Parametric Integrals for Binomial Series with Harmonic Polynomials

Abstract

Binomial series involving harmonic polynomials are expressed in terms of parametric integrals. By evaluating these parametric integrals, we establish several remarkable closed formulae for the infinite series containing both central binomial coefficients and harmonic numbers. Most of the values for binomial series found in this paper concern the dilogarithm and trilogarithm functions.

1. Introduction and Outline

By introducing a variable x, we extend the harmonic numbers to harmonic polynomials

$${\mathcal{H}}_n^{\langle m\rangle }\left(x\right)={\displaystyle \sum _{k=1}^n}{\displaystyle \frac{x^{k-1}}{k^m}},\mathrm{where}m,n\in {\mathbb{N}}_0.$$

When $x=\pm 1$, they become the usual and skew harmonic numbers

$${\mathbf{H}}_n^{\langle m\rangle }={\displaystyle \sum _{k=1}^n}{\displaystyle \frac{1}{k^m}}\mathrm{and}{\overline{\mathbf{H}}}_n^{\langle m\rangle }={\displaystyle \sum _{k=1}^n}{\displaystyle \frac{{(-1)}^{k-1}}{k^m}}.$$

In the case of $m=1$, we shall omit it from these notations. Recently, finite sums and infinite series around harmonic numbers have become active research topics (see for example [1,2,3,4,5,6]) for their applications in mathematics (particularly number theory [7,8,9], analytic combinators [10,11]), physics (standing waves in strings [12]), and computer science (algorithmic analysis [13]), just for examples.

The main approaches are Abel’s lemma on summation by parts (cf. [14,15,16]), hypergeometric series method (cf. [17,18,19,20,21]) and the logarithmic integrals (cf. [22,23,24,25,26]). Different from the above-mentioned methods, this paper will investigate analytically closed formulae of binomial series associated with these harmonic polynomials exclusively by evaluating the parametric integrals. Concretely, this will be done by first expressing these binomial series in terms of parametric integrals and then evaluating them (manually or by “Wolfram Mathematica Version 11”) explicitly in closed form. Several remarkable summation formulae will be derived as applications. As a showcase, we illustrate a few representatives as follows:

• Corollary 5 (d) and (i):

  $$\begin{array}{ll}& {\Phi }_1\left(1,\frac{1}{32}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{32}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }=-{\displaystyle \frac{4\sqrt{2}}{\sqrt{7}}}{\mathrm{Li}}_2\left(4\sqrt{14}-15\right),\hfill \\ & {\Phi }_1\left(1,\frac{-1}{32}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{32}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }=-{\displaystyle \frac{4\sqrt{2}}{3}}{\mathrm{Li}}_2\left(17-12\sqrt{2}\right).\hfill \end{array}$$
• Proposition 1 (b) and (g):

  $$\begin{array}{ll}& {\Phi }_2(1,\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left({\displaystyle \frac{\sqrt{5}-3}{2}}\right),\hfill \\ & {\Phi }_2(1,\frac{-1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left({\displaystyle \frac{7-3\sqrt{5}}{2}}\right).\hfill \end{array}$$
• Proposition 2 (i) and (d):

  $$\begin{array}{ll}& {\Phi }_3(-1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n^{\langle 2\rangle }}{n+1}}=2{ln}^2\left({\displaystyle \frac{1+\sqrt{2}}{2}}\right)-4{\mathrm{Li}}_2\left({\displaystyle \frac{1-\sqrt{2}}{2}}\right),\hfill \\ & {\Phi }_3(1,\frac{3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}={\displaystyle \frac{8}{3}}{\mathrm{Li}}_2\left({\displaystyle \frac{1}{4}}\right)-4{ln}^2\left({\displaystyle \frac{4}{3}}\right).\hfill \end{array}$$

The rest of the paper will be organized as follows. In Section 2 and Section 3, we shall investigate the binomial series involving harmonic polynomials of the first order ${\mathcal{H}}_n\left(x\right)$ and the second order ${\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)$, respectively. Then the bisection series will be examined in Section 4, where a solution to a conjectured formula by Sun [27] will be presented. Finally, the paper will end with a brief comment about the remaining questions on the subject.

The informed reader will notice that many series examined in this paper are expressed in terms of the polylogarithmic function (cf. [28]), that can be defined by repeated integrals

$${\mathrm{Li}}_{m+1}\left(z\right)={\int }_0^z{\displaystyle \frac{{\mathrm{Li}}_m\left(y\right)}{y}}dy\mathrm{with}{\mathrm{Li}}_1\left(z\right)=-ln(1-z).$$

Two initial ones ${\mathrm{Li}}_2\left(z\right)$ and ${\mathrm{Li}}_3\left(z\right)$ are called dilogarithm and trilogarithm functions, respectively. The Hurwitz zeta function and the Riemann zeta series are defined for $m,x\in \mathbb{C}$ with $\Re \left(m\right)>1$ and $x\notin \mathbb{Z}\setminus \mathbb{N}$ by

$${\zeta }_m\left(x\right)={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{1}{{(x+n)}^m}}\mathrm{and}{\zeta }_m={\zeta }_m\left(1\right)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{1}{n^m}}.$$

In order to assure the accuracy, all the displayed expressions are experimentally checked using “Wolfram Mathematica version 11”. Since most of the integral computations and series manipulations are cumbersome and tedious, we confine ourselves to give the finally simplified formulae without producing long and complicated intermediate reductions.

2. Binomial Series with Harmonic Polynomials ${\mathcal{H}}_{\mathit{n}}$(x)

It is well-known that the usual harmonic numbers admit the generating function and the integral representation as follows:

$${\displaystyle \sum _{n=1}^{\infty }}{\mathbf{H}}_n{y}^n={\displaystyle \frac{ln(1-y)}{y-1}}\mathrm{and}{\mathbf{H}}_n={\int }_0^1{\displaystyle \frac{1-T^n}{1-T}}dT.$$

Likewise, we have no difficulty in establishing their bivariate counterparts

$${\displaystyle \sum _{n=1}^{\infty }}{y}^n{\mathcal{H}}_n\left(x\right)={\displaystyle \frac{ln(1-xy)}{x(y-1)}}\mathrm{and}{\mathcal{H}}_n\left(x\right)={\int }_0^1{\displaystyle \frac{1-{\left(Tx\right)}^n}{1-Tx}}dT.$$

In this section, we shall examine four classes of binomial series associated with harmonic polynomials of the first order and show several infinite series identities as consequences.

2.1. Series ${\varphi }_1(x,y)$ with ${\mathcal{H}}_n\left(x\right)$

Boyadzhiev [29] derived the following univariate generating function:

$${\displaystyle \sum _{n=1}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n={\displaystyle \frac{2}{\sqrt{1-4y}}}ln{\displaystyle \frac{1+\sqrt{1-4y}}{2\sqrt{1-4y}}}.$$

By combining the series rearrangement with the generating function for central binomial coefficients

$${\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)y^n={\displaystyle \frac{1}{\sqrt{1-4y}}},$$

(1)

the authors found its bivariate form (cf. Li and Chu [30]), which is reproduced, as the first series ${\varphi }_1(x,y)$ for integrity, in the following lemma.

Lemma 1

($-{\displaystyle \frac{1}{4}}\le y<{\displaystyle \frac{1}{4}}$).

$$\begin{array}{ll}\hfill {\varphi }_1(x,y):={\displaystyle \sum _{n=0}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n\left(x\right)& ={\int }_0^1{\displaystyle \frac{dT}{1-Tx}}\left\{{\displaystyle \frac{1}{\sqrt{1-4y}}}-{\displaystyle \frac{1}{\sqrt{1-4Txy}}}\right\}\hfill \\ & ={\displaystyle \frac{2}{x\sqrt{1-4y}}}ln\left({\displaystyle \frac{1+\sqrt{1-4y}}{\sqrt{1-4y}+\sqrt{1-4xy}}}\right).\hfill \end{array}$$

The above series contains several numerical series as special cases. In addition to those given in [30], further examples are displayed below.

Corollary 1.

Binomial series with ${\mathbf{H}}_n$ or ${\overline{\mathbf{H}}}_n$ classified according to their convergence rates:

• Series of convergence rate ${\displaystyle \frac{1}{4}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{a}\right)& {\varphi }_1(1,-\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n=\sqrt{2}ln\left({\displaystyle \frac{1+\sqrt{2}}{2\sqrt{2}}}\right),\hfill \\ \hfill \left(\mathrm{b}\right)& {\varphi }_1(-1,-\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n=\sqrt{2}ln\left({\displaystyle \frac{\sqrt{2}}{1+\sqrt{2}}}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{1}{5}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{c}\right)& {\varphi }_1(1,\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n=2\sqrt{5}ln\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right),\hfill \\ \hfill \left(\mathrm{d}\right)& {\varphi }_1(-1,\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n=2\sqrt{5}ln\left(\sqrt{5}-1\right),\hfill \\ \hfill \left(\mathrm{e}\right)& {\varphi }_1(1,-\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n={\displaystyle \frac{2}{3}}\sqrt{5}ln\left({\displaystyle \frac{3+\sqrt{5}}{6}}\right),\hfill \\ \hfill \left(\mathrm{f}\right)& {\varphi }_1(-1,-\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n={\displaystyle \frac{2}{3}}\sqrt{5}ln\left(3-\sqrt{5}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{1}{7}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{g}\right)& {\varphi }_1(1,\frac{1}{7})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{7}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n=2\sqrt{{\displaystyle \frac{7}{3}}}ln\left({\displaystyle \frac{3+\sqrt{21}}{6}}\right),\hfill \\ \hfill \left(\mathrm{h}\right)& {\varphi }_1(1,-\frac{1}{7})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{7}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n=2\sqrt{{\displaystyle \frac{7}{11}}}ln\left({\displaystyle \frac{11+\sqrt{77}}{22}}\right),\hfill \\ \hfill \left(\mathrm{i}\right)& {\varphi }_1(-1,\frac{1}{7})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{7}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n=2\sqrt{{\displaystyle \frac{7}{3}}}ln\left({\displaystyle \frac{\sqrt{21}+\sqrt{77}}{7+\sqrt{21}}}\right),\hfill \\ \hfill \left(\mathrm{j}\right)& {\varphi }_1(-1,-\frac{1}{7})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{7}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n=2\sqrt{{\displaystyle \frac{7}{11}}}ln\left({\displaystyle \frac{\sqrt{21}+\sqrt{77}}{7+\sqrt{77}}}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{\sqrt{3}}{8}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{k}\right)& {\varphi }_1(1,\frac{\sqrt{3}}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{\sqrt{3}}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n=2(1+\sqrt{3})ln\left({\displaystyle \frac{2+\sqrt{3}}{2}}\right),\hfill \\ \hfill \left(\mathrm{l}\right)& {\varphi }_1(1,\frac{-\sqrt{3}}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-\sqrt{3}}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n=(1-\sqrt{3})ln\left({\displaystyle \frac{4}{3}}\right),\hfill \\ \hfill \left(\mathrm{m}\right)& {\varphi }_1(-1,\frac{\sqrt{3}}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{\sqrt{3}}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n=(1+\sqrt{3})ln\left(12-6\sqrt{3}\right),\hfill \\ \hfill \left(\mathrm{n}\right)& {\varphi }_1(-1,\frac{-\sqrt{3}}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-\sqrt{3}}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n=(\sqrt{3}-1)ln\left(4-2\sqrt{3}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{3}{16}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{o}\right)& {\varphi }_1(1,\frac{3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n=4ln\left({\displaystyle \frac{3}{2}}\right),\hfill \\ \hfill \left(\mathrm{p}\right)& {\varphi }_1(1,\frac{-3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n={\displaystyle \frac{4}{\sqrt{7}}}ln\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{\sqrt{7}}}\right),\hfill \\ \hfill \left(\mathrm{q}\right)& {\varphi }_1(-1,\frac{3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n=4ln\left({\displaystyle \frac{1+\sqrt{7}}{3}}\right),\hfill \\ \hfill \left(\mathrm{r}\right)& {\varphi }_1(-1,\frac{-3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n={\displaystyle \frac{4}{\sqrt{7}}}ln\left({\displaystyle \frac{5-\sqrt{7}}{3}}\right).\hfill \end{array}$$

2.2. Series ${\varphi }_2(x,y)$ with ${\displaystyle \frac{{\mathcal{H}}_n\left(x\right)}{n}}$

Now, we are going to examine the second series

$${\varphi }_2(x,y):={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{y^n}{n}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n\left(x\right).$$

The closed formula is given as in the following theorem.

Theorem 1

($\left|x\right|\le 1$ and $\left|y\right|\le {\displaystyle \frac{1}{4}}$).

$$\begin{array}{ll}\hfill {\varphi }_2(x,y)& ={\displaystyle \frac{2}{x}}\{ln(1-x)ln{\displaystyle \frac{1+\sqrt{1-4y}}{1+\sqrt{1-4xy}}}-{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{1-4xy}}{1+\sqrt{1-4y}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{1-4xy}}{1-\sqrt{1-4y}}}\right)\hfill \\ & -2\pi \mathbf{i}\chi (y>0)ln\left({\displaystyle \frac{1+\sqrt{1-4xy}}{2}}\right)+{\mathrm{Li}}_2\left({\displaystyle \frac{2}{1+\sqrt{1-4y}}}\right)+{\mathrm{Li}}_2\left({\displaystyle \frac{2}{1-\sqrt{1-4y}}}\right)\}.\hfill \end{array}$$

Proof. 

Recalling the generating function (cf. [31])

$${\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{y^n}{n}}=2ln\left({\displaystyle \frac{2}{1+\sqrt{1-4y}}}\right),$$

(2)

we can express the binomial series with harmonic polynomials as

$$\begin{array}{ll}\hfill {\varphi }_2(x,y)& ={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{y^n}{n}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\int }_0^1{\displaystyle \frac{1-{\left(Tx\right)}^n}{1-Tx}}dT\hfill \\ & ={\int }_0^1{\displaystyle \frac{2dT}{1-Tx}}ln{\displaystyle \frac{1+\sqrt{1-4Txy}}{1+\sqrt{1-4y}}}.\hfill \end{array}$$

The indefinite integral in question equals

$$\begin{array}{ll}\hfill \int {\displaystyle \frac{2dT}{1-Tx}}ln{\displaystyle \frac{1+\sqrt{1-4Txy}}{1+\sqrt{1-4y}}}={\displaystyle \frac{2}{x}}\{& ln\left({\displaystyle \frac{1+\sqrt{1-4y}}{1+\sqrt{1-4Txy}}}\right)ln\left({\displaystyle \frac{\sqrt{1-4y}-\sqrt{1-4Txy}}{\sqrt{1-4y}+1}}\right)\hfill \\ \hfill +& ln\left({\displaystyle \frac{1+\sqrt{1-4y}}{1+\sqrt{1-4Txy}}}\right)ln\left({\displaystyle \frac{\sqrt{1-4y}+\sqrt{1-4Txy}}{\sqrt{1-4y}-1}}\right)\hfill \\ \hfill -& {\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{1-4Txy}}{1+\sqrt{1-4y}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{1-4Txy}}{1-\sqrt{1-4y}}}\right)\}+C,\hfill \end{array}$$

which can be verified by computing the derivative of the expression on the right. By substituting the integral limits, we find that

$$\begin{array}{ll}\hfill {\varphi }_2(x,y)& ={\displaystyle \frac{2}{x}}\{ln\left({\displaystyle \frac{2}{1+\sqrt{1-4y}}}\right)ln\left({\displaystyle \frac{\sqrt{1-4y}-1}{\sqrt{1-4y}+1}}\right)+ln\left({\displaystyle \frac{1+\sqrt{1-4y}}{1+\sqrt{1-4xy}}}\right)ln\left({\displaystyle \frac{\sqrt{1-4y}-\sqrt{1-4xy}}{1+\sqrt{1-4y}}}\right)\hfill \\ \hfill +& ln\left({\displaystyle \frac{2}{1+\sqrt{1-4y}}}\right)ln\left({\displaystyle \frac{\sqrt{1-4y}+1}{\sqrt{1-4y}-1}}\right)+ln\left({\displaystyle \frac{1+\sqrt{1-4y}}{1+\sqrt{1-4xy}}}\right)ln\left({\displaystyle \frac{\sqrt{1-4y}+\sqrt{1-4xy}}{\sqrt{1-4y}-1}}\right)\hfill \\ \hfill +& {\mathrm{Li}}_2\left({\displaystyle \frac{2}{1+\sqrt{1-4y}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{2}{1-\sqrt{1-4y}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{1-4xy}}{1+\sqrt{1-4y}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{1-4xy}}{1-\sqrt{1-4y}}}\right)\}.\hfill \end{array}$$

By reducing the above four terms of logarithmic products to

$$ln(1-x)ln\left({\displaystyle \frac{1+\sqrt{1-4y}}{1+\sqrt{1-4xy}}}\right)-2\pi \mathbf{i}\chi (y>0)ln\left({\displaystyle \frac{1+\sqrt{1-4xy}}{2}}\right),$$

we confirm the formula stated in Theorem 1.  □

By assigning particular values for x and y, we deduce, from Theorem 1, the following infinite series identities. Among them, (a) and (b) can be found in [32] (§4.8.1 and §4.8.2), while (k) and (l) in [33].

Corollary 2.

Binomial series with ${\mathbf{H}}_n$ or ${\overline{\mathbf{H}}}_n$ classified according to their convergence rates:

• Series of convergence rate ${\displaystyle \frac{1}{4}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{a}\right)& {\varphi }_2(1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}={\displaystyle \frac{{\pi }^2}{3}},\hfill \\ \hfill \left(\mathrm{b}\right)& {\varphi }_2(1,-\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}=2{\mathrm{Li}}_2\left(2\sqrt{2}-3\right),\hfill \\ \hfill \left(\mathrm{c}\right)& {\varphi }_2(-1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n}}=-{\displaystyle \frac{{\pi }^2}{3}}-4{\mathrm{Li}}_2(-\sqrt{2}),\hfill \\ \hfill \left(\mathrm{d}\right)& {\varphi }_2(-1,-\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n}}=2ln(1+\sqrt{2})ln\left({\displaystyle \frac{1+\sqrt{2}}{2}}\right)-{\displaystyle \frac{{\pi }^2}{12}}.\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{1}{5}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{e}\right)& {\varphi }_2(1,\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}={\displaystyle \frac{2{\pi }^2}{15}}-2{ln}^2\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right),\hfill \\ \hfill \left(\mathrm{f}\right)& {\varphi }_2(1,\frac{-1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}=2{\mathrm{Li}}_2\left({\displaystyle \frac{3\sqrt{5}-7}{2}}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{1}{8}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{g}\right)& {\varphi }_2(1,\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}=2{\mathrm{Li}}_2\left(3-2\sqrt{2}\right),\hfill \\ \hfill \left(\mathrm{h}\right)& {\varphi }_2(1,-\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}=2{\mathrm{Li}}_2\left(2\sqrt{6}-5\right),\hfill \\ \hfill \left(\mathrm{i}\right)& {\varphi }_2(-1,\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n}}={\displaystyle \frac{{\pi }^2}{3}}-2{\mathrm{Li}}_2\left({\displaystyle \frac{\sqrt{2}-1}{\sqrt{2}+\sqrt{3}}}\right)\hfill \\ & -2ln\left({\displaystyle \frac{\sqrt{2}+\sqrt{3}}{1+\sqrt{2}}}\right)ln\left({\displaystyle \frac{\sqrt{2}+\sqrt{3}}{2\sqrt{2}-2}}\right)-2{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{2}}{\sqrt{2}+\sqrt{3}}}\right),\hfill \\ \hfill \left(\mathrm{j}\right)& {\varphi }_2\left(-1,-\frac{1}{8}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n}}=4{ln}^2\left(\sqrt{2}+\sqrt{3}\right)\hfill \\ & +2{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{2}}{\sqrt{2}+\sqrt{3}}}\right)+2{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{2}}{\sqrt{2}-\sqrt{3}}}\right)-2ln2ln\left({\displaystyle \frac{\sqrt{2}+\sqrt{3}}{1+\sqrt{2}}}\right).\hfill \end{array}$$
• Four series with difference convergence rates:

  $$\begin{array}{ll}\hfill \left(\mathrm{k}\right)& {\varphi }_2(1,\frac{2}{9})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{2}{9}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}={\displaystyle \frac{{\pi }^2}{6}}-{ln}^{2}2,\hfill \\ \hfill \left(\mathrm{l}\right)& {\varphi }_2(1,\frac{3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}=2{\mathrm{Li}}_2\left({\displaystyle \frac{1}{3}}\right),\hfill \\ \hfill \left(\mathrm{m}\right)& {\varphi }_2(1,\frac{1}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}=2{\mathrm{Li}}_2\left(7-4\sqrt{3}\right),\hfill \\ \hfill \left(\mathrm{n}\right)& {\varphi }_2(1,\frac{-1}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n}}=2{\mathrm{Li}}_2\left(4\sqrt{5}-9\right).\hfill \end{array}$$

2.3. Series ${\varphi }_3(x,y)$ with ${\displaystyle \frac{{\mathcal{H}}_n\left(x\right)}{n+1}}$

The third series is defined by

$${\varphi }_3(x,y):={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n\left(x\right).$$

This series can be evaluated in closed form as in the following simpler formula.

Theorem 2

($\left|x\right|\le 1$ and $\left|y\right|\le {\displaystyle \frac{1}{4}}$).

$$\begin{array}{ll}\hfill {\varphi }_3(x,y)& ={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{y^n}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n\left(x\right)\hfill \\ & ={\displaystyle \frac{1}{xy}}ln\left({\displaystyle \frac{2}{1+\sqrt{1-4xy}}}\right)+{\displaystyle \frac{\sqrt{1-4y}}{xy}}ln\left({\displaystyle \frac{\sqrt{1-4y}+\sqrt{1-4xy}}{1+\sqrt{1-4y}}}\right).\hfill \end{array}$$

Proof. 

Applying the generating function for Catalan numbers (cf. [29])

$${\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{y^n}{n+1}}={\displaystyle \frac{2}{1+\sqrt{1-4y}}},$$

(3)

we can reformulate the series into a definite integral

$$\begin{array}{ll}\hfill {\varphi }_3(x,y)& ={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\int }_0^1{\displaystyle \frac{1-{\left(Tx\right)}^n}{1-Tx}}dT\hfill \\ & ={\int }_0^1{\displaystyle \frac{2dT}{1-Tx}}\left\{{\displaystyle \frac{1}{1+\sqrt{1-4y}}}-{\displaystyle \frac{1}{1+\sqrt{1-4Txy}}}\right\}.\hfill \end{array}$$

It is not difficult to check that the above integrand has the following antiderivative

$${\displaystyle \frac{1}{2xy}}ln\left({\displaystyle \frac{(1-Tx)(\sqrt{1-4Txy}-1)}{Tx(\sqrt{1-4Txy}+1)}}\right)-{\displaystyle \frac{2ln(Tx-1)}{x(1+\sqrt{1-4y})}}-{\displaystyle \frac{\sqrt{1-4y}}{2xy}}ln\left({\displaystyle \frac{\sqrt{1-4y}-\sqrt{1-4Txy}}{\sqrt{1-4y}+\sqrt{1-4Txy}}}\right).$$

By substituting the integral limits and then simplifying the resulting expression, we arrive at the closed formula stated in Theorem 2.  □

By specifying particular values for x and y in Theorem 2, numerous infinite series identities can be derived as consequences. We confine ourselves to show the following sample ones.

Corollary 3.

Binomial series with ${\mathbf{H}}_n$ or ${\overline{\mathbf{H}}}_n$ classified according to their convergence rates:

• Series of convergence rate ${\displaystyle \frac{1}{4}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{a}\right)& {\varphi }_3(1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}=4ln2,\hfill \\ \hfill \left(\mathrm{b}\right)& {\varphi }_3(-1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}=4ln\left({\displaystyle \frac{1+\sqrt{2}}{2}}\right),\hfill \\ \hfill \left(\mathrm{c}\right)& {\varphi }_3(1,-\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}=4ln\left({\displaystyle \frac{1+\sqrt{2}}{2}}\right)+4\sqrt{2}ln\left({\displaystyle \frac{1+\sqrt{2}}{2\sqrt{2}}}\right),\hfill \\ \hfill \left(\mathrm{d}\right)& {\varphi }_3(-1,-\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}=4ln2+4\sqrt{2}ln\left(2-\sqrt{2}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{1}{5}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{e}\right)& {\varphi }_3(1,\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}=5ln\left({\displaystyle \frac{5-\sqrt{5}}{2}}\right)-\sqrt{5}ln\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right),\hfill \\ \hfill \left(\mathrm{f}\right)& {\varphi }_3(1,\frac{-1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}=5ln\left({\displaystyle \frac{3+\sqrt{5}}{2\sqrt{5}}}\right)+3\sqrt{5}ln\left({\displaystyle \frac{3+\sqrt{5}}{6}}\right),\hfill \\ \hfill \left(\mathrm{g}\right)& {\varphi }_3(-1,\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}=5ln\left({\displaystyle \frac{3+\sqrt{5}}{2\sqrt{5}}}\right)-\sqrt{5}ln\left(\sqrt{5}-1\right),\hfill \\ \hfill \left(\mathrm{h}\right)& {\varphi }_3(-1,\frac{-1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}=5ln\left({\displaystyle \frac{5-\sqrt{5}}{2}}\right)+3\sqrt{5}ln\left(3-\sqrt{5}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{1}{7}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{i}\right)& {\varphi }_3(1,\frac{1}{7})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{7}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}=7ln\left({\displaystyle \frac{7-\sqrt{21}}{2}}\right)-\sqrt{21}ln\left({\displaystyle \frac{3+\sqrt{21}}{6}}\right),\hfill \\ \hfill \left(\mathrm{j}\right)& {\varphi }_3(1,\frac{-1}{7})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{7}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}=7ln\left({\displaystyle \frac{7+\sqrt{77}}{14}}\right)+\sqrt{77}ln\left({\displaystyle \frac{11+\sqrt{77}}{22}}\right),\hfill \\ \hfill \left(\mathrm{k}\right)& {\varphi }_3(-1,\frac{1}{7})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{7}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}=7ln\left({\displaystyle \frac{7+\sqrt{77}}{14}}\right)-\sqrt{21}ln\left({\displaystyle \frac{\sqrt{21}+\sqrt{77}}{7+\sqrt{21}}}\right),\hfill \\ \hfill \left(\mathrm{l}\right)& {\varphi }_3(-1,\frac{-1}{7})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{7}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}=\sqrt{77}ln\left({\displaystyle \frac{\sqrt{21}+\sqrt{77}}{7+\sqrt{77}}}\right)-7ln\left({\displaystyle \frac{7+\sqrt{21}}{14}}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{1}{8}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{m}\right)& {\varphi }_3(1,\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}=8ln\left({\displaystyle \frac{2\sqrt{2}}{1+\sqrt{2}}}\right)+4\sqrt{2}ln\left({\displaystyle \frac{2}{1+\sqrt{2}}}\right),\hfill \\ \hfill \left(\mathrm{n}\right)& {\varphi }_3(-1,\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}=8ln\left({\displaystyle \frac{\sqrt{2}+\sqrt{3}}{2\sqrt{2}}}\right)+4\sqrt{2}ln\left({\displaystyle \frac{1+\sqrt{2}}{1+\sqrt{3}}}\right),\hfill \\ \hfill \left(\mathrm{o}\right)& {\varphi }_3(1,-\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}=8ln\left({\displaystyle \frac{\sqrt{2}+\sqrt{3}}{2\sqrt{2}}}\right)+4\sqrt{6}ln\left({\displaystyle \frac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}}\right),\hfill \\ \hfill \left(\mathrm{p}\right)& {\varphi }_3(-1,-\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}=8ln\left({\displaystyle \frac{2\sqrt{2}}{1+\sqrt{2}}}\right)+4\sqrt{6}ln\left({\displaystyle \frac{1+\sqrt{3}}{\sqrt{2}+\sqrt{3}}}\right).\hfill \end{array}$$
• Series of convergence rate ${\displaystyle \frac{1}{16}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{q}\right)& {\varphi }_3(1,\frac{3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n+1}}={\displaystyle \frac{8}{3}}ln\left({\displaystyle \frac{32}{27}}\right),\hfill \\ \hfill \left(\mathrm{r}\right)& {\varphi }_3(-1,\frac{3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{n+1}}={\displaystyle \frac{8}{3}}ln\left({\displaystyle \frac{17+7\sqrt{7}}{32}}\right).\hfill \end{array}$$

2.4. Series ${\varphi }_4(x,y)$ with ${\displaystyle \frac{{\mathcal{H}}_n\left(x\right)}{2n+1}}$

The fourth series ${\varphi }_4(x,y)$ below is more difficult than previous ones

$${\varphi }_4(x,y):={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{2n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n\left(x\right).$$

Nevertheless, we did succeed, with a help of “Wolfram Mathematica version 11”, in delving into the following compact formula involving eight terms of dilogarithm function.

