Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent
Jes\'{u}s Guillera
Jonathan Sondow
The two-fold aim of the paper is to unify and generalize on the one hand the double integrals of Beukers for $\zeta(2)$ and $\zeta(3),$ and of the second author for Euler's constant $\gamma$ and its alternating analog $\ln(4 / \pi),$ and on the other hand the infinite products of the first author for $e$, of the second author for $\pi,$ and of Ser for $e^\gamma.$ We obtain new double integral and infinite product representations of many classical constants, as well as a generalization to Lerch's transcendent of Hadjicostas's double integral formula for the Riemann zeta function, and logarithmic series for the digamma and Euler beta functions. The main tools are analytic continuations of Lerch's function, including Hasse's series. We also use Ramanujan's polylogarithm formula for the sum of a particular series involving harmonic numbers, and his relations between certain dilogarithm values.
math0506319.tex
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\begin{document}

\title[Double integrals and infinite products]{Double integrals and infinite products for  some classical
constants via analytic continuations of Lerch's transcendent}

\keywords{Lerch transcendent, zeta function, infinite product,
double integral, polylogarithm}

\date{11 June 2005}

\subjclass[2000]{Primary 11M06, 11M35, 11Y60. Secondary 33B15,
33B30}

\author{Jes\'{u}s Guillera \and Jonathan Sondow}

\address{Av. Ces\'{a}reo Alierta 31, esc. izda $\rm 4^{o}A$ \vskip 0cm Zaragoza 50008 Spain \vskip 0cm}

\email{jguillera@able.es \vskip 0cm \textit{URL}\!:
http://personal.auna.com/jguillera}

\address{209 West 97th Street Apt 6F \vskip 0cm New York, NY 10025 USA \vskip 0cm}

\email{jsondow@alumni.princeton.edu \vskip 0cm \textit{URL}\!:
http://home.earthlink.net/\~jsondow/}

\maketitle

\newtheorem{teor}{Theorem}[section]
\newtheorem{lema}{Lemma}[section]
\newtheorem{coro}{Corollary}[section]
\newtheorem{ejem}{Example}[section]

\begin{abstract}
The two-fold aim of the paper is to unify and generalize on the one
hand the double integrals of Beukers for $\zeta(2)$ and $\zeta(3),$
and of the second author for Euler's constant $\gamma$ and its
alternating analog $\ln(4 / \pi),$ and on the other hand the
infinite products of the first author for $e$, of the second author
for $\pi,$ and of Ser for $e^\gamma.$ We obtain new double integral
and infinite product representations of many classical constants, as
well as a generalization to Lerch's transcendent of Hadjicostas's
double integral formula for the Riemann zeta function, and
logarithmic series for the digamma and Euler beta functions. The
main tools are analytic continuations of Lerch's function, including
Hasse's series. We also use Ramanujan's polylogarithm formula for
the sum of a particular series involving harmonic numbers, and his
relations between certain dilogarithm values.
\end{abstract}

\section*{Contents}
\begin{center}
\begin{tabular}{c l l l r}
  \ref{intro} & Introduction & & & 1 \\
  \ref{analy} & The Lerch Transcendent & & & 2 \\
  \ref{doubl} & Double Integrals & & & 7 \\
  \ref{hadji} & A Generalization of Hadjicostas's Formula & & & 14 \\
  \ref{infin} & Infinite Products & & & 16 \\
              & References & & & 20
\end{tabular}
\end{center}

\section{Introduction} \label{intro}
This paper is primarily about double integrals and infinite
products related to the Lerch transcendent $\Phi(z,s,u)$.

Concerning double integrals, our aim is to unify and generalize the
integrals over the unit square $[0,1]^2$ of Beukers \cite{beukers}
and Hadjicostas \cite{hadjicostas0}, \cite{hadjicostas} for values
of the Riemann zeta function (Examples \ref{int2beukers} and
\ref{doublehadji}), and those of the second author \cite{sondow0},
\cite{sondow1} for Euler's constant $\gamma$ and its alternating
analog $\ln (4 / \pi)$ (Examples \ref{eulersondow} and
\ref{intlnpi}). We do this in Theorems \ref{teorintegrals1} and
\ref{teorintegrals2}, using a classical integral analytic
continuation of $\Phi$ (Lemmas \ref{philemma1} and \ref{philemma2}).
As applications, we obtain new double integral representations of
many constants, including  $\ln 2$, $\ln \varphi$, $\pi^3$, $\ln
\sigma$, $\Gamma(3/4)$,  $\zeta(5),$ and $G/\pi$ (Examples
\ref{int2ln2}, \ref{int2lngolden}, \ref{int2picub}, \ref{intsomos},
\ref{int2GAMMA3over4}, \ref{intzeta5}, and \ref{int2catoverpi}),
where $\varphi$ is the golden ratio, $\sigma$ is one of Somos's
quadratic recurrence constants, and $G$ is Catalan's constant.
(Single integrals for them follow by the change of variables in the
proof of Theorem \ref{teorintegrals1}.) Corollary \ref{coro3.1}
gives a double integral for the polylogarithm function; examples use
Ramanujan's relations between certain values of the dilogarithm
(Example \ref{ramanujandylogs}). Corollaries \ref{corointgammas},
\ref{coro3.5} and \ref{corointpsi} evaluate certain types of double
integrals in terms of the function $\ln \Gamma(x)$ and its
derivative, the digamma function; Examples \ref{intlnpi},
\ref{int2pioversqrt3} and \ref{eulersondow} for $\ln \pi$,
$\pi/\sqrt{3}$ and $\gamma$ are interesting special cases. Theorem
\ref{corointrama} evaluates a class of double integrals with the
help of Ramanujan's summation of a series involving harmonic
numbers; applications use classical values of the di- and
trilogarithm (Example \ref{di-trilog}).
\par Regarding infinite products, our aim is extend those of the first author (see
\cite{sondow2}) for $e$ (Example \ref{raize1}), of the second author
\cite{sondow2} for $\pi$ (Example \ref{prodpi}), and of Ser
\cite{ser} (see also \cite{sondow3}, \cite{sondow2}) for $e^\gamma$
(Example \ref{prodeuler}). To do this we use two analytic
continuations of $\Phi$ by series, one due to Hasse \cite{hasse}
(Theorems \ref{phidoubleseries} and \ref{hasse}). We find new
infinite products for many constants, among them $\Gamma(1/4)
\Gamma(3/4)^{-1}$, $e^{G/\pi}$, $\pi / e$, $e^{\pi}$, $\sigma$, and
the Glaisher-Kinkelin constant $A$ (Examples \ref{prodGammas},
\ref{prodcat}, \ref{pipare}, \ref{prodexppi}, \ref{prodsigma}, and
\ref{prodgla}). The product for $\sigma$ has rational factors.

Finally, we mention other results obtained: explicit formulas for
the Bernoulli and Euler polynomials (Examples \ref{bernoullipol} and
\ref{eulerpol}), and logarithmic series for the digamma and Euler
beta functions (Theorems \ref{teordigamma} and \ref{teorbeta}). The
latter series leads to infinite products for the exponential
function of the same shape as other products in the paper, but which
we construct without using Lerch's function (Theorem \ref{expx} and
Examples \ref{raize1} and \ref{prode2}).

