32\right)^{-2z}\\+\,&\left(\tfrac52\right)^{-z}-{\small\log_2 3}\cdot {\small3^{-z}}\Big)+\mathcal O\Big(\left(\tfrac32\right)^{-3z}\Big)\color{gray}{,\,\,z\to\infty.}\end{align} $$

Can we compute more terms of this series or find a general formula for its terms? Does this series converge?

Answer 1

$$F(t)=\zeta(-\log_2 t)-1=t(1+ \sum_{n\ge 3} t^{\log_2 (n) - 1})=t (1+g(t))$$

It is an element of the ring of formal series $$A=\Bbb{R}[[\{t^{1+e}, e\in E\}]], \qquad E= \{ \sum_{l=1}^L m_l (\log_2(n_l)-1), m_l\ge 0, n_l\ge 2\}$$

The main points are that

• the set of exponents $1+E$ is well-ordered, ie. any strictly increasing sequence $\to \infty$,

• for $H(t)=c_0+\sum_{e\in E} c_e t^{1+e}\in A$ then $H(F(t))$ is in $A$:

$$H(F(t))=c_0+\sum_{e\in E} c_e t^{1+e} \sum_{k\ge 0} {1+e\choose k} g(t)^k \in A$$

Whence $F(t)$ has an inverse in $A$.

Start with $G_1=t$, then for $j=1,2,\ldots$

• find $c_j,e_j$ such that $G_j(F(t))= t+c_j t^{1+e_j}(1+o(1))$ and set $G_{j+1}(t) = G_j(t)-c_j t^{1+e_j}$.

We get that $e_j$ is a strictly increasing sequence $\in E$ so $\lim_{j\to \infty} e_j=\infty$ and hence $\lim_{j\to \infty} G_j(t)$ converges in $A$, to $F^{-1}(t)$, giving your asymptotic expansion of $F^{-1}(t)$ at $t=0$ ie. your asymptotic expansion of the inverse of $\zeta(s)$ as $s\to +\infty$.

I don't think the formal series $F^{-1}(t)$ converges for any $t\ne 0$.