Theorem 3

($\left|x\right|\le 1$ and $\left|y\right|\le {\displaystyle \frac{1}{4}}$). With the sign function $\eta (x,y)$ being defined by

$$\eta (x,y)=\left\{\begin{array}{c}-1,x<0\mathit{and}y<0;\hfill \\ 1,\mathit{otherwise};\hfill \end{array}\right.$$

the following summation formula holds:

$$\begin{array}{ll}{\varphi }_4(x,y)& ={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{2n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n\left(x\right)={\displaystyle \frac{\mathbf{i}}{2\sqrt{x}\sqrt{xy}}}\{2\mathbf{i}arcsin\left(2\sqrt{xy}\right)ln(1+\sqrt{x})\\ & -\pi \eta (x,y)arcsin\left(2\sqrt{xy}\right)+\mathbf{i}ln(1-x)\left[\eta (x,y)arcsin\left(2\sqrt{y}\right)-arcsin\left(2\sqrt{xy}\right)\right]\\ & +{\mathrm{Li}}_2\left({\displaystyle \frac{2\sqrt{xy}+\sqrt{x(4y-1)}}{\mathbf{i}\sqrt{x}}}\right)+{\mathrm{Li}}_2\left({\displaystyle \frac{(2\sqrt{xy}+\sqrt{x(4y-1)})(2\mathbf{i}\sqrt{xy}-\sqrt{1-4xy})}{\mathbf{i}\sqrt{x}}}\right)\\ & +{\mathrm{Li}}_2\left({\displaystyle \frac{2\sqrt{xy}-\sqrt{x(4y-1)}}{\mathbf{i}\sqrt{x}}}\right)+{\mathrm{Li}}_2\left({\displaystyle \frac{(2\sqrt{xy}-\sqrt{x(4y-1)})(2\mathbf{i}\sqrt{xy}-\sqrt{1-4xy})}{\mathbf{i}\sqrt{x}}}\right)\\ & -{\mathrm{Li}}_2\left({\displaystyle \frac{-2\sqrt{xy}+\sqrt{x(4y-1)}}{\mathbf{i}\sqrt{x}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{(2\sqrt{xy}+\sqrt{x(4y-1)})(\sqrt{1-4xy}-2\mathbf{i}\sqrt{xy})}{\mathbf{i}\sqrt{x}}}\right)\\ & -{\mathrm{Li}}_2\left({\displaystyle \frac{-2\sqrt{xy}-\sqrt{x(4y-1)}}{\mathbf{i}\sqrt{x}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{(2\sqrt{xy}-\sqrt{x(4y-1)})(\sqrt{1-4xy}-2\mathbf{i}\sqrt{xy})}{\mathbf{i}\sqrt{x}}}\right)\}.\end{array}$$

Proof. 

First, dividing across (1) by $y^{1/2}$ and integrating with respect to y over $[0,y]$, we can compute the binomial series

$${\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{y^n}{2n+1}}={\displaystyle \frac{arcsin\left(2\sqrt{y}\right)}{2\sqrt{y}}}.$$

(4)

It is utilized to reformulate the bivariate series as follows:

$$\begin{array}{ll}\hfill {\varphi }_4(x,y)& ={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{2n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)\left\{{\displaystyle \frac{{\mathrm{Li}}_1\left(x\right)}{x}}-{\int }_0^1{\displaystyle \frac{{\left(Tx\right)}^n}{1-Tx}}dT\right\}\hfill \\ & ={\displaystyle \frac{arcsin\left(2\sqrt{y}\right){\mathrm{Li}}_1\left(x\right)}{2x\sqrt{y}}}-{\int }_0^1{\displaystyle \frac{arcsin\left(2\sqrt{Txy}\right)}{2\sqrt{Txy}(1-Tx)}}dT\begin{array}{|c|}\hline T\to T^2\\ \hline\end{array}\hfill \\ & ={\displaystyle \frac{arcsin\left(2\sqrt{y}\right){\mathrm{Li}}_1\left(x\right)}{2x\sqrt{y}}}-{\int }_0^1{\displaystyle \frac{arcsin\left(2T\sqrt{xy}\right)}{(1-T^{2}x)\sqrt{xy}}}dT.\hfill \end{array}$$

Then the explicit formula in Theorem 3 follows after having evaluated the rightmost integral by “Wolfram Mathematica version 11” and performed a number of complex manipulations and simplifications for the resulting expression.  □

Even though the above expression looks ugly, we did succeed in determining, under long and tedious reductions, the following closed formulae for appropriately selected values of x and y.

Corollary 4.

Binomial series with ${\mathbf{H}}_n$ or ${\overline{\mathbf{H}}}_n$ classified according to their convergence rates:

• Series of convergence rate ${\displaystyle \frac{1}{4}}$:

  $$\begin{array}{ll}\hfill \left(\mathrm{a}\right)& {\varphi }_4(1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{2n+1}}=4G-\pi ln2,\hfill \\ \hfill \left(\mathrm{b}\right)& {\varphi }_4(-1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{2n+1}}={\displaystyle \frac{\pi }{2}}ln2-5G+4\Im \left\{{\mathrm{Li}}_2\left({\displaystyle \frac{1+\mathbf{i}}{\sqrt{2}}}\right)\right\},\hfill \\ \hfill \left(\mathrm{c}\right)& {\varphi }_4(1,-\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{2n+1}}=4{\mathrm{Li}}_2(1-\sqrt{2})\hfill \\ & \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} +{\displaystyle \frac{{\pi }^2}{4}}-2ln2ln(1+\sqrt{2})-{\mathrm{Li}}_2(2\sqrt{2}-3),\hfill \\ \hfill \left(\mathrm{d}\right)& {\varphi }_4(-1,-\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n}{2n+1}}=ln2ln(1+\sqrt{2})\hfill \\ & \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}-{\displaystyle \frac{{\pi }^2}{4}}+2{\mathrm{Li}}_2(\sqrt{2}-1)-2{\mathrm{Li}}_2(1-\sqrt{2}).\hfill \end{array}$$
• Further four series with different convergence rates:

  $$\begin{array}{ll}\hfill \left(\mathrm{e}\right)& {\varphi }_4(1,\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{2n+1}}=-{\displaystyle \frac{{\pi }^2}{2}}-{\displaystyle \frac{\pi ln2}{\sqrt{2}}}-2G\sqrt{2}+{\displaystyle \frac{1}{8}}{\zeta }_2\left({\displaystyle \frac{1}{8}}\right)+{\displaystyle \frac{1}{8}}{\zeta }_2\left({\displaystyle \frac{3}{8}}\right)\hfill \\ & ={\displaystyle \frac{{\pi }^2}{2}}-{\displaystyle \frac{\pi ln2}{\sqrt{2}}}-2G\sqrt{2}-{\displaystyle \frac{1}{8}}{\zeta }_2\left({\displaystyle \frac{5}{8}}\right)-{\displaystyle \frac{1}{8}}{\zeta }_2\left({\displaystyle \frac{7}{8}}\right),\hfill \\ \hfill \left(\mathrm{f}\right)& {\varphi }_4(1,\frac{-1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{2n+1}}={\displaystyle \frac{{\pi }^2}{2\sqrt{2}}}+4\sqrt{2}{\mathrm{Li}}_2\left(-\sqrt{2-\sqrt{3}}\right)\hfill \\ & -\sqrt{2}{\mathrm{Li}}_2(\sqrt{3}-2)-\sqrt{2}ln2ln(2+\sqrt{3}),\hfill \\ \hfill \left(\mathrm{g}\right)& {\varphi }_4(1,-\frac{1}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{2n+1}}=4ln\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right)ln\left({\displaystyle \frac{1+\sqrt{5}}{4}}\right)-2{\mathrm{Li}}_2\left({\displaystyle \frac{\sqrt{5}-3}{2}}\right)-{\displaystyle \frac{{\pi }^2}{30}},\hfill \\ \hfill \left(\mathrm{h}\right)& {\varphi }_4(1,\frac{-1}{32})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{32}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{2n+1}}={\displaystyle \frac{5{\pi }^2}{3\sqrt{2}}}-4\sqrt{2}{ln}^{2}2-2\sqrt{2}{\mathrm{Li}}_2\left({\displaystyle \frac{-1}{2}}\right)-8\sqrt{2}{\mathrm{Li}}_2\left({\displaystyle \frac{1}{\sqrt{2}}}\right).\hfill \end{array}$$

3. Binomial Series with Harmonic Polynomials ${\mathcal{H}}_{\mathit{n}}^{\mathbf{\langle }\mathbf{2}\mathbf{\rangle }}\mathbf{\left(}\mathit{x}\mathbf{\right)}$

For the harmonic polynomials of the second order ${\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)$, it is not hard to derive their ordinary generating function

$${\displaystyle \sum _{n=1}^{\infty }}{y}^n{\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)={\displaystyle \frac{{\mathrm{Li}}_2\left(xy\right)}{x(1-y)}},\mathrm{where}{\mathrm{Li}}_m\left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{z^n}{n^m}}.$$

The harmonic polynomials of the second order can also be expressed by the integral below

$${\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)=-{\int }_0^{1}lnT{\displaystyle \frac{1-{\left(Tx\right)}^n}{1-Tx}}dT={\displaystyle \frac{{\mathrm{Li}}_2\left(x\right)}{x}}+{\int }_0^1{\displaystyle \frac{{\left(Tx\right)}^{n}lnT}{1-Tx}}dT.$$

This will be crucial for us to examine four further classes of binomial series associated with ${\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)$.