\section{The Lerch Transcendent} \label{analy}
The \textit{Lerch transcendent} $\Phi$ (see \cite[section
1.11]{bateman}, \cite[section 64:12]{spanier}) is the analytic
continuation of the series
\begin{equation}\label{lerch1}
\Phi(z,s,u)={1 \over u^s}+{z \over (u+1)^s}+{z^2 \over
(u+2)^s}+\cdots,
\end{equation}
which converges for any real number $u>0$ if $z$ and $s$ are any
complex numbers with either $|z|<1$, or $|z|=1$ and $\Re(s)>1$.
Special cases include the analytic continuations of the
\textit{Riemann zeta function}
\begin{equation}\label{fzeta}
\zeta(s)=\sum_{k=1}^{\infty} {1 \over k^s}=\Phi(1,s,1),
\end{equation}
the \textit{Hurwitz zeta function}
\begin{equation}\nonumber
\zeta(s,u)=\sum_{k=0}^{\infty} {1 \over (u+k)^s}=\Phi(1,s,u),
\end{equation}
the \textit{alternating zeta function} (also known as
\textit{Dirichlet's eta function $\eta(s)$})
\begin{equation}\label{alter}
\zeta^{*}(s)= \sum_{k=1}^{\infty} {(-1)^{k-1} \over k^s} =
\Phi(-1,s,1),
\end{equation}
the \textit{Dirichlet beta function}
\begin{equation}\label{dirichletbeta}
\beta(s)=\sum_{k=0}^{\infty} {(-1)^k \over (2k+1)^s}= 2^{-s} \Phi
\left(-1,s,{1 \over 2} \right),
\end{equation}
the \textit{Legendre chi function}
\begin{equation}\label{legendrechi}
\chi_s(z)=\sum_{k=0}^{\infty} {z^{2k+1} \over (2k+1)^s}= 2^{-s} z
\Phi \left(z^2,s,{1 \over 2} \right),
\end{equation}
the \textit{polylogarithm}
\begin{equation}\label{polyloga}
{\textrm{Li}_n(z)}=\sum_{k=1}^{\infty} {z^k \over k^n}=z
\Phi(z,n,1),
\end{equation}
and the \textit{Lerch zeta function}
$L(\lambda,\alpha,s)=\Phi(\exp(2 \pi i \lambda),s,\alpha)$.
\par From (\ref{lerch1}), Lerch's transcendent satisfies the
following identities:
\begin{align}
\Phi(z,s,u+1) &= {1 \over z} \left(  \Phi(z,s,u)- {1 \over u^s} \right), \label{induu} \\
\Phi(z,s-1,u) &= \left( u+ z {\partial \over \partial z} \right)
\Phi(z,s,u), \label{indulerch} \\
\Phi(z,s+1,u) &= -{1 \over s} {\partial \Phi \over \partial u}
(z,s,u). \label{lerchpar}
\end{align}

\par {\it Henceforth, we assume that $u$ and $v$ are positive real numbers.} \rm The
following two lemmas are classical. (See \cite{bateman},
\cite{milnor}, \cite{spanier}, and \cite{whittaker}.)

\begin{lema} \label{philemma1}
If either $|z|<1$ and $\Re(s)>0$, or $z=1$ and $\Re(s)>1$, then
\begin{equation} \label{intelerch}
\Phi(z,s,u)={1 \over \Gamma(s)} \int_{0}^{\infty} {e^{-(u-1)t}
\over e^t-z} t^{s-1} dt.
\end{equation}
\end{lema}

\begin{proof}
Replacing ${e^{-(u-1)t} \over e^t-z}$ with $\sum_{k \geq 0} z^k
e^{-(u+k)t}$, and noting that
\begin{equation}\label{intint}
\int_{0}^{\infty} e^{-(u+k)t} t^{s-1} dt= {1 \over (u+k)^s}
\int_{0}^{\infty} e^{-x}x^{s-1}dx={\Gamma(s) \over (u+k)^s},
\end{equation}
the lemma follows.
\end{proof}

\begin{lema} \label{philemma2}
The series {\rm (\ref{lerch1})} extends to a function $\Phi(z,s,u)$
which is defined, and holomorphic in $z$ and $s$, for $z \in
\mathbf{C}-[1, \infty)$ and all complex $s$, and which is given by
the integral formula {\rm (\ref{intelerch})} if $\Re(s)>0$.
\end{lema}

\begin{proof}
If $z \in \mathbf{C} -[1, \infty)$, the denominator of the
integrand in (\ref{intelerch}) does not vanish. It follows that
for $\Re(s)>0$ the right side of (\ref{intelerch}) defines a
function which is holomorphic in $z$ and $s$. The extension to
other values of $s$ follows inductively using the identity
(\ref{indulerch}).
\end{proof}

\par For $z \in \mathbf{C}-[1,\infty)$ and $s=0,-1,-2, \dots$, there is
a closed formula for the function $\Phi(z,s,u)$. In fact, summing
the series $\Phi(z,0,u)=1+z+z^2+ \cdots$ and using
(\ref{indulerch}), we see inductively that the functions
\begin{align}
\Phi(z,0,u) &= {1 \over 1-z}, \label{racionales0} \\
\Phi(z,-1,u) &= {u \over 1-z}+ {z \over (1-z)^2}, \qquad \cdots
\label{racionales}
\end{align}
are rational, with rational coefficients, holomorphic in $z$
except for a pole at $z=1$. In particular, setting $z=-1$ and
$m=-s$, we define the $m$th {\it Euler polynomial}
\begin{equation}\label{polieuler}
E_m(x)=2 \Phi(-1,-m,x).
\end{equation}

On the other hand, if $s$ and $u$ are positive integers, then
$\Phi(z,s,u)$ is a transcendental function. For example, when
$s=u=1$ series (\ref{lerch1}) is $1+{z \over 2}+{z^2 \over
3}+\cdots$, which evidently sums to
\begin{equation}\label{z11}
\Phi(z,1,1)=-{1 \over z} \ln(1-z)={1 \over z} \textrm{Li}_1(z).
\end{equation}
For $s=2$, setting $u=1$ in (\ref{indulerch}) and writing the
right side as ${ \partial  \over  \partial z} \; z \Phi(z,2,1)$,
we find that the integral
$$\Phi(z,2,1)={1 \over z} \int_0^z \Phi(w,1,1)dw={1 \over z} \textrm{Li}_2(z)$$
is $z^{-1}$ times the classical {\it dilogarithm function}. The
identity (\ref{induu}) leads by induction to similar formulas for
$\Phi(z,1,u)$ and $\Phi(z,2,u)$ when $u=2,3, \dots$.
\par The following three examples give evaluations of $
\partial \Phi / \partial s $ that we will need.

\begin{ejem} \rm
Differentiating the relation (see, for example, \cite{sondowzeta})
\begin{equation}\label{alterzeta}
\zeta^{*}(s)=\left( 1-{2 \over 2^s} \right) \zeta(s)
\end{equation}
at $s=-1$, and using (\ref{alter}) and the values
$\zeta(-1)=-1/12$ and
\begin{equation}\label{gla}
\zeta'(-1)={1 \over 12}-\ln A,
\end{equation}
where
\begin{equation} \nonumber
A=\lim_{n \to \infty} {1^1 2^2 3^3 \cdots n^n \over n^{ {1 \over
2} n^2+ {1 \over 2} n + {1 \over 12}}  e^{ -{1 \over 4} n^2}}
\end{equation}
is the \textit{Glaisher-Kinkelin constant} (see \cite[section
2.15]{finch}), we get
\begin{equation}
{\partial \Phi \over \partial s}(-1,-1,1) = \ln {A^3 \over 2^{1/3}
e^{1/4}}.  \label{valor1}
\end{equation}
\end{ejem}

\begin{ejem} \rm
Differentiating (\ref{alterzeta}) and the functional equation of the
zeta function \cite[p. 27]{spanier}
$$ \zeta(s)=2 (2 \pi)^{s-1} \Gamma(1-s) \sin {\pi s \over 2} \zeta(1-s)$$
at $s=-2,$ and using (\ref{alter}) and the value $\zeta(-2)=0$, we
deduce that
\begin{equation}
{\partial \Phi \over \partial s}(-1,-2,1) ={7 \zeta(3) \over 4
\pi^2}, \label{valor3}
\end{equation}
where $\zeta(3)$ is \textit{Ap\'{e}ry's constant}.
\end{ejem}

\begin{ejem} \rm
Differentiating (\ref{dirichletbeta}) and the functional equation of
the Dirichlet beta function \cite[p. 27]{spanier},
$$ \beta(s)=\left( \pi \over 2 \right)^{s-1} \Gamma(1-s) \cos {\pi s \over 2} \beta(1-s)$$
at $s=-1,$ and using the value $\beta(-1)=0$, we obtain the
evaluation
\begin{equation}
{\partial \Phi \over \partial s} \left(-1,-1,{1 \over 2} \right) =
{G \over \pi}, \label{valor2}
\end{equation}
where
\begin{equation}\label{catalan}
G=\beta(2)={1 \over 4} \Phi\left( -1,2,{1 \over 2} \right)=\Re
(\Phi(i,2,1))
\end{equation}
is \textit{Catalan's constant}.
\end{ejem}

\begin{teor} \label{phidoubleseries}
For all complex $s$, and complex $z$ with $\Re(z)<1/2$,
\begin{equation}\label{gs3}
(1-z)\Phi(z,s,u)=\sum_{n=0}^{\infty} \left({-z \over
1-z}\right)^{n} \sum_{k=0}^{n} (-1)^k {n \choose k} (u+k)^{-s}.
\end{equation}
\end{teor}

\begin{proof} The right side of (\ref{gs3}) defines a holomorphic function of
$z$ on the half-plane $\Re(z)<1/2$, because the inequality implies
that $|{-z \over 1-z}|<1.$ We prove (\ref{gs3}) when $|z|<1/2,$ and
the result follows by analytic continuation.

