1 Introduction↩︎

Multiple zeta values are defined by \[\zeta(k_1,\cdots,k_r)=\sum_{n_1>\cdots>n_r>0}\frac{1}{n_1^{k_1}\cdots n_r^{k_r}},\qquad k_1\geq 2,k_2,\cdots, k_{r}\geq 1.\] For a multiple zeta value \(\zeta(k_1,\cdots,k_r)\), denote by \(N=k_1+\cdots+k_r\) and \(r\) its weight and depth respectively.

For \(k_1\geq 2,k_2,\cdots, k_{r}\geq 1\), the multiple zeta-star value \(\zeta^{\star}(k_1,\cdots,k_r)\) is defined by \[\zeta^{\star}(k_1,\cdots,k_r)=\sum_{n_1\geq\cdots\geq n_r\geq 1}\frac{1}{n_1^{k_1}\cdots n_r^{k_r}}.\] The theory of multiple zeta-star values has been developed in parallel with, but not merely as a byproduct of, the theory of multiple zeta values. Its algebraic foundations are closely related to Hoffman’s work on multiple harmonic series and harmonic algebras [1],[2]. Important structural and identity results were obtained by Ohno-Wakabayashi [3], Ohno-Zudilin [4], Ihara-Kajikawa-Ohno-Okuda [5], Muneta [6], Kaneko-Ohno [7], Yamamoto [8], Zhao [9], and others.

In [10], the author discovered the following bijective map:
Denote by \[\mathbb{Z}_{+}=\{1,2,\cdots, n,\cdots\}\] and \[\mathcal{T}=\big{\{} (k_1,\cdots,k_r,\cdots)\;\big{|}\; k_i\in \mathbb{Z}_{+} , i\geq 1; k_1\geq 2; \mathrm{if}\; k_1=2,\, \mathrm{then}\; k_i\geq 2 \;\mathrm{for\;some} \;i\geq 2 \big{\}}.\] There is a bijective map \[\eta: \mathcal{T}\rightarrow (1,+\infty),\] \[{\boldsymbol{k}}=(k_1,\cdots,k_r,\cdots)\mapsto \eta({\boldsymbol{k}})=\mathop{\mathrm{lim}}_{r\rightarrow +\infty} \zeta^{\star}(k_1,\cdots,k_r).\] We call the map \(\eta\) zeta-star correspondence. In fact, one can compare the theory of zeta-star correspondence to the theory of continued fractions. For more related topics on zeta-star correspondence, see[11], [12], [13],[14],[10]. Hirose, Murahara and Onozuka [15] discovered the above bijective map independently.

For \((k_1,\cdots,k_r)\in (\mathbb{Z}^+)^r\) and \(n\in \mathbb{Z}^+\), define \[H_n^\star( \varnothing )=1,\;\;\;\;H_n^\star(k_1, \cdots,k_r)=\sum_{n\geq m_1\geq \cdots \geq m_r\geq 1} \frac{1}{m_1^{k_1}\cdots m_r^{k_r}}.\] For a fixed \(n\geq 2\), is there any kind of zeta-star correspondence map for \(H_n^\star\)?

In fact, for \(n=2\), Theorem 1 is essentially another reformulation of the binary expansion. For fixed \((k_1,\cdots,k_r)\in (\mathbb{Z}^+)^r\), the multiple harmonic sum \[H_n^{\star}(k_1,\cdots,k_r)\] can be viewed as a function of the integer variable \(n\). More generally, is there any kind of zeta-star correspondence map for \(\mathfrak{h}_s^\star\) when \(s\) is a complex variable instead of a positive integer?

To achieve the above goals, we have to first define \(H_s^\star(k_1,\cdots,k_r)\). For \(\mathrm{Re}\;s>0\) and \(k_1\in \mathbb{Z}^+\), define \[G_n^\star(s;k_1)=\sum_{m=1}^n\frac{1}{(m+s)^{k_1}} .\]

\[H_s^\star(\varnothing)=1,\quad H_s^\star(k_1)=\mathop{\mathrm{lim}}_{M\rightarrow +\infty} \bigg{(}H_M^\star(k_1)-G_M^\star(s;k_1) \bigg{)}.\] It is clear that \[H_s^\star(k_1)\big{|}_{s=n}=H_n^\star(k_1)\]for \(n\in \mathbb{Z}^+\).

For \(r\geq 2\),\((k_1,\cdots,k_r)\in(\mathbb{Z}^+)^r\) and \(\mathrm{Re}(s)>0\), by induction, define \[G_n^\star(s;k_1,\cdots,k_r)=\sum_{n\geq m_1\geq \cdots \geq m_r\geq 1} \frac{1}{(m_1+s)^{k_1}\cdots (m_r+s)^{k_r}}.\] \[H_s^\star(k_1, \cdots,k_r) = \mathop{\mathrm{lim}}_{M\rightarrow +\infty} \Big{(} H_M^\star(k_1,\cdots,k_r)- \sum_{j=1}^r G_M^\star(s;k_1,\cdots, k_j)H_s^\star (k_{j+1},\cdots, k_r) \Big{)} .\]

In fact, we have

A divided difference is a generalization of the ordinary difference quotient to several points. It is a fundamental tool in interpolation, approximation theory and numerical analysis. For a survey about divided differences, see [16]. We will give a brief introduction in Section 3.

Formula in Theorem 1, \((ii)\) for \(s=n\) appears to be an elementary divided-difference reformulation of the known integral representations of finite multiple harmonic star sums. The existence of integral representations for such sums is classical in this context, especially in Yamamoto’s work [17], while the identity connecting divided differences of monomials with complete homogeneous symmetric polynomials is standard. We have not found an explicit occurrence of the exact divided-difference cube-integral form in the literature.

For \(\mathrm{Re}(s)>0\), does the limit \[\mathop{\mathrm{lim}}_{r \rightarrow +\infty} H_s^\star(k_1, \cdots,k_r)\] exist for any \((k_1,\cdots, k_r,\cdots)\in (\mathbb{Z}^+)^\infty\)?

When \(s\) is a complex number instead of positive integer, we have

We call \(\mathfrak{h}_s\) the finite Zeta-star correspondence. By the Grothendieck period conjecture or the Kontsevich-Zagier conjecture [18], it is widely believed that all the multiple zeta-star values are transcendental. Thus one of the fundamental questions in the theory of Zeta-star correspondence is to determine the infinite sequence \(\eta^{-1}(x)\in \mathcal{T}\) for a rational or algebraic number \(x>1\). As \[\mathop{\mathrm{lim}}_{\mathrm{Re} (s)\rightarrow +\infty} \mathfrak{h}_s\Big{|}_{\mathcal{T}}=\eta\] Theorem 1 may gives us a potential way to achieve this goal by the theory of complex analysis.

Theorem 1 gives us another way to interpret the theory of multiple zeta-star values. Roughly speaking, zeta-star correspondence can be viewed as a complex deformation of binary expansion at the infinite point.

In general, we believe that the following conjecture is true.

2 Finite zeta-star correspondence in integer cases↩︎

In this section, we will give a total order structure on the set of finite multiple star harmonic sums. Theorem 1 is proved as an application. The analysis in this section is similar to the analysis of multiple zeta-star values in [10]. The main references are [10], [13].

Denote by \[\widehat{\mathcal{S}}=\Big{\{}(k_1,\cdots,k_r)\,\Big{|}\,k_1,\cdots,k_r\in\left(\mathbb{Z}^{+}\right)^r, r\geq 1\Big{\}}.\] There is a total order structure \(\succ\) on the set of \(\widehat{\mathcal{S}}\) which is defined by: \[(k_1,\cdots,k_r,k_{r+1})\succ (k_1,\cdots,k_r )\] and \[(k_1,\cdots,k_p)\succ (l_1,\cdots,l_q)\] if \((k_1,\cdots,k_{i-1})=(l_1,\cdots,l_{i-1})\) and \(k_i<l_i\) for some \(i\leq \mathrm{min}\{p,q\}\). If \[(k_1,\cdots,k_p)\succ (l_1,\cdots,l_q),\] we also write as \[(l_1,\cdots,l_q)\; \rotatebox[origin=c]{180}{\text{ \succ}} (k_1,\cdots,k_p).\]

It is clear that \[\begin{align} &\;\;\;\;H_n^\star(\{1\}^r)\\ &<\sum_{n\geq m_1\geq \cdots \geq m_r\geq 1} \frac{1}{m_1\cdots m_{r-1}}=\sum_{n\geq m_1\geq \cdots\geq m_{r-1}\geq 1}\frac{m_{r-1}}{ m_1\cdots m_{r-1} }=\cdots=n. \\ \end{align}\]

\(\hfill\Box\)

\((i)\) For \(n=1\), this is trivial. For \(n\geq 2\), as \[\begin{align} &\;\;\;\; H_n^\star(\{1\}^r) \\ &= \sum_{n\geq m_1\geq \cdots \geq m_r\geq 1} \frac{1}{m_1\cdots m_r} \\ &= \left( \sum_{\substack{n\geq m_1\geq \cdots \geq m_r\geq 1\\ m_1=1 }} + \sum_{\substack{n\geq m_1\geq \cdots \geq m_r\geq 1\\ m_1\geq 2,m_2=1 }} +\cdots+ \sum_{\substack{n\geq m_1\geq \cdots \geq m_r\geq 1\\ m_r\geq 2 }} \right) \frac{1}{m_1\cdots m_r} \\ &=1+\sum_{n\geq m_1\geq 2}\frac{1}{m_1}+\cdots+\sum_{n\geq m_1\geq \cdots\geq m_r\geq 2}\frac{1}{m_1\cdots m_r} , \\ \end{align}\] it follows that \[\begin{align} &\;\;\;\; \mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_n^\star(\{1\}^r) \\ &=1+\sum_{n\geq m_1\geq 2}\frac{1}{m_1}+\cdots+\sum_{n\geq m_1\geq \cdots\geq m_r\geq 2}\frac{1}{m_1\cdots m_r} +\cdots \\ &= \prod_{n\geq m\geq 2} \frac{1}{1-\frac{1}{m}} \\ &=n. \end{align}\] \((ii)\) By definition, \[\begin{align} &\;\;\;\; H_n^\star(k_1,\cdots,k_{l-1}, k_l+1,\{1\}^r) \\ &= \sum_{n\geq m_1\geq \cdots\geq m_{l}\geq m_{l+1}\geq \cdots \geq m_{l+r}\geq 1} \frac{1}{m_1^{k_1}\cdots m_{l-1}^{k_{l-1}}m_l^{k_l+1} m_{l+1}\cdots m_{l+r}} \\ &= \sum_{n\geq m_1\geq \cdots\geq m_{l}\geq 1} \frac{1}{m_1^{k_1}\cdots m_l^{k_l} } \cdot \frac{1}{m_l} \sum_{m_l\geq m_{l+1}\geq \cdots \geq m_{l+r}\geq 1} \frac{1}{m_{l+1}\cdots m_{l+r}}. \\ \end{align}\] By \((i)\), \[\mathop{\mathrm{lim}}_{r\rightarrow +\infty} \frac{1}{m_l} \sum_{m_l\geq m_{l+1}\geq \cdots \geq m_{l+r}\geq 1} \frac{1}{m_{l+1}\cdots m_{l+r}}=1.\] Thus \[\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_n^\star(k_1,\cdots,k_{l-1}, k_l+1,\{1\}^r) =H_n^\star(k_1,\cdots,k_{l-1}, k_l) .\] \(\hfill\Box\)

For \(n\geq 2\), by definition, \[\begin{align} &\;\;\;\; H_n^\star(k_1,\cdots,k_r,k_{r+1}) \\ &= \sum_{n\geq m_1\geq \cdots\geq m_r\geq m_{r+1}\geq 1} \frac{1}{m_1^{k_1}\cdots m_r^{k_r}m_{r+1}^{k_{r+1}} } \\ &= \left(\sum_{\substack{n\geq m_1\geq \cdots\geq m_r\geq m_{r+1}\geq 1\\ m_{r+1}=1}} +\sum_{\substack{n\geq m_1\geq \cdots\geq m_r\geq m_{r+1}\geq 1\\ m_{r+1}\geq 2}} \right)\frac{1}{m_1^{k_1}\cdots m_r^{k_r}m_{r+1}^{k_{r+1}} } \\ &=H_n^{\star}(k_1,\cdots,k_r)+\sum_{n\geq m_1\geq \cdots\geq m_r\geq m_{r+1}\geq 2} \frac{1}{m_1^{k_1}\cdots m_r^{k_r}m_{r+1}^{k_{r+1}} } . \end{align}\] So we have \[H_n^\star(k_1,\cdots,k_r,k_{r+1}) >H_n^{\star}(k_1,\cdots,k_r).\] For \[(k_1,\cdots,k_p)\succ (l_1,\cdots,l_q),\] we assume that \((k_1,\cdots,k_{i-1})=(l_1,\cdots,l_{i-1})\) and \(k_i<l_i\) for some \(i\leq \mathrm{min}\{p,q\}\). Then by Lemma 1, \[\begin{align} &\;\;\;\; H_n^\star(l_1,\cdots,l_q) \\ &\leq H_n^\star(l_1,\cdots,l_{i-1},l_i,\{1\}^{q-i}) \\ &\leq \sum_{n\geq m_1\geq \cdots\geq m_i\geq 1} \frac{1}{m_1^{l_1}\cdots m_i^{l_i}} \cdot \sum_{m_i\geq m_{i+1}\geq \cdots \geq m_q\geq 1} \frac{1}{m_{i+1}\cdots m_q}\\ &<\sum_{n\geq m_1\geq \cdots\geq m_i\geq 1} \frac{m_i}{m_1^{l_1}\cdots m_i^{l_i}} =H_n^\star(l_1,\cdots,l_{i-1},l_i-1). \\ \end{align}\] Since \(H_n^\star(l_1,\cdots,l_{i-1},l_i-1)\leq H_n^\star(k_1,\cdots,k_i)\leq H_n^\star(k_1,\cdots,k_p)\), we have \[H_n^\star(k_1,\cdots,k_p)>H_n^\star(l_1,\cdots,l_q).\] \(\hfill\Box\)
Now we are ready to prove Theorem 1. For \((k_1,\cdots,k_r,\cdots)\in \widehat{\mathcal{T}}=(\mathbb{Z}^+)^\infty\) and \(n\geq 2\), define \[a_r=H_n^\star(k_1,\cdots,k_r).\] By Proposition 1, we have \(a_r<a_{r+1}\). By Lemma 1, \[a_r\leq H_n^\star(\{1\}^r)<n.\] As a result, the sequence \(\{a_r\,|\,r\geq 1\}\) is an increasing bounded sequence, thus the limit \[\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_n^{\star}(k_1,\cdots,k_r)\] exists and the map \[\mathfrak{h}_n: \widehat{\mathcal{T}}\rightarrow (1, n],\] \[{\boldsymbol{k}}=(k_1,\cdots,k_r,\cdots)\mapsto \mathfrak{h}_n({\boldsymbol{k}})=\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_n^{\star}(k_1,\cdots,k_r).\] is well-defined. It suffices to show that \(\mathfrak{h}_n\) is bijective.