3.1. Series ${\Phi }_1(x,y)$ with ${\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)$

According to (1), we can rewrite the binomial series below as a parametric integral

$$\begin{array}{ll}\hfill {\Phi }_1(x,y):={\displaystyle \sum _{n=0}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)& =-{\displaystyle \sum _{n=0}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\int }_0^1{\displaystyle \frac{1-{\left(Tx\right)}^n}{1-Tx}}lnTdT\hfill \\ & ={\int }_0^1{\displaystyle \frac{lnTdT}{1-Tx}}\left\{{\displaystyle \frac{1}{\sqrt{1-4Txy}}}-{\displaystyle \frac{1}{\sqrt{1-4y}}}\right\}.\hfill \end{array}$$

Computing it and then simplifying the result using “Wolfram Mathematica version 11”, we find the closed formula as in the theorem below. Because the process just described is cumbersome and tedious, we confine ourselves to give the finally simplified result without producing long and complicated intermediate reductions.

Theorem 4

($\left|x\right|<1$ and $-{\displaystyle \frac{1}{4}}\le y<{\displaystyle \frac{1}{4}}$).

$$\begin{array}{ll}\hfill {\Phi }_1(x,y)& ={\displaystyle \frac{2\pi \mathbf{i}\chi (y<0)}{x\sqrt{1-4y}}}ln\left({\displaystyle \frac{1-\sqrt{1-4xy}}{2xy\sqrt{1-4xy}}}\right)+{\displaystyle \frac{ln\left(\frac{1+\sqrt{1-4y}}{1-\sqrt{1-4y}}\right)}{x\sqrt{1-4y}}}ln\left(xy{\displaystyle \frac{1+\sqrt{1-4xy}}{1-\sqrt{1-4xy}}}\right)\hfill \\ & +{\displaystyle \frac{1}{x\sqrt{1-4y}}}\{2{\mathrm{Li}}_2(-\sqrt{1-4y})-2{\mathrm{Li}}_2\left(\sqrt{1-4y}\right)-2{\mathrm{Li}}_2\left(-\sqrt{{\displaystyle \frac{1-4y}{1-4xy}}}\right)\hfill \\ & +{\mathrm{Li}}_2\left(x\right)+{\mathrm{Li}}_2\left({\displaystyle \frac{2\sqrt{1-4y}}{1+\sqrt{1-4y}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{-2\sqrt{1-4y}}{1-\sqrt{1-4y}}}\right)+2{\mathrm{Li}}_2\left(\sqrt{{\displaystyle \frac{1-4y}{1-4xy}}}\right)\hfill \\ & +{\mathrm{Li}}_2\left(-{\displaystyle \frac{1-\sqrt{1-4xy}}{1+\sqrt{1-4y}}}\sqrt{{\displaystyle \frac{1-4y}{1-4xy}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{1+\sqrt{1-4xy}}{1+\sqrt{1-4y}}}\sqrt{{\displaystyle \frac{1-4y}{1-4xy}}}\right)\hfill \\ & +{\mathrm{Li}}_2\left(-{\displaystyle \frac{1+\sqrt{1-4xy}}{1-\sqrt{1-4y}}}\sqrt{{\displaystyle \frac{1-4y}{1-4xy}}}\right)-{\mathrm{Li}}_2\left({\displaystyle \frac{1-\sqrt{1-4xy}}{1-\sqrt{1-4y}}}\sqrt{{\displaystyle \frac{1-4y}{1-4xy}}}\right)\}.\hfill \end{array}$$

As applications, we record the following interesting summation formulae, where formulae (a) and (g) have been obtained previously by the authors [30] (Theorems 9 & 10).

Corollary 5.

Binomial series (positive and alternating) with ${\mathbf{H}}_n^{\langle 2\rangle }$ or ${\overline{\mathbf{H}}}_n^{\langle 2\rangle }$:

• Positive series with ${\mathbf{H}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{a}\right)& {\Phi }_1\left(1,\frac{1}{8}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }=-2\sqrt{2}{\mathrm{Li}}_2(2\sqrt{2}-3),\hfill \\ \hfill \left(\mathrm{b}\right)& {\Phi }_1\left(1,\frac{1}{16}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }={\displaystyle \frac{-4}{\sqrt{3}}}{\mathrm{Li}}_2\left(4\sqrt{3}-7\right),\hfill \\ \hfill \left(\mathrm{c}\right)& {\Phi }_1\left(1,\frac{3}{16}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }={\displaystyle \frac{2{\pi }^2}{3}}-2{ln}^{2}3-8{\mathrm{Li}}_2\left({\displaystyle \frac{1}{3}}\right),\hfill \\ \hfill \left(\mathrm{d}\right)& {\Phi }_1\left(1,\frac{1}{32}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{32}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }=-4\sqrt{{\displaystyle \frac{2}{7}}}{\mathrm{Li}}_2\left(4\sqrt{14}-15\right),\hfill \\ \hfill \left(\mathrm{e}\right)& {\Phi }_1\left(1,\frac{1}{64}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{64}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }={\displaystyle \frac{-8}{\sqrt{15}}}{\mathrm{Li}}_2\left(8\sqrt{15}-31\right).\hfill \end{array}$$
• Alternating series with ${\mathbf{H}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{f}\right)& {\Phi }_1\left(1,-\frac{1}{4}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }=-\sqrt{2}{\mathrm{Li}}_2(3-2\sqrt{2}),\hfill \\ \hfill \left(\mathrm{g}\right)& {\Phi }_1\left(1,\frac{-1}{8}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }=-83{\mathrm{Li}}_2(5-2\sqrt{6}),\hfill \\ \hfill \left(\mathrm{h}\right)& {\Phi }_1\left(1,\frac{-1}{16}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }={\displaystyle \frac{-4}{\sqrt{5}}}{\mathrm{Li}}_2\left(9-4\sqrt{5}\right),\hfill \\ \hfill \left(\mathrm{i}\right)& {\Phi }_1\left(1,\frac{-1}{32}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{32}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }=-{\displaystyle \frac{4\sqrt{2}}{3}}{\mathrm{Li}}_2\left(17-12\sqrt{2}\right),\hfill \\ \hfill \left(\mathrm{j}\right)& {\Phi }_1\left(1,\frac{-1}{64}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{64}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n^{\langle 2\rangle }={\displaystyle \frac{-8}{\sqrt{17}}}{\mathrm{Li}}_2\left(33-8\sqrt{17}\right).\hfill \end{array}$$
• Two series with skew harmonic number ${\overline{\mathbf{H}}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{k}\right)& {\Phi }_1\left(-1,\frac{-1}{4}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n^{\langle 2\rangle }=\sqrt{2}{ln}^2(1+\sqrt{2})+\sqrt{2}{\mathrm{Li}}_2(3-2\sqrt{2})-{\displaystyle \frac{{\pi }^2}{4\sqrt{2}}},\hfill \\ \hfill \left(\mathrm{l}\right)& {\Phi }_1\left(-1,\frac{1}{5}\right)={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n^{\langle 2\rangle }={\displaystyle \frac{\sqrt{5}}{2}}\{4{\mathrm{Li}}_2\left({\displaystyle \frac{5-\sqrt{5}}{3}}\right)-4{\mathrm{Li}}_2\left({\displaystyle \frac{-2-\sqrt{5}}{3}}\right)+4{\mathrm{Li}}_2\left({\displaystyle \frac{\sqrt{5}-3}{2}}\right)\hfill \\ & -{\displaystyle \frac{{\pi }^2}{6}}-2ln3ln\left({\displaystyle \frac{20}{9}}\right)-4ln\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right)ln\left({\displaystyle \frac{20}{81}}(5+\sqrt{5})\right)-8{\mathrm{Li}}_2\left({\displaystyle \frac{1}{3}}\right)+{\mathrm{Li}}_2\left({\displaystyle \frac{1}{81}}\right)\}.\hfill \end{array}$$

3.2. Series ${\Phi }_2(x,y)$ with ${\displaystyle \frac{{\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)}{n}}$

By means of (2), we can express the binomial series as

$$\begin{array}{ll}\hfill {\Phi }_2(x,y)& :={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{y^n}{n}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)\hfill \\ & ={\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{y^n}{n}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\int }_0^1{\displaystyle \frac{1-{\left(Tx\right)}^n}{1-Tx}}lnTdT\hfill \\ & ={\int }_0^1{\displaystyle \frac{2lnTdT}{1-Tx}}ln{\displaystyle \frac{1+\sqrt{1-4y}}{1+\sqrt{1-4Txy}}}.\hfill \end{array}$$

For this parametric integral, “Wolfram Mathematica version 11” fails to produce an analytic formula in closed form. However, for specific numerical values of x and y, we find several elegant identities. Some of them are shown below as examples. Among them, the first one (a) can be found in Olaikhan [32] (§4.8.4).