Reversing the order of summation (for justification, see
\cite{emery}), the right side of (\ref{gs3}) is equal to
$$
\sum_{k=0}^{\infty} {(-1)^k \over (u+k)^s} \sum_{n=k}^{\infty} {n
\choose k} \left( {-z \over 1-z} \right)^n = \sum_{k=0}^{\infty}
{(-1)^k \over (u+k)^s} (1-z) (-z)^k =(1-z) \Phi(z,s,u),
$$ and the theorem follows.
\end{proof}

\subsection*{Remark} Theorem \ref{phidoubleseries} excludes the Hurwitz zeta function
$\zeta(s,u)=\Phi(1,s,u).$ However, the two cases $u=1$ and $u=1/2$
can be obtained via the relations (\ref{fzeta}), (\ref{alter}),
(\ref{alterzeta}), and $\zeta \left(s, {1/2} \right)=(2^s-1)
\zeta(s,1).$ In the case $u=1$, formula (\ref{gs3}) with $z=-1$
gives a globally convergent series for the alternating zeta function
(\ref{alter}). Then (\ref{alterzeta}) yields a formula for the
Riemann zeta function, valid for all $s \neq 1$,
$$
\zeta(s)={1 \over 1-2^{1-s}} \sum_{n=0}^{\infty} {1 \over 2^{n+1}}
\sum_{k=0}^{n} (-1)^k {n \choose k} (k+1)^{-s},
$$
which was conjectured by Knopp, proved by Hasse \cite[p.
464]{hasse}, and rediscovered by the second author
\cite{sondowzeta}.

\begin{teor}[Hasse] \label{hasse}
For $z=1$, the series $(\ref{lerch1})$ extends to a function
\begin{equation}\label{hur1}
\Phi(1,s,u)={1 \over s-1} \sum_{n=0}^{\infty} {1 \over n+1}
\sum_{k=0}^{n} (-1)^k {n \choose k} (u+k)^{1-s}
\end{equation}
defined for all complex $s \neq 1,$ with a simple pole at $s=1.$
Moreover, the series in {\rm (\ref{hur1})} converges uniformly on
compact sets in the \textit{s}-plane to an entire function.
\end{teor}
\begin{proof} We  give a sketch; for details, see \cite{hasse}.
\par First assume that $\Re(s)>1.$ Setting $z=1$ in
(\ref{intelerch}), we rewrite it as
$$
\Phi(1,s,u)={1 \over \Gamma(s)} \int_{0}^{\infty} {t \over 1-e^{-t}}
e^{-ut} t^ {s-2}dt={1 \over \Gamma(s)} \int_{0}^{\infty}
\sum_{n=0}^{\infty} {(1-e^{-t})^n \over n+1} e^{-ut} t^{s-2} dt.
$$
Interchanging the integral and sum, we expand the binomial and get
$$
\Phi(1,s,u)={1 \over \Gamma(s)} \sum_{n=0}^{\infty} {1 \over n+1}
\sum_{k=0}^{n} (-1)^k {n \choose k} \int_{0}^{\infty} e^{-(u+k)t}
t^{s-2} dt.
$$
Now (\ref{intint}) (with $k$ in place of $n$, and $s-1$ in place of
$s$) and the relation $\Gamma(s)=(s-1) \Gamma(s-1)$ yield
(\ref{hur1}) when $\Re(s)>1$.

\par For any $s \in \mathbf{C}$, the inner sum in (\ref{hur1}) with $n>0$
evidently is equal to
\begin{equation}\label{innersum}
\sum_{k=0}^{n} (-1)^k {n \choose k} (u+k)^{1-s}=(s-1)_n \int_0^1
\cdots \int_0^1 (u+x_1+ \cdots x_n)^{1-s-n} dx_1 \cdots dx_n,
\end{equation}
where the Pochhammer symbol $(a)_n$ stands for the product $a(a+1)
\cdots (a+n-1)$. It follows (see \cite{hasse} or \cite{sondow0})
that the series in (\ref{hur1}) converges as required. By analytic
continuation, this proves the theorem.
\end{proof}

\begin{ejem} \rm \label{bernoullipol}
Note that when $s$ is zero or a negative integer, the factor
$(s-1)_n$ in (\ref{innersum}) vanishes for $n>m:=1-s$, so that
\begin{equation}\label{eva0}
\sum_{k=0}^n (-1)^k {n \choose k} (u+k)^m=0 \qquad (n>m=0,1,2,
\dots).
\end{equation}
In this case the series in (\ref{hur1}) becomes a finite sum, in
fact, a polynomial with rational coefficients. Defining the $m$th
\textit{Bernoulli polynomial} by $B_0(x)=1$ and $B_m(x)=-m
\Phi(1,1-m,x)$ for $m=1,2, \dots$, we thus have the explicit formula
\begin{equation} \nonumber
B_m(x)= \sum_{n=0}^{m} {1 \over n+1} \sum_{k=0}^{n} (-1)^k {n
\choose k} (x+k)^m.
\end{equation}
In particular,
\begin{equation}\label{b10x}
B_1(x)=-\Phi(1,0,x)=x-{1 \over 2}.
\end{equation}
\end{ejem}

\begin{ejem} \rm \label{eulerpol}
Taking $z=-1$ and $s=-m$ in (\ref{gs3}), we use (\ref{eva0}) to get
the explicit formula for the $m$th Euler polynomial
(\ref{polieuler})
\begin{equation} \nonumber
E_m(x)= \sum_{n=0}^{m} {1 \over 2^n} \sum_{k=0}^{n} (-1)^k {n
\choose k} (x+k)^m.
\end{equation}
\end{ejem}

The next two results relate the analytic continuation of
$\Phi(z,s,u)$ for $z \in \mathbf{C}-[1,\infty)$ (Lemmas
\ref{philemma1} and \ref{philemma2}) to that for $z=1$ (Theorem
\ref{hasse}). (Note that $\Phi(z,s,u)$ \textit{is not continuous in
$z$ at} $z=1$: see (\ref{racionales0}), (\ref{racionales}) and
\cite[p. 30, equation (12)]{bateman}.)

\begin{coro} \label{corointdephi}
For all complex $s$,
\begin{equation}\nonumber
\int_{0}^{\infty} {\Phi \left(-z ,s,u \right) \over 1+z} dz=s
\Phi(1,s+1,u).
\end{equation}
\end{coro}

\begin{proof} Replace $z$ with $-z$ in (\ref{gs3}), multiply by
$(1+z)^{-2}$, and integrate from $z=0$ to $\infty$. Comparing the
result with (\ref{hur1}) gives the corollary.
\end{proof}

\begin{lema} \label{convertir}
If either $z \in \mathbf{C}- \{ \: (-\infty,-1] \: \bigcup \:
[1,\infty) \: \}$ and $s \in \mathbf{C}$, or $z=\pm 1$ and $s \in
\mathbf{C}- \{ 1 \}$, then
\begin{equation}\label{convertir1}
\Phi(z,s,u)=2^{-s} \left[ \Phi \left(z^2,s, {u \over 2} \right) +
z \Phi \left(z^2,s, {u+1 \over 2} \right) \right].
\end{equation}
\end{lema}
\begin{proof} If $|z|<1$  (respectively, $z= \pm 1$ and $\Re(s)>1$),
formula (\ref{convertir1}) is proved by splitting the series
(\ref{lerch1}) into two sums, one over even numbers and the other
over odd numbers, and factoring out $2^{-s}.$ The result then
follows by analytic continuation in $z$ (respectively, in $s).$
\end{proof}

\par As an application, (\ref{convertir1}) with $z=\pm 1$ and
$s=0,-1,-2, \dots$ implies the classical relations among the
Bernoulli and Euler polynomials \cite[equations 19:5:6 and
20:3:3]{spanier}.

