\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm{\LARGE\bf Vacca-Type Series for Values of the\\ \vskip 0in Generalized Euler Constant Function \\ \vskip .13in and its Derivative } \vskip 1cm \large Khodabakhsh Hessami Pilehrood\footnote{Author's current address: Mathematics Department, Faculty of Basic Sciences, Shahrekord University, Shahrekord, P.O. Box 115, Iran. Research was in part supported by a grant from IPM (No.\ 87110018).} and Tatiana Hessami Pilehrood\footnote{Author's current address: Mathematics Department, Faculty of Basic Sciences, Shahrekord University, Shahrekord, P.O. Box 115, Iran. Research was in part supported by a grant from IPM (No.\ 87110019).} \\ Institute for Studies in Theoretical Physics and Mathematics \\ Tehran, Iran \\ \href{mailto:hessamik@ipm.ir}{\tt hessamik@ipm.ir}\\ \href{mailto:hessamik@gmail.com}{\tt hessamik@gmail.com} \\ \href{mailto:hessamit@ipm.ir}{\tt hessamit@ipm.ir} \\ \href{mailto:hessamit@gmail.com}{\tt hessamit@gmail.com} \end{center} \vskip .2 in \begin{abstract} We generalize well-known Catalan-type integrals for Euler's constant to values of the ge\-ne\-ra\-li\-zed Euler constant function and its derivatives. Using generating functions appearing in these integral representations, we give new Vacca and Ramanujan-type series for values of the generalized Euler constant function and Ad\-dison-type series for values of the generalized Euler constant function and its derivative. As a consequence, we get base-$B$ rational series for $\log\frac{4}{\pi},$ $\frac{G}{\pi}$ (where $G$ is Catalan's constant), $\frac{\zeta'(2)}{\pi^2}$, and also for logarithms of the Somos and Glaisher-Kinkelin constants. \end{abstract} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \section{Introduction} \label{intro} J.~Sondow \cite{s} proved the following two formulas: \begin{equation} \qquad\gamma=\sum_{n=1}^{\infty}\frac{N_{1,2}(n)+N_{0,2}(n)}{2n(2n+1)}, \label{eq1} \end{equation} \begin{equation} \log\frac{4}{\pi}=\sum_{n=1}^{\infty}\frac{N_{1,2}(n)-N_{0,2}(n)}{2n(2n+1)}, \label{eq2} \end{equation} where $\gamma$ is Euler's constant and $N_{i,2}(n)$ is the number of $i$'s in the binary expansion of $n$ (see sequences \seqnum{A000120} and \seqnum{A023416} in Sloane's database \cite{oeis}). The series (\ref{eq1}) is equivalent to the well-known Vacca series \cite{va} \begin{equation} \gamma=\sum_{n=1}^{\infty}(-1)^n\frac{\lfloor\log_2n\rfloor}{n}=\sum_{n=1}^{\infty} (-1)^n\frac{N_{1,2}\bigl(\lfloor\frac{n}{2}\rfloor\bigr)+ N_{0,2}\bigl(\lfloor\frac{n}{2}\rfloor\bigr)}{n} \label{eq3} \end{equation} and both series (\ref{eq1}) and (\ref{eq3}) may be derived from Catalan's integral \cite{ca} \begin{equation} \gamma=\int_0^1\frac{1}{1+x}\sum_{n=1}^{\infty}x^{2^n-1}\,dx. \label{eq4} \end{equation} To see this it suffices to note that $$ G(x)=\frac{1}{1-x}\sum_{n=0}^{\infty}x^{2^n}=\sum_{n=1}^{\infty} (N_{1,2}(n)+N_{0,2}(n))x^n $$ is a generating function of the sequence $N_{1,2}(n)+N_{0,2}(n),$ (see \seqnum{A070939}), which is the binary length of $n,$ rewrite (\ref{eq4}) as $$ \gamma=\int_0^1(1-x)\frac{G(x^2)}{x}\,dx $$ and integrate the power series termwise. In view of the equality \begin{equation*} 1=\int_0^1\sum_{n=1}^{\infty}x^{2^n-1}\,dx, \end{equation*} which is easily verified by termwise integration, (\ref{eq4}) is equivalent to the formula \begin{equation} \gamma=1-\int_0^1\frac{1}{1+x}\sum_{n=1}^{\infty}x^{2^n}\,dx \label{eq05} \end{equation} obtained independently by Ramanujan (see \cite[Corollary 2.3]{bebo}). Catalan's integral (\ref{eq05}) gives the following rational series for $\gamma:$ \begin{equation} \gamma=1-\int_0^1(1-x)G(x^2)\,dx=1-\sum_{n=1}^{\infty} \frac{N_{1,2}(n)+N_{0,2}(n)}{(2n+1)(2n+2)}. \label{eq06} \end{equation} Averaging (\ref{eq1}), (\ref{eq06}) and (\ref{eq4}), (\ref{eq05}), respectively, we get Addison's series for $\gamma$ \cite{ad} $$ \gamma=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{N_{1,2}(n)+N_{0,2}(n)}% {2n(2n+1)(2n+2)} $$ and its corresponding integral \begin{equation} \gamma=\frac{1}{2}+\frac{1}{2}\int_0^1\frac{1-x}{1+x}\sum_{n=1}^{\infty} x^{2^n-1}\,dx, \label{eq065} \end{equation} respectively. Integrals (\ref{eq05}), (\ref{eq4}) were generalized to an arbitrary integer base $B>1$ by S.~Ramanujan and by B.~C.~Berndt and D.~C.~Bowman (see \cite{bebo}): \begin{equation} \gamma =1-\int_0^1\left(\frac{1}{1-x}-\frac{Bx^{B-1}}{1-x^B}\right) \sum_{n=1}^{\infty}x^{B^n}\,dx \qquad\qquad (\text{Ramanujan}), \label{eq07} \end{equation} \begin{equation} \gamma =\int_0^1\left(\frac{B}{1-x^B}-\frac{1}{1-x}\right)\sum_{n=1}^{\infty} x^{B^n-1}\,dx \qquad\qquad (\text{Berndt-Bowman}). \label{eq08} \end{equation} Formula (\ref{eq08}) implies the generalized Vacca series for $\gamma$ (see \cite[Theorem 2.6]{bebo}) proposed by L.