Proposition 1.

Binomial series (positive and alternating) with ${\mathbf{H}}_n^{\langle 2\rangle }$ or ${\overline{\mathbf{H}}}_n^{\langle 2\rangle }$:

• Positive series with ${\mathbf{H}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{a}\right)& {\Phi }_2(1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}={\displaystyle \frac{3{\zeta }_3}{2}},\hfill \\ \hfill \left(\mathrm{b}\right)& {\Phi }_2(1,\frac{1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left({\displaystyle \frac{\sqrt{5}-3}{2}}\right),\hfill \\ \hfill \left(\mathrm{c}\right)& {\Phi }_2(1,\frac{3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left({\displaystyle \frac{-1}{3}}\right),\hfill \\ \hfill \left(\mathrm{d}\right)& {\Phi }_2(1,\frac{6}{49})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{6}{49}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left({\displaystyle \frac{-1}{6}}\right),\hfill \\ \hfill \left(\mathrm{e}\right)& {\Phi }_2(1,\frac{\sqrt{3}-1}{3})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{\sqrt{3}-1}{3}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left(1-\sqrt{3}\right).\hfill \end{array}$$
• Alternating series with ${\mathbf{H}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{f}\right)& {\Phi }_2(1,\frac{-1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left(3-2\sqrt{2}\right),\hfill \\ \hfill \left(\mathrm{g}\right)& {\Phi }_2(1,\frac{-1}{5})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{5}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left({\displaystyle \frac{7-3\sqrt{5}}{2}}\right),\hfill \\ \hfill \left(\mathrm{h}\right)& {\Phi }_2(1,\frac{-6}{25})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-6}{25}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left({\displaystyle \frac{1}{6}}\right),\hfill \\ \hfill \left(\mathrm{i}\right)& {\Phi }_2(1,\frac{-11}{100})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-11}{100}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left({\displaystyle \frac{1}{11}}\right),\hfill \\ \hfill \left(\mathrm{j}\right)& {\Phi }_2(1,\frac{2\sqrt{3}-4}{3})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{2\sqrt{3}-4}{3}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n}}=-2{\mathrm{Li}}_3\left(1-{\displaystyle \frac{\sqrt{3}}{2}}\right).\hfill \end{array}$$
• Two series with skew harmonic number ${\overline{\mathbf{H}}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{k}\right)& {\Phi }_2(-1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n^{\langle 2\rangle }}{n}}={\displaystyle \frac{10}{3}}{ln}^3(1+\sqrt{2})-{\displaystyle \frac{{\pi }^2}{6}}ln(1+\sqrt{2})+4{\mathrm{Li}}_3(\sqrt{2}-1)\hfill \\ & -5{\zeta }_3+4ln(1+\sqrt{2})\left\{3{\mathrm{Li}}_2(3-2\sqrt{2})-2{\mathrm{Li}}_2(2\sqrt{2}-3)\right\}-4{\mathrm{Li}}_3(1-\sqrt{2}),\hfill \\ \hfill \left(\mathrm{l}\right)& {\Phi }_2(-1,\frac{-1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n^{\langle 2\rangle }}{n}}={\displaystyle \frac{4}{3}}{ln}^3(1+\sqrt{2})+{\displaystyle \frac{{\pi }^2}{2}}ln(1+\sqrt{2})+8{\mathrm{Li}}_3(3-2\sqrt{2})\hfill \\ & -9{\zeta }_3+4ln(1+\sqrt{2})\left\{3{\mathrm{Li}}_2(3-2\sqrt{2})-2{\mathrm{Li}}_2(2\sqrt{2}-3)\right\}-4{\mathrm{Li}}_3(2\sqrt{2}-3).\hfill \end{array}$$

3.3. Series ${\Phi }_3(x,y)$ with ${\displaystyle \frac{{\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)}{n+1}}$

In view of (3), we can manipulate the series

$$\begin{array}{ll}\hfill {\Phi }_3(x,y)& :={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)\left\{{\displaystyle \frac{{\mathrm{Li}}_2\left(x\right)}{x}}+{\int }_0^1{\displaystyle \frac{{\left(Tx\right)}^{n}lnT}{1-Tx}}dT\right\}\hfill \\ & ={\displaystyle \frac{2{\mathrm{Li}}_2\left(x\right)}{x(1+\sqrt{1-4y})}}+{\int }_0^1{\displaystyle \frac{2lnTdT}{(1-Tx)(1+\sqrt{1-4Txy})}}\hfill \\ & ={\int }_0^1{\displaystyle \frac{2lnTdT}{1-Tx}}\left\{{\displaystyle \frac{1}{1+\sqrt{1-4Txy}}}-{\displaystyle \frac{1}{1+\sqrt{1-4y}}}\right\}.\hfill \end{array}$$

Even though “Wolfram Mathematica version 11” is unable to compute this parametric integral in closed expression, it does produce a few remarkable formulae for specifically assigned values of x and y as exhibited below.

Proposition 2.

Binomial series (positive and alternating) with ${\mathbf{H}}_n^{\langle 2\rangle }$ or ${\overline{\mathbf{H}}}_n^{\langle 2\rangle }$:

• Positive series with ${\mathbf{H}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{a}\right)& {\Phi }_3(1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}={\displaystyle \frac{{\pi }^2}{3}}-4{ln}^{2}2,\hfill \\ \hfill \left(\mathrm{b}\right)& {\Phi }_3(1,\frac{1}{8})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{8}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}=4(2-\sqrt{2}){\mathrm{Li}}_2\left({\displaystyle \frac{2-\sqrt{2}}{4}}\right)-{\displaystyle \frac{2+\sqrt{2}}{2}}{ln}^2\left({\displaystyle \frac{3+2\sqrt{2}}{8}}\right),\hfill \\ \hfill \left(\mathrm{c}\right)& {\Phi }_3(1,\frac{1}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}=8(2-\sqrt{3}){\mathrm{Li}}_2\left({\displaystyle \frac{2-\sqrt{3}}{4}}\right)-4(2+\sqrt{3}){ln}^2{\displaystyle \frac{2+\sqrt{3}}{4}},\hfill \\ \hfill \left(\mathrm{d}\right)& {\Phi }_3(1,\frac{3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}={\displaystyle \frac{8}{3}}{\mathrm{Li}}_2\left({\displaystyle \frac{1}{4}}\right)-4{ln}^2\left({\displaystyle \frac{4}{3}}\right).\hfill \end{array}$$
• Alternating series with ${\mathbf{H}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{e}\right)& {\Phi }_3(1,\frac{-1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}=2(1+\sqrt{2}){ln}^2\left({\displaystyle \frac{1+\sqrt{2}}{2}}\right)+{\displaystyle \frac{4}{1+\sqrt{2}}}{\mathrm{Li}}_2\left({\displaystyle \frac{1-\sqrt{2}}{2}}\right),\hfill \\ \hfill \left(\mathrm{f}\right)& {\Phi }_3(1,\frac{-1}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}=4(2+\sqrt{5}){ln}^2\left({\displaystyle \frac{2+\sqrt{5}}{4}}\right)-8(2-\sqrt{5}){\mathrm{Li}}_2\left({\displaystyle \frac{2-\sqrt{5}}{4}}\right),\hfill \\ \hfill \left(\mathrm{g}\right)& {\Phi }_3(1,\frac{-3}{16})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}={\displaystyle \frac{16}{3}}{ln}^2\left({\displaystyle \frac{2+\sqrt{7}}{4}}\right)+{\displaystyle \frac{8}{3}}(2-\sqrt{7}){\mathrm{Li}}_2\left({\displaystyle \frac{11-4\sqrt{7}}{3}}\right),\hfill \\ \hfill \left(\mathrm{h}\right)& {\Phi }_3(1,\frac{-1}{64})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{64}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^{\langle 2\rangle }}{n+1}}={\displaystyle \frac{8{\pi }^2}{3}}(4-\sqrt{17})-16(4-\sqrt{17}){\mathrm{Li}}_2\left({\displaystyle \frac{8}{4+\sqrt{17}}}\right)\hfill \\ & -32ln\left({\displaystyle \frac{4+\sqrt{17}}{8}}\right)\left\{6ln2+(2-\sqrt{17})ln(4+\sqrt{17})\right\}.\hfill \end{array}$$
• Two series with skew harmonic number ${\overline{\mathbf{H}}}_n^{\langle 2\rangle }$:

  $$\begin{array}{ll}\hfill \left(\mathrm{i}\right)& {\Phi }_3(-1,\frac{1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n^{\langle 2\rangle }}{n+1}}=2{ln}^2\left({\displaystyle \frac{1+\sqrt{2}}{2}}\right)-4{\mathrm{Li}}_2\left({\displaystyle \frac{1-\sqrt{2}}{2}}\right),\hfill \\ \hfill \left(\mathrm{j}\right)& {\Phi }_3(-1,\frac{-1}{4})={\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{-1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\overline{\mathbf{H}}}_n^{\langle 2\rangle }}{n+1}}={\displaystyle \frac{{\pi }^2}{3}}-4{ln}^{2}2\hfill \\ & \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}-{\displaystyle \frac{{\pi }^2}{\sqrt{2}}}+4\sqrt{2}{ln}^2(1+\sqrt{2})+4\sqrt{2}{\mathrm{Li}}_2\left(3-2\sqrt{2}\right).\hfill \end{array}$$