\begin{ejem} \rm
Differentiating (\ref{convertir1}) with respect to $s$ at $s=0$,
and using (\ref{b10x}) and the formula ${\partial \Phi \over
\partial s}(1,0,u)=\ln {\Gamma(u) \over \sqrt {2 \pi}}$ (see \cite[p. 26]{bateman}), we get
\begin{equation}\label{valor0}
{\partial \Phi \over \partial s} (-1,0,u)=\ln {\Gamma \left( {u
\over 2} \right) \over \Gamma \left( {u+1 \over  2} \right)
\sqrt{2}}.
\end{equation}
\end{ejem}

\section{Double Integrals} \label{doubl}
Using the integral analytic continuation of Lerch's function in
Lemmas \ref{philemma1} and \ref{philemma2}, we evaluate certain
types of double integrals.

\begin{teor} \label{teorintegrals1}
If either $z \in \mathbf{C}-[1,\infty)$ and $\Re(s)>-2$, or $z=1$
and $ \Re(s)>-1$, then
\begin{align}
\int_0^1 \int_0^1 {x^{u-1} y^{v-1} \over 1-xyz} (-\ln xy)^{s}dxdy
&= \Gamma(s+1)
{\Phi(z,s+1,v)-\Phi(z,s+1,u) \over u-v}, \label{teor11} \\
\int_0^1 \int_0^1 {(xy)^{u-1} \over 1-xyz} (-\ln xy)^{s}dxdy &=
{\Gamma(s+2)} \Phi(z,s+2,u). \label{teor13}
\end{align}
\end{teor}
\begin{proof} The integrals define holomorphic functions of $z$ and/or $s$
under the conditions stated. Making the change of variables $x=X/Y$,
$y=Y$ in integral (\ref{teor11}), it is equal to
$$
\int_0^1 {X^{u-1} \over 1-Xz} (-\ln X)^{s} \int_{X}^{1}
Y^{v-u-1}dY dX ={1 \over u-v} \int_0^1 {X^{v-1}-X^{u-1} \over
1-Xz} (-\ln X)^{s}dX.
$$
Substituting $t=-\ln X$ and using (\ref{intelerch}) (with $s+1$ in
place of $s$), we obtain equality (\ref{teor11}) when $\Re(s)>0$ and
either $|z|<1$ or $z=1$; the result then follows by analytic
continuation in $z$ or $s$. To prove (\ref{teor13}), let $v \to u$
in (\ref{teor11}), and use the identity (\ref{lerchpar}).
\end{proof}

\begin{ejem} \rm \label{ejem4.1}
Setting $z=-i$, $u=1$ and $s=0$ in (\ref{teor13}), we use
(\ref{lerch1}), (\ref{alterzeta}), (\ref{catalan}), and Euler's
formula $\zeta(2)=\pi^2/6$ to get
$$\int_0^1 \int_0^1 {1 \over 1+xyi} dx dy=G-{\pi^2 i \over 48}.$$
\end{ejem}

\par The next three examples use the substitution $x \to x^2$,
$y \to y^2$.

\begin{ejem} \rm \label{ejem4.2}
Setting $z=-1$, $u=1$, $v=1/2$ and $s=1$ in (\ref{teor11}), we get
$$\int_0^1 \int_0^1 {-x \ln xy \over 1+x^2y^2} dx dy=
G - {\pi^2 \over 48}.$$
\end{ejem}

\begin{ejem} \rm \label{int2pi2}
Setting $z=1$, $u=1$, $v=1/2$ and $s=1$ in (\ref{teor11}) gives
$$\int_0^1 \int_0^1 {-x \ln xy \over 1-x^2y^2} dx dy=
{\pi^2 \over 12}.$$
\end{ejem}

\begin{coro} \label{coro3.1}
If either $z \in \mathbb{C}-[1,\infty)$ and $n \geq -1$, or $z=1$
and $n \geq 0,$ then
$$\int_0^1 \int_0^1 {(-\ln xy)^{n} \over 1-xyz} dx dy= {(n+1)!  {\rm Li}_{n+2}(z) \over z}.$$
In particular, for $z \in \mathbb{C}-[1,\infty)$
$$\int_0^1 \int_0^1 {-1 \over (1-xyz) \ln xy} dx dy= {-\ln(1-z) \over z}.$$
\end{coro}
\begin{proof} Let $u=1$ in (\ref{teor13}) and set $s=n$. Then use
(\ref{polyloga}) and (\ref{z11}).
\end{proof}

\begin{ejem} \rm \label{int2ln2}
Take $z=1/2$. Using Euler's and Landen's values of the di- and
trilogarithm at $1/2$ (see \cite[pp. 1-2]{lewin}), we see that
(compare Example \ref{intsomos})
\begin{align*}
\int_0^1 \int_0^1 {-1 \over (2-xy) \ln xy} dx dy &= \ln 2, &
\int_0^1 \int_0^1 {1 \over 2-xy} dx dy &= {\pi^2 \over 12}-{\ln^2
2 \over 2},
\end{align*}
\begin{equation*}
\int_0^1 \int_0^1 {-\ln xy \over 2-xy} dx dy = {7 \zeta (3) \over
4}-{\pi^2 \ln 2 \over 6}+{\ln^3 2 \over 3}.
\end{equation*}
\end{ejem}

\begin{ejem} \rm \label{int2lngolden}
Take $z=\varphi^{-1}$, $\varphi^{-2}$, $-\varphi$, and
$-\varphi^{-1}$, where $\varphi=(1+\sqrt{5})/2$ is the {\it golden
ratio}. Using the relation $1+\varphi=\varphi^2$ to simplify
$\ln(1-z)$, and using Landen's values for $\textrm{Li}_2(z)$ and
$\textrm{Li}_3(\varphi^{-2})$ (see \cite[equations (1.20), (1.21),
and (6.13)]{lewin2}), we obtain
\begin{align*}
\int_0^1 \int_0^1 {-1 \over (\varphi-xy) \ln xy} dx dy &= 2 \ln
\varphi, & \int_0^1 \int_0^1 {1 \over \varphi-xy } dx dy &= {\pi^2
\over 10}-\ln^2 \varphi, \\
\int_0^1 \int_0^1 {-1 \over (\varphi^2-xy) \ln xy} dx dy &= \ln
\varphi, &  \int_0^1 \int_0^1 {1 \over \varphi^2-xy } dx dy  &=
{\pi^2 \over 15} -\ln^2 \varphi, \\
\int_0^1 \int_0^1 {-1 \over (1+\varphi xy) \ln xy} dx dy &= \frac{2
\ln \varphi}{\varphi}, & \int_0^1 \int_0^1 {1 \over 1+\varphi xy }
dx dy &= {\pi^2
\over 10 \varphi}+\frac{\ln^2 \varphi}{\varphi}, \\
\int_0^1 \int_0^1 {-1 \over (\varphi+xy) \ln xy} dx dy &= \ln
\varphi, &  \int_0^1 \int_0^1 {1 \over \varphi+xy } dx dy  &= {\pi^2
\over 15} -\frac{\ln^2 \varphi}{2},
\end{align*}
\begin{equation*}
\int_0^1 \int_0^1 {-\ln xy \over \varphi^2-xy} dx dy = {8 \zeta
(3) \over 5}-{4 \pi^2 \ln \varphi \over 15}+{4 \ln^3 \varphi \over
3}.
\end{equation*}
\end{ejem}

\begin{ejem}\label{ramanujandylogs} \rm
Using Ramanujan's values \cite[Part IV, p. 324]{ramanujan} for ${\rm
Li}_2(-1/8)+{\rm Li}_2(1/9)$, ${\rm Li}_2(-1/2)+\frac{1}{6} {\rm
Li}_2(1/9)$, and ${\rm Li}_2(1/3)-\frac{1}{6}{\rm Li}_2(1/9)$, and
substituting $x,y \to x^2, y^2$ in our double integral for the last
${\rm Li}_2(1/9)$, we get, respectively,
\begin{align}
\int_0^1 \int_0^1 \frac{1-2xy}{(8+xy)(9-xy)} dx dy &=\frac{1}{2}
\ln^2 \frac{9}{8}, \nonumber \\
\int_0^1 \int_0^1 \frac{52-7xy}{(2+xy)(9-xy)} dx dy &=\frac{\pi^2}{3}+3 \ln^2 2 +2 \ln^2 3-6 \ln 2 \ln 3, \nonumber \\
\int_0^1 \int_0^1 \frac{9+xy}{9-x^2 y^2} dx dy &=
\frac{\pi^2}{6}-\frac{\ln^2 3}{2}. \nonumber
\end{align}
\end{ejem}