~Carlitz \cite{carl}: \begin{equation} \gamma=\sum_{n=1}^{\infty}\frac{\varepsilon(n)}{n}\lfloor\log_Bn\rfloor, \label{eq09} \end{equation} where \begin{equation} \varepsilon(n)=\begin{cases} B-1, & \text{if $B$ divides $n$}; \\ -1, & \text{otherwise}; \end{cases} \label{eq095} \end{equation} and the averaging integral of (\ref{eq07}) and (\ref{eq08}) produces the generalized Addison series for $\gamma$ found by Sondow \cite{s}: \begin{equation} \gamma=\frac{1}{2}+\sum_{n=1}^{\infty}\frac{\lfloor\log_BBn\rfloor P_B(n)}{Bn(Bn+1)\cdots (Bn+B)}, \label{eq010} \end{equation} where $P_B(x)$ is a polynomial of degree $B-2$ defined by \begin{equation} P_B(x)=(Bx+1)(Bx+2)\cdots (Bx+B-1)\sum_{m=1}^{B-1}\frac{m(B-m)}{Bx+m}. \label{polyn} \end{equation} In this paper, we consider the generalized Euler constant function \begin{equation} \gamma_{a,b}(z)=\sum_{n=0}^{\infty}\left(\frac{1}{an+b}- \log\left(\frac{an+b+1}{an+b}\right)\right)z^n, \qquad a,b \in{\mathbb N}, \qquad |z|\le 1, \label{eq7} \end{equation} which is related to the constants in (\ref{eq1}), (\ref{eq2}) as $\gamma_{1,1}(1)=\gamma,$ $\gamma_{1,1}(-1)=\log\frac{4}{\pi}.$ Basic properties of a special case of this function, $\gamma_{1,1}(z),$ were studied earlier in \cite{sh,he0}. In Section~\ref{S2}, we show that $\gamma_{a,b}(z)$ admits an analytic continuation to the domain ${\mathbb C}\setminus [1,+\infty)$ in terms of the Lerch transcendent. In Sections~\ref{S3}--\ref{S4}, we generalize Catalan-type integrals (\ref{eq07}), (\ref{eq08}) to values of the generalized Euler constant function and its derivatives. Using generating functions appearing in these integral representations, we give new Vacca- and Ramanujan-type series for values of $\gamma_{a,b}(z)$ and Addison-type series for values of $\gamma_{a,b}(z)$ and its derivative. In Section \ref{S5}, we get base-$B$ rational series for $\log\frac{4}{\pi},$ $\frac{G}{\pi},$ (where $G$ is Catalan's constant), $\frac{\zeta'(2)}{\pi^2}$ and also for logarithms of the Somos and Glaisher-Kinkelin constants. We also mention a connection of our approach to summation of series of the form $$\ \sum_{n=1}^{\infty}N_{\omega,B}(n)Q(n,B) \quad\text{and}\quad \sum_{n=1}^{\infty}\frac{N_{\omega,B}(n)P_B(n)}{Bn(Bn+1)\cdots (Bn+B)}, $$ where $Q(n,B)$ is a rational function of $B$ and $n$ \begin{equation} Q(n,B)=\frac{1}{Bn(Bn+1)}+\frac{2}{Bn(Bn+2)}+\cdots+\frac{B-1}{Bn(Bn+B-1)}, \label{eq5} \end{equation} and $N_{\omega,B}(n)$ is the number of occurrences of a word $\omega$ over the alphabet $\{0,1,\ldots, B-1\}$ in the $B$-ary expansion of $n,$ considered in \cite{assh}. Moreover, we answer some questions posed in \cite{assh} concerning possible generalizations of the series (\ref{eq1}) and (\ref{eq2}) to any integer base $B>1.$ Note that in the above notation, the generalized Vacca series (\ref{eq09}) can be written as follows: \begin{equation} \gamma= \sum_{k=1}^{\infty}L_B(k)Q(k,B), \label{eq6} \end{equation} where $L_B(k):=\lfloor\log_BBk\rfloor =\sum_{\alpha=0}^{B-1}N_{\alpha,B}(k)$ is the $B$-ary length of $k.$ Indeed, representing $n=Bk+r,$ $0\le r\le B-1$ and summing in (\ref{eq09}) over $k\ge 1$ and $0\le r\le B-1$ we get $$ \gamma=\sum_{k=1}^{\infty}\lfloor\log_BBk\rfloor\left( \frac{B-1}{Bk}-\frac{1}{Bk+1}-\cdots-\frac{1}{Bk+B-1}\right) =\sum_{k=1}^{\infty}\lfloor\log_BBk\rfloor Q(k,B). $$ Using the same notation, the generalized Addison series (\ref{eq010}) gives another base-$B$ expansion of Euler's constant \begin{equation} \gamma=\frac{1}{2}+\sum_{n=1}^{\infty} \frac{L_B(n)P_B(n)}{Bn(Bn+1)\cdots (Bn+B)} =\frac{1}{2}+\sum_{n=1}^{\infty}L_B(n) \left(Q(n,B)-\frac{B-1}{2Bn(n+1)}\right) \label{eu} \end{equation} which converges faster than (\ref{eq6}) to $\gamma.$ Here we used the fact that $$ \sum_{n=1}^{\infty}\sum_{\alpha=0}^{B-1}\frac{N_{\alpha, B}(n)}{n(n+1)}=\frac{B}{B-1}, $$ which can be easily checked by \cite[Section 3]{as}. On the other hand, \begin{equation*} \begin{split} Q&(n,B)-\frac{B-1}{2Bn(n+1)}=\frac{1}{2}\sum_{m=1}^{B-1}\left(\frac{1}{Bn}- \frac{2}{Bn+m}+\frac{1}{Bn+B}\right) \\ &=\frac{1}{Bn(Bn+B)}\sum_{m=1}^{B-1} \left(2m-B+\frac{2m(B-m)}{Bn+m}\right)=\frac{P_B(n)}{Bn(Bn+1)\cdots(Bn+B)}. \end{split} \end{equation*} Finally, we give a brief description of some other generalized Euler constants that have appeared in the literature in Section \ref{S6}. \section{Analytic continuation} \label{S2} We consider the generalized Euler constant function $\gamma_{a,b}(z)$ defined in (\ref{eq7}), where $a, b$ are positive real numbers, $z\in {\mathbb C},$ and the series converges when $|z|\le 1.$ We show that $\gamma_{a,b}(z)$ admits an analytic continuation to the domain ${\mathbb C}\setminus [1,+\infty).