For \({\boldsymbol{k}},{\boldsymbol{l}}\in \widehat{\mathcal{T}}\) and \({\boldsymbol{k}}\neq {\boldsymbol{l}}\), without loss of generality, we assume that \[(k_1,\cdots,k_i)=(l_1,\cdots,l_i), \quad k_{i+1}>l_{i+1}.\] By Proposition 1, for \(r>i+1\), we have \[H_n^\star(k_1,\cdots, k_i, k_{i+1},\cdots k_r) <H_n^\star(k_1,\cdots, k_i, k_{i+1}-1) \leq H_n^\star(l_1,\cdots,l_i,l_{i+1}).\] Therefore \[\mathfrak{h}_n({\boldsymbol{k}})=\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_n^{\star}(k_1,\cdots,k_r)\leq H_n^\star(l_1,\cdots,l_i,l_{i+1}).\] On the other hand \[H_n^\star(l_1,\cdots,l_i,l_{i+1})< \mathfrak{h}_n({\boldsymbol{l}})=\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_n^{\star}(l_1,\cdots,l_r) .\] In conclusion, we have \[\mathfrak{h}_n({\boldsymbol{k}}) < \mathfrak{h}_n({\boldsymbol{l}}).\] So the map \(\mathfrak{h}_n\) is injective.

To show that \(\mathfrak{h}_n\) is surjective, for \(r\geq 1\), define \[Z_{\{1\}^r}(n)=\Big{(}H_n^\star(\{1\}^r), n\Big],\] \[Z_{k_1,\cdots,k_r}(n)=\Big{(}H_n^\star(k_1,\cdots,k_r), H_n^\star(k_1,\cdots ,k_{i-1},k_i-1) \Big]\] for \((k_1,\cdots,k_r)\neq (\{1\}^r)\) and \[k_i\geq 2,k_{i+1}=\cdots=k_r=1.\] From the total order structure of multiple star harmonic sums, we have \[Z_{k_1,\cdots,k_r}(n)\bigcap Z_{l_1,\cdots,l_r}(n)=\varnothing,\quad (k_1,\cdots,k_r)\neq (l_1,\cdots,l_r),\] \[(1,n]=\bigcup_{(k_1,\cdots,k_r)\in (\mathbb{Z}^+)^r} Z_{k_1,\cdots,k_r}(n), \quad r\geq 1,\] \[Z_{k_1,\cdots,k_r}(n)=\bigcup_{(k_{r+1},\cdots,k_{r+p})\in (\mathbb{Z}^+)^p} Z_{k_1,\cdots,k_r,k_{r+1},\cdots,k_{r+p}}(n) .\]

By definition, it suffices to show that \[H_n^\star(k_1,\cdots,k_r)+m\left( Z_{k_1,\cdots,k_r}(n) \right)= \sum_{n\geq m_1\geq \cdots \geq m_r\geq 1}\frac{m_r}{m_1^{k_1}\cdots m_r^{k_r}}.\] For \((k_1,\cdots,k_r)=(\{1\}^r)\), \[\begin{align} &\;\;\;\; \sum_{n\geq m_1\geq \cdots \geq m_r\geq 1}\frac{m_r}{m_1\cdots m_r} \\ &= \sum_{n\geq m_1\geq \cdots \geq m_r\geq 1}\frac{1}{m_1\cdots m_{r-1}} \\ &=\sum_{n\geq m_1\geq \cdots \geq m_{r-1}\geq 1}\frac{m_{r-1}}{m_1\cdots m_{r-1}} \\ &\qquad\cdots\\ &=\sum_{n\geq m_1\geq 1}\frac{m_1}{m_1}\\ &=n\\ &= H_n^\star(\{1\}^r)+m\left( Z_{\{1\}^r}(n) \right). \end{align}\] For \((k_1,\cdots,k_r)\neq (\{1\}^r)\) and \[k_i\geq 2,k_{i+1}=\cdots=k_r=1,\] similarly, we have \[\begin{align} &\;\;\;\; \sum_{n\geq m_1\geq \cdots \geq m_r\geq 1}\frac{m_r}{m_1^{k_1}\cdots m_i^{k_i} m_{i+1}\cdots m_r} \\ &= \sum_{n\geq m_1\geq \cdots \geq m_{r-1}\geq 1}\frac{m_{r-1}}{m_1^{k_1}\cdots m_i^{k_i} m_{i+1}\cdots m_{r-1}} \\ &\qquad\cdots \\ &= \sum_{n\geq m_1\geq \cdots \geq m_{r-1}\geq 1}\frac{m_i}{m_1^{k_1}\cdots m_i^{k_i} } \\ &= H_n^\star(k_1,\cdots,k_r)+m\left( Z_{k_1,\cdots,k_r}(n) \right). \end{align}\] \(\hfill\Box\)
For \(x\in (1,n]\), if \(x=n\), by Lemma 1, one has \[x=n=\mathfrak{h}_n(({\{1\}^{\infty}})) .\] If \(x=H_n^\star(k_1,\cdots,k_r)\), by Lemma 1, one has \[x=H_n^\star(k_1,\cdots,k_r) =\mathfrak{h}_n\left( (k_1,\cdots, k_{r-1}, k_r+1,\{1\}^{\infty}) \right) .\] If \[x\neq H_n^\star(k_1,\cdots,k_r)\] for all \((k_1,\cdots,k_r)\in \left( \mathbb{Z}^+ \right)^r\), \(r\geq 1\), from the basic properties of \[Z_{k_1,\cdots,k_r}(n),\] there is a \[{\boldsymbol{k}}=(k_1,\cdots,k_r,\cdots)\in \widehat{\mathcal{T}}\] such that \[x\in Z_{k_1,\cdots,k_r}(n)\] for all \(r\geq 1\). As a result, \[H_n^\star (k_1,\cdots,k_r) <x\leq H_n^\star (k_1,\cdots,k_r) + m\left( Z_{k_1,\cdots,k_r}(n) \right).\] Since \[\begin{align} &\;\;\;\; m\left( Z_{k_1,\cdots,k_r}(n) \right)\\ &= \sum_{n\geq m_1\geq \cdots \geq m_r\geq 1}\frac{m_r-1}{m_1^{k_1}\cdots m_r^{k_r}} \\ &\leq \sum_{n\geq m_1\geq \cdots \geq m_r\geq 1}\frac{m_r-1}{m_1\cdots m_r}=n-H_n^\star\left(\{1\}^r\right) , \\ \end{align}\] we have \[\mathop{\mathrm{lim}}_{r\rightarrow +\infty} m\left( Z_{k_1,\cdots,k_r}(n) \right) = \mathop{\mathrm{lim}}_{r\rightarrow +\infty} \left( n-H_n^\star\left(\{1\}^r\right) \right) =0.\] Therefore, \[x= \mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_n^\star (k_1,\cdots,k_r) =\mathfrak{h}_n({\boldsymbol{k}}).\] In a word, the map \(\mathfrak{h}_n: \widehat{\mathcal{T}}\rightarrow (1,n]\) is bijective.

3 Analytic continuation of finite multiple star harmonic sums↩︎

In this section, we will review the theory of divided differences. For more references, see [22], [16]. By using the divided differences, we will give the integral representation of \[H_s^\star(k_1,\cdots,k_r).\] As an application, we will show that \(H_s^\star(k_1,\cdots,k_r)\) is well-defined and analytic on \(\mathrm{Re}(s)>0\). In this paper, for \(t>0\) and \(s\in\mathbb{C}\), the function \(t^s\) is defined by \[t^s=e^{s\,\mathrm{log}\,t}.\]

For the function with sufficient differentiability, the divided difference has the following geometric explanation, which is known as Hermite-Genocchi formula. For the history of Hermite-Genocchi formula, one can consult the survey [16].

For \(n=1\), it is clear that \[[x_0,x_1]f=\frac{f(x_1)-f(x_0)}{x_1-x_0}=\int^1_0f^\prime \left( (1-u)x_0+ux_1 \right)du.\] For \(n\geq 2\), by induction, \[[x_0,\cdots, x_{n-1}]f=\int_{S_{n-1}} f^{(n-1)}(u_0x_0+\cdots +u_{n-1}x_{n-1})du_1\cdots du_{n-1}.\] Here \[u_0=1-u_1-\cdots-u_{n-1}.\] By the definition and the symmetry of divided difference, \[\begin{align} &\;\;\;\;[x_0,\cdots,x_n]f \\ & =\frac{ [x_0,\cdots,x_{n-2},x_n]f- [x_0,\cdots,x_{n-2},x_{n-1}]f }{x_n-x_{n-1}} \\ &= \int_{S_{n-1}} \frac{ F(u_1,\cdots, u_{n-2},x_n)- F(u_1,\cdots, u_{n-2},x_{n-1}) }{x_n-x_{n-1}} du_1\cdots du_{n-1} , \\ \end{align}\] which \[F(u_1,\cdots, u_{n-2}, y)=f^{(n-1)}\left( u_0x_0+\cdots +u_{n-2}x_{n-2}+u_{n-1}y \right).\] For fixed \((u_0,u_1,\cdots, u_{n-1})\), denote by \[A=u_0x_0+\cdots +u_{n-2}x_{n-2}+u_{n-1}x_{n-1},\] \[B=u_0x_0+\cdots+u_{n-2}x_{n-2}+u_{n-1} x_n,\] then \[B-A=u_{n-1}(x_n-x_{n-1}).\] By the Newton-Leibniz formula, one has \[f^{(n-1)}(B)-f^{(n-1)}(A)=(B-A)\int^1_0 f^{(n)}\left( (1-s)A+sB \right)ds.\] Thus \[\frac{f^{(n-1)}(B)-f^{(n-1)}(A)}{x_n-x_{n-1}}=u_{n-1} \int^1_0 f^{(n)}\left( (1-s)A+sB \right)ds.\] As a result, \[\begin{align} &\;\;\;\; [x_0,\cdots, x_n]f\\ &= \int_{S_{n-1}} \left( \int^1_0 u_{n-1} f^{(n)}\left( \sum_{i=0}^{n-2} u_ix_i+u_{n-1}(1-s)s_{n-1}+u_{n-1}sx_n \right) ds \right) du_1\cdots du_{n-1} . \\ \end{align}\] By changing of variables: \[y_i=u_i, \quad 0\leq i\leq n-2,\] \[y_{n-1}=u_{n-1}(1-s),\quad y_n=u_{n-1}s,\] then \[y_0+\cdots+y_n=1,\quad y_i\geq 0.\] Therefore \[[x_0,\cdots, x_n]f=\int_{S_n} f^{(n)}(y_0x_0+y_1x_1+\cdots+y_nx_n) dy_1\cdots dy_n.\] \(\hfill\Box\)

One can prove the above statement by the explicit formula of divided difference in Definition 1. Here we will show that the above statement is a direct corollary of the Leibniz formula for divided differences.

By [22], we have the following Leibniz formula \[[y_0,\cdots, y_m]\left(fg\right)=\sum_{k=0}^m [y_0,\cdots, y_k]f\cdot [y_k,\cdots,y_m]g. \] By the Leibniz formula \((1)\), it follows that \[[y_0, \cdots, y_m]\left(tg(t) \right)=y_0[y_0,\cdots, y_m]g+[y_1,\cdots,y_m]g. \] By the symmetry of divided difference with respect to \[y_0,y_1,\cdots, y_m,\] it is easy to check that \[[y_0,\cdots, y_m]g= \frac{[y_0,y_2,\cdots, y_m]g-[y_1,y_2,\cdots,y_m]g }{y_0-y_1} . \] As a result, \[\begin{align} & \;\;\;\; [1,x_1,x_2,\cdots,x_r]\left(tg(t)\right) \\ &= [1,x_1,x_2,\cdots,x_r]g + [x_1,x_2,\cdots,x_r]g \\ &= \frac{ [1,x_2,\cdots,x_r]g - [x_1,x_2,\cdots,x_r]g }{1-x_1} + [x_1,x_2,\cdots,x_r]g \\ &= \frac{ [1,x_2,\cdots,x_r]g-x_1[x_1,x_2,\cdots,x_r]g }{1-x_1} . \end{align}\] \(\hfill\Box\)

For \(u_i\neq u_j, i\neq j\), we have the following explicit formula \[[u_0,u_1,\cdots, u_r]t^{M+r}=\sum_{i=0}^r \frac{u_i^{M+r}}{\prod_{j\neq i} (u_i-u_j)}\] Thus \[\sum_{M=0}^{+\infty}[u_0,u_1,\cdots, u_r]t^{M+r}\cdot X^M=\sum_{i=0}^r \frac{u_i^{r}}{\prod_{j\neq i} (u_i-u_j)}\cdot \frac{1}{1-u_iX}.\] By the partial-fraction decomposition, it is easy to check that \[\sum_{i=0}^r \frac{u_i^{r}}{\prod_{j\neq i} (u_i-u_j)}\cdot \frac{1}{1-u_iX}=\prod_{i=0}^r\frac{1}{1-u_iX}.\] As a result, \[\begin{align} & \sum_{M=0}^{+\infty}[u_0,u_1,\cdots, u_r]t^{M+r}\cdot X^M = \prod_{i=0}^r\frac{1}{1-u_iX} =\sum_{N=0}^{+\infty} h_M(u_0,\cdots,u_r) X^M. \end{align}\] Thus the Lemma is proved for \(u_i\neq u_j,i\neq j\). In general cases, the Lemma follows from continuous extension. \(\hfill\Box\)
Let \[X_0=1,X_j=x_1x_2\cdots x_{k_1+\cdots +k_j}, \;\;\; 1\leq j\leq r.\] By Lemma 1, for \(M\in \mathbb{Z}^+\), one has \[\begin{align} &\;\;\;\; \int_{(0,1)^{k_1+\cdots+k_r}}[X_0,X_1,\cdots,X_r]t^{M+r-1}dx_1dx_2\cdots dx_{k_1+\cdots+k_r} \\ &= \int_{(0,1)^{k_1+\cdots+k_r}} h_{M-1}(X_0,\cdots,X_r) dx_1dx_2\cdots dx_{k_1+\cdots+k_r} \\ &= \int_{(0,1)^{k_1+\cdots+k_r}} \sum_{\substack{ a_0+\cdots +a_r=M-1 \\ a_0,\,\cdots,a_r\geq 0 }} X_0^{a_0}\cdots X_r^{a_r} dx_1dx_2\cdots dx_{k_1+\cdots+k_r} \\ &= \int_{(0,1)^{k_1+\cdots+k_r}} \sum_{\substack{ a_0+\cdots +a_r=M-1 \\ a_0,\,\cdots,a_r\geq 0 }} \\ &\;\;\;\; (x_1\cdots x_{k_1})^{a_1+\cdots a_r}(x_{k_1+1}\cdots x_{k_1+k_2} )^{a_2+\cdots +a_r} \cdots (x_{k_1+\cdots +k_{r-1}+1}\cdots x_{k_1+\cdots +k_r})^{a_r}\\ &\;\;\;\;dx_1dx_2\cdots dx_{k_1+\cdots+k_r} \\ &=\int_{(0,1)^{k_1+\cdots+k_r}} \sum_{M\geq m_1\geq m_2\geq\cdots\geq m_r\geq 1} \\ &\;\;\;\; (x_1\cdots x_{k_1})^{m_1-1}(x_{k_1+1}\cdots x_{k_1+k_2} )^{m_2-1} \cdots (x_{k_1+\cdots +k_{r-1}+1}\cdots x_{k_1+\cdots +k_r})^{m_r-1} \\ &\;\;\;\; dx_1dx_2\cdots dx_{k_1+\cdots+k_r} \\ &= \sum_{M\geq m_1\geq\cdots\geq m_r\geq 1} \frac{1}{m_1^{k_1}\cdots m_r^{k_r}}\\ &=H_M^\star(k_1,\cdots,k_r). \end{align} \] By the same analysis, one has \[\begin{align} &\;\;\;\; \int_{(0,1)^{k_1+\cdots+k_r}}[X_0,X_1,\cdots,X_r]t^{M+r-1}\cdot (x_1x_2\cdots x_{k_1+\cdots +k_r})^sdx_1dx_2\cdots dx_{k_1+\cdots+k_r} \\ &= \int_{(0,1)^{k_1+\cdots+k_r}} h_{M-1}(X_0,\cdots,X_r) \cdot (x_1x_2\cdots x_{k_1+\cdots +k_r})^s dx_1dx_2\cdots dx_{k_1+\cdots+k_r} \\ &= G_M^\star(s;k_1,\cdots,k_r). \\ \end{align} \] In conclusion, we have \[H_M^\star(k_1,\cdots,k_r)= \int_{(0,1)^{k_1+\cdots+k_r}}[X_0,X_1,\cdots,X_r]t^{M+r-1}dx_1\cdots dx_{k_1+\cdots+k_r},\] \[G_M^\star(s;k_1,\cdots,k_r)=\int_{(0,1)^{k_1+\cdots+k_r}}[X_0,X_1,\cdots,X_r]t^{M+r-1}\cdot (x_1\cdots x_{k_1+\cdots +k_r})^sdx_1\cdots dx_{k_1+\cdots+k_r} .\]