\begin{ejem} \rm
In Bailey, Borwein, and Plouffe's \cite{bailey} dilogarithm "ladder"
$6 {\rm Li}_2(1/2)-6{\rm Li}_2(1/4)-2{\rm Li}_2(1/8)+{\rm
Li}_2(1/64)=\zeta(2)$, we replace ${\rm Li}_2(1/2^k)$ with a double
integral in which we substitute $x,y \to x^k, y^k,$ for $k=1,2,3,$
and $6$, obtaining
\[ \int_0^1 \int_0^1 \frac{(4+2xy+x^2 y^2)(4-8xy+x^2 y^2)}{64-x^6y^6} dx dy=\frac{\pi^2}{72}. \]
\end{ejem}

\begin{coro} \label{coro3.2}
We have
\begin{align}
\int_0^1 \int_0^1 {(-\ln xy)^s \over 1-xy} dxdy &= \Gamma(s+2)
\zeta(s+2) \qquad & (\Re(s)>-1), \label{cozeta} \\
\int_0^1 \int_0^1 {(-\ln xy)^s \over 1+xy} dxdy &= \Gamma(s+2)
\zeta^{*}(s+2) \qquad & (\Re(s)>-2), \label{coalter} \\
\int_0^1 \int_0^1 {(-\ln xy)^s \over 1+x^2y^2} dxdy &= \Gamma(s+2)
\beta(s+2) \qquad & (\Re(s)>-2). \label{cobeta}
\end{align}
\end{coro}

\begin{proof} In (\ref{teor13}) take $z=1$, $u=1$; $z=-1$, $u=1$; and $z=-1$, $u=1/2$; then
use relations (\ref{fzeta}), (\ref{alter}) and
(\ref{dirichletbeta}), respectively. After using
(\ref{dirichletbeta}), substitute $x,y \to x^2,y^2$.
\end{proof}

\par For integer $s \geq 0,$ formula (\ref{cozeta}) is due to Hadjicostas
\cite{hadjicostas0}.

\begin{ejem} \rm \label{int2beukers}
Taking $s=0$ and $1$ in (\ref{cozeta}), we recover Beukers's
integrals \cite{beukers}
\begin{equation}\label{beukers}
\int_0^1 \int_0^1 {1 \over 1-xy} dxdy= \zeta(2), \qquad \int_0^1
\int_0^1 {-\ln xy \over 1-xy} dxdy=2 \zeta(3).
\end{equation}
\end{ejem}

\begin{ejem} \rm \label{int2picub}
Taking $s=-1,0$, and $1$ in (\ref{cobeta}), and using
(\ref{catalan}) and the values $\beta(1)=\pi/4$ and
$\beta(3)=\pi^3/32$ (see \cite[p. 29]{spanier}), we get
$$ \int_0^1 \int_0^1 {-1 \over (1+x^2y^2) \ln xy} dx dy =
{\pi \over 4}, \qquad \int_0^1 \int_0^1 {1 \over 1+x^2y^2} dx dy=G,
$$
$$
\int_0^1 \int_0^1 {-\ln xy \over 1+x^2y^2} dx dy=\frac{\pi^3}{16}.
$$
\end{ejem}

\begin{ejem}\label{ejemplozetazeros} \rm
Let $\rho$ be a non-trivial zero of the zeta function (\ref{fzeta}).
Differentiate (\ref{coalter}) at $s=\rho-2$ and use
(\ref{alterzeta}) to get (compare Example \ref{non-trivial})
$$\int_0^1 \int_0^1 {(-\ln xy)^{\rho-2} \over 1+xy}  \ln (-\ln xy) dxdy=
(1-2^{3-\rho}) \Gamma(\rho) \zeta'(\rho).$$
\end{ejem}

\begin{coro}\label{corolegendre-chi}
If either $z \in \mathbf{C}-((-\infty,-1] \bigcup [1,\infty))$ and
$\Re(s)>-2$, or $z \pm 1$ and $ \Re(s)>-1$, then
\[ \int_0^1 \int_0^1 \frac{(-\ln xy)^s}{1-x^2 y^2 z^2}  dx dy=\Gamma(s+2) \frac{\chi_{s+2}(z)}{z}, \]
where $\chi_s(z)$ is Legendre's chi function {\rm
(\ref{legendrechi})}.
\end{coro}
\begin{proof}
In (\ref{teor13}), take $u=1/2$ and substitute $x, y, z \to x^2,
y^2, z^2$. Then use (\ref{legendrechi}).
\end{proof}

\begin{ejem}\label{ejemlegendre-chi} \rm
Set $s=0$, and take $z=\tan \frac{\pi}{8}=\sqrt{2}-1$ and
$z=\varphi^{-3}=\sqrt{5}-2$. Using Landen's values \cite[section
1.8.2]{lewin2} for $\chi_2(\tan \frac{\pi}{8})$ and
$\chi_2(\varphi^{-3})$, we obtain, respectively,
\begin{align}
\int_0^1 \int_0^1 \frac{1}{1-x^ 2 y^2 \tan^2 \frac{\pi}{8}}dx dy
&=\frac{\pi^2}{16 \tan \frac{\pi}{8}}- \frac{\ln^2 \tan
\frac{\pi}{8}}{4 \tan \frac{\pi}{8}}, \nonumber \\
\int_0^1 \int_0^1 \frac{1}{\varphi^6-x^2 y^2} dx dy &=
\frac{\pi^2}{24 \varphi^3}-\frac{3 \ln^2 \varphi}{4 \varphi^3}.
\nonumber
\end{align}
\end{ejem}

\begin{coro} \label{coroteor12}
For $z \in \mathbf{C} -[1,\infty)$,
\begin{align}
\int_0^1 \int_0^1 {-x^{u-1} y^{v-1} \over (1-xyz) \ln xy} dxdy &=
{1 \over u-v} \left[{\partial \Phi \over
\partial s}(z,0,v)- {\partial \Phi \over  \partial s} (z,0,u)
\right], \label{teor12} \\
\int_0^1 \int_0^1 {-(xy)^{u-1} \over(1-xyz) \ln xy} dxdy &= \Phi
(z,1,u). \label{teore12}
\end{align}
\end{coro}

\begin{proof} Substitute $(s+1)^{-1} \Gamma(s+2)$ for $\Gamma(s+1)$ in
(\ref{teor11}) and let $s \to -1^{+}$. Since $z \in
\mathbf{C}-[1,\infty)$, in view of (\ref{racionales0}) the result is
(\ref{teor12}). Letting $v \to u$ in (\ref{teor12}), and using
identity (\ref{lerchpar}), we arrive at $\rm (\ref{teore12})$.
\end{proof}

\begin{ejem} \rm
Set $z=0$. Using (\ref{lerch1}), we get
$$\int_0^1 \int_0^1 {x^{u-1} y^{v-1} \over -\ln xy}
dxdy={1 \over u-v} \ln {u \over v}, \qquad \qquad \int_0^1
\int_0^1 {(xy)^{u-1} \over -\ln xy} dxdy={1 \over u}.
$$
\end{ejem}

\begin{ejem} \rm \label{intsomos}
Set $z=1/2$, $u=2$, and $v=1$ in (\ref{teor12}). Using the
derivative of (\ref{induu}) with respect to $s$ at $s=0$, we get
(compare Example \ref{int2ln2})
$$
\int_0^1 \int_0^1 {-x \over (2-xy) \ln xy} dx dy= \ln \sigma,
$$
where
\begin{equation}\label{somos}
\sigma= \sqrt{1 \sqrt {2 \sqrt {3 \cdots }}}=1^{1/2} 2^{1/4} 3^{1/8}
\cdots = \exp \left[ -{1 \over 2} {\partial \Phi \over
\partial s} \left( {1 \over 2},0,1 \right) \right]
\end{equation}
is one of \textit{Somos's quadratic recurrence constants}
\cite{somos} (see \cite[p. 446]{finch}, which uses the notation
$\gamma$ instead of $\sigma;$ see also \cite[p. 348]{ramanujan2}).
\end{ejem}