$ The following theorem is a slight modification of \cite[Theorem 3]{sh}. \begin{theorem} Let $a, b$ be positive real numbers, $z\in {\mathbb C},$ $|z|\le 1.$ Then \begin{equation} \gamma_{a,b}(z)=\int_0^1\int_0^1\frac{(xy)^{b-1}(1-x)}{(1-zx^ay^a)(-\log xy)}\,dxdy=\int_0^1\frac{x^{b-1}(1-x)}{1-zx^a}\left(\frac{1}{1-x}+ \frac{1}{\log x}\right)\,dx. \label{eq8} \end{equation} The integrals converge for all $z\in {\mathbb C}\setminus (1,+\infty)$ and give the analytic continuation of the generalized Euler constant function $\gamma_{a,b}(z)$ for $z\in {\mathbb C}\setminus [1,+\infty).$ \end{theorem} \begin{proof} Denoting the double integral in (\ref{eq8}) by $I(z)$ and for $|z|\le 1,$ expanding $(1-zx^ay^a)^{-1}$ in a geometric series we have \begin{equation*} \begin{split} I(z)&=\sum_{k=0}^{\infty}z^k\int_0^1\int_0^1\frac{(xy)^{ak+b-1}(1-x)}{(-\log xy)}\,dxdy \\ &=\sum_{k=0}^{\infty}z^k\int_0^1\int_0^1\int_0^{+\infty}(xy)^{t+ak+b-1}(1-x)\,dxdydt \\ &=\sum_{k=0}^{\infty}z^k\int_0^{+\infty}\left(\frac{1}{(t+ak+b)^2}-\Bigl( \frac{1}{t+ak+b}-\frac{1}{t+ak+b+1}\Bigr)\right)\,dt=\gamma_{a,b}(z). \end{split} \end{equation*} On the other hand, making the change of variables $u=x^a,$ $v=y^a$ in the double integral we get $$ I(z)=\frac{1}{a}\int_{0}^1\int_{0}^1\frac{(uv)^{\frac{b}{a}-1}(1-u^{\frac{1}{a}})}% {(1-zuv)(-\log uv)}\,dudv. $$ Now by \cite[Corollary 3.3]{gs}, for $z\in {\mathbb C}\setminus [1,+\infty)$ we have $$ I(z)=\frac{1}{a}\Phi\Bigl(z,1,\frac{b}{a}\Bigr)-\frac{\partial\Phi}{\partial s}\Bigl(z,0,\frac{b}{a}\Bigr)+\frac{\partial\Phi}{\partial s}\Bigl(z,0,\frac{b+1}{a}\Bigr), $$ where $\Phi(z,s,u)$ is the Lerch transcendent, a holomorphic function in $z$ and $s,$ for $z\in {\mathbb C}\setminus [1,+\infty)$ and all complex $s$ (see \cite[Lemma 2.2]{gs}), which is the analytic continuation of the series $$ \Phi(z,s,u)=\sum_{n=0}^{\infty}\frac{z^n}{(n+u)^s}, \qquad u>0. $$ To prove the second equality in (\ref{eq8}), make the change of variables $X=xy,$ $Y=y$ and integrate with respect to $Y.$ \end{proof} \begin{corollary} \label{c1} Let $a,b$ be positive real numbers, $l\in {\mathbb N},$ $z\in {\mathbb C}\setminus [1,+\infty).$ Then for the $l$-th derivative we have $$ \gamma_{a,b}^{(l)}(z)=\int_0^1\int_0^1\frac{(xy)^{al+b-1}(x-1)}% {(1-zx^ay^a)^{l+1}\log xy}\,dxdy= \int_0^1\frac{x^{la+b-1}(1-x)}{(1-zx^a)^{l+1}} \left(\frac{1}{1-x}+\frac{1}{\log x}\right)\,dx. $$ \end{corollary} From Corollary \ref{c1}, \cite[Cor.3.3, 3.8, 3.9]{gs} and \cite[Lemma 4]{assh} we get \begin{corollary} \label{c1.1} Let $a, b$ be positive real numbers, $z\in {\mathbb C}\setminus [1, +\infty).$ Then the following equalities hold: $$ \gamma_{a,b}(1)=\log\Gamma\Bigl(\frac{b+1}{a}\Bigr)-\log\Gamma\Bigl( \frac{b}{a}\Bigr)-\frac{1}{a}\psi\Bigl(\frac{b}{a}\Bigr), $$ $$ \gamma_{a,b}(z)=\frac{1}{a}\Phi\Bigl(z,1,\frac{b}{a}\Bigr)- \frac{\partial\Phi}{\partial s}\Bigl(z,0,\frac{b}{a}\Bigr) +\frac{\partial\Phi}{\partial s}\Bigl(z,0,\frac{b+1}{a}\Bigr), $$ \begin{equation*} \begin{split} \gamma'_{a,b}(z)=&-\frac{b}{a^2}\Phi\Bigl(z,1,\frac{b}{a}+1\Bigr)+ \frac{1}{a(1-z)}+\frac{b}{a}\frac{\partial\Phi}{\partial s}\Bigl(z,0,\frac{b}{a}+1\Bigr)- \frac{\partial\Phi}{\partial s}\Bigl(z,-1,\frac{b}{a}+1\Bigr)- \\[10pt] &\frac{b+1}{a}\frac{\partial\Phi}{\partial s}\Bigl(z,0,\frac{b+1}{a}+1\Bigr)+\frac{\partial\Phi}{\partial s} \Bigl(z,-1,\frac{b+1}{a}+1\Bigr), \end{split} \end{equation*} where $\Phi(z,s,u)$ is the Lerch transcendent and $\psi(x)=\frac{d}{dx}\log\Gamma(x)$ is the logarithmic derivative of the gamma function. \end{corollary} \section{Catalan-type integrals for $\gamma_{a,b}^{(l)}(z).$} \label{S3} Berndt and Bowman \cite{bebo} demonstrated that for $x>0$ and any integer $B>1,$ \begin{equation} \frac{1}{1-x}+\frac{1}{\log x}=\sum_{k=1}^{\infty}\frac{(B-1)+ (B-2)x^{\frac{1}{B^k}}+(B-3)x^{\frac{2}{B^k}}+\cdots+x^{\frac{B-2}{B^k}}}% {B^k(1+x^{\frac{1}{B^k}}+x^{\frac{2}{B^k}}+\cdots+x^{\frac{B-1}{B^k}})}. \label{eq9} \end{equation} The special cases $B=2,3$ of this equality can be found in Ramanujan's third note book \cite[p.\ 364]{ra}. Using this key formula we prove the following generalization of the integral (\ref{eq08}). \begin{theorem} \label{t2} Let $a, b, B$ be positive integers with $B>1,$ $l$ a non-negative integer. If either $z\in {\mathbb C}\setminus [1,+\infty)$ and $l\ge 1,$ or $z\in {\mathbb C}\setminus (1,+\infty)$ and $l=0,$ then \begin{equation} \gamma_{a,b}^{(l)}(z)=\int_0^1\left(\frac{B}{1-x^B}-\frac{1}{1-x}\right) F_l(z,x)\,dx \label{eq10} \end{equation} where \begin{equation} F_l(z,x)=\sum_{k=1}^{\infty}\frac{x^{(b+al)B^k-1}(1-x^{B^k})}% {(1-zx^{aB^k})^{l+1}}. \label{eq017} \end{equation} \end{theorem} \begin{proof} First we note that the series of variable $x$ on the right-hand side of (\ref{eq9}) converges uniformly on $[0,1],$ since the absolute value of its general term does not exceed $\frac{B-1}{2B^{k-1}}.