\((i)\) By Lemma 1, for \(r\geq 1\), \[[1,X_1,\cdots,X_r]t^{s+r-1}=\frac{ [1,X_2,\cdots ,X_r]t^{s+r-2}-X_1[X_1,X_2,\cdots, X_r]t^{s+r-2} }{1-X_1}\] By the definition of divided difference, we have \[[X_1,X_2,\cdots, X_r]t^{s+r-2}=X_1^{s-1} \left[1, \frac{X_2}{X_1},\cdots, \frac{X_r}{X_1} \right] t^{s+r-2}.\] Thus \[\begin{align} &\;\;\;\;A_r(s; X_1, \cdots, X_r) \\ &= \frac{1}{1-X_1} [1,X_2,\cdots,X_r]t^{s+r-2}-\frac{X_1^s}{1-X_1} \left[1, \frac{X_2}{X_1},\cdots, \frac{X_r}{X_1} \right] t^{s+r-2} \\ &= \frac{1}{1-X_1}A_{r-1}(s; X_2,\cdots,X_r)-\frac{X_1^s}{1-X_1}A_{r-1}\left( s; \frac{X_2}{X_1},\cdots, \frac{X_r}{X_1} \right). \\ \end{align}\] \((ii)\) For \(r=1\), the statement \((i)\) is proved in \((i)\). By induction, one has \[\begin{align} &\;\;\;\;A_r(s; X_1, \cdots, X_r) \\ &= \frac{1}{1-X_1}A_{r-1}(s; X_2,\cdots,X_r)-\frac{X_1^s}{1-X_1}A_{r-1}\left( s; \frac{X_2}{X_1},\cdots, \frac{X_r}{X_1} \right) \\ &= \frac{1}{1-X_1}\left[\frac{1}{(1-X_2)\cdots (1-X_r)}- \sum_{j=2}^r \frac{X_j^s}{(1-X_2)\cdots (1-X_j)} A_{r-j}\left(s; \frac{X_{j+1}}{X_j},\cdots, \frac{X_r}{X_j} \right) \right] \\ &\;\;\;\;-\frac{X_1^s}{1-X_1}A_{r-1}\left( s; \frac{X_2}{X_1},\cdots, \frac{X_r}{X_1} \right) \\ &= \frac{1}{P_r}-\sum_{j=1}^r \frac{X_j^s}{P_j}A_{r-j}\left(s; \frac{X_{j+1}}{X_j},\cdots, \frac{X_r}{X_j} \right) . \end{align}\] \((iii)\) The formula follows immediately from the Leibniz formula \((1)\) of divided differences and the formula \[[y_1,y_2,\cdots, y_r]t^{s+r-2}=y_1^{s-1} \left[1, \frac{y_2}{y_1},\cdots, \frac{y_r}{y_1} \right] t^{s+r-2}.\] \((iv)\) By Lemma 1, for \(1\leq j\leq r\) and \(M\geq 1\), \[\begin{align} &\;\;\;\; A_j(M; X_1,\cdots, X_j) \\ &= [1,X_1, \cdots, X_j] t^{M+j-1} \\ &=h_{M-1}(1,X_1,\cdots,X_j) \\ &= \sum_{q=0}^{M-1} h_q(X_1,\cdots, X_j). \\ \end{align}\] As \[0<X_r<\cdots<X_1<1,\] one has \[\mathop{\mathrm{lim}}_{M\rightarrow +\infty} A_j(M; X_1,\cdots, X_j) = \sum_{q=0}^{+\infty} h_q(X_1,\cdots, X_j)=\frac{1}{\prod_{i=1}^j (1-X_i) }=\frac{1}{P_j}.\] So we have \[\begin{align} &\;\;\;\; \mathop{\mathrm{lim}}_{M\rightarrow +\infty} P_{M,r}(s; X_1,\cdots,X_r) \\ &= \frac{1}{P_r}-\sum_{j=1}^r X_j^s \cdot \frac{1}{P_j} A_{r-j}\left( s; \frac{X_{j+1}}{X_j},\cdots, \frac{X_r}{X_j} \right) \\ &= A_r(s; X_1,\cdots, X_r). \end{align}\] \(\hfill\Box\)

By the Hermite-Genocchi formula in Theorem 1 for \(f(t)= t^{s+r-1}\). \[\begin{align} &\;\;\;\; I_s(k_1,\cdots,k_r) \\ &= \int_{(0,1)^{K_r}} \left( \int_{S_r} f^{(r)} \left( u_0+u_1X_1+\cdots+u_rX_r \right)du_1\cdots du_r \right)dx_1dx_2\cdots dx_{K_r}.\\ \end{align}\] Here \(f^{(r)}(t)=(s+r-1)\cdots (s+1)s \cdot t^{s-1}\) and \[S_r=\big{\{}(u_1,\cdots,u_r)\in \mathbb{R}^r\,\big{|}\, u_j\geq 0, \sum_{j=1}^r u_j\leq 1\big{\}},\qquad u_0=1-\sum_{j=1}^r u_j.\] Let \[\sigma_0=\mathop{\mathrm{inf}}_{s\in\Omega}\mathrm{Re}\,s>0.\] As \[0\leq u_0+u_1X_1+\cdots+u_rX_r \leq 1,\] there is a constant \(C_{\Omega,r}^\prime\) that \[\Big{|}[1,X_1,\cdots ,X_r] t^{s+r-1}\Big{|}\leq C_{\Omega,r}^\prime \int_{S_r} \left( u_0+u_1X_1+\cdots+u_rX_r \right)^{\sigma_0-1} du_1\cdots du_r.\] Since \(0\leq u_0\leq 1\) and \(X_r\leq X_j\leq 1\) for all \(j\), \[u_0+u_1X_1+\cdots +u_rX_r\geq u_0+(1-u_0)X_r\geq \frac{u_0+X_r }{2}.\] Thus \[\Big{|}[1,X_1,\cdots ,X_r] t^{s+r-1}\Big{|}\leq D_{\Omega,r} \int_{S_r} \left( u_0+X_r \right)^{\sigma_0-1} du_1\cdots du_r.\] By direct calculation, \[\begin{align} &\;\;\;\; \int_{S_r} \left( u_0+X_r \right)^{\sigma_0-1} du_1\cdots du_r \\ &= \int_{S_r} \left( 1+X_r -u_1-\cdots-u_r \right)^{\sigma_0-1} du_1\cdots du_r \\ &=\int_{1\geq v_1\geq \cdots \geq v_r\geq 0} \left(1+X_r-v_1 \right)^{\sigma_0-1}dv_1\cdots dv_r \\ &\leq \int^1_0(1+X_r-v_1)^{\sigma_0-1}dv_1\\ &\leq \frac{ (X_r+1)^{\sigma_0}-X_r^{\sigma_0} }{\sigma_0} \\ &\leq \frac{2^{\sigma_0}}{\sigma_0}. \end{align}\] Therefore \[\Big{|} [1,X_1,\cdots, X_r]t^{s+r-1} \Big{|}\leq C_{\Omega,r}\] for some \(C_{\Omega,r}\). For each fixed \((x_1,\cdots,x_{K_r})\) in the open cube \((0,1)^{K_r}\), the function \(s\mapsto [1,X_1,\cdots, X_r]t^{s+r-1}\) is entire. By the Morera’s theorem and dominated convergence, \(I_s(k_1,\cdots,k_r)\) is holomorphic on \(\mathrm{Re}\,s>0\). \(\hfill\Box\)

Define \[\sigma_0=\mathop{\mathrm{inf}}_{s\in \Omega} \mathrm{Re}\,s>0,\quad B=\mathop{\mathrm{sup}}_{s\in \Omega}\,|s|.\] For \(r=1\), \[P_{M,1}(s;X_1)=A_1(M; X_1)-X_1^sA_1(M;X_1).\] Thus \[\begin{align} &\;\;\;\; P_{M,1}(s;X_1)-A_1(s;X_1) \\ &= [1,X_1]t^M-X_1^s [1,X_1]t^M-[1,X_1]t^s \\ &= \frac{1-X_1^{M}}{1-X_1}-X_1^s \frac{1-X_1^{M}}{1-X_1} -\frac{1-X_1^s}{1-X_1} \\ &=-\frac{ X_1^M(1-X_1^s ) }{1-X_1} . \\ \end{align}\] Now we have \[\begin{align} &\;\;\;\; \int_{(0,1)^{K_1}} \Big{|} P_{M,1}(s; X_1 ) -A_1(s;X_1) \Big{|}dx_1\cdots dx_{K_1} \\ &= \int_{(0,1)^{K_1}} \Bigg{|} \frac{ X_1^M(1-X_1^s ) }{1-X_1} \Bigg{|}dx_1\cdots dx_{K_1} \\ &\leq \int_{(0,1)^{K_1}} \sum_{m\geq 0} \Big{|}X_1^{M+m}(1-X_1^s) \Big{|} dx_1\cdots dx_{K_1}. \\ \end{align}\] Since \[1-x^s=s \int^1_x t^{s-1}dt,\quad x\in(0,1),\] one has \[\big{|}1-x^s\big{|}\leq \big{|} s \big{|} \int^1_x t^{\sigma_0-1}dt\] For \(s\in \Omega\), if \(\sigma_0\geq 1\), then \[\big{|}1-x^s\big{|}\leq B(1-x), \quad x\in(0,1).\] If \(0<\sigma_0<1\), \[\big{|}1-x^s\big{|}\leq \big{|} s \big{|} \int^1_x t^{\sigma_0-1}dt \leq B\cdot \frac{1-x^{\sigma_0} }{\sigma_0} \leq B\cdot \frac{1-x}{\sigma_0},\quad x\in(0,1).\] In both cases, for \(s\in\Omega\), we have \[\big{|}1-x^s\big{|}\leq C(1-x), \quad x\in (0,1), \] which \(C=B\,\mathrm{max}\Big{\{} 1,\frac{1}{\sigma_0} \Big{\}}\). Thus \[\begin{align} &\;\;\;\; \int_{(0,1)^{K_1}} \Big{|} P_{M,1}(s; X_1 ) -A_r(s;X_1) \Big{|}dx_1\cdots dx_{K_1} \\ &\leq \int_{(0,1)^{K_1}} \sum_{m\geq 0} \Big{|}X_1^{M+m}(1-X_1^s) \Big{|} dx_1\cdots dx_{K_1} \\ &\leq C \int_{(0,1)^{K_1}} \sum_{m\geq 0} \Big{|}X_1^{M+m}(1-X_1) \Big{|} dx_1\cdots dx_{K_1} \\ &\leq C \int_{(0,1)^{K_1}} X_1^{M} dx_1\cdots dx_{K_1} \\ &\leq \frac{C}{(M+1)^{K_1}}. \end{align}\] By induction, we assume that the statement is proved for \(r<n\). For \(r=n\), by Lemma 1 and Lemma 1, we have \[\begin{align} &\;\;\;\; \int_{(0,1)^{K_r}} \Big{|} P_{M,r}(s; X_1,\cdots,X_r ) -A_r(s;X_1,\cdots, X_r) \Big{|}dx_1\cdots dx_{K_r}\\ &= \int_{(0,1)^{K_r}} \Big{|} A_r(M;X_1,\cdots, X_r) - A_r(M+s;X_1,\cdots, X_r) \Big{|}dx_1\cdots dx_{K_r}\\ &= \int_{(0,1)^{K_r}} \Big{|} [1,X_1,\cdots, X_r](t^{M+r-1}-t^{M+s+r-1}) \Big{|}dx_1\cdots dx_{K_r}.\\ \end{align}\] By the explicit formula of divided differences in Definition 1, one has \[[1,y_1,\cdots,y_r]\left( (t-1)g(t) \right)= [y_1,\cdots,y_r]\left( g(t) \right). \] Define \[\phi_s(t)=\frac{1-t^s}{1-t}, G_{M,r}(t)=t^{M+r-1}\phi_s(t),\] then \[t^{M+r-1}-t^{M+s+r-1} =(1-t) G_{M,r}(t).\] As \[\phi_s(t)=s\int^1_0 \left( t+u(1-t) \right)^{s-1}du,\] for \(0\leq j\leq r-1\) we have \[\phi_s^{(j)}(t)= s(s-1)\cdots (s-j)\int^1_0(1-u)^j \left( t+u(1-t) \right)^{s-1-j}du.\] Thus for \(s\in \Omega\) and \(t\in (0,1)\), \[\big{|} \phi_s^{(j)}(t) \big{|}\leq L_{\Omega,r} \int^1_0 (t+u(1-t))^{\mathrm{Re}\,s-1-j}du,\quad 0\leq j\leq r-1.\] For \(u\in [0,1]\), one has \(0<t+u(1-t)\leq 1\). Therefore \[\big{|} \phi_s^{(j)}(t) \big{|}\leq L_{\Omega,r} \int^1_0 (t+u(1-t))^{\sigma_0-1-j}du.\] For \(0<t<1\), \(t+u(1-t)\geq t\). It follows that \[\left(t+u(1-t)\right)^{-r}\leq t^{-r}.\] Thus \[\big{|} \phi_s^{(j)}(t) \big{|}\leq L_{\Omega,r}t^{-j} \int^1_0 (t+u(1-t))^{\sigma_0-1}du= L_{\Omega,r}t^{-j} \frac{ 1-t^{\sigma_0}}{\sigma_0(1-t)} .\] If \(0<\sigma_0\leq 1\), it is clear that \[0\leq \frac{1-t^{\sigma_0}}{1-t}\leq 1.\] If \(\sigma_0\geq 1\), then \[\frac{1-t^{\sigma_0}}{1-t}=\sigma_0\xi^{\sigma_0-1}\leq \sigma_0,\] which \(\xi\in (0,1)\). In both cases, we have \[\big{|} \phi_s^{(j)}(t) \big{|}\leq C_{\Omega,r} t^{-j}, \quad t\in(0,1). \] Here \(C_{\Omega,r}=L_{\Omega,r} \mathop{\mathrm{max}}\big{\{}1,\frac{1}{\sigma_0} \big{\}}\). By the Hermite-Genocchi formula in Theorem 1, we have \[[X_1,\cdots, X_r]G_{M,r}(t)= \int_{S_{r-1}} G_{M,r}^{(r-1)}(u_0X_1+\cdots+u_{r-1}X_{r} ) du_1\cdots du_{r-1} . \] Here \[S_{r-1}=\big{\{}(u_1,\cdots,u_{r-1})\in \mathbb{R}^{r-1}\,\big{|}\, u_j\geq 0, \sum_{j=1}^{r-1}u_j\leq 1\big{\}},\] \[u_0=1-u_1-\cdots -u_{r-1}.\] Since \[\begin{align} &\;\;\;\;G_{M,r}^{(r-1)}(t)\\ &=\sum_{i=0}^{r-1} \binom{r-1}{i} \left( t^{M+r-1} \right)^{(r-1-i)} \phi^{(i)}_s(t)\\ &= \sum_{i=0}^{r-1} \binom{r-1}{i} \frac{ (M+r-1)! }{ (M+i)! }t^{M+i}\phi^{(i)}_s(t),\\ \\ \end{align}\] by the formula \((8)\) we have \[\begin{align} &\;\;\;\;\big{|}G_{M,r}^{(r-1)}(t)\big{|}\\ &\leq C_{\Omega,r} \sum_{i=0}^{r-1} \binom{r-1}{i} (M+r-1)^{r-1-i} t^M \\ &\leq C_{\Omega,r} (M+r)^{r-1}t^M. \end{align} \] From the formula \((9)\) and the inequality \((10)\), one has \[\begin{align} &\;\;\;\;\Big{|}[X_1,\cdots, X_r]G_{M,r}(t)\Big{|}\\ & \leq C_{\Omega,r} (M+r)^{r-1} \int_{S_{r-1}} \left( u_0X_1+\cdots+u_{r-1}X_r \right)^M du_1\cdots du_{r-1}. \\ \end{align} \]