\begin{coro} \label{corointgammas}
For $u>0$ and $v>0$,
\begin{align}
\int_0^1 \int_0^1 {-x^{u-1}y^{v-1} \over (1+xy) \ln xy} dx dy &=
{1 \over u-v} \ln { \Gamma ( {v \over 2} ) \; \Gamma({u+1 \over2})
\over {\Gamma({u \over 2}) \; \Gamma({v+1 \over 2})}}, \label{icoro1} \\
\int_0^1 \int_0^1 {-(x y)^{u-1} \over (1+xy) \ln xy} dx dy &= {1
\over 2} \left[ \psi \left( {u+1 \over 2} \right)- \psi \left( {u
\over 2} \right) \right], \label{icoro2}
\end{align}
where $\psi(u)= {d \over du} \ln \Gamma(u)=\Gamma'(u) / \Gamma(u)$
is the {\rm digamma function}.
\end{coro}
\begin{proof}
To prove (\ref{icoro1}), set  $z=-1$ in (\ref{teor12}) and use
(\ref{valor0}). To prove (\ref{icoro2}), let $v \to u$ in
(\ref{icoro1}).
\end{proof}

\begin{ejem} \rm \label{intlnpi}
Setting $u=1$ in (\ref{icoro2}) and $u=2$ and $v=1$ in
(\ref{icoro1}) gives integrals for $\ln 2$ and $\ln (\pi/2),$
respectively. Their sum and difference are integrals for $\ln \pi$
and the "alternating Euler constant" $\ln (4/\pi)$ (compare Example
\ref{eulersondow} and see \cite{sondow1} and \cite{sondow2}):
\[
\int_0^1 \int_0^1 {1+x \over (1+xy) (-\ln xy)} dx dy = \ln \pi,
\qquad \int_0^1 \int_0^1 {1-x \over (1+xy) (-\ln xy)} dx dy =\ln
\frac{4}{\pi}.
\]
\end{ejem}

\begin{ejem} \rm \label{int2lnGAMMA3over4}
Set $u=1$ and $v=1/2$ in (\ref{icoro1}), and substitute $x \to
x^2$, $y \to y^2$. Using the reflection formula $\Gamma(t)
\Gamma(1-t) =\pi / \sin \pi t$ with $t=1/4$, we get
$$\int_0^1 \int_0^1 {-x \over (1+x^2 y^2) \ln xy} dx dy
=\ln {\sqrt{2 \pi} \over \Gamma \left( 3/4 \right) ^2}.
$$
\end{ejem}

\begin{coro} \label{coro3.5}
For $u>0$ and $v>0$,
\begin{align}
\int_0^1 \int_0^1 {x^{u-1} y^{v-1} \over 1-xy} dxdy
&= {\psi(u)-\psi(v) \over u-v}, \label{teo} \\
\int_0^1 \int_0^1 {(xy)^{u-1} \over 1-xy} dxdy &= \psi'(u).
\label{teoo}
\end{align}
\end{coro}

\begin{proof}
To prove (\ref{teo}), set $z=1$ in (\ref{teor11}), let $s \to 0$,
and use the formula \cite[p. 271]{whittaker}
\begin{equation}\label{ww}
\lim_{s \to 1} \left( \Phi(1,s,u)-{1 \over s-1} \right)=-\psi(u).
\end{equation}
To prove (\ref{teoo}), let $v \to u$ in (\ref{teo}).
\end{proof}

\begin{ejem} \rm \label{int2pioversqrt3}
Take $u=1/3$ and $v=2/3$ in (\ref{teo}). Since
$\psi(2/3)-\psi(1/3)=\pi/\sqrt{3}$ (see \cite[equation
44:7:1]{spanier}), the change of variables $x \to x^3$ and $y \to
y^3$ yields
$$\int_0^1 \int_0^1 {y \over 1-x^3 y^3} dxdy = {\pi \over 3 \sqrt{3}}.$$
\end{ejem}

\begin{ejem} \rm
Taking $u=1/2$ in (\ref{teoo}) and using the value $\psi'(1/2)=\pi^2
/2$ (see \cite[p. 427]{spanier}), we substitute $x \to x^2$, $y \to
y^2$ and obtain
$$\int_0^1 \int_0^1 {1 \over 1-x^2 y^2} dxdy = {\pi^2 \over 8}.$$
\end{ejem}

\begin{coro}
If $\Re(s)>-2$, then
\begin{align}
\int_0^1 \int_0^1 x^{u-1}y^{v-1} (-\ln xy)^s dxdy &=\Gamma(s+1)
{v^{-s-1}-u^{-s-1} \over u-v}, \label{coraz0} \\
\int_0^1 \int_0^1 (xy)^{u-1} (-\ln xy)^s dxdy
&=\Gamma(s+2)u^{-s-2}. \label{coroz0}
\end{align}
\end{coro}

\begin{proof}
Set $z=0$ in (\ref{teor11}) and (\ref{teor13}), and use
(\ref{lerch1}).
\end{proof}

\begin{ejem} \rm
Taking $s=-3/2$, $u=2$ and $v=1$ in (\ref{coraz0}) gives
$$\int_0^1 \int_0^1 {x \over (-\ln xy)^{3/2}} dxdy=2(\sqrt 2 -1) \sqrt \pi.$$
\end{ejem}

\begin{ejem} \rm \label{int2GAMMA3over4}
Taking $u=1$ and $s=-3/2$ or $s=-5/4$ in (\ref{coroz0}) gives
$$\int_0^1 \int_0^1 {1 \over  \left( -\ln xy
\right)^{3/2}} dx dy={\sqrt \pi}, \qquad \int_0^1 \int_0^1{1 \over
(-\ln xy)^{5/4}} dxdy= \Gamma \left( 3/4 \right).$$
\end{ejem}

\begin{coro}
If  $z \in \mathbf{C} -[1,\infty)$ and $\Re(s)>-2$, then
\begin{multline}\label{teor21}
\int_0^1 \int_0^1 {x^{u-1} y^{v-1} \over (1-xyz)^2} (-\ln
xy)^{s}dxdy \\ ={\Gamma(s+1) \over u-v} [ (1-v)
\Phi(z,s+1,v)+\Phi(z,s,v)+(u-1) \Phi(z,s+1,u)-\Phi(z,s,u)],
\end{multline}
\vskip -0.6cm
\begin{multline}\label{teor23}
\int_0^1 \int_0^1 {(xy)^{u-1} \over (1-xyz)^2} (-\ln xy)^{s}dxdy =
\Gamma(s+2) [(1-u) \Phi(z,s+2,u)+\Phi(z,s+1,u)].
\end{multline}
\end{coro}
\begin{proof} Differentiate (\ref{teor11}) with respect to $z$ and apply identity (\ref{indulerch}).
Replacing $u$ with $u-1$, and $v$ with $v-1$, we arrive at
(\ref{teor21}). To prove (\ref{teor23}), let $v \to u$ in
(\ref{teor21}) and use identity (\ref{lerchpar}).
\end{proof}

\begin{ejem} \rm
Set $z=-1$, $u=1/2$, $v=1$, and $s=2$ in (\ref{teor21}).
Substituting $x \to x^2$, $y \to y^2$, and using (\ref{alter}),
(\ref{dirichletbeta}), (\ref{catalan}), and the values
$\zeta^{*}(2)=\pi^2/12$ and $\beta(3)=\pi^3/32$, we get
$$\int_0^1 \int_0^1 {x \ln^2 xy \over (1+x^2 y^2)^2} dx dy=G-{\pi^2 \over 48}+{\pi^3 \over 32}.$$
\end{ejem}

\begin{ejem} \rm \label{intzeta5}
Set $z=-1$, $u=1$ and $s=4$ in (\ref{teor23}) to get
$$\int_0^1 \int_0^1 {\ln^4 xy \over (1+x y)^2} dx dy={225 \over 2} \zeta(5).$$
\end{ejem}

\begin{ejem} \rm
Setting $z=-1$, $u=1/2$ and $s=-1$ in (\ref{teor23}), and
substituting $x \to x^2$, $y \to y^2$, we get
$$\int_0^1 \int_0^1 {-1 \over (1+x^2 y^2)^2 \ln xy} dx dy={\pi+2 \over 8}.$$
\end{ejem}

\begin{coro}
If  $z \in \mathbf{C} -[1,\infty)$, then
\begin{multline} \nonumber
\int_0^1 \int_0^1 {-x^{u-1} y^{v-1} \over (1-xyz)^2 \ln xy} dxdy
\\ ={1\over u-v} \left[ (1-v) {\partial \Phi \over \partial
s}(z,0,v)+ {\partial \Phi \over \partial s}(z,-1,v)-(1-u)
{\partial \Phi \over \partial s}(z,0,u)- {\partial \Phi \over
\partial s}(z,-1,u) \right].
\end{multline}
\end{coro}
\begin{proof} Substitute $(s+1)^{-1} \Gamma(s+2)$ for $\Gamma(s+1)$ in $\rm (\ref{teor21})$
and let $s \to -1^{+}$ . \end{proof}

\begin{ejem} \rm
Setting $z=-1$, $u=2$ and $v=1$, we use (\ref{valor0}),
(\ref{valor1}), and (\ref{induu}) to get (compare \cite[equation
(9)]{sondow1})
$$\int_0^1 \int_0^1 {-x \over (1+xy)^2 \ln xy} dx dy=
\ln {A^6 \over 2^{1/6} \sqrt{\pi e}},
$$
where $A$ is the Glaisher-Kinkelin constant.
\end{ejem}

\begin{ejem} \rm \label{int2catoverpi}
Set $z=-1$, $u=2$ and $v=1/2$. Using (\ref{valor2}) and
substituting $x \to x^2$ and $y \to y^2$, we get
$$\int_0^1 \int_0^1 {-x^2 \over (1+x^2 y^2)^2 \ln xy} dx dy={G \over \pi}.$$
\end{ejem}