$ Then for $l\ge 0,$ multiplying both sides of (\ref{eq9}) by $\frac{x^{la+b-1}(1-x)}{(1-zx^a)^{l+1}}$ and integrating over $0\le x\le 1$ we get $$ \gamma_{a,b}^{(l)}(z)=\sum_{k=1}^{\infty}\int_0^1\frac{x^{la+b-1}(1-x)}{(1-zx^a)^{l+1}} \cdot\frac{(B-1)+(B-2)x^{\frac{1}{B^k}}+\cdots+x^{\frac{B-2}{B^k}}}% {B^k(1+x^{\frac{1}{B^k}}+x^{\frac{2}{B^k}}+\cdots+x^{\frac{B-1}{B^k}})}\,dx. $$ Replacing $x$ by $x^{B^k}$ in each integral we find \begin{equation*} \begin{split} \gamma_{a,b}^{(l)}(z)&=\sum_{k=1}^{\infty}\int_0^1\frac{x^{(la+b)B^k-1}(1-x^{B^k})}% {(1-zx^{aB^k})^{l+1}}\cdot\frac{(B-1)+(B-2)x+\cdots+x^{B-2}}{1+x+x^2+\cdots+x^{B-1}}\,dx \\ &=\int_0^1\left(\frac{B}{1-x^B}-\frac{1}{1-x}\right)F_l(z,x)\,dx, \end{split} \end{equation*} as required. \end{proof} From Theorem \ref{t2} we readily get a generalization of Ramanujan's integral. \begin{corollary} \label{c2} Let $a, b, B$ be positive integers with $B>1,$ $l$ a non-negative integer. If either $z\in {\mathbb C}\setminus [1,+\infty)$ and $l\ge 1,$ or $z\in {\mathbb C}\setminus (1,+\infty)$ and $l=0,$ then \begin{equation} \gamma_{a,b}^{(l)}(z)=\int_0^1\frac{x^{b+al-1}(1-x)}{(1-zx^a)^{l+1}}\,dx +\int_0^1\left(\frac{Bx^{B}}{1-x^B}-\frac{x}{1-x}\right)F_l(z,x)\,dx. \label{eq018} \end{equation} \end{corollary} \begin{proof} First we note that the series (\ref{eq017}), considered as a sum of functions of the variable $x$ converges uniformly on $[0, 1-\varepsilon]$ for any $\varepsilon>0.$ Then integrating termwise we have $$ \int_0^{1-\varepsilon}F_l(z,x)\,dx=\sum_{k=1}^{\infty} \int_0^{1-\varepsilon}\frac{x^{(b+al)B^k-1}(1-x^{B^k})}% {(1-zx^{aB^k})^{l+1}}\,dx. $$ Making the change of variable $y=x^{B^k}$ in each integral we get $$ \int_0^{1-\varepsilon}F_l(z,x)\,dx=\sum_{k=1}^{\infty} \frac{1}{B^k} \int_0^{(1-\varepsilon)^{B^k}}\frac{y^{b+al-1}(1-y)}% {(1-zy^a)^{l+1}}\,dy. $$ Since the last series, considered as a series in the variable $\varepsilon,$ converges uniformly on $[0,1],$ letting $\varepsilon$ tend to zero we get \begin{equation} \int_0^1F_l(z,x)\,dx=\frac{1}{B-1}\int_0^1 \frac{y^{b+al-1}(1-y)}{(1-zy^a)^{l+1}}\,dy. \label{eq019} \end{equation} Now from (\ref{eq10}) and (\ref{eq019}) it follows that $$ \gamma_{a,b}^{(l)}(z)-\int_0^1 \frac{y^{b+al-1}(1-y)}{(1-zy^a)^{l+1}}\,dy= \int_0^1\left( \frac{Bx^B}{1-x^B}-\frac{x}{1-x}\right)F_l(z,x)\,dx, $$ and the proof is complete. \end{proof} Averaging the formulas (\ref{eq10}) and (\ref{eq018}), we get the following generalization of the integral (\ref{eq065}). \begin{corollary} \label{c3} Let $a, b, B$ be positive integers with $B>1,$ $l$ a non-negative integer. If either $z\in {\mathbb C}\setminus [1,+\infty)$ and $l\ge 1,$ or $z\in {\mathbb C}\setminus (1,+\infty)$ and $l=0,$ then $$ \gamma_{a,b}^{(l)}(z)=\frac{1}{2}\int_0^1\frac{x^{b+al-1}(1-x)}{(1-zx^a)^{l+1}}\,dx +\frac{1}{2}\int_0^1\left(\frac{B(1+x^B)}{1-x^B}-\frac{1+x}{1-x}\right)F_l(z,x)\,dx. $$ \end{corollary} \section{Vacca-type series for $\gamma_{a,b}(z)$ and $\gamma'_{a,b}(z).$} \label{S4} \begin{theorem} \label{t3} Let $a,b,B$ be positive integers with $B>1,$ $z\in {\mathbb C},$ $|z|\le 1.$ Then for the generalized Euler constant function $\gamma_{a,b}(z)$, the following expansion is valid: $$ \gamma_{a,b}(z)=\sum_{k=1}^{\infty}a_kQ(k,B)= \sum_{k=1}^{\infty}a_{\lfloor\frac{k}{B}\rfloor}\frac{\varepsilon(k)}{k}, $$ where $Q(k,B)$ is a rational function given by {\rm (\ref{eq5})}, $\{a_k\}_{k=0}^{\infty}$ is a sequence defined by the generating function \begin{equation} G(z,x)=\frac{1}{1-x}\sum_{k=0}^{\infty}\frac{x^{bB^k}(1-x^{B^k})}% {1-zx^{aB^k}}=\sum_{k=0}^{\infty}a_kx^k \label{eq11} \end{equation} and $\varepsilon(k)$ is defined in {\rm (\ref{eq095})}. \end{theorem} \begin{proof} For $l=0,$ rewrite (\ref{eq10}) in the form $$ \gamma_{a,b}(z)=\int_0^1\frac{1-x^B}{x}\left(\frac{B}{1-x^B}-\frac{1}{1-x}\right) G(z,x^B)\,dx $$ where $G(z,x)$ is defined in (\ref{eq11}). Then, since $a_0=0,$ we have \begin{equation} \gamma_{a,b}(z)=\int_0^1(B-1-x-x^2-\cdots-x^{B-1})\sum_{k=1}^{\infty}a_kx^{Bk-1}\, dx. \label{eq12} \end{equation} Expanding $G(z,x)$ in a power series of $x,$ $$ G(z,x)=\sum_{k=0}^{\infty}\sum_{m=0}^{\infty}z^mx^{(am+b)B^k}(1+x+\cdots+x^{B^k-1}), $$ we see that $a_k=O(\ln_B k).$ Therefore, by termwise integration in (\ref{eq12}), which can be easily justified by the same way as in the proof of Corollary \ref{c2}, we get \begin{equation*} \begin{split} \gamma_{a,b}(z) &=\sum_{k=1}^{\infty}a_k\int_0^1[(x^{Bk-1}-x^{Bk})+(x^{Bk-1}-x^{Bk+1})+\cdots+ (x^{Bk-1}-x^{Bk+B-2})]\,dx \\ &=\sum_{k=1}^{\infty}a_kQ(k,B). \end{split} \end{equation*} \end{proof} \begin{theorem} Let $a,b,B$ be positive integers with $B>1,$ $z\in {\mathbb C},$ $|z|\le 1.