3.4. Series ${\Phi }_4(x,y)$ with ${\displaystyle \frac{{\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)}{2n+1}}$

Keeping in mind (4), we can reformulate the binomial series

$$\begin{array}{ll}\hfill {\Phi }_4(x,y)& :={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{2n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n^{\langle 2\rangle }\left(x\right)\hfill \\ & ={\displaystyle \sum _{n=0}^{\infty }}{\displaystyle \frac{y^n}{2n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)\left\{{\displaystyle \frac{{\mathrm{Li}}_2\left(x\right)}{x}}+{\int }_0^1{\displaystyle \frac{{\left(Tx\right)}^{n}lnT}{1-Tx}}dT\right\}\hfill \\ & ={\displaystyle \frac{arcsin\left(2\sqrt{y}\right){\mathrm{Li}}_2\left(x\right)}{2x\sqrt{y}}}+{\int }_0^1{\displaystyle \frac{lnTarcsin\left(2\sqrt{Txy}\right)dT}{2(1-Tx)\sqrt{Txy}}}\hfill \\ & ={\int }_0^1{\displaystyle \frac{lnTdT}{1-Tx}}\left\{{\displaystyle \frac{arcsin\left(2\sqrt{Txy}\right)}{2\sqrt{Txy}}}-{\displaystyle \frac{arcsin\left(2\sqrt{y}\right)}{2\sqrt{y}}}\right\}.\hfill \end{array}$$

Unfortunately, with “Wolfram Mathematica version 11”, we are able to evaluate neither this parametric integral in closed form, nor the corresponding numerical integrals when x and y are fixed by particular values.

4. Generating Functions for Bisection Series

Given a formal power series

$$\Lambda \left(y\right)={\displaystyle \sum _{n=0}^{\infty }}{\lambda }_n{y}^n.$$

There are two associated bisection series

$${\displaystyle \frac{\Lambda \left(y\right)+\Lambda (-y)}{2}}={\displaystyle \sum _{n=0}^{\infty }}{\lambda }_{2n}{y}^{2n}\mathrm{and}{\displaystyle \frac{\Lambda \left(y\right)-\Lambda (-y)}{2}}={\displaystyle \sum _{n=0}^{\infty }}{\lambda }_{2n+1}{y}^{2n+1}.$$

As showcases, we shall confine ourselves to examine only the bisection series related to ${\varphi }_1(x,y)$ in this section. Similar studies can be carried out analogously for the remaining binomial series, which will be more complex. In order to facilitate the subsequent applications, we shall need the following generating function for ${\mathcal{H}}_{2n}\left(x\right)$, which resembles that ${\varphi }_1(x,y)$ as in Lemma 1 and can be verified without difficulty.

Theorem 5

($\left|x\right|\le 1$ and $-{\displaystyle \frac{1}{4}}\le y<{\displaystyle \frac{1}{4}}$).

$$\begin{array}{ll}\hfill \varphi (x,y):={\displaystyle \sum _{n=0}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_{2n}\left(x\right)& ={\int }_0^1{\displaystyle \frac{dT}{1-Tx}}\left\{{\displaystyle \frac{1}{\sqrt{1-4y}}}-{\displaystyle \frac{1}{\sqrt{1-4T^2{x}^{2}y}}}\right\}\hfill \\ & ={\displaystyle \frac{1}{x\sqrt{1-4y}}}ln\left({\displaystyle \frac{1-4xy-\sqrt{1-4y}\sqrt{1-4x^{2}y}}{{(1-x)}^2(1-\sqrt{1-4y})}}\right).\hfill \end{array}$$

This may be considered to be a bivariate analog of its limiting case $x\to 1$ obtained by Chen [34] (cf. [22]):

$$\varphi (1,y)=\underset{x\to 1}{lim}\varphi (x,y)={\displaystyle \sum _{n=0}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_{2n}={\displaystyle \frac{1}{\sqrt{1-4y}}}ln\left({\displaystyle \frac{1+\sqrt{1-4y}}{2(1-4y)}}\right).$$

When $x=-1$, the corresponding function becomes

$$\varphi (-1,y)={\displaystyle \sum _{n=0}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_{2n}={\displaystyle \frac{1}{\sqrt{1-4y}}}ln\left({\displaystyle \frac{1-\sqrt{1-4y}}{2y}}\right).$$

For these two series, we can immediately write down their bisection series:

$$\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right){\mathbf{H}}_{4n}{y}^{2n}={\displaystyle \frac{\varphi (1,y)+\varphi (1,-y)}{2}}={\displaystyle \frac{ln\left(\frac{1+\sqrt{1-4y}}{2(1-4y)}\right)}{2\sqrt{1-4y}}}+{\displaystyle \frac{ln\left(\frac{1+\sqrt{1+4y}}{2(1+4y)}\right)}{2\sqrt{1+4y}}},\hfill \end{array}$$

(5)

$$\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n+2}{2n+1}}\right){\mathbf{H}}_{4n+2}{y}^{2n+1}={\displaystyle \frac{\varphi (1,y)-\varphi (1,-y)}{2}}={\displaystyle \frac{ln\left(\frac{1+\sqrt{1-4y}}{2(1-4y)}\right)}{2\sqrt{1-4y}}}-{\displaystyle \frac{ln\left(\frac{1+\sqrt{1+4y}}{2(1+4y)}\right)}{2\sqrt{1+4y}}},\hfill \end{array}$$

(6)

$$\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right){\overline{\mathbf{H}}}_{4n}{y}^{2n}={\displaystyle \frac{\varphi (-1,y)+\varphi (-1,-y)}{2}}={\displaystyle \frac{ln\left(\frac{1-\sqrt{1-4y}}{2y}\right)}{2\sqrt{1-4y}}}-{\displaystyle \frac{ln\left(\frac{1+\sqrt{1+4y}}{2}\right)}{2\sqrt{1+4y}}},\hfill \end{array}$$

(7)

$$\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n+2}{2n+1}}\right){\overline{\mathbf{H}}}_{4n+2}{y}^{2n+1}={\displaystyle \frac{\varphi (-1,y)-\varphi (-1,-y)}{2}}={\displaystyle \frac{ln\left(\frac{1-\sqrt{1-4y}}{2y}\right)}{2\sqrt{1-4y}}}+{\displaystyle \frac{ln\left(\frac{1+\sqrt{1+4y}}{2}\right)}{2\sqrt{1+4y}}}.\hfill \end{array}$$

(8)

4.1. Four Bisection Series from ${\varphi }_1(x,y)$

When $x=\pm 1$ in Lemma 1, we have two univariate series (cf. [22,29]):

$$\begin{array}{ll}\hfill {\varphi }_1(1,y)& ={\displaystyle \sum _{n=0}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathbf{H}}_n={\displaystyle \frac{2}{\sqrt{1-4y}}}ln{\displaystyle \frac{1+\sqrt{1-4y}}{2\sqrt{1-4y}}},\hfill \\ \hfill {\varphi }_1(-1,y)& ={\displaystyle \sum _{n=0}^{\infty }}{y}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\overline{\mathbf{H}}}_n={\displaystyle \frac{2}{\sqrt{1-4y}}}ln\left({\displaystyle \frac{\sqrt{1+4y}+\sqrt{1-4y}}{1+\sqrt{1-4y}}}\right).\hfill \end{array}$$

Then, four related bisection series can be displayed as follows:

$$\begin{array}{l}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right){\mathbf{H}}_{2n}{y}^{2n}={\displaystyle \frac{{\varphi }_1(1,y)+{\varphi }_1(1,-y)}{2}}\hfill \\ ={\displaystyle \frac{1}{\sqrt{1-4y}}}ln{\displaystyle \frac{1+\sqrt{1-4y}}{2\sqrt{1-4y}}}+{\displaystyle \frac{1}{\sqrt{1+4y}}}ln{\displaystyle \frac{1+\sqrt{1+4y}}{2\sqrt{1+4y}}},\hfill \end{array}$$

(9)

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n+2}{2n+1}}\right){\mathbf{H}}_{2n+1}{y}^{2n+1}={\displaystyle \frac{{\varphi }_1(1,y)-{\varphi }_1(1,-y)}{2}}\hfill \\ ={\displaystyle \frac{1}{\sqrt{1-4y}}}ln{\displaystyle \frac{1+\sqrt{1-4y}}{2\sqrt{1-4y}}}-{\displaystyle \frac{1}{\sqrt{1+4y}}}ln{\displaystyle \frac{1+\sqrt{1+4y}}{2\sqrt{1+4y}}},\hfill \end{array}$$

(10)

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right){\overline{\mathbf{H}}}_{2n}{y}^{2n}={\displaystyle \frac{{\varphi }_1(-1,y)+{\varphi }_1(-1,-y)}{2}}\hfill \\ ={\displaystyle \frac{1}{\sqrt{1-4y}}}ln\left({\displaystyle \frac{\sqrt{1+4y}+\sqrt{1-4y}}{1+\sqrt{1-4y}}}\right)+{\displaystyle \frac{1}{\sqrt{1+4y}}}ln\left({\displaystyle \frac{\sqrt{1-4y}+\sqrt{1+4y}}{1+\sqrt{1+4y}}}\right),\hfill \end{array}$$