By the formulas \((7)\) and \((11)\), we have \[\begin{align} &\;\;\;\; \int_{(0,1)^{K_r}} \Big{|} P_{M,r}(s; X_1,\cdots,X_r ) -A_r(s;X_1,\cdots, X_r) \Big{|}dx_1\cdots dx_{K_r}\\ &= \int_{(0,1)^{K_r}} \Big{|} [1,X_1,\cdots, X_r]((1-t)G_{M,r}(t)) \Big{|}dx_1\cdots dx_{K_r}\\ &= \int_{(0,1)^{K_r}} \Big{|} [X_1,\cdots, X_r]G_{M,r}(t) \Big{|}dx_1\cdots dx_{K_r}\\ &\leq C_{\Omega,r}(M+r)^{r-1} \int_{(0,1)^{K_r}} \left(\int_{S_{r-1}} \left( u_0X_1+\cdots+u_{r-1}X_r \right)^Mdu_1\cdots du_{r-1}\right) dx_1\cdots dx_{K_r} . \end{align}\] Define \[A_{M,r}= \int_{(0,1)^{K_r}} \left(\int_{S_{r-1}} \left( u_0X_1+\cdots+u_{r-1}X_r \right)^Mdu_1\cdots du_{r-1}\right) dx_1\cdots dx_{K_r} .\] By the Hermite-Genocchi formula in Theorem 1, \[\int_{S_{r-1}} \left( u_0X_1+\cdots+u_{r-1}X_r \right)^Mdu_1\cdots du_{r-1} =\frac{M!}{(M+r-1)!} [X_1,\cdots, X_r] t^{M+r-1}.\] From the integral representation \((4)\) of the multiple harmonic sums and \[[X_1,\cdots,X_r]t^{M+r-1}=X_1^M \left[1,\frac{X_2}{X_1},\cdots,\frac{X_r}{X_1}\right]t^{M+r-1} ,\] one has \[\begin{align} &\;\;\;\; A_{M,r} \\ &= \frac{M!}{(M+r-1)!} \int_{S_{r-1}} [X_1,\cdots, X_r] t^{M+r-1} dx_1\cdots dx_{K_r} \\ &=\frac{M!}{(M+r-1)!} \int_{S_{r-1}} X_1^M \left[1,\frac{X_2}{X_1},\cdots,\frac{X_r}{X_1}\right]t^{M+r-1} dx_1\cdots dx_{K_r} \\ &=\frac{M!}{(M+r-1)!} \frac{1}{(M+1)^{k_1}} H_{M+1}^\star(k_2,\cdots,k_r).\\ \end{align}\] As a result, \[A_{M,r}\leq \frac{1}{(M+1)^r}\left(1+\frac{1}{2}+\cdots+\frac{1}{M+1} \right)^{r-1} \leq \frac{(2\mathrm{log}\;(M+2))^{r-1}}{(M+1)^r}.\] So we have \[\begin{align} &\;\;\;\; \int_{(0,1)^{K_r}} \Big{|} P_{M,r}(s; X_1,\cdots,X_r ) -A_r(s;X_1,\cdots, X_r) \Big{|}dx_1\cdots dx_{K_r}\\ &\leq C_{\Omega,r}(M+r)^{r-1} \frac{(2\mathrm{log}\;(M+2))^{r-1}}{(M+1)^r}\\ &\leq C_{{\boldsymbol{k}},\Omega} \frac{ \left(\mathrm{log}\,(M+2)\right)^{r-1} }{M+1}. \end{align}\] In conclusion, the Lemma is proved. \(\hfill\Box\)

For \(r=1\) and \(k_1\in \mathbb{Z}^+\), it is clear that \[\begin{align} &H_s^\star(k_1)=\mathop{\mathrm{lim}}_{M\rightarrow +\infty} \bigg{(}H_M^\star(k_1)-G_M^\star(s;k_1) \bigg{)}=\sum_{m=1}^{+\infty} \left( \frac{1}{m^{k_1}}-\frac{1}{(m+s)^{k_1}} \right) \end{align}\] is absolutely convergent on \(\mathrm{Re}(s)>0\). Thus \(H_s^\star(k_1)\) is well-defined and analytic on \(\mathrm{Re}(s)>0\). Since \[\frac{1}{m^{k_1}}-\frac{1}{(m+s)^{k_1}}=\int_{(0,1)^{k_1}} (x_1\cdots x_{k_1})^{m-1}\left[ 1- (x_1\cdots x_{k_1})^{s} \right]dx_1\cdots dx_{k_1},\] one has \[\begin{align} &\;\;\;\; H_s^\star(k_1)\\ &= \int_{(0,1)^{k_1}}\sum_{m=1}^{+\infty} (x_1\cdots x_{k_1})^{m-1}\left[ 1- (x_1\cdots x_{k_1})^{s} \right]dx_1\cdots dx_{k_1}\\ &=\int_{(0,1)^{k_1}}\frac{ 1- (x_1\cdots x_{k_1})^{s} }{1-x_1\cdots x_{k_1} } dx_1\cdots dx_{k_1}\\ &= \int_{(0,1)^{k_1}} [1,X_1]t^s dx_1\cdots dx_{k_1} . \\ \end{align}\] By induction, assuming that \(H_s^\star(k_1,\cdots,k_r)\) is well-defined and analytic on \(\mathrm{Re}(s)>0\) for \(r<n\) and \[H_s^\star(k_1,\cdots,k_r)=\int_{(0,1)^{k_1+\cdots+k_r}} [1,X_1,\cdots, X_r]t^{s+r-1}dx_1\cdots dx_{k_1+\cdots+k_r}\] on \(\mathrm{Re}(s)>0\) . For \(r=n\) , by Lemma 1, \[I_s(k_1,\cdots,k_r):=\int_{(0,1)^{K_r}} [1,X_1,\cdots ,X_r] t^{s+r-1} dx_1dx_2\cdots dx_{K_r}\] is holomorphic on \(\mathrm{Re}\,s>0\). For \(r=n, M\geq 1\), by Lemma 1, \((iv)\), Lemma 1 and Lemma 1, the function \[\begin{align} &\;\;\;\;H_s^\star(k_1,\cdots,k_r)\\ &=\mathop{\mathrm{lim}}_{M\rightarrow +\infty} \Big{(} H_M^\star(k_1,\cdots,k_r)- \sum_{j=1}^r G_M^\star(s;k_1,\cdots, k_j)H_s^\star (k_{j+1},\cdots, k_r) \Big{)} \\ &= \mathop{\mathrm{lim}}_{M\rightarrow +\infty} \int_{(0,1)^{K_r}} P_{M,r}(s;X_1,\cdots,X_r)dx_1\cdots dx_{K_r} \\ &= \int_{(0,1)^{K_r}}A_r(s;X_1,\cdots, X_r) dx_1\cdots dx_{K_r} \end{align}\] is well-defined and holomorphic on \(\mathrm{Re}\,s>0\). \(\hfill\Box\)

4 Finite zeta-star correspondence in general cases↩︎

In this section, we will give a complete proof of Theorem 1. Many basic properties of finite zeta-star correspondence are also discussed in this section.

Define \(\sigma=\mathrm{Re}(s)>0\). By Theorem 1, \[H_s^\star(k_1,\cdots,k_r,k_{r+1})=\int_{(0,1)^{K_{r+1}}} [1,X_1,\cdots, X_r,X_{r+1}]t^{s+r}dx_1\cdots dx_{K_{r+1}},\] \[H_s^\star(k_1,\cdots,k_r)=\int_{(0,1)^{K_r}} [1,X_1,\cdots, X_r]t^{s+r-1}dx_1\cdots dx_{K_r}.\] Here \[K_i=k_1+\cdots+k_i,\qquad X_i=x_1x_2\cdots x_{K_i}, \qquad i=1,\cdots,r+1.\] By the symmetry of divided differences and Lemma 1, \[\left[1,X_1,\cdots, X_r,X_{r+1}\right] t^{s+r} =\frac{[1,X_1,\cdots,X_r] t^{s+r-1}-X_{r+1}[X_1,\cdots, X_r,X_{r+1}] t^{s+r-1} }{1-X_{r+1}}.\] It follows that \[\begin{align} &\;\;\;\; H_s^\star(k_1,\cdots,k_r,k_{r+1})- H_s^\star(k_1,\cdots,k_r) \\ &= \int_{(0,1)^{K_{r+1}}} \left( \left[1,X_1,\cdots, X_r,X_{r+1}\right] t^{s+r} - [1,X_1,\cdots, X_r]t^{s+r-1} \right) dx_1\cdots dx_{K_{r+1}} \\ &= \int_{(0,1)^{K_{r+1}}} \left( \frac{[1,X_1,\cdots,X_r] t^{s+r-1}-X_{r+1}[X_1,\cdots, X_r,X_{r+1}] t^{s+r-1} }{1-X_{r+1}}- [1,X_1,\cdots, X_r]t^{s+r-1} \right) \\ &\qquad dx_1\cdots dx_{K_{r+1}} \\ &= \int_{(0,1)^{K_{r+1}}} \left( \frac{X_{r+1}\left([1,X_1,\cdots,X_r] t^{s+r-1}-[X_1,\cdots, X_r,X_{r+1}] t^{s+r-1} \right) }{1-X_{r+1}} \right) dx_1\cdots dx_{K_{r+1}} \\ &=\int_{(0,1)^{K_{r+1}}} \frac{X_{r+1}}{1-X_{r+1}} \left([1,X_1,\cdots,X_r] t^{s+r-1}-[X_1,\cdots, X_r,X_{r+1}] t^{s+r-1} \right) dx_1\cdots dx_{K_{r+1}} \end{align}\]