$ Then for the generalized Euler constant function, the following expansion is valid: $$ \gamma_{a,b}(z)=\int_0^1\frac{x^{b-1}(1-x)}{1-zx^a}\,dx-\sum_{k=1}^{\infty}a_k \widetilde{Q}(k,B), $$ where \begin{equation*} \begin{split} \widetilde{Q}(k,B)&=\frac{B-1}{Bk(k+1)}-Q(k,B) \\ &= \frac{B-1}{(Bk+B)(Bk+1)}+\frac{B-2}{(Bk+B)(Bk+2)}+\cdots+ \frac{1}{(Bk+B)(Bk+B-1)} \end{split} \end{equation*} and the sequence $\{a_k\}_{k=1}^{\infty}$ is defined in Theorem {\rm\ref{t3}}. \end{theorem} \begin{proof} From Corollary \ref{c2} with $l=0,$ using the same method as in the proof of Theorem \ref{t3}, we get \begin{equation*} \begin{split} &\int_0^1\left(\frac{Bx^B}{1-x^B}-\frac{x}{1-x}\right)F_0(z,x)= \int_0^1\frac{1-x^B}{x}\left(\frac{Bx^B}{1-x^B}-\frac{x}{1-x}\right)G(z,x^B)\,dx \\ &=\int_0^1(Bx^{B-1}-(1+x+\cdots+x^{B-1}))\sum_{k=1}^{\infty}a_kx^{Bk}\,dx \\ &=\sum_{k=1}^{\infty}a_k\int_0^1[(x^{Bk+B-1}-x^{Bk+B-2})+\cdots+ (x^{Bk+B-1}-x^{Bk+1})+(x^{Bk+B-1}-x^{Bk})]\,dx \\ &=-\sum_{k=1}^{\infty}a_k\widetilde{Q}(k,B). \end{split} \end{equation*} \end{proof} \begin{theorem} \label{t5} Let $a,b,B$ be positive integers with $B>1,$ $z\in {\mathbb C},$ $|z|\le 1.$ Then for the generalized Euler constant function $\gamma_{a,b}(z)$ and its derivative, the following expansion is valid: $$ \gamma_{a,b}^{(l)}(z)=\frac{1}{2}\int_0^1\frac{x^{b+al-1}(1-x)}{(1-zx^a)^{l+1}}\,dx +\sum_{k=1}^{\infty}a_{k,l}\frac{P_B(k)}{Bk(Bk+1)\cdots(Bk+B)}, \qquad l=0,1, $$ where $P_B(k)$ is a polynomial of degree $B-2$ given by {\rm (\ref{polyn})}, $z\ne 1$ if $l=1,$ and the sequence $\{a_{k,l}\}_{k=0}^{\infty}$ is defined by the generating function \begin{equation} G_l(z,x)=\frac{1}{1-x}\sum_{k=0}^{\infty}\frac{x^{(b+al)B^k}(1-x^{B^k})}% {(1-zx^{aB^k})^{l+1}}=\sum_{k=0}^{\infty}a_{k,l}x^k, \qquad l=0, 1. \label{eq24} \end{equation} \end{theorem} \begin{proof} Expanding $G_l(z,x)$ in a power series of $x,$ $$ G_l(z,x)=\sum_{k=0}^{\infty}\sum_{m=0}^{\infty}\binom{m+l}{l}z^mx^{(b+al+am)B^k}(1+x+x^2+ \cdots+x^{B^k-1}), $$ we see that $a_{k,l}=O(k^l\ln_Bk).$ Therefore, for $l=0,1,$ by termwise integration we get \begin{equation*} \begin{split} &\int_0^1\left(\frac{B(1+x^B)}{1-x^B}-\frac{1+x}{1-x}\right)F_l(z,x)dx =\int_0^1\frac{1-x^B}{x}\left(\frac{B(1+x^B)}{1-x^B}-\frac{1+x}{1-x}\right)G_l(z,x^B)dx \\ &=\int_0^1[(B-1)-2x-2x^2-\cdots-2x^{B-1}+(B-1)x^B]\sum_{k=1}^{\infty}a_{k,l}x^{Bk-1}\,dx \\ &=\sum_{k=1}^{\infty}a_{k,l}\left(\frac{B-1}{Bk}-\frac{2}{Bk+1}-\frac{2}{Bk+2}-\cdots -\frac{2}{Bk+B-1}+\frac{B-1}{Bk+B}\right) \\ &=2\sum_{k=1}^{\infty}a_{k,l}\frac{P_B(k)}{Bk(Bk+1)\cdots(Bk+B)}, \end{split} \end{equation*} where $P_B(k)$ is defined in (\ref{polyn}) and the last series converges since $\frac{P_B(k)}{Bk(Bk+1)\cdots(Bk+B)}=O(k^{-3}).$ Now our theorem easily follows from Corollary \ref{c3}. \end{proof} \section{Examples of rational series} \label{S5} It is easily seen that the generating function (\ref{eq24}) satisfies the following functional equation: \begin{equation} G_l(z,x)-\frac{1-x^B}{1-x}G_l(z,x^B)=\frac{x^{b+al}}{(1-zx^a)^{l+1}}, \label{eq25} \end{equation} which is equivalent to the following identity for series: $$ \sum_{k=0}^{\infty}a_{k,l}x^k-(1+x+\cdots+x^{B-1})\sum_{k=0}^{\infty}a_{k,l}x^{Bk} =\sum_{k=l}^{\infty}\binom{k}{l}z^{k-l}x^{ak+b}. $$ Comparing coefficients of powers of $x$ we get an alternative definition of the sequence $\{a_{k,l}\}_{k=0}^{\infty}$ by means of the recursion $$ a_{0,l}=a_{1,l}=\cdots=a_{al+b-1,l}=0 $$ and for $k\ge al+b,$ \begin{equation} a_{k,l}=\begin{cases} a_{\lfloor\frac{k}{B}\rfloor,l}, & \qquad\text{if} \qquad k\not\equiv b \pmod a; \\ a_{\lfloor\frac{k}{B}\rfloor,l}+\binom{(k-b)/a}{l}z^{\frac{k-b}{a}-l}, & \qquad\text{if} \qquad k\equiv b\pmod a. \end{cases} \label{re} \end{equation} On the other hand, in view of Corollary \ref{c1.1}, $\gamma_{a,b}(z)$ and $\gamma'_{a,b}(z)$ can be explicitly expressed in terms of the Lerch transcendent, $\psi$-function and logarithm of the gamma function. This allows us to sum the series in Theorems \ref{t3}--\ref{t5} in terms of these functions. \begin{example} Suppose that $\omega$ is a non-empty word over the alphabet $\{0,1,\ldots,B-1\}.$ Then obviously $\omega$ is uniquely defined by its length $|\omega|$ and its size $v_B(\omega)$ which is the value of $\omega$ when interpreted as an integer in base $B.$ Let $N_{\omega,B}(k)$ be the number of (possibly overlapping) occurrences of the block $\omega$ in the $B$-ary expansion of $k.$ Note that for every $B$ and $\omega,$ $N_{\omega,B}(0)=0,$ since the $B$-ary expansion of zero is the empty word. If the word $\omega$ begins with $0,$ but $v_B(\omega)\ne 0,$ then in computing $N_{\omega,B}(k)$ we assume that the $B$-ary expansion of $k$ starts with an arbitrary long prefix of $0$'s. If $v_B(\omega)=0$ we take for $k$ the usual shortest $B$-ary expansion of $k.