\par For the proof of the final theorem in this section, we require
a lemma.

\begin{lema}
If $H_{n,r}=1+{1 \over 2^r}+ \cdots +{1 \over n^r}$ is the $n$th
{\rm harmonic number of order} $r$ and $H_n=H_{n,1}$, and if
$|z|<1$, then
\begin{equation}\label{ramalema}
\sum_{n=1}^{\infty} \frac{H_{n,2}}{n}z^n+2 \sum_{n=1}^{\infty}
\frac{H_n}{n^2} z^n =3 {\rm Li}_3(z)-{\rm Li}_2(z) \ln(1-z).
\end{equation}
\end{lema}
\begin{proof}
We have
\begin{align}
\sum_{n=1}^{\infty} \frac{H_n}{n^2} z^n-{\rm Li}_3(z)=&
\sum_{n=2}^{\infty} \sum_{k=1}^{n-1} \frac{z^n}{n^2 k}
=\sum_{k=1}^{\infty} \sum_{n=k+1}^{\infty} \frac{z^n}{n^2 k}=
\sum_{k=1}^{\infty} \sum_{j=1}^{\infty} \frac{z^{k+j}}{(k+j)^2 k}
\nonumber \\ =& \sum_{k=1}^{\infty} \frac{1}{k} \sum_{j=1}^{\infty}
\int_0^z \frac{1}{t} \int_0^t u^{k+j-1} du dt= \sum_{k=1}^{\infty}
\frac{1}{k}\int_0^z \frac{1}{t} \int_0^t  \frac{u^k}{1-u} du dt \nonumber \\
=& \int_0^z \frac{1}{t} \int_0^t  \frac{-\ln(1-u)}{1-u} du dt =
\frac{1}{2} \int_0^z \frac{\ln^2(1-t)}{t}dt \nonumber
\end{align}
and in  a similar way
\[
\sum_{n=1}^{\infty} \frac{H_{n,2}}{n} z^n-{\rm Li}_3(z) = \int_0^z
\frac{{\rm Li}_2(t)}{1-t}dt. \nonumber
\]
Adding twice the first identity to the second, we obtain
\[ \sum_{n=1}^{\infty} \frac{H_{n,2}}{n}z^n+2 \sum_{n=1}^{\infty} \frac{H_n}{n^2} z^n-3 {\rm Li}_3(z)=
\int_0^z \left[ \frac{{\rm Li}_2(t)}{1-t}+\frac{\ln^2(1-t)}{t}
\right] dt=-{\rm Li}_2(z) \ln(1-z), \] the evaluation of the
integral being easily checked by differentiation.
\end{proof}

\begin{teor} \label{corointrama}
For $z \in \mathbf{C}-[1, \infty)$, the following analytic
continuations hold:
\begin{multline}\label{corohar1}
\int_0^1 \int_0^1 {-\ln (1-xz) \over 1-xy} dx dy \\
= \left[{1 \over 2}\ln z \ln(1-z)+{\rm Li}_2(1-z)
\right]\ln(1-z)+{\rm Li}_3(z)-{\rm Li}_3(1-z)+\zeta(3),
\end{multline}
\vskip -0.6cm
\begin{multline}\label{corohar2}
\int_0^1 \int_0^1 {-\ln (1-xyz) \over 1-xy} dx dy = {\rm Li}_2(1-z)
\ln(1-z)-{\rm Li}_3(z)-2 {\rm Li}_3(1-z)+2 \zeta(3).
\end{multline}
\end{teor}

\begin{proof}
By analytic continuation, it suffices to prove (\ref{corohar1}) and
(\ref{corohar2}) when $|z|<1.$ \par Expanding $-\ln(1-xz)$ in a
series, and using (\ref{teo}) and the identity
$\psi(n+1)-\psi(1)=H_n$ (see \cite[equation 44:5:5]{spanier}), we
see that
\begin{equation}\nonumber
\int_0^1 \int_0^1 {-\ln (1-xz) \over 1-xy} dx dy=\sum_{n=1}^{\infty}
{z^n \over n} \int_0^1 \int_0^1 {x^n \over 1-xy} dx dy
=\sum_{n=1}^{\infty} {H_n \over n^2} z^n.
\end{equation}
Ramanujan's summation of the last series \cite[p. 251, Entry 9(i)
and p. 259, equation (12.1)]{ramanujan} completes the proof of
(\ref{corohar1}).
\par Similarly, using (\ref{teoo}) and the identity
$\psi'(n+1)={1 \over 6} \pi^2-H_{n,2}$ (see \cite[equation
44:12:6]{spanier}), we find that
\begin{equation}\nonumber
\int_0^1 \int_0^1 {-\ln (1-xyz) \over 1-xy} dx dy=-{\pi^2 \over 6}
\ln(1-z)-\sum_{n=1}^{\infty} {H_{n,2} \over n} z^n.
\end{equation}
Now apply (\ref{ramalema}) with $\sum_{n=1}^{\infty} H_n n^{-2} z^n$
replaced by the right side of (\ref{corohar1}). Simplifying with
Euler's identity ${\rm Li}_2(z)+{\rm Li}_2(1-z)=\frac{1}{6}\pi^2-\ln
z \ln(1-z)$ (see \cite[p. 1]{lewin}), we arrive at (\ref{corohar2}).
\end{proof}

\begin{ejem}\label{di-trilog} \rm
First take $z=1/2$; write $\ln\left(1-{x \over 2} \right)$ as
$\ln(2-x)-\ln 2$ in (\ref{corohar1}), rewrite (\ref{corohar2})
similarly, and apply the first equation in (\ref{beukers}). Then
take $z=-1$. Using the values of ${\rm Li}_2(1/2)$ and ${\rm
Li}_3(1/2)$ (see Example \ref{int2lngolden}), and the inversion
formulas for ${\rm Li}_2(y)+ {\rm Li}_2(1/y)$ and ${\rm
Li}_3(y)-{\rm Li}_3(1/y)$ (see \cite[equations (1.10) and
(6.7)]{lewin2}) with $y=2$, we obtain
\begin{align}
\int_0^1 \int_0^1 {\ln(2-x) \over 1-xy} dxdy &=\int_0^1 \int_0^1
{\ln(1+xy) \over 1-xy} dx dy={\pi^2 \ln 2 \over 4}-\zeta(3), \nonumber \\
\int_0^1 \int_0^1 {\ln(2-xy) \over 1-xy} dxdy &=\int_0^1 \int_0^1
{\ln (1+x) \over 1-xy} dx dy ={5 \over 8} \zeta(3). \nonumber
\end{align}
\end{ejem}

\begin{ejem} \rm
Letting $z \to 1^{-}$, we deduce that
\begin{equation} \nonumber
\int_0^1 \int_0^1 {-\ln (1-x) \over 1-xy} dx dy  = 2 \int_0^1
\int_0^1 {-\ln (1-xy) \over 1-xy} dx dy  = 2 \zeta(3).
\end{equation}
\end{ejem}

\section{A Generalization of Hadjicostas's Formula} \label{hadji}
Hadjicostas \cite{hadjicostas2} asked for a generalization of the
double integral formulas in \cite{hadjicostas} and \cite{sondow1}
for the Riemann and alternating zeta functions (Examples
\ref{doublehadji} and \ref{doublesondow}). To fill the bill, we
multiply the integrand of (\ref{teor13}) by $1-x$, and obtain a
formula valid on a larger half-plane of the complex $s$-plane.

\begin{teor} \label{teorintegrals2}
If either $z \in \mathbf{C}-[1, \infty)$ and $\Re(s)>-3$, or $z=1$
and $\Re(s)>-2$, then
\begin{multline} \label{hadjiteor}
\int_0^1 \int_0^1 {1-x \over 1-xyz} (xy)^{u-1} (-\ln xy)^{s}dxdy
\\ =\Gamma(s+2) \left[ \Phi(z,s+2,u)+{(1-z)
\Phi(z,s+1,u)-u^{-s-1} \over z (s+1) } \right].
\end{multline}
\end{teor}

\begin{proof} The integral in (\ref{hadjiteor}) defines a function which
is holomorphic in $s$, when $\Re(s)>-3$ if $z \in \mathbf{C}-[1,
\infty)$, and when $\Re(s)>-2$ if $z=1$. We prove (\ref{hadjiteor})
with $\Re(s)>0$ and the result then follows by analytic
continuation.