By the Hermite-Genocchi formula, we have \[[1,X_1,\cdots, X_r]f=\int_{S_r} f^{(r)}(u_0+u_1X_1+\cdots+u_rX_r) du_1\cdots du_r,\] \[[X_1,\cdots, X_r,X_{r+1}]f=\int_{S_r} f^{(r)}(u_0X_{r+1}+u_1X_1+\cdots+u_rX_r) du_1\cdots du_r\] for \(f=t^{s+r-1}\). Here \[S_r=\big{\{}(u_1,\cdots,u_r)\in \mathbb{R}^r\,\big{|}\, u_j\geq 0, \sum_{j=1}^r u_j\leq 1\big{\}},\quad u_0=1-\sum_{j=1}^r u_j.\] By direct calculation, we have \[\begin{align} &\;\;\;\; \Big{|} [1,X_1,\cdots, X_r]f - [X_1,\cdots, X_r,X_{r+1}]f \Big{|} \\ &\leq \int_{S_r} \Big{|} f^{(r)}(u_0+u_1X_1+\cdots+u_rX_r) -f^{(r)}(u_0X_{r+1}+u_1X_1+\cdots+u_rX_r) \Big{|} du_1\cdots du_r\\ &\leq \Bigg{|} \frac{ \Gamma(s+r) }{\Gamma(s)} \Bigg{|} \int_{S_r} \Bigg{|} \Big[(u_0+u_1X_1+\cdots+u_rX_r)^{s-1}\\ &\;\;\;\; -(u_0X_{r+1}+u_1X_1+\cdots+u_rX_r)^{s-1} \Big] \Bigg{|} du_1\cdots du_r\\ &\leq \Bigg{|} \frac{ \Gamma(s+r) }{\Gamma(s)} \Bigg{|} \Bigg{(} \int_{S_r} (u_0+u_1X_1+\cdots+u_rX_r)^{\sigma-1} du_1\cdots du_r\\ &\;\;\;\; +\int_{S_r} (u_0X_{r+1}+u_1X_1+\cdots+u_rX_r)^{\sigma-1} du_1\cdots du_r \Bigg{)} \\ \end{align}\] If \(\sigma\geq 1\), then \[\int_{S_r} (u_0+u_1X_1+\cdots+u_rX_r)^{\sigma-1} du_1\cdots du_r\leq \frac{1}{r!},\] \[\int_{S_r} (u_0X_{r+1}+u_1X_1+\cdots+u_rX_r)^{\sigma-1} du_1\cdots du_r \leq \frac{1}{r!}.\] If \(0<\sigma<1\), then \[\begin{align} &\;\;\;\; \int_{S_r} (u_0+u_1X_1+\cdots+u_rX_r)^{\sigma-1} du_1\cdots du_r \\ &\leq \int_{S_r} (u_0+u_1X_r+\cdots+u_rX_r)^{\sigma-1} du_1\cdots du_r \\ &\leq \int_{S_r} [1-(u_1+\cdots+u_r)(1-X_r)] ^{\sigma-1} du_1\cdots du_r \\ &\leq \int_{S_r} [1-(u_1+\cdots+u_r)] ^{\sigma-1} du_1\cdots du_r \\ &\leq \int_{0<v_1<\cdots<v_r<1} (1-v_r)^{\sigma-1} dv_1\cdots dv_r \\ &\leq \int_0^1 \frac{ v_r^{r-1}(1-v_r)^{\sigma-1} }{(r-1)!} dv_r \\ &\leq \frac{1}{(r-1)!}\frac{ \Gamma(r)\Gamma(\sigma) }{\Gamma(r+\sigma)}\\ &\leq \frac{\Gamma(\sigma)}{\Gamma(r+\sigma)} \end{align}\] and similarly \[\begin{align} &\;\;\;\; \int_{S_r} (u_0X_{r+1}+u_1X_1+\cdots+u_rX_r)^{\sigma-1} du_1\cdots du_r \\ &= \int_{S_r} (u_0X_1+u_1X_2+\cdots+u_rX_{r+1})^{\sigma-1} du_1\cdots du_r \\ &\leq X_1^{\sigma-1} \int_{S_r} \left(u_0+u_1\frac{X_2}{X_1}+\cdots+u_r\frac{X_{r+1}}{X_1}\right)^{\sigma-1} du_1\cdots du_r \\ &\leq X_1^{\sigma-1} \int_{S_r} \left(u_0+u_1\frac{X_{r+1}}{X_1}+\cdots+u_r\frac{X_{r+1}}{X_1}\right)^{\sigma-1} du_1\cdots du_r \\ &\leq X_1^{\sigma-1} \int_{S_r} \left(1-(u_1+\cdots+u_r)\left(1-\frac{X_{r+1}}{X_1}\right)\right)^{\sigma-1} du_1\cdots du_r\\ &\leq \frac{\Gamma(\sigma)}{\Gamma(r+\sigma)} X_1^{\sigma-1} \\ &\leq \frac{\Gamma(\sigma)}{\Gamma(r+\sigma)} X_{r+1}^{\sigma-1} \end{align}\]

In conclusion, for \(\sigma\geq 1\), \[\Big{|} [1,X_1,\cdots, X_r]f - [X_1,\cdots, X_r,X_{r+1}]f \Big{|}\leq \Bigg{|} \frac{ \Gamma(s+r) }{\Gamma(s)} \Bigg{|} \frac{2}{r!};\] For \(0<\sigma<1\), \[\Big{|} [1,X_1,\cdots, X_r]f - [X_1,\cdots, X_r,X_{r+1}]f \Big{|}\leq \Bigg{|} \frac{ \Gamma(s+r) }{\Gamma(s)} \Bigg{|} \frac{\Gamma(\sigma)}{\Gamma(r+\sigma)} \left( 1+X_{r+1}^{\sigma-1} \right).\] By the asymptotic formula of Gamma function, it follows that \[\frac{\Gamma(r+s) }{r!}=\frac{\Gamma(r+s)}{\Gamma(r+1)}\sim r^{s-1},\qquad \frac{\Gamma(r+s)}{\Gamma(r+\sigma)}\sim r^{s-\sigma}\] for \(r\rightarrow +\infty\). In both cases, there is a constant \(C_1\) (depends only on \(s\)) such that \[\Big{|} [1,X_1,\cdots, X_r]f - [X_1,\cdots, X_r,X_{r+1}]f \Big{|}\leq C_1 r^{\sigma} (1+X_{r+1}^{\sigma-1} )\] Thus \[\begin{align} &\;\;\;\; \Big{|} H_s^\star(k_1,\cdots,k_r,k_{r+1})- H_s^\star(k_1,\cdots,k_r) \Big{|} \\ &\leq \int_{(0,1)^{K_{r+1}}} \frac{X_{r+1}}{1-X_{r+1}}\Big{|}[1,X_1,\cdots,X_r] f-[X_1,\cdots, X_r,X_{r+1}] f \Big{|} dx_1\cdots dx_{K_{r+1}} \\ &\leq C_1r^{\sigma} \int_{(0,1)^{K_{r+1}}} \frac{X_{r+1}}{1-X_{r+1}} \left(1+X_{r+1} ^{\sigma-1}\right) dx_1\cdots dx_{K_{r+1}} \\ &\leq C_1r^{\sigma} \int_{(0,1)^{K_{r+1}}} \sum_{m\geq 1} \left(X_{r+1}^m+X_{r+1} ^{m+\sigma-1}\right) dx_1\cdots dx_{K_{r+1}} \\ &\leq C_1r^{\sigma} \sum_{m\geq 1} \left( \frac{1}{(m+1)^{K_{r+1}}} + \frac{1}{(m+\sigma)^{K_{r+1}}} \right)\\ &\leq C_1r^{\sigma} \sum_{m\geq 1} \left( \frac{1}{(m+1)^{{r+1}}} + \frac{1}{(m+\sigma)^{{r+1}}} \right)\\ &\leq C_1 r^{\sigma} \left( \frac{1}{2^{r+1}}+\frac{1}{(1+\sigma)^{r+1}}+ \int_{2}^{+\infty} \left( \frac{1}{x^{r+1}} +\frac{1}{(x+\sigma-1)^{r+1}} \right) dx \right) \\ &\leq C_1 r^{\sigma} \left( \frac{1}{2^{r+1}}+\frac{1}{(1+\sigma)^{r+1}}\right) \left(1+\frac{1}{r}\right) \\ &\leq 2\, C_1 r^{\sigma} \left( \frac{1}{2^{r+1}}+\frac{1}{(1+\sigma)^{r+1}}\right) \end{align}\] As a result, the Lemma is proved. \(\hfill\Box\)
By Lemma 1, for \(\mathrm{Re}\,(s)>0\) and \({\boldsymbol{k}}=(k_1,\cdots,k_r,\cdots)\), the sequence \[H_s^\star(k_1,\cdots,k_r), r\geq 1\] is a Cauchy sequence in \(\mathbb{C}\). So the limit \[\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_s^\star(k_1,\cdots,k_r)\] exists. Thus Theorem 1 \((i)\) is proved.

For the proof of Theorem 1, \((ii)\), we use the total order structure on \[\widehat{\mathcal{S}}=\Big{\{}(k_1,\cdots,k_r)\,\Big{|}\,k_1,\cdots,k_r\in\left(\mathbb{Z}^{+}\right)^r, r\geq 1\Big{\}}\] which is introduced in Section 2.

To deal with the order structure of \(H_s^\star(k_1,\cdots,k_r)\) for \(s>1\), define \[\left(\mathcal{F}_kh\right)(z):=\int_{(0,z)\times (0,1)^{k-1}} \frac{h(1)-h( u_1\cdots u_k ) }{1-u_1\cdots u_k} du_1\cdots du_k, \quad 0\leq z\leq 1,\quad k\in \mathbb{Z}^+\] for any function \(h\). For \[(k_1,\cdots,k_r)\in \left(\mathbb{Z}^+\right)^r\] and a function \(h\), let \[\mathcal{L}_{\varnothing}(h)=h(1),\] \[\mathcal{L}_{k_1,\cdots,k_r}(h):= \int_{(0,1)^{K_r}} [1,X_1,\cdots,X_r] \left( t^{r-1}h(t)\right)dx_1\cdots dx_{K_r},\] which \[K_j=k_1+\cdots +k_j,\qquad X_j=x_1x_2\cdots x_{K_j}, \qquad 1\leq j\leq r.\] It is clear that \[H_s^\star(k_1,\cdots,k_r)=\mathcal{L}_{k_1,\cdots,k_r}(t^s).\]

For \(r=1\), \[\mathcal{L}_{k_1}(h)=\int_{(0,1)^{K_1}} [1,X_1](h(t))dx_1\cdots dx_{K_1}=(\mathcal{F}_{k_1}(h))(1).\] For \(r\geq 2\) and fixed \(x_{K_1+1},\cdots,x_{K_r}\), denote by \[z_0=1,\qquad z_i=x_{K_1+1}\cdots x_{K_{i+1}},\qquad 1\leq i\leq r-1.\] Then \[X_1=X_1 z_0,\quad X_2=X_1z_1,\quad,\cdots, \quad X_r=X_1z_{r-1}.\] Since \(t^{r-1}h(1)\) is a polynomial of degree \(\leq r-1\), its \(r\)-th order divided difference is \(0\). Therefore \[\begin{align} &\;\;\;\; [1,X_1z_0,\cdots, X_1z_{r-1}] t^{r-1}h(t) \\ &= [1,X_1z_0,\cdots, X_1z_{r-1}] t^{r-1}\left(h(t)-h(1) \right) \\ &= \sum_{i=0}^{r-1} \frac{ z_i^{r-1}\left( h(X_1z_i)-h(1) \right) }{(X_1z_i-1) \prod_{\substack{0\leq j\leq r-1 \\ j\neq i }} (z_i-z_j) }.\\ \end{align}\] As \[\begin{align} &\;\;\;\; \int_{(0,1)^{k_1}} \frac{ h(1)-h(X_1z_i) }{1-X_1z_i}dx_1\cdots dx_{k_1}\\ &=\frac{1}{z_i} \int_{(0,z_i)\times (0,1)^{k-1}} \frac{h(1)-h( x_1\cdots x_{k_1}) }{1-x_1\cdots x_{k_1}} dx_1\cdots dx_{k_1}\\ &=\frac{(\mathcal{F}_{k_1}h)(z_i) }{z_i}, \\ \end{align}\] we have \[\int_{(0,1)^{k_1}} [1,X_1z_0,\cdots, X_1z_{r-1}] t^{r-1}h(t) dx_1\cdots dx_{k_1}=[z_0,\cdots, z_{r-1}]\left( t^{r-2}(\mathcal{F}_{k_1}h)(t) \right).\] By integrating with respect to \(x_{K_1+1},\cdots, x_{K_r}\), we have \[\mathcal{L}_{k_1,\cdots,k_r}(h)=\mathcal{L}_{k_2,\cdots,k_r}\left( \mathcal{F}_{k_1}h \right).\] By induction, one has \[H_s^\star(k_1,\cdots,k_r)= \mathcal{L}_{k_1,\cdots,k_r}(t^s)=\left(\mathcal{F}_{k_r}\cdots \mathcal{F}_{k_1} t^s\right)(1)\] \(\hfill\Box\)

By Lemma 1, we have \[H_s^\star(\{1\}^m)=\mathcal{L}_{ \{1\}^ m}(t^s)=(\mathcal{F}_1^mt^s)(1).\] By definition and repeated using of integration by parts, \[\begin{align} &\;\;\;\; (\mathcal{F}_1^mt^s)(1)\\ & = \int^1_0 \frac{(\mathcal{F}_1^{m-1}t^s)(1)- (\mathcal{F}_1^{m-1}t^s)(u) }{1-u}du \\ &= \int^1_0 [ (\mathcal{F}_1^{m-1}t^s)(1)- (\mathcal{F}_1^{m-1}t^s)(u) ] d\,\mathrm{log}\frac{1}{1-u} \\ &=\int^1_0 \mathrm{log}\frac{1}{1-u} d[ (\mathcal{F}_1^{m-1}t^s)(u)] \\ &= \int^1_0 \mathrm{log}\frac{1}{1-u} \cdot \frac{(\mathcal{F}_1^{m-2}t^s)(1)- (\mathcal{F}_1^{m-2}t^s)(u) }{1-u}du\\ &= \frac{1}{2!} \int^1_0 [ (\mathcal{F}_1^{m-2}t^s)(1)- (\mathcal{F}_1^{m-2}t^s)(u) ] d\,\mathrm{log}^2\frac{1}{1-u} \\ &=\qquad \cdots \\ &=\frac{1}{(m-1)!} \int^1_0 \left[(\mathcal{F}_1t^s)(1)-(\mathcal{F}_1t^s)(u)\right] d\,\mathrm{log}^{m-1}\frac{1}{1-u} \\ &= \frac{1}{(m-1)!} \int^1_0 \mathrm{log}^{m-1}\frac{1}{1-u} d\,[(\mathcal{F}_1t^s)(u)] \\ &=\frac{1}{(m-1)!} \int^1_0 \mathrm{log}^{m-1}\frac{1}{1-u} \cdot \frac{ 1-u^s }{1-u}du \\ &=\frac{1}{m!} \int^1_0 (1-u^s)d\, \mathrm{log}^{m}\frac{1}{1-u} \\ &= \frac{s}{m!}\int^1_0 \left( -\mathrm{log}\,(1-u) \right)^m u^{s-1}du.\\ \\ \end{align}\] By changing of variables \(u=1-t\), we have \[H_s^\star(\{1\}^m)=\frac{s}{m!}\int^1_0\left( -\mathrm{log}\;t \right)^m(1-t)^{s-1}dt.\] \(\hfill\Box\)

By changing of variables, we have \[\frac{ ( \mathcal{F}_kh )(z) }{z}=\int_{(0,1)^k} Q_h(zu_1\cdots u_k) du_1\cdots du_k\] and \[( \mathcal{F}_kh )(1) =\int_{(0,1)^k} Q_h(u_1\cdots u_k) du_1\cdots du_k.\] For \(z,u_1,\cdots,u_k\in (0,1)\), one has \[zu_1\cdots u_k<u_1\cdots u_k.\] From the fact that \(Q_h\) is strictly increasing, we have \[Q_h(zu_1\cdots u_k)<Q_h(u_1\cdots u_k).\] Thus \[\frac{(\mathcal{F}_kh)(z)}{z}<(\mathcal{F}_kh)(1), \qquad 0<z<1.\]

Denote by \(\xi=\mathcal{F}_kh\). If \(k=1\), then \[\xi^\prime(z)=Q_h(z).\] Thus \(\xi^\prime\) is strictly increasing. If \(k\geq 2\), by definition \[\xi^\prime(z)=\int^1_0Q_h(zv_1\cdots v_{k-1} )dv_1\cdots dv_{k-1} .\] As \(Q_h\) is strictly increasing, \(\xi^\prime\) is also strictly increasing. In both cases, the function \(\xi\) is a strictly convex function and \(Q_\xi(z)>0\) for \(z\in (0,1)\). Then for \(1>z>z^\prime>0\), \[\begin{align} &\;\;\;\; Q_{\xi}(z)-Q_{\xi}(z^\prime)\\ &=\frac{ \xi(1)-\xi(z) }{1-z}-\frac{\xi(1)-\xi(z^\prime) }{1-z^\prime} \\ &= \frac{1}{(1-z)(1-z^\prime)} \left[ (z-z^\prime) \xi(1) +(1-z)\xi(z^\prime)-(1-z^\prime) \xi(z) \right] \\ &= \frac{1}{1-z} \left[ \frac{z-z^\prime}{1-z^\prime} \xi(1) +\frac{1-z}{1-z^\prime}\xi(z^\prime)- \xi(z) \right] >0 . \\ \end{align}\] In conclusion, \(Q_{\mathcal{F}_kh}\) is positive and strictly increasing on \((0,1)\). \(\hfill\Box\)

\((i)\) By Lemma 1, for \(s>1\), one has \[0<(1-t)^{s-1}<1,\qquad t\in (0,1).\] So we have \[H_s^\star(\{1\}^m)<\frac{s}{m!}\int^1_0\left( -\mathrm{log}\;t \right)^mdt.\] As \[\int^1_0\left( -\mathrm{log}\;t \right)^mdt=\int^{+\infty}_0 u^m e^{-u}du=\Gamma(m+1)=m!,\] one has \[H_s^\star(\{1\}^m)< s.\] \((ii)\) For fixed \(s>1\), define \(h(s)=t^s, t\in (0,1)\). Since \[Q_h(z)=\frac{ 1-z^s }{1-z}=s\int^1_0 \left(z+(1-z)u \right)^{s-1} du,\] \(Q_h(z)\) is positive and strictly increasing on \(0<z<1\). Define \[g=\mathcal{F}_{k_{r-1}}\cdots \mathcal{F}_{k_1} t^s.\] For \(r=1\), \(g\) is interpreted as \(g=t^s\). By Lemma 1, \[Q_g(z)=\frac{g(1)-g(z) }{1-z}\] is positive and strictly increasing on \((0,1)\).