$ Now we consider equation (\ref{eq25}) with $l=0,$ $z=1$ \begin{equation} G(1,x)-\frac{1-x^B}{1-x}G(1,x^B)=\frac{x^b}{1-x^a} \label{eq26} \end{equation} and for a given non-empty word $\omega,$ set $a=B^{|\omega|}$ in (\ref{eq26}) and \begin{equation*} b=\begin{cases} B^{|\omega|}, & \quad \text{if}\quad v_B(\omega)=0; \\ v_B(\omega), & \quad \text{if}\quad v_B(\omega)\ne 0. \end{cases} \end{equation*} Then by (\ref{re}), it is easily seen that $a_k:=a_{k,0}=N_{\omega, B}(k),$ $k=1,2,\ldots,$ and by Theorem \ref{t3}, we get another proof of the following statement (see \cite[Sections 3, 4.2]{assh}). \end{example} \begin{corollary} \label{c4} Let $\omega$ be a non-empty word over the alphabet $\{0,1,\ldots, B-1\}.$ Then \begin{equation*} \sum_{k=1}^{\infty}N_{\omega, B}(k)Q(k,B)=\begin{cases} \gamma_{B^{|\omega|},v_B(\omega)}(1), & \quad \text{if}\quad v_B(\omega)\ne 0; \\ \gamma_{B^{|\omega|}, B^{|\omega}|}(1), & \quad \text{if}\quad v_B(\omega)= 0. \end{cases} \end{equation*} \end{corollary} By Corollary \ref{c1.1}, the right-hand side of the last equality can be calculated explicitly and we have \begin{equation} \sum_{k=1}^{\infty}N_{\omega, B}(k)Q(k,B)=\begin{cases} \log\Gamma\left(\frac{v_B(\omega)+1}{B^{|\omega|}}\right)- \log\Gamma\left(\frac{v_B(\omega)}{B^{|\omega|}}\right)- \frac{1}{B^{|\omega|}}\psi\left(\frac{v_B(\omega)}{B^{|\omega|}}\right), & \text{if}\,\, v_B(\omega)\ne 0; \\ \log\Gamma\left(\frac{1}{B^{|\omega|}}\right)+ \frac{\gamma}{B^{|\omega|}}-|\omega|\log B, & \text{if}\,\, v_B(\omega)= 0. \end{cases} \label{eq28} \end{equation} \begin{corollary} \label{c54} Let $\omega$ be a non-empty word over the alphabet $\{0,1,\ldots, B-1\}.$ Then \begin{equation*} \begin{split} &\sum_{k=1}^{\infty}\frac{N_{\omega, B}(k)P_B(k)}{Bk(Bk+1)\cdots (Bk+B)} \\ &=\begin{cases} \gamma_{B^{|\omega|},v_B(\omega)}(1)-\frac{1}{2B^{|\omega|}} \left(\psi\Bigl(\frac{v_B(\omega)+1}{B^{|\omega|}}\Bigr) -\psi\Bigl(\frac{v_B(\omega)}{B^{|\omega|}}\Bigr)\right), & \quad \text{if}\quad v_B(\omega)\ne 0; \\ \gamma_{B^{|\omega|}, B^{|\omega}|}(1)- \frac{1}{2B^{|\omega|}}\psi\Bigl(\frac{1}{B^{|\omega|}}\Bigr)- \frac{\gamma}{2B^{|\omega|}}-\frac{1}{2}, & \quad \text{if}\quad v_B(\omega)= 0. \end{cases} \end{split} \end{equation*} \end{corollary} \begin{proof} The required statement easily follows from Theorem \ref{t5}, Corollary \ref{c4} and the equality $$ \int_0^1\frac{x^{b-1}(1-x)}{1-x^a}\,dx=\sum_{k=0}^{\infty} \left(\frac{1}{ak+b}-\frac{1}{ak+b+1}\right)=\frac{1}{a} \left(\psi\Bigl(\frac{b+1}{a}\Bigr)-\psi\Bigl(\frac{b}{a}\Bigr)\right). $$ \end{proof} From Theorem \ref{t3}, (\ref{eq25}) and (\ref{re}) with $a=1,$ $l=0$ we have \begin{corollary} \label{c5} Let $b, B$ be positive integers with $B>1,$ $z\in {\mathbb C},$ $|z|\le 1.$ Then $$ \gamma_{1,b}(z)=\sum_{k=1}^{\infty}a_kQ(k,B)=\sum_{k=1}^{\infty}a_{\lfloor\frac{k}{B}\rfloor} \frac{\varepsilon(k)}{k}, $$ where $a_0=a_1=\cdots=a_{b-1}=0,$ $a_k=a_{\lfloor\frac{k}{B}\rfloor}+z^{k-b},$ $k\ge b.$ \end{corollary} Similarly, from Theorem \ref{t5} we have \begin{corollary} \label{c65} Let $b,B$ be positive integers with $B>1,$ $z\in {\mathbb C},$ $|z|\le 1.$ Then $$ \gamma_{1,b}(z)=\frac{1}{2}\sum_{k=0}^{\infty}\frac{z^k}{(k+b)(k+b+1)} +\sum_{k=1}^{\infty}a_k\frac{P_B(k)}{Bk(Bk+1)\cdots (Bk+B)}, $$ where $a_0=a_1=\cdots=a_{b-1}=0,$ $a_k=a_{\lfloor\frac{k}{B}\rfloor}+z^{k-b},$ $k\ge b.$ \end{corollary} \begin{example} If in Corollary \ref{c5} we take $z=1,$ then we get that $a_k$ is equal to the $B$-ary length of $\lfloor\frac{k}{b}\rfloor,$ i. e., $$ a_k=\sum_{\alpha=0}^{B-1}N_{\alpha,B}\left(\Bigl\lfloor\frac{k}{b}\Bigr\rfloor\right) =L_B\left(\Bigl\lfloor\frac{k}{b}\Bigr\rfloor\right). $$ On the other hand, $$ \gamma_{1,b}(1)=\log b-\psi(b)=\log b-\sum_{k=1}^{b-1}\frac{1}{k}+\gamma $$ and hence we get \begin{equation} \log b-\psi(b)=\sum_{k=1}^{\infty}L_B\left(\Bigl\lfloor\frac{k}{b}\Bigr\rfloor \right)Q(k,B). \label{eq29} \end{equation} If $b=1,$ formula (\ref{eq29}) gives (\ref{eq6}). If $b>1,$ then from (\ref{eq29}) and (\ref{eq6}) we get \begin{equation} \log b=\sum_{k=1}^{b-1}\frac{1}{k}+\sum_{k=1}^{\infty}\left(L_B\left(\Bigl\lfloor \frac{k}{b}\Bigr\rfloor\right)-L_B(k)\right)Q(k,B), \label{eq30} \end{equation} which is equivalent to \cite[Theorem 2.8]{bebo}. Similarly, from Corollary \ref{c65} we obtain (\ref{eu}) and \begin{equation} \log b=\sum_{k=1}^{b-1}\frac{1}{k}-\frac{b-1}{2b}+\sum_{k=1}^{\infty} \frac{\left(L_B(\lfloor \frac{k}{b}\rfloor)-L_B(k)\right)P_B(k)}% {Bk(Bk+1)\cdots (Bk+B)}. \label{eq30.5} \end{equation} \end{example} \begin{example} Using the fact that for any integer $B>1,$ $$ L_B\left(\Bigl\lfloor\frac{k}{B}\Bigr\rfloor\right)-L_B(k)=-1, $$ from (\ref{eq28}), (\ref{eq6}) and (\ref{eq30}) we get the following rational series for $\log\Gamma(1/B):$ $$ \log\Gamma\left(\frac{1}{B}\right)=\sum_{k=1}^{B-1}\frac{1}{k} +\sum_{k=1}^{\infty}\Bigl(N_{0,B}(k)-\frac{1}{B}L_B(k)-1\Bigr)Q(k,B). $$ \end{example} \begin{example} Substituting $b=1,$ $z=-1$ in Corollary \ref{c5} we get the generalized Vacca series for $\log\frac{4}{\pi}.