(11)

$$\begin{array}{c}{\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n+2}{2n+1}}\right){\overline{\mathbf{H}}}_{2n+1}{y}^{2n+1}={\displaystyle \frac{{\varphi }_1(-1,y)-{\varphi }_1(-1,-y)}{2}}\hfill \\ ={\displaystyle \frac{1}{\sqrt{1-4y}}}ln\left({\displaystyle \frac{\sqrt{1+4y}+\sqrt{1-4y}}{1+\sqrt{1-4y}}}\right)-{\displaystyle \frac{1}{\sqrt{1+4y}}}ln\left({\displaystyle \frac{\sqrt{1-4y}+\sqrt{1+4y}}{1+\sqrt{1+4y}}}\right).\hfill \end{array}$$

(12)

4.2. Two Interesting Series about ${\mathbf{H}}_n$

According to the equations

$${\mathbf{H}}_n={\mathbf{H}}_{2n}-{\overline{\mathbf{H}}}_{2n}={\mathbf{H}}_{2n+1}-{\overline{\mathbf{H}}}_{2n+1},$$

we deduce from (9)–(12) the following pair of unusual identities:

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right){\mathbf{H}}_n{y}^{2n}& ={\displaystyle \frac{1}{\sqrt{1-4y}}}ln\left({\displaystyle \frac{1-2y+\sqrt{1-4y}}{1-4y+\sqrt{1-16y^2}}}\right)+{\displaystyle \frac{1}{\sqrt{1+4y}}}ln\left({\displaystyle \frac{1+2y+\sqrt{1+4y}}{1+4y+\sqrt{1-16y^2}}}\right),\hfill \\ \hfill {\displaystyle \sum _{n=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4n+2}{2n+1}}\right){\mathbf{H}}_n{y}^{2n+1}& ={\displaystyle \frac{1}{\sqrt{1-4y}}}ln\left({\displaystyle \frac{1-2y+\sqrt{1-4y}}{1-4y+\sqrt{1-16y^2}}}\right)-{\displaystyle \frac{1}{\sqrt{1+4y}}}ln\left({\displaystyle \frac{1+2y+\sqrt{1+4y}}{1+4y+\sqrt{1-16y^2}}}\right).\hfill \end{array}$$

4.3. Four Series About Differences of Harmonic Numbers

By Combining (5)–(8) with (9)–(12), we derive four closed formulae below for differences of harmonic numbers:

$$\begin{array}{cc}& {\displaystyle \sum _{n=0}^{\infty }}{y}^{2n}\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right)\left\{2{\mathbf{H}}_{4n}-{\mathbf{H}}_{2n}\right\}=-{\displaystyle \frac{ln\left(1+4y\right)}{2\sqrt{1+4y}}}-{\displaystyle \frac{ln\left(1-4y\right)}{2\sqrt{1-4y}}},\hfill \\ & {\displaystyle \sum _{n=0}^{\infty }}{y}^{2n}\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right)\left\{2{\overline{\mathbf{H}}}_{4n}-{\overline{\mathbf{H}}}_{2n}\right\}=ln\left({\displaystyle \frac{2}{\sqrt{1+4y}+\sqrt{1-4y}}}\right)\left\{{\displaystyle \frac{1}{\sqrt{1+4y}}}+{\displaystyle \frac{1}{\sqrt{1-4y}}}\right\},\hfill \\ & {\displaystyle \sum _{n=0}^{\infty }}{y}^{2n+1}\left({\displaystyle \genfrac{}{}{0pt}{}{4n+2}{2n+1}}\right)\left\{2{\mathbf{H}}_{4n+2}-{\mathbf{H}}_{2n+1}\right\}={\displaystyle \frac{ln\left(1+4y\right)}{2\sqrt{1+4y}}}-{\displaystyle \frac{ln\left(1-4y\right)}{2\sqrt{1-4y}}},\hfill \\ & {\displaystyle \sum _{n=0}^{\infty }}{y}^{2n+1}\left({\displaystyle \genfrac{}{}{0pt}{}{4n+2}{2n+1}}\right)\left\{2{\overline{\mathbf{H}}}_{4n+2}-{\overline{\mathbf{H}}}_{2n+1}\right\}=ln\left({\displaystyle \frac{\sqrt{1+4y}+\sqrt{1-4y}}{2}}\right)\left\{{\displaystyle \frac{1}{\sqrt{1+4y}}}-{\displaystyle \frac{1}{\sqrt{1-4y}}}\right\}.\hfill \end{array}$$

In view of the equality

$$2{\overline{\mathbf{H}}}_{4n}-{\overline{\mathbf{H}}}_{2n}=2{\mathbf{H}}_{4n}-3{\mathbf{H}}_{2n}+{\mathbf{H}}_n,$$

the second series displayed above is equivalent to the following conjectured one made recently by Sun [27] (Conjecture 2.8)

$${\displaystyle \sum _{k=0}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{4k}{2k}}\right){\left({\displaystyle \frac{x(1-x)}{4}}\right)}^k\left\{2{\mathbf{H}}_{4k}-3{\mathbf{H}}_{2k}+{\mathbf{H}}_k\right\}={\displaystyle \frac{\sqrt{1-x}}{2x-1}}ln(1-x),$$

where the series on the left converges for ${\displaystyle \frac{1-\sqrt{2}}{2}}<x<{\displaystyle \frac{1}{2}}$ so that $\left|4x\right(1-x\left)\right|<1$.

4.4. Two Series About Crossing Differences of Harmonic Numbers

Combining the four series in (5), (6), (11) and (12), we can obtain two further closed formulae below for crossing differences of harmonic numbers:

$$\begin{array}{cc}& {\displaystyle \sum _{n=0}^{\infty }}{y}^{2n}\left({\displaystyle \genfrac{}{}{0pt}{}{4n}{2n}}\right)\left\{2{\mathbf{H}}_{4n}+{\overline{\mathbf{H}}}_{2n}\right\}={\displaystyle \frac{ln\left(\frac{\sqrt{1+4y}+\sqrt{1-4y}}{2-8y}\right)}{\sqrt{1-4y}}}+{\displaystyle \frac{ln\left(\frac{\sqrt{1+4y}+\sqrt{1-4y}}{2+8y}\right)}{\sqrt{1+4y}}},\hfill \\ & {\displaystyle \sum _{n=0}^{\infty }}{y}^{2n+1}\left({\displaystyle \genfrac{}{}{0pt}{}{4n+2}{2n+1}}\right)\left\{2{\mathbf{H}}_{4n+2}+{\overline{\mathbf{H}}}_{2n+1}\right\}={\displaystyle \frac{ln\left(\frac{\sqrt{1+4y}+\sqrt{1-4y}}{2-8y}\right)}{\sqrt{1-4y}}}-{\displaystyle \frac{ln\left(\frac{\sqrt{1+4y}+\sqrt{1-4y}}{2+8y}\right)}{\sqrt{1+4y}}}.\hfill \end{array}$$

5. Conclusions and Further Comments

By evaluating parametric integrals, we have shown numerous closed formulae for binomial series containing harmonic numbers. These values in terms of $\pi$, logarithmic, and polylogarithmic functions may serve the reader as a comprehensive source for further exploration.

Due to the tremendous complexity and difficulty, most of the parametric integrals can not be evaluated manually. Thanks to the modern computer algebra systems, it becomes possible for us to realize related computations by employing “Wolfram Mathematica version 11”. In addition to the precedent series, we may also consider the next one

$${\displaystyle \sum _{n=1}^{\infty }}{\displaystyle \frac{y^n}{n^2}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\mathcal{H}}_n\left(x\right)={\int }_0^y{\varphi }_2(x,u){\displaystyle \frac{du}{u}}={\int }_0^y{\displaystyle \frac{du}{u}}{\int }_0^1{\displaystyle \frac{2dv}{1-vx}}ln{\displaystyle \frac{1+\sqrt{1-4uvx}}{1+\sqrt{1-4u}}}.$$

Due to the presence of the free variable “x”, we actually fail to compute this double integral either manually or by using “Wolfram Mathematica version 11”. Nevertheless, we did detect values for the following two numerical series

$$\begin{array}{cc}& {\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{1}{4}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n^2}}={\displaystyle \frac{9{\zeta }_3}{2}}-{\displaystyle \frac{2{\pi }^2}{3}}ln2,\hfill \\ & {\displaystyle \sum _{n=1}^{\infty }}{\left({\displaystyle \frac{3}{16}}\right)}^n\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n}{n^2}}={\displaystyle \frac{ln2}{3}}\left\{3{ln}^{2}3-24ln2ln3+26{ln}^{2}2\right\}-2{\mathrm{Li}}_3\left({\displaystyle \frac{1}{3}}\right)\hfill \\ & -{\displaystyle \frac{{\zeta }_3}{2}}-6ln2{\mathrm{Li}}_2\left({\displaystyle \frac{1}{3}}\right)+ln\left({\displaystyle \frac{128}{81}}\right){\mathrm{Li}}_2\left({\displaystyle \frac{3}{4}}\right)+4{\mathrm{Li}}_3\left({\displaystyle \frac{3}{4}}\right);\hfill \end{array}$$

where the former series can be found in Olaikhan [32] (§4.8.3) and Dasireddy [35]. We record two further series due to Olaikhan [32] (§4.8.5 and §4.8.6) (with the latter being corrected)

$${\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_n^2}{4^n\times n}}={\displaystyle \frac{21{\zeta }_3}{2}}\mathrm{and}{\displaystyle \sum _{n=1}^{\infty }}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right){\displaystyle \frac{{\mathbf{H}}_{2n}^{\langle 2\rangle }}{4^n\times n}}=2\pi G-{\displaystyle \frac{25{\zeta }_3}{8}}.$$

Alternative proofs via the approaches devised in this paper may also be worthwhile.