Define \(F=\mathcal{F}_{k_r} g\). By Lemma 1, we have \[F(1)=\left(\mathcal{F}_{k_r} g \right)(1)=H_s^\star(k_1,\cdots,k_{r-1},k_r) \] and \[H_s^\star(k_1,\cdots,k_{r-1}, k_r+1,\{1\}^m)= \left(\mathcal{F}_1^m \mathcal{F}_{k_r+1}g \right)(1). \]

For any differentiable function \(\varphi\), by the essentially same analysis as Lemma 1, we have \[\begin{align} &\;\;\;\; \left( \mathcal{F}_1^m \varphi \right)(1) \\ &= \int^1_0 \frac{ \left( \mathcal{F}_1^{m-1} \varphi \right)(1)- \left( \mathcal{F}_1^{m-1} \varphi \right)(u) }{1-u} du \\ &= \int^1_0 \left[ \left( \mathcal{F}_1^{m-1} \varphi \right)(1)- \left( \mathcal{F}_1^{m-1} \varphi \right)(u)\right] d\,\mathrm{log}\frac{1}{1-u} \\ &= \int^1_0 \mathrm{log}\frac{1}{1-u} d\left[ \left( \mathcal{F}_1^{m-1} \varphi \right)(u) \right] \\ &\cdots \\ &= \frac{1}{(m-1)!} \int^1_0 \mathrm{log}^{m-1}\frac{1}{1-u} d\,[(\mathcal{F}_1\varphi)(u)] \\ &= \frac{1}{(m-1)!} \int^1_0 \mathrm{log}^{m-1}\frac{1}{1-u} \cdot \frac{1-\varphi(u)}{1-u} du \\ &=\frac{1}{m!}\int^1_0 \varphi^\prime(u)\, \mathrm{log}^{m}\frac{1}{1-u} du. \\ \end{align}\] Let \(\varphi= \mathcal{F}_{k_r+1}g\). By definition, it follows that \[\varphi^\prime(u)= \int_{(0,1)^{k_r}} \frac{1-g(uv_1\cdots v_{k_1}) }{1-uv_1\cdots v_{k_1}} dv_1\cdots dv_{k_1}=\frac{\left(\mathcal{F}_{k_r}g\right)(u) }{u}=\frac{F(u)}{u}.\] So we have \[H_s^\star(k_1,\cdots,k_{r-1}, k_r+1,\{1\}^m)= \left(\mathcal{F}_1^m \mathcal{F}_{k_r+1}g \right)(1)=\frac{1}{m!}\int^1_0\frac{F(u)}{u} \mathrm{log}^{m}\frac{1}{1-u} du.\] By Lemma 1, one has \[\frac{F(u)}{u}<F(1),\qquad 0<u<1.\] Therefore, \[H_s^\star(k_1,\cdots,k_{r-1}, k_r+1,\{1\}^m) <\frac{F(1)}{m!} \int^1_0 \mathrm{log}^{m}\frac{1}{1-u} du.\] Since \[\int^1_0 \mathrm{log}^{m}\frac{1}{1-u} du= \int^1_0 \mathrm{log}^{m}\frac{1}{u} du=\int^{+\infty}_0 x^m e^{-x}dx=m!\] and \(F(1)=H_s^\star(k_1,\cdots,k_{r-1},k_r)\), we have \[H_s^\star(k_1,\cdots,k_{r-1}, k_r+1,\{1\}^m)< H_s^\star(k_1,\cdots,k_{r-1},k_r).\] \((iii)\) It suffices to prove that the statement in special cases \[(i_1,\cdots,i_r)=(1,0,\cdots,0), (0,1,\cdots,0),\cdots,(0,0,\cdots,1).\] The general cases follows from the special cases by induction. For \[(i_1,\cdots,i_r)=(1,0,\cdots,0),\] we give the detailed proof for this case. The other cases are analogous. By Theorem 1, we have \[H_s^\star(k_1,k_2, \cdots,k_r)=\int_{(0,1)^{k_1+\cdots+k_r}} [1,X_1,\cdots, X_r]t^{s+r-1}dx_1\cdots dx_{k_1+\cdots+k_r},\] \[H_s^\star(k_1+1,k_2, \cdots,k_r)=\int^1_0\left(\int_{(0,1)^{k_1+\cdots+k_r}} [1,yX_1,\cdots, yX_r]t^{s+r-1}dx_1\cdots dx_{k_1+\cdots+k_r}\right)dy.\] Here \[X_j=x_1\cdots x_{K_j},\qquad 1\leq j\leq r.\] By the Hermite-Genocchi formula, \[[1,X_1,\cdots,X_r]t^{s+r-1}=\frac{\Gamma(s+r)}{\Gamma(s)} \int_{S_r} \left(u_0+u_1X_1+\cdots+u_rX_r \right)^{s-1} du_1\cdots du_r,\] \[[1,yX_1,\cdots,yX_r]t^{s+r-1}=\frac{\Gamma(s+r)}{\Gamma(s)} \int_{S_r} \left(u_0+u_1yX_1+\cdots+u_ryX_r \right)^{s-1} du_1\cdots du_r .\] From the above expressions, for \(s>1\), it is clear that \[[1,X_1,\cdots,X_r]t^{s+r-1}> [1,yX_1,\cdots,yX_r]t^{s+r-1},\qquad y\in (0,1).\] So \[[1,X_1,\cdots,X_r]t^{s+r-1}> \int^1_0 [1,yX_1,\cdots,yX_r]t^{s+r-1} dy.\] By integrating with respect to \(x_1,x_2,\cdots,x_{k_1+\cdots+k_r}\) over \((0,1)^{k_1+\cdots+k_r}\), one has \[H_s^\star(k_1,k_2, \cdots,k_r)> H_s^\star(k_1+1,k_2, \cdots,k_r).\] \(\hfill\Box\)

By Lemma 1, \[H_s^\star(\{1\}^m)=\frac{s}{m!}\int^1_0\left( -\mathrm{log}\;t \right)^m(1-t)^{s-1}dt.\] For \(s>1\), it is clear that \[\Big{|} \frac{ (1-t)^{s-1}-1 }{t} \Big{|}\leq c_1, \quad t\in (0,1)\] for some constant \(c_1>0\). Thus \[\begin{align} &\;\;\;\; \Big{|} H_s^\star(\{1\}^m)-s \Big{|} \\ &= \Big{|} \frac{s}{m!}\int^1_0\left( -\mathrm{log}\;t \right)^m(1-t)^{s-1}dt- \frac{s}{m!}\int^1_0\left( -\mathrm{log}\;t \right)^mdt \Big{|} \\ & \leq \frac{s}{m!}\int^1_0 \left( -\mathrm{log}\;t \right)^m \Big{|} (1-t)^{s-1}-1 \Big{|} dt \\ & \leq \frac{c_1 s}{m!}\int^1_0 \left( -\mathrm{log}\;t \right)^m t\, dt . \\ \end{align}\] By changing of variables \(t=e^{-u}=e^{-\frac{v}{2}}\), we have \[\int^1_0 \left( -\mathrm{log}\;t \right)^m t\, dt = \int^{+\infty}_0 u^m e^{-2u} du= \frac{1}{2^{m+1}}\int^{+\infty}_0 v^m e^{-v} dv=\frac{m!}{2^{m+1}}.\] As a result, \[\Big{|} H_s^\star(\{1\}^m)-s \Big{|}<\frac{c_1}{2^{m+1}}s.\] So we have \[\mathop{\mathrm{lim}}_{m\rightarrow +\infty} H_s^\star(\{1\}^m)= s.\] \((ii)\) Here we use the same analysis as the proof of Lemma 1, \((ii)\). Define \[F=\mathcal{F}_{k_r}\mathcal{F}_{k_{r-1}}\cdots \mathcal{F}_{k_1} t^s.\] Then \[H_s^\star(k_1,\cdots,k_{r-1},k_r)=F(1),\] \[\begin{align} &\;\;\;\;H_s^\star(k_1,\cdots,k_{r-1}, k_r+1,\{1\}^m)\\ &=\frac{1}{m!}\int^1_0\frac{F(u)}{u} \mathrm{log}^{m}\frac{1}{1-u} du\\ &= \frac{1}{m!}\int^1_0\frac{F(1-u)}{1-u} \mathrm{log}^{m}\frac{1}{u} du. \\ \end{align}\] It follows that \[\begin{align} & \;\;\;\;H_s^\star(k_1,\cdots,k_{r-1}, k_r+1,\{1\}^m)- H_s^\star(k_1,\cdots,k_{r-1},k_r) \\ &= \frac{1}{m!}\int^1_0\frac{F(1-u)}{1-u} \mathrm{log}^{m}\frac{1}{u} du-F(1) \\ &=\frac{1}{m!}\int^1_0\left( \frac{F(1-u)}{1-u}-F(1)\right) \mathrm{log}^{m}\frac{1}{u} du. \end{align}\] By the proof of Lemma 1, one has \[\frac{F(1-u)}{1-u}=\int_{(0,1)^{k_r}} Q_h\left((1-u)u_1\cdots u_{k_r}\right)du_1\cdots du_{k_r} ,\] \[F(1)=\int_{(0,1)^{k_r}} Q_h(u_1\cdots u_{k_r})du_1\cdots du_{k_r}\] for \[h=\mathcal{F}_{k_{r-1}}\cdots \mathcal{F}_{k_1}t^s,\quad Q_h(z)=\frac{ h(1)-h(z)}{1-z},\quad 0<z<1.\] For \(s>1\), by induction one can show that \(F^\prime(u)\) is bounded on \((0,1)\). Moreover, for \(s>1\), by repeated using of Lemma 1, \(Q_h(z)\) is positive and strictly increasing on \((0,1)\). For \(u\in \left(\frac{1}{2},1\right)\), \[\Bigg{|} \frac{F(1-u)}{1-u}-F(1)\Bigg{|} \leq 2F(1).\] For \(u\in \left(0,\frac{1}{2}\right)\), there is a constant \(c_2>0\) such that \[\Bigg{|} \frac{F(1-u)}{1-u}-F(1)\Bigg{|} \leq c_2u.\] So we have \[\begin{align} & \;\;\;\;\Big{|}H_s^\star(k_1,\cdots,k_{r-1}, k_r+1,\{1\}^m)- H_s^\star(k_1,\cdots,k_{r-1},k_r) \Big{|} \\ &= \Bigg{|}\frac{1}{m!}\int^1_0\frac{F(1-u)}{1-u} \mathrm{log}^{m}\frac{1}{u} du-F(1)\Bigg{|} \\ &\leq \frac{1}{m!}\int^1_0\Bigg{|} \frac{F(1-u)}{1-u}-F(1)\Bigg{|} \mathrm{log}^{m}\frac{1}{u} du\\ &\leq \frac{c_2}{m!} \int^{\frac{1}{2}}_0 u\, \mathrm{log}^{m}\frac{1}{u} du +\frac{ 2F(1)}{m!}\int^1_{\frac{1}{2}} \mathrm{log}^{m}\frac{1}{u} du\\ &\leq \frac{c_2}{m!} \int^{1}_0 u\, \mathrm{log}^{m}\frac{1}{u} du +\frac{ 2F(1)}{m!} \\ &\leq \frac{c_2}{2^{m+1}} +\frac{2F(1)}{m!}. \end{align}\] As a result, \[\mathop{\mathrm{lim}}_{m\rightarrow +\infty} H_s^\star(k_1,\cdots,k_{r-1}, k_r+1,\{1\}^m)=H_s^\star(k_1,\cdots,k_{r-1}, k_r).\] \(\hfill\Box\)

By the proof of Lemma 1, we have \[\begin{align} &\;\;\;\; H_s^\star(k_1,\cdots,k_p,k_{p+1})- H_s^\star(k_1,\cdots,k_p) \\ &= \int_{(0,1)^{K_{p+1}}} \left( \left[1,X_1,\cdots, X_p,X_{p+1}\right] t^{s+p} - [1,X_1,\cdots, X_p]t^{s+p-1} \right) dx_1\cdots dx_{K_{p+1}} \\ &=\int_{(0,1)^{K_{p+1}}} \frac{X_{p+1}}{1-X_{p+1}} \left([1,X_1,\cdots,X_p] t^{s+p-1}-[X_1,\cdots, X_p,X_{p+1}] t^{s+p-1} \right) dx_1\cdots dx_{K_{p+1}} . \end{align}\] By the Hermite-Genocchi formula, for \(s>1\) and \[0<X_{p+1}<X_p<\cdots<X_1<1,\] one has \[[1,X_1,\cdots,X_p] t^{s+p-1}-[X_1,\cdots, X_p,X_{p+1}] t^{s+p-1} >0.\] Thus \[H_s^\star(k_1,\cdots,k_p,k_{p+1})>H_s^\star(k_1,\cdots,k_p).\] If \((k_1,\cdots,k_p)\succ (l_1,\cdots,l_q)\), we assume that \[(k_1,\cdots,k_{i-1})=(l_1,\cdots,l_{i-1}), \quad k_i<l_i\] for some \(i\leq \mathrm{min}\{ p,q\}\). By induction, we have \[H_s^\star(k_1,\cdots,k_p)\geq H_s^\star(k_1,\cdots,k_{i-1},k_i).\] From Lemma 1, \((ii),(iii)\), it follows that \[H_s^\star(k_1,\cdots,k_{i-1},k_i)>H_s^\star(k_1,\cdots,k_{i-1}, k_i+1, \{1\}^{q-i})\] and \[H_s^\star(k_1,\cdots,k_{i-1}, k_i+1, \{1\}^{q-i})\geq H_s^\star(l_1,\cdots,l_q).\] As a result, \[H_s^\star(k_1,\cdots,k_p)> H_s^\star(l_1,\cdots,l_q).\] \(\hfill\Box\)
Now we are ready to prove Theorem 1, \((ii)\). For \[(k_1,\cdots,k_r,\cdots)\in \widehat{\mathcal{T}}=(\mathbb{Z}^+)^\infty\] and \(s>1\), define \[a_r=H_s^\star(k_1,\cdots,k_r).\] By Theorem 1, we have \(a_r<a_{r+1}\). By Lemma 1, \[a_r\leq H_s^\star(\{1\}^r)<s.\] As a result, the sequence \(\{a_r\,|\,r\geq 1\}\) is an increasing bounded sequence, thus the limit \[\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_s^{\star}(k_1,\cdots,k_r)\] exists and the map \[\mathfrak{h}_s: \widehat{\mathcal{T}}\rightarrow (1, s],\] \[{\boldsymbol{k}}=(k_1,\cdots,k_r,\cdots)\mapsto \mathfrak{h}_s({\boldsymbol{k}})=\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_s^{\star}(k_1,\cdots,k_r).\] is well-defined. It suffices to show that \(\mathfrak{h}_s\) is bijective.