$ \end{example} \begin{corollary} \label{c6} Let $B\in {\mathbb N},$ $B>1.$ Then $$ \log\frac{4}{\pi}=\sum_{k=1}^{\infty}a_kQ(k,B)=\sum_{k=1}^{\infty}a_{\lfloor\frac{k}{B}\rfloor} \frac{\varepsilon(k)}{k}, $$ where \begin{equation} a_0=0, \qquad a_k=a_{\lfloor\frac{k}{B}\rfloor}+(-1)^{k-1}, \quad k\ge 1. \label{eq31} \end{equation} In particular, if $B$ is even, then \begin{equation} \log\frac{4}{\pi}=\sum_{k=1}^{\infty}(N_{odd,B}(k)-N_{even,B}(k))Q(k,B) =\sum_{k=1}^{\infty} \frac{\left(N_{odd,B}(\lfloor\frac{k}{B}\rfloor)- N_{even,B}(\lfloor\frac{k}{B}\rfloor)\right)}{k} \varepsilon(k), \label{eq32} \end{equation} where $N_{odd,B}(k)$ (respectively $N_{even,B}(k)$) is the number of occurrences of the odd (respectively even) digits in the $B$-ary expansion of $k.$ \end{corollary} \begin{proof} To prove (\ref{eq32}), we notice that if $B$ is even, then the sequence $\widetilde{a}_k:=N_{odd,B}(k)-N_{even,B}(k)$ satisfies recurrence (\ref{eq31}). \end{proof} Substituting $b=1, z=-1$ in Corollary \ref{c65} with the help of (\ref{eq30.5}) we get the generalized Addison series for $\log\frac{4}{\pi}.$ \begin{corollary} \label{c69} Let $B>1$ be a positive integer. Then $$ \log\frac{4}{\pi}=\frac{1}{4}+\sum_{k=1}^{\infty}\frac{\left(L_B(\lfloor \frac{k}{2}\rfloor)-L_B(k)+a_k\right)P_B(k)}{Bk(Bk+1)\cdots (Bk+B)}, $$ where the sequence $a_k$ is defined in Corollary {\rm\ref{c6}}. In particular, if $B$ is even, then $$ \log\frac{4}{\pi}=\frac{1}{4}+\sum_{k=1}^{\infty}\frac{\left(L_B(\lfloor \frac{k}{2}\rfloor)-2N_{even,B}(k)\right)P_B(k)}{Bk(Bk+1)\cdots (Bk+B)}. $$ \end{corollary} \begin{example} For $t>1,$ the generalized Somos constant $\sigma_t$ is defined by $$ \sigma_t=\sqrt[t]{1\sqrt[t]{2\sqrt[t]{3\cdots}}}= 1^{1/t}2^{1/t^2}3^{1/t^3}\cdots=\prod_{n=1}^{\infty}n^{1/t^n} $$ (see \cite[Section 3]{sh}). In view of the relation \cite[Theorem 8]{sh} \begin{equation} \gamma_{1,1}\left(\frac{1}{t}\right)=t\log\frac{t}% {(t-1)\sigma_t^{t-1}}, \label{somos} \end{equation} by Corollary \ref{c5} and formula (\ref{eq30}) we get \end{example} \begin{corollary} \label{c7} Let $B\in {\mathbb N},$ $B>1,$ $t\in {\mathbb R},$ $t>1.$ Then $$ \log\sigma_t=\frac{1}{(t-1)^2}+\frac{1}{t-1}\sum_{k=1}^{\infty} \left(L_B\Bigl(\Bigl\lfloor\frac{k}{t}\Bigr\rfloor\Bigr)-L_B\Bigl(\Bigl\lfloor \frac{k}{t-1}\Bigr\rfloor\Bigr)-\frac{a_k}{t}\right)Q(k,B), $$ where $a_0=0,$ $a_k=a_{\lfloor\frac{k}{B}\rfloor}+t^{1-k},$ $k\ge 1.$ \end{corollary} In particular, setting $B=t=2$ we get the following rational series for Somos's quadratic recurrence constant: $$ \log\sigma_2=1-\frac{1}{2}\sum_{k=1}^{\infty}\frac{b_k}{2k(2k+1)}, $$ where $b_1=3,$ $b_k=b_{\lfloor\frac{k}{2}\rfloor}+\frac{1}{2^{k-1}},$ $k\ge 2.$ From (\ref{somos}), (\ref{eq30.5}) and Theorem \ref{t5} we find \begin{corollary} \label{c8} Let $B\in {\mathbb N},$ $B>1,$ $t\in {\mathbb R},$ $t>1.$ Then \begin{equation*} \begin{split} &\log\sigma_t=\frac{3t-1}{4t(t-1)^2} \\ &+\frac{t+1}{2(t-1)}\sum_{k=1}^{\infty} \left(L_B\Bigl(\Bigl\lfloor\frac{k}{t}\Bigr\rfloor\Bigr)-L_B\Bigl(\Bigl\lfloor \frac{k}{t-1}\Bigr\rfloor\Bigr)-\frac{2a_k}{t(t+1)}\right) \frac{P_B(k)}{Bk(Bk+1)\cdots(Bk+B)}, \end{split} \end{equation*} where the sequence $a_k$ is defined in Corollary {\rm \ref{c7}}. \end{corollary} In particular, if $B=t=2$ we get $$ \log\sigma_2=\frac{5}{8}-\frac{1}{2}\sum_{k=1}^{\infty} \frac{c_k}{2k(2k+1)(2k+2)}, $$ where $c_1=4,$ $c_k=c_{\lfloor\frac{k}{2}\rfloor}+\frac{1}{2^{k-1}},$ $k\ge 2.$ \begin{example} The Glaisher-Kinkelin constant is defined by the limit \cite[p.135]{fi} $$ A:=\lim_{n\to\infty}\frac{1^22^2\cdots n^n}{n^{\frac{n^2+n}{2}+\frac{1}{12}}e^{-\frac{n^2}{4}}}=1.28242712\cdots. $$ Its connection to the generalized Euler constant function $\gamma_{a,b}(z)$ is given by the formula \cite[Corollary 4]{sh} \begin{equation} \gamma'_{1,1}(-1)=\log\frac{2^{11/6}A^6}{\pi^{3/2}e}. \label{glaisher} \end{equation} By Theorem \ref{t5}, since $$ \int_0^1\frac{x(1-x)}{(1+x)^2}\,dx=3\log 2-2, $$ we have $$ \log A=\frac{4}{9}\log 2-\frac{1}{4}\log\frac{4}{\pi}+ \frac{1}{6}\sum_{k=1}^{\infty}a_{k,1}\frac{P_B(k)}{Bk(Bk+1)\cdots(Bk+B)}, $$ where the sequence $a_{k,1}$ is defined by the generating function (\ref{eq24}) with $a=b=l=1,$ $z=-1,$ or using the recursion (\ref{re}): $$ a_{0,1}=a_{1,1}=0, \qquad a_{k,1}=a_{\lfloor\frac{k}{B}\rfloor,1}+(-1)^k(k-1), \quad k\ge 2. $$ Now by Corollary \ref{c69} and (\ref{eq30.5}) we get \end{example} \begin{corollary} \label{c12} Let $B>1$ be a positive integer. Then $$ \log A=\frac{13}{48}-\frac{1}{36}\sum_{k=1}^{\infty} \left(7L_B(k)-7L_B\Bigl(\Bigl\lfloor\frac{k}{2}\Bigr\rfloor\Bigr) +b_k\right)\frac{P_B(k)}{Bk(Bk+1)\cdots(Bk+B)}, $$ where $b_0=0,$ $b_k=b_{\lfloor\frac{k}{B}\rfloor}+(-1)^{k-1}(6k+3),$ $k\ge 1.