For \({\boldsymbol{k}},{\boldsymbol{l}}\in \widehat{\mathcal{T}}\) and \({\boldsymbol{k}}\neq {\boldsymbol{l}}\), without loss of generality, we assume that \[(k_1,\cdots,k_i)=(l_1,\cdots,l_i), \quad k_{i+1}>l_{i+1}.\] By Lemma 1, for \(r>i+1\), we have \[H_s^\star(k_1,\cdots, k_i, k_{i+1},\cdots k_r) <H_s^\star(k_1,\cdots, k_i, k_{i+1}-1) \leq H_s^\star(l_1,\cdots,l_i,l_{i+1}).\] Therefore \[\mathfrak{h}_s({\boldsymbol{k}})=\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_s^{\star}(k_1,\cdots,k_r)\leq H_s^\star(l_1,\cdots,l_i,l_{i+1}).\] On the other hand \[H_s^\star(l_1,\cdots,l_i,l_{i+1})< \mathfrak{h}_s({\boldsymbol{l}})=\mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_s^{\star}(l_1,\cdots,l_r) .\] In conclusion, we have \[\mathfrak{h}_s({\boldsymbol{k}}) < \mathfrak{h}_s({\boldsymbol{l}}).\] So the map \(\mathfrak{h}_s\) is injective.

To show that \(\mathfrak{h}_s\) is surjective, for \(r\geq 1\), define \[Z_{\{1\}^r}(s)=\Big{(}H_s^\star(\{1\}^r), s\Big],\] \[Z_{k_1,\cdots,k_r}(s)=\Big{(}H_s^\star(k_1,\cdots,k_r), H_s^\star(k_1,\cdots ,k_{i-1},k_i-1) \Big]\] for \((k_1,\cdots,k_r)\neq (\{1\}^r)\) and \[k_i\geq 2,k_{i+1}=\cdots=k_r=1.\] From the total order structure of multiple star harmonic sums, we have \[Z_{k_1,\cdots,k_r}(s)\bigcap Z_{l_1,\cdots,l_r}(s)=\varnothing,\quad (k_1,\cdots,k_r)\neq (l_1,\cdots,l_r),\] \[(1,s]=\bigcup_{(k_1,\cdots,k_r)\in (\mathbb{Z}^+)^r} Z_{k_1,\cdots,k_r}(s), \quad r\geq 1,\] \[Z_{k_1,\cdots,k_r}(s)=\bigcup_{(k_{r+1},\cdots,k_{r+p})\in (\mathbb{Z}^+)^p} Z_{k_1,\cdots,k_r,k_{r+1},\cdots,k_{r+p}}(s) .\]

Define \[f(x)=[a_0,\cdots,a_m,cx]t^{\alpha+1}.\] By definition, \[\begin{align} &\;\;\;\; [a_0,\cdots,a_m,cx,cx]t^{\alpha+1} \\ &=\mathop{\mathrm{lim}}_{\varepsilon\rightarrow\, 0} \frac{ [a_0,\cdots,a_m,c(x+\varepsilon)]t^{\alpha+1}- [a_0,\cdots,a_m,cx]t^{\alpha+1} }{c(x+\varepsilon)-cx}\\ &=\frac{ f^\prime(x) }{c}. \end{align}\] Thus \[\begin{align} &\;\;\;\;\int^1_0[a_0,\cdots,a_m,cx,cx]t^{\alpha+1} dx\\ &=\frac{1}{c}\int^1_0 f^\prime(x) dx \\ &=\frac{1}{c}\left(f(1)-f(0) \right)\\ &=\frac{1}{c}\left([a_0,\cdots, a_m,c]t^{\alpha+1}-[a_0,\cdots,a_m,0]t^{\alpha+1} \right)\\ &=[a_0,\cdots,a_m,c,0]t^{\alpha+1}. \\ \end{align}\] By the symmetry of divided difference and Lemma 1, for \(a_0=1\) one has \[[a_0,\cdots,a_m,c,0]t^{\alpha+1}=\frac{ [a_0,\cdots,a_m,c]t^{\alpha} -0\cdot [ a_0,\cdots,a_m,c ]t^{\alpha} }{a_0-0}= [a_0,\cdots,a_m,c]t^{\alpha}.\] For \(a_0>0\), it follows that \[\begin{align} &\;\;\;\; [a_0,\cdots,a_m,c,0]t^{\alpha+1} \\ &= a_0^{\alpha+1-m-2} \left[1,\frac{a_1}{a_0},\cdots, \frac{a_m}{a_0},\frac{c}{a_0},0 \right]t^{\alpha+1} \\ &= a_0^{\alpha-m-1} \left[1,\frac{a_1}{a_0},\cdots, \frac{a_m}{a_0},\frac{c}{a_0} \right]t^{\alpha} \\ &= [a_0,\cdots,a_m,c]t^{\alpha}. \end{align}\] \(\hfill\Box\)

By Lemma 1, we have \[\begin{align} &\;\;\;\; R_{\{1\}^r} \\ &=\int_{(0,1)^r} [1,X_1,\cdots,X_{r-1}, X_r,X_r]t^{s+r}dx_1\cdots dx_r \\ &= \int_{(0,1)^{r-1}} [1,X_1,\cdots,X_{r-2}, X_{r-1},X_{r-1}]t^{s+r-1}dx_1\cdots dx_{r-1} \\ &=\int^1_0[1,x_1,x_1]t^{s+1} dx_1 \\ &= [1,1]t^s\\ &=s. \end{align}\] For \((k_1,\cdots,k_r)\neq (\{1\}^r)\) satisfying \[k_i\geq 2,k_{i+1}=\cdots=k_r=1,\] by the exactly same analysis, we have \[R_{k_1,\cdots,k_r} = H_s^\star(k_1,\cdots ,k_{i-1},k_i-1).\]

\(\hfill\Box\)

For \((k_1,\cdots,k_r)\in \left(\mathbb{Z}^+\right)^r\), define \[X_j=x_1\cdots x_{k_1+\cdots +k_j}, \quad 1\leq j\leq r.\] Let \[\mathcal{M}_s(X_1,\cdots,X_r)=[1,X_1,\cdots,X_{r-1},X_r,X_r]t^{s+r}-[1,X_1,\cdots,X_{r-1}, X_r]t^{s+r-1}.\] By the symmetry of divided differences and Leibniz formula \((1),(2)\), we have \[\begin{align} &\;\;\;\; [1,X_1,\cdots,X_{r-1}, X_r, X_r]\left(tg(t) \right)\\ &= [X_r, 1,X_1,\cdots,X_{r-1}, X_r]\left(tg(t) \right) \\ &= X_r [X_r, 1,X_1,\cdots, X_{r-1},X_r] \left(g(t)\right)+[1,X_1,\cdots, X_{r-1},X_r]\left(g(t)\right)\\ &=X_r[1,X_1,\cdots,X_{r-1}, X_r, X_r]\left(g(t) \right)+ [1,X_1,\cdots, X_{r-1},X_r]\left(g(t)\right). \end{align}\] Thus \[[1,X_1,\cdots,X_{r-1}, X_r, X_r]\left((t-X_r)g(t) \right) =[1,X_1,\cdots, X_{r-1},X_r]\left(g(t)\right).\] From the identity \[t^{s+r}=(t-X_r)t^{s+r-1}+X_rt^{s+r-1},\] it follows that \[\mathcal{M}_s(X_1,\cdots,X_r)=X_r [1,X_1,\cdots,X_{r-1},X_r,X_r]t^{s+r-1}.\] By Theorem 1 and Lemma 1, we have \[m\left(Z_{k_1,\cdots,k_r}(s)\right)=\int_{(0,1)^{K_r}} \mathcal{M}_s(X_1,\cdots,X_r) dx_1\cdots dx_{K_r}. \] In particular, if \(k_1=\cdots=k_r=1\), then \[m\left(Z_{\{1\}^r}(s)\right)=\int_{(0,1)^{r}} \mathcal{M}_s(x_1,\cdots,x_1\cdots x_r) dx_1\cdots dx_{r}. \]

By the Hermite-Genocchi formula, for \(s>1\), one has \[\begin{align} &\;\;\;\; \mathcal{M}_s(X_1,\cdots,X_r)\\ &=X_r\frac{\Gamma(s+r)}{\Gamma(s-1)}\int_{S_{r+1}} \left(u_0+u_1X_1+\cdots+u_{r-1}X_{r-1}+ (u_r+u_{r+1}) X_r \right)^{s-2} du_1\cdots du_{r+1}.\\ \end{align}\] Define \(\Phi_s(y_1,y_2, \cdots,y_r)=\mathcal{M}_s(y_1,y_1y_2, \cdots,y_1\cdots y_r)\). Then \[\begin{align} &\;\;\;\; \Phi_s(y_1,y_2, \cdots,y_r)\\ &=y_1\cdots y_r\frac{\Gamma(s+r)}{\Gamma(s-1)}\int_{S_{r+1}} \\ &\;\;\;\;\left[u_0+u_1y_1+\cdots+u_{r-1}y_1\cdots y_{r-1}+ (u_r+u_{r+1}) y_1\cdots y_r \right]^{s-2} du_1\cdots du_{r+1}.\\ \end{align}\] Since \[\begin{align} &\;\;\;\; \frac{\partial\Phi_s }{\partial y_j} \\ &= \frac{\Gamma(s+r)}{\Gamma(s-1)} \cdot \frac{y_1\cdots y_r}{y_j} \int_{S_{r+1}} \mathcal{L}^{s-2}\left( 1+(s-2)\frac{\mathcal{P}_j }{\mathcal{L}} \right) du_1\cdots du_{r+1} . \\ \end{align}\] Here \[\mathcal{L}=u_0+u_1y_1+\cdots+u_{r-1}y_1\cdots y_{r-1}+ (u_r+u_{r+1}) y_1\cdots y_r\] and \[\mathcal{P}_j=y_j\frac{\partial \mathcal{L} }{\partial y_j}=u_jy_1\cdots y_j+\cdots+u_{r-1}y_1\cdots y_{r-1}+(u_r+u_{r+1}) y_1\cdots y_r.\] If \(s\geq 2\), it is clear that \[\frac{\partial\Phi_s }{\partial y_j} >0, \quad y_j\in(0,1),\quad 1\leq j\leq r.\] If \(1<s<2\), since \({\mathcal{P}_j}/{\mathcal{L}}\leq 1\), \[1+(s-2)\frac{\mathcal{P}_j }{\mathcal{L}} \geq 0.\] In conclusion, for \(s>1\), \[\frac{\partial\Phi_s }{\partial y_j} >0, \quad y_j\in(0,1),\quad 1\leq j\leq r.\] As a result, \[\begin{align} &\;\;\;\; m\left(Z_{k_1,\cdots,k_r}(s)\right) \\ &= \int_{(0,1)^{K_r}} \Phi_s(x_1\cdots x_{K_1}, x_{K_1+1}\cdots x_{K_2},\cdots, x_{K_{r-1}+1}\cdots x_{K_r} ) dx_1\cdots dx_{K_r} \\ &\leq \int_{(0,1)^{K_r}} \Phi_s( x_{K_1}, x_{K_2},\cdots, x_{K_r} ) dx_1dx_2\cdots dx_{K_r} \\ &\leq \int_{(0,1)^{r}} \Phi_s( x_{K_1}, x_{K_2},\cdots, x_{K_r} ) dx_{K_1}dx_{K_2}\cdots dx_{K_r} \\ &\leq m\left(Z_{\{1\}^r}(s)\right). \end{align}\] \(\hfill\Box\)
For \(s>1\) and \(x\in (1,s]\), if \(x=s\), by Lemma 1, one has \[x=s=\mathfrak{h}_s(({\{1\}^{\infty}})) .\] If \(x=H_s^\star(k_1,\cdots,k_r)\), by Lemma 1, one has \[x=H_s^\star(k_1,\cdots,k_r) =\mathfrak{h}_s\left( (k_1,\cdots, k_{r-1}, k_r+1,\{1\}^{\infty}) \right) .\] If \[x\neq H_s^\star(k_1,\cdots,k_r)\] for all \((k_1,\cdots,k_r)\in \left( \mathbb{Z}^+ \right)^r\), \(r\geq 1\), from the basic properties of \[Z_{k_1,\cdots,k_r}(s),\] there is a \[{\boldsymbol{k}}=(k_1,\cdots,k_r,\cdots)\in \widehat{\mathcal{T}}\] such that \[x\in Z_{k_1,\cdots,k_r}(s)\] for all \(r\geq 1\). As a result, \[H_s^\star (k_1,\cdots,k_r) <x\leq H_s^\star (k_1,\cdots,k_r) + m\left( Z_{k_1,\cdots,k_r}(s) \right).\]

Since by Lemma 1 \[\begin{align} &\;\;\;\; m\left( Z_{k_1,\cdots,k_r}(s) \right)\leq m\left( Z_{\{1\}^r}(s) \right), \end{align}\] we have \[\mathop{\mathrm{lim}}_{r\rightarrow +\infty} m\left( Z_{k_1,\cdots,k_r}(s) \right) = \mathop{\mathrm{lim}}_{r\rightarrow +\infty} \left( s-H_s^\star\left(\{1\}^r\right) \right) =0.\] Therefore, \[x= \mathop{\mathrm{lim}}_{r\rightarrow +\infty} H_s^\star (k_1,\cdots,k_r) =\mathfrak{h}_s({\boldsymbol{k}}).\] In a word, the map \(\mathfrak{h}_s: \widehat{\mathcal{T}}\rightarrow (1,s]\) is bijective for \(s>1\). Theorem 1 \((ii)\) is proved.