$ \end{corollary} In particular, if $B=2$ we get $$ \log A=\frac{13}{48}-\frac{1}{36}\sum_{k=1}^{\infty} \frac{c_k}{2k(2k+1)(2k+2)}, $$ where $c_1=16,$ $c_k=c_{\lfloor\frac{k}{2}\rfloor}+(-1)^{k-1}(6k+3),$ $k\ge 2.$ Using the formula expressing $\frac{\zeta'(2)}{\pi^2}$ in terms of the Glaisher-Kinkelin constant \cite[p.\ 135]{fi}, $$ \log A=-\frac{\zeta'(2)}{\pi^2}+\frac{\log 2\pi+\gamma}{12}, $$ by Corollaries \ref{c65}, \ref{c69} and \ref{c12}, we get \begin{corollary} Let $B>1$ be a positive integer. Then $$ \frac{\zeta'(2)}{\pi^2}=-\frac{1}{16}+\frac{1}{36}\sum_{k=1}^{\infty} \left(4L_B(k)-L_B\Bigl(\Bigl\lfloor\frac{k}{2}\Bigr\rfloor\Bigr)+c_k\right) \frac{P_B(k)}{Bk(Bk+1)\cdots(Bk+B)}, $$ where $c_0=0,$ $c_k=c_{\lfloor\frac{k}{B}\rfloor}+(-1)^{k-1}6k,$ $k\ge 1.$ \end{corollary} \begin{example} First we evaluate $\gamma_{2,1}^{(l)}(-1)$ for $l=0, 1.$ From Corollaries \ref{c1}, \ref{c1.1} and \cite[Examples 3.9, 3.15]{gs} we have $$ \gamma_{2,1}(-1)=\int_0^1\int_0^1\frac{(x-1)\,dxdy}{(1+x^2y^2)\log xy}=\frac{\pi}{4}-2\log\Gamma\Bigl(\frac{1}{4}\Bigr)+\log\sqrt{2\pi^3} $$ and \begin{equation*} \begin{split} \gamma'_{2,1}(-1)=&-\frac{1}{4}\Phi(-1,1,3/2)+\frac{1}{2} \Phi(-1,0,3/2) +\frac{1}{2}\frac{\partial\Phi}{\partial s}(-1,0,3/2) \\ &-\frac{\partial\Phi}{\partial s}(-1,-1,3/2)- \frac{\partial\Phi}{\partial s}(-1,0,2)+\frac{\partial\Phi}{\partial s}(-1,-1,2). \end{split} \end{equation*} The last expression can be evaluated explicitly (see \cite[Section 2]{gs}) and we get $$ \gamma'_{2,1}(-1)=\frac{G}{\pi}+\frac{\pi}{8}-\log\Gamma\Bigl( \frac{1}{4}\Bigr)-3\log A+\log\pi+\frac{1}{3}\log 2, $$ or \begin{equation} \frac{G}{\pi}=\gamma'_{2,1}(-1)-\frac{1}{2}\gamma_{2,1}(-1) +\frac{1}{4}\log\frac{4}{\pi}+3\log A-\frac{7}{12}\log 2. \label{eq101} \end{equation} On the other hand, by Theorem \ref{t5} and (\ref{re}) we have \begin{equation} \gamma_{2,1}(-1)=\frac{\pi}{8}-\frac{1}{4}\log 2+\sum_{k=1}^{\infty}a_{k,0}\frac{P_B(k)}{Bk(Bk+1)\cdots (Bk+B)}, \label{eq102} \end{equation} where $a_{0,0}=0,$ $a_{2k,0}=a_{\lfloor\frac{2k}{B}\rfloor,0},$ $k\ge 1,$ $a_{2k+1,0}=a_{\lfloor\frac{2k+1}{B}\rfloor,0}+(-1)^k,$ $k\ge 0,$ and \begin{equation} \gamma'_{2,1}(-1)=\frac{\pi}{16}-\frac{1}{4}\log 2+\sum_{k=1}^{\infty}a_{k,1}\frac{P_B(k)}{Bk(Bk+1)\cdots (Bk+B)}, \label{eq103} \end{equation} where $a_{0,1}=0,$ $a_{2k,1}=a_{\lfloor\frac{2k}{B}\rfloor,1},$ $k\ge 1,$ $a_{2k+1,1}=a_{\lfloor\frac{2k+1}{B}\rfloor,1}+(-1)^{k-1}k,$ $k\ge 0.$ Now from (\ref{eq101}) -- (\ref{eq103}), (\ref{eq30.5}) and Corollary \ref{c69} we get the following expansion for $G/\pi.$ \end{example} \begin{corollary} Let $B>1$ be a positive integer. Then $$ \frac{G}{\pi}=\frac{11}{32}+\sum_{k=1}^{\infty}\left( \frac{1}{8}L_B\Bigl(\Bigl\lfloor\frac{k}{2}\Bigr\rfloor\Bigr) -\frac{1}{8}L_B(k)+c_k\right)\frac{P_B(k)}{Bk(Bk+1)\cdots (Bk+B)}, $$ where $c_0=0,$ $c_{2k}=c_{\lfloor\frac{2k}{B}\rfloor}+k,$ $k\ge 1,$ $c_{2k+1}=c_{\lfloor\frac{2k+1}{B}\rfloor}+ \frac{(-1)^{k-1}-1}{2}(2k+1),$ $k\ge 0.$ \end{corollary} In particular, if $B=2$ we get $$ \frac{G}{\pi}=\frac{11}{32}+\sum_{k=1}^{\infty}\frac{b_k}{2k(2k+1)(2k+2)}, $$ where $b_1=-\frac{9}{8},$ $b_{2k}=b_k+k,$ $b_{2k+1}=b_k+ \frac{(-1)^{k-1}-1}{2}(2k+1),$ $k\ge 1.$ \section{Other generalized Euler constants} \label{S6} The purpose of this section is to draw attention to different generalizations of Euler's constant for which many interesting results remain to be discovered. The simplest way to generalize Euler's constant \begin{equation} \gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\log n\right), \label{6.1} \end{equation} which is related to the digamma function by the equality $\gamma=-\psi(1),$ is to consider for $0<\alpha\le 1,$ $$ \gamma(\alpha)=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k+\alpha}-\log n\right)= \lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k+\alpha}-\log(n+\alpha)\right). $$ Tasaka \cite{ta} proved that $\gamma(\alpha)=-\psi(\alpha).$ Its connection to the generalized Euler constant function $\gamma_{a,b}(z)$ is given by the formula $$ \gamma(\alpha)+\log\alpha=\gamma_{1,\alpha}(1). $$ Briggs \cite{br} and Lehmer \cite{l} studied the analog of $\gamma$ corresponding to the arithmetical progression of positive integers $r,$ $r+m,$ $r+2m, \ldots,$ $(r\le m):$ $$ \gamma(r,m)=\lim_{n\to\infty}\left(H(n,r,m)-\frac{1}{m}\log n\right), $$ where $H(n,r,m)=\underset{k\equiv r\!\!\!\pmod m}{\sum\limits_{0