By the explicit formula of divided differences, the left hand side is \[\frac{1}{\prod_{i=1}^r (1-Xy_i) }+X^s\sum_{i=1}^r \frac{ y_i^{s+r-1}}{(Xy_i-1) \prod_{j\neq i} (y_i-y_j) }.\] By Lemma 1, \[\frac{1}{\prod_{i=1}^r (1-Xy_i) }=\sum_{n=1}^{+\infty} X^{n-1} [y_1,\cdots,y_r] t^{n+r-2}.\] From the explicit formula of divided differences, it follows that \[\begin{align} &\;\;\;\; \sum_{n=1}^{+\infty} X^{s+n-1}[y_1,\cdots,y_r]t^{s+n+r-2} \\ &=\sum_{n=1}^{+\infty} X^{s+n-1} \sum_{i=1}^r \frac{ y_i^{s+n+r-2} }{\prod_{j\neq i}(y_i-y_j)} \\ &=X^s \sum_{i=1}^r \frac{ y_i^{s+r-2} }{\prod_{j\neq i}(y_i-y_j)} \sum_{n=1}^{+\infty} \left( Xy_i \right)^{n-1} \\ &=X^s \sum_{i=1}^r \frac{ y_i^{s+r-2} }{ (1-Xy_i) \prod_{j\neq i}(y_i-y_j)}. \end{align}\] As \((Xy_i-1)^{-1}=-(1-Xy_i)^{-1}\), we have \[\begin{align} &\;\;\;\; [1,Xy_1,Xy_2,\cdots, Xy_r]t^{s+r-1}\\ &=\sum_{n=1}^{+\infty} \left( X^{n-1}[y_1,\cdots,y_r]t^{n+r-2} -X^{s+n-1}[y_1,\cdots,y_r]t^{s+n+r-2}\right).\\ \end{align}\] \(\hfill\Box\)

By definition, \[H_n^\star(k_1,\cdots,k_r)- H_{s+n}^\star(k_1,\cdots,k_r) =\int_{(0,1)^{K_r}} [1,X_1,\cdots,X_r] \left(t^{s+r-1}-t^{s+n+r-1} \right) dx_1\cdots dx_{K_r}.\] By Lemma 1, \((iii)\), \[\begin{align} &\;\;\;\;[1,X_1,\cdots,X_r] \left(t^{s+r-1}-t^{s+n+r-1} \right) \\ &= A_r(n; X_1,\cdots, X_r) - A_r(s+n; X_1,\cdots, X_r) \\ &= P_{n,r}(s; X_1,\cdots,X_r)-A_r(s; X_1,\cdots, X_r). \end{align}\] From Lemma 1, it follows that \[\bigg{|} H_n^\star(k_1,\cdots,k_r)- H_{s+n}^\star(k_1,\cdots,k_r) \bigg{|}\leq \frac{ c_{s,{\boldsymbol{k}}} \left( \mathrm{log}\,(M+2) \right)^{r-1}}{n+1},\qquad n\geq 1.\] for some \(c_s>0\). \(\hfill\Box\)
By Lemma 1, we have \[\begin{align} &\;\;\;\; H_s^\star(k_1,\cdots,k_r) \\ &= \int_{(0,1)^{K_r}} [1,X_1,\cdots,X_r] t^{s+r-1} dx_1\cdots dx_{K_r} \\ &= \int_{(0,1)^{K_r}} \sum_{n=1}^{+\infty} \left( X_1^{n-1}\left[1,\frac{X_2}{X_1},\cdots,\frac{X_r}{X_1}\right]t^{n+r-2} -X_1^{s+n-1} \left[1,\frac{X_2}{X_1},\cdots,\frac{X_r}{X_1}\right]t^{s+n+r-2}\right) dx_1\cdots dx_{K_r} \\ &= \sum_{n=1}^{+\infty} \left(\frac{H_n^\star(k_2,\cdots,k_r)}{n^{k_1}} - \frac{H_{s+n}^\star(k_2,\cdots,k_r)}{(s+n)^{k_1}} \right) . \\ \end{align}\] Since \[\begin{align} &\;\;\;\;\Bigg{|} \frac{H_n^\star(k_2,\cdots,k_r)}{n^{k_1}} - \frac{H_{s+n}^\star(k_2,\cdots,k_r)}{(s+n)^{k_1}} \Bigg{|}\\ &\leq H_n^\star(k_2,\cdots,k_r) \Bigg{|} \frac{1}{n^{k_1}} - \frac{1}{(s+n)^{k_1}} \Bigg{|} + \Bigg{|} \frac{H_n^\star(k_2,\cdots,k_r)- H_{s+n}^\star(k_2,\cdots,k_r)}{(s+n)^{k_1}} \Bigg{|}, \\ \end{align}\] by Lemma 1, the series \[\sum_{n=1}^{+\infty} \left(\frac{H_n^\star(k_2,\cdots,k_r)}{n^{k_1}} - \frac{H_{s+n}^\star(k_2,\cdots,k_r)}{(s+n)^{k_1}} \right)\] is absolutely convergent. To prove that \[\mathop{\mathrm{lim}}_{\mathrm{Re} (s)\rightarrow +\infty} \mathfrak{h}_s\Big{|}_{\mathcal{T}}=\eta\] with bounded imaginary part \(\mathrm{Im}\,(s)\), it suffices to show that for any \(m\geq 1\) \[\mathop{\mathrm{lim}}_{\mathrm{Re} (s)\rightarrow +\infty} H_s^\star(k_1,\cdots,k_r)=\zeta^\star(k_1,\cdots,k_r)\] uniformly for every \((k_1,\cdots,k_r) \; \rotatebox[origin=c]{180}{\text{ \succ}} (2,\{1\}^{m})\).

By the inductive definition of \(H_s^\star(k)\), one has \[H_s^\star(k_1)=\mathop{\mathrm{lim}}_{M\rightarrow +\infty} \bigg{(}H_M^\star(k_1)-G_M^\star(s;k_1) \bigg{)}=\sum_{m=1}^{+\infty} \left( \frac{1}{m^{k_1}}-\frac{1}{ (m+s)^{k_1} } \right). \] For \(\mathrm{Re}(z)\geq 1\), we have \[\frac{1}{z^{k_1}}=\frac{1}{\Gamma(k_1)} \int^{+\infty}_0 e^{-zu} u^{k_1-1}du.\] From the formula \((16)\), it follows that \[H_s^\star(k_1)=\frac{1}{\Gamma(k_1)} \int^{+\infty}_0 \left( 1-e^{-su} \right)\frac{ u^{k_1-1}du }{e^u-1 }.\] We assume that the Lemma is proved for \(r<m\). For \(r=m\), we have \[\begin{align} &\;\;\;\; H_s^\star (k_1,\cdots,k_r) \\ &= \sum_{n=1}^{+\infty} \left(\frac{H_n^\star(k_2,\cdots,k_r)}{n^{k_1}} - \frac{H_{s+n}^\star(k_2,\cdots,k_r)}{(s+n)^{k_1}} \right) \\ &= \sum_{n=1}^{+\infty} \left[ \int^{+\infty}_0 \left(\frac{ 1-e^{-nu } }{ n^{k_1} } - \frac{ 1-e^{-(s+n)u } }{ (s+n)^{k_1} } \right) f_{k_2,\cdots,k_r}(u)du \right]. \\ \end{align}\] For \(k\in\mathbb{Z}^+\) and \(\mathrm{Re}(z)\geq 1\), it is easy to check that \[\frac{1-e^{-zu}}{z^k}=\frac{1}{(k-1)!}\int^{+\infty}_0 e^{-zx}\left(x^{k-1}-(x-u)_+^{k-1} \right)dx.\] Here \[(x-u)_+= \begin{cases} x-u,&x>u;\\ 0,&x\leq u.\\ \end{cases}\] Thus \[\begin{align} &\;\;\;\; H_s^\star (k_1,\cdots,k_r) \\ &= \sum_{n=1}^{+\infty} \left[ \int^{+\infty}_0 \left( \frac{1}{(k-1)!}\int^{+\infty}_0 (e^{-nx} -e^{-(s+n)x} )\left(x^{k-1}-(x-u)_+^{k-1} \right)dx \right) f_{k_2,\cdots,k_r}(u)du \right] \\ &= \int^{+\infty}_0 \left( \frac{1}{(k-1)!}\int^{+\infty}_0 \frac{ 1-e^{-sx} }{e^x-1} \left(x^{k-1}-(x-u)_+^{k-1} \right)dx \right) f_{k_2,\cdots,k_r}(u)du \\ &= \int^{+\infty}_0 \left(1-e^{-sx}\right) f_{k_1,\cdots,k_r}(x) dx. \end{align}\] Here \[f_{k_1,\cdots,k_r}(x)=\frac{1}{(k_1-1)!}\int^{+\infty}_0 \frac{ x^{k-1}-(x-u)_+^{k-1} }{ e^x-1 } f_{k_2,\cdots,k_r}(u) du.\] \(\hfill\Box\)

For \(s\in\mathbb{R}\) and \(s>1\), by the total order structure in Theorem 1, we have \[0 <H_s^\star(k_1,\cdots,k_r) \leq H_s^\star(\{1\}^{m})\] for \((k_1,\cdots,k_r) \; \rotatebox[origin=c]{180}{\text{ \succ}} (\{1\}^m)\). Thus for \(s\in \mathbb{R}\), it suffices to show that \[H_s^\star (\{1\}^m)\leq \left(1+ \mathrm{log}\,(1+s) \right)^m\] for all \(s>1\). Denote by \([s]\) the integer part of \(s\). Define \(n=[s]+1\). By Lemma 1, one has \[0< H_s^\star(\{1\}^m)<H_n^\star(\{1\}^m)<\left( H_n^\star(1) \right)^m\] Since \[H_n^\star(1)=1+\frac{1}{2}+\cdots +\frac{1}{n}<1+\int^n_1\frac{dx}{x}=1+\mathrm{log}\,n,\] we have \[\Big{|} H_s^\star(k_1,\cdots,k_r) \Big{|}\leq \left(1+\mathrm{log}\,n \right)^m\leq \left(1+\mathrm{log}(1+s)\right)^m\] for \(s>1\). For \(\mathrm{Re}(s)> 1\), by Lemma 1, \[\Big{|} H_s^\star(k_1,\cdots,k_r) \Big{|}\leq \int^{+\infty}_0 \Big{|} 1-e^{-su } \Big{|} f_{k_1,\cdots,k_r}(u)du.\] We have \[\Big{|} 1-e^{-w}\Big{|}\leq 3\left(1-e^{-|w|} \right),\qquad \mathrm{Re}(w)>0. \] In fact, the formula \((17)\) can be proved by the following simple observation: \[\Big{|} 1-e^{-w}\Big{|}\leq \mathrm{min}\{|w|,2\}\leq 3\left(1-e^{-|w|} \right), \quad \mathrm{Re}(w)>0\] By the formula \((17)\), we have \[\Big{|} H_s^\star(k_1,\cdots,k_r) \Big{|}\leq 3 H_{|s|}^\star(k_1,\cdots,k_r) \leq 3\left(1+ \mathrm{log}\,(1+|s|) \right)^m\] for \(\mathrm{Re}(s)> 1\). Similarly, \[\Big{|} H_s^\star(\{1\}^r) \Big{|}<3 H_{|s|}^\star(\{1\}^r) <3 |s|\] for any \(\mathrm{Re}(s)>1\) and \(r\geq 1\). \(\hfill\Box\)
By Lemma 1 and Lemma 1, we have \[\Big{|} H_s^\star(k_1,\cdots,k_r) -\zeta^\star(k_1,\cdots,k_r) \Big{|} \leq \sum_{n=1}^{+\infty} \Bigg{|} \frac{H_{s+n}^\star(k_2,\cdots,k_r)}{(s+n)^{k_1}} \Bigg{|}\] for \(k_1\geq 2, k_2,\cdots,k_r\geq 1\). By Lemma 1, for \[(k_1,\cdots,k_r) \; \rotatebox[origin=c]{180}{\text{ \succ}} (2, \{1\}^{m})\] and \(\sigma=\mathrm{Re}(s)> 1\), one has \[\Big{|} H_s^\star(k_1,\cdots,k_r) -\zeta^\star(k_1,\cdots,k_r) \Big{|} \leq \begin{cases} \sum_{n=1}^{+\infty} \frac{3\left(1+ \mathrm{log}\,(1+|s+n|) \right)^m }{(\sigma+n)^2}&k_1=2;\\ \sum_{n=1}^{+\infty} \frac{3|s+n|}{(\sigma+n)^3} &k_1\geq 3.\\ \end{cases}\] By the above formula, we have \[\Big{|} \mathop{\mathrm{lim}}_{r\rightarrow+\infty} H_s^\star(k_1,\cdots,k_r) - \mathop{\mathrm{lim}}_{r\rightarrow+\infty} \zeta^\star(k_1,\cdots,k_r) \Big{|} \leq B_s\] for any \({\boldsymbol{k}}=(k_1,\cdots,k_r,\cdots)\in \mathcal{T}\subseteq \widehat{\mathcal{T}}\) with \({\boldsymbol{k}} \; \rotatebox[origin=c]{180}{\text{ \succ}} (2,\{1\}^m\). Here \[B_s=\begin{cases} \sum_{n=1}^{+\infty} \frac{3\left(1+ \mathrm{log}\,(1+|s+n|) \right)^m }{(\sigma+n)^2}&k_1=2;\\ \sum_{n=1}^{+\infty} \frac{3|s+n|}{(\sigma+n)^3} &k_1\geq 3.\\ \end{cases}\] So we have \[\mathop{\mathrm{lim}}_{\mathrm{Re}(s)\rightarrow+\infty} \left( \mathop{\mathrm{lim}}_{r\rightarrow+\infty} H_s^\star(k_1,\cdots,k_r) \right)= \mathop{\mathrm{lim}}_{r\rightarrow+\infty} \zeta^\star(k_1,\cdots,k_r)\] with bounded imaginary part \(\mathrm{Im}(s)\) for any \({\boldsymbol{k}}=(k_1,\cdots,k_r,\cdots)\in \mathcal{T}\subseteq \widehat{\mathcal{T}}\). As a result, \[\mathop{\mathrm{lim}}_{\mathrm{Re} (s)\rightarrow +\infty} \mathfrak{h}_s\Big{|}_{\mathcal{T}}=\eta\] with bounded imaginary part \(|\mathrm{Im}(s)|\). Theorem 1 \((iii)\) is proved.

Acknowledgements↩︎

This project is supported by the National Natural Science Foundation of China (Grant No.12571009) and the Natural Science Foundation of Hunan Province, China (Grant No.2026JJ40003). The author wants to thank ChatGPT for very helpful conversations about the theory of divided differences.

References↩︎