Replace $u$ with $u+1$ in (\ref{teor11}) and set $v=u$.
Subtracting the result from (\ref{teor13}) gives
\begin{multline}
\int_0^1 \int_0^1 {(xy)^{u-1}-x^u y^{u-1} \over 1-xyz} (-\ln
xy)^{s}dxdy  \\ = \Gamma(s+2) \Phi(z,s+2,u)+ \Gamma(s+1) \left[
\Phi(z,s+1,u+1)- \Phi(z,s+1,u) \right]. \label{pru1}
\end{multline}
Using identity (\ref{induu}), we obtain (\ref{hadjiteor}), and the
theorem follows.
\end{proof}

\begin{ejem} \rm \label{doublehadji}
The case $z=u=1$ is Hadjicostas's formula \cite{chapman},
\cite{hadjicostas} (see also \cite{sondow1})
\begin{equation}\label{hadj}
\int_0^1 \int_0^1 {1-x \over 1-xy} (-\ln xy)^s dxdy=\Gamma(s+2)
\left [ \zeta(s+2)-{1 \over s+1} \right] \quad (\Re(s)>-2).
\end{equation}
\end{ejem}

\begin{ejem} \rm \label{doublesondow}
Taking $z=-1$ and $u=1$ gives the analogous formula \cite{sondow1}
$$\int_0^1 \int_0^1 {1-x \over 1+xy} (-\ln xy)^s dxdy=
\Gamma(s+2) \left [ \zeta^{*}(s+2)+{1-2 \zeta^{*}(s+1) \over s+1}
\right] \quad (\Re(s)>-3).$$
\end{ejem}

\begin{ejem} \label{non-trivial} \rm
Let $\rho$ be a non-trivial zero of $\zeta(s)$. Differentiating
(\ref{hadj}) with respect to $s$ at $s=\rho-2$, we get (compare
Example \ref{ejemplozetazeros})
$$ \int_0^1 \int_0^1 {1-x \over 1-xy} (-\ln xy)^{\rho-2} \ln (-\ln
xy) dx dy=\Gamma(\rho) \left[ \zeta'(\rho) + {1 \over (\rho-1)^2}
- {\psi(\rho) \over \rho-1} \right].$$
\end{ejem}

\begin{coro} \label{corointpsi}
For $u>0$,
\begin{equation} \nonumber
 \int_0^1 \int_0^1 {1-x \over (1-xy) (-\ln xy)} (xy)^{u-1}
 dxdy =\ln u -\psi(u).
\end{equation}
\end{coro}
\begin{proof} Set $z=1$ in (\ref{hadjiteor}), let $s \to -1^{+},$ and use
(\ref{ww}). \end{proof}

\begin{ejem} \rm \label{eulersondow}
Setting $u=1$ and using the value \cite[p. 427]{spanier}
\begin{equation}\label{psi1}
\psi(1)=-\gamma,
\end{equation}
where $\gamma=\lim_{n \to \infty} \left(1+{1 \over 2}+ \cdots + {1
\over n} -\ln n \right)$ is \textit{Euler's constant}, we recover
the formula in \cite{sondow0}, \cite{sondow1}, and \cite{sondow2}
$$\int_0^1 \int_0^1 {1-x \over (1-xy) (-\ln xy)} dxdy =\gamma.$$
\end{ejem}

\begin{coro}
For $z \in \mathbf{C}-[1, \infty)$,
\begin{multline} \nonumber
\int_0^1 \int_0^1 {1-x \over (1-xyz) \ln^2 xy}  (xy)^{u-1} dxdy
\\ = {\partial  \Phi \over \partial s}(z,0,u)+ {1 \over z} \left[ (z-1)
{\partial \Phi \over \partial s} (z,-1,u)-u \ln u \right]+{1 \over
z-1}.
\end{multline}
\end{coro}
\begin{proof} Let $s \to -2^{+}$ in (\ref{hadjiteor}) and use
(\ref{racionales0}). \end{proof}

\begin{ejem} \rm
Take $u=1/2$ and $z=-1$. Substituting $x \to x^2$, $y \to y^2$ and
using (\ref{valor0}) and (\ref{valor2}), we get
$$\int_0^1 \int_0^1 {1-x^2 \over (1+x^2 y^2) \ln^2 xy}dxdy=
\ln {\Gamma \left( 1/4 \right) \over 2 \Gamma \left( 3/4 \right)}+
{2 G \over \pi} -{1 \over 2}.$$
\end{ejem}

\begin{ejem} \rm
Replacing $\Phi$ with the series (\ref{lerch1}), we let $z \to 0$
and deduce that
$$
\int_0^1 \int_0^1 {1-x \over \ln^2 xy} (xy)^{u-1} dx dy = (u+1)
\ln \left(1 + {1 \over u} \right)-1.
$$
\end{ejem}

\begin{coro}
If $\Re(s)>-3$, then
\begin{equation} \nonumber
\int_0^1 \int_0^1 (1-x) (xy)^{u-1} (-\ln xy)^s dxdy= \Gamma(s+1)
\left[(s+1-u)u^{-s-2}+(u+1)^{-s-1} \right].
\end{equation}
\end{coro}
\begin{proof} Set $z=0$ in (\ref{pru1}) and use (\ref{lerch1}).
\end{proof}

\begin{ejem} \rm
Taking $s=-5/2$ and $u=1$ gives
$$\int_0^1 \int_0^1 {1-x \over (-\ln xy)^{5/2}} dxdy={\sqrt{\pi} \over 3} (8 \sqrt 2 - 10).$$
\end{ejem}

\section{Infinite Products} \label{infin}

Using corollaries of Theorems \ref{phidoubleseries} and \ref{hasse},
together with logarithmic series for the digamma and Euler beta
functions (Theorems \ref{teordigamma} and \ref{teorbeta}), we derive
infinite products for many constants.

\begin{coro} \label{coroinf1}
For $m=0, 1, 2, \dots$ and complex $z$ with $\Re(z) < 1/2$,
\begin{equation} \nonumber
{\partial \Phi \over \partial s} \left(z,-m,u \right) = \left( u +
z {\partial \over \partial z} \right)^m \sum_{n=0}^{\infty} {1
\over 1-z} \left( {-z \over 1-z} \right)^n
\sum_{k=0}^{n}(-1)^{k+1} {n \choose k} \ln(u+k).
\end{equation}
\end{coro}

\begin{proof}
Differentiate (\ref{gs3}) with respect to $s$ at $s=0$ and
multiply by $(1-z)^{-1}$. Then apply (\ref{indulerch}) \textit{m}
times.
\end{proof}

\begin{ejem} \rm \label{prodpi}
Setting $m=0$, $z=-1$ and $u=1$ and multiplying by $2$ gives
$$
2 {\partial \Phi \over \partial s} (-1,0,1)=\sum_{n=0}^{\infty} {1
\over 2^{n}} \sum_{k=0}^{n} (-1)^{k+1} {n \choose k} \ln (k+1).
$$
Using (\ref{valor0}), we recover the product from \cite{sondow2}
$$
{\pi \over 2}=\left( {2 \over 1} \right)^ {1/2} \left( {2^2 \over
1 \cdot 3} \right) ^{1/4} \left( {2^3 \cdot  4 \over 1 \cdot 3^3}
\right)^{1/8} \left( {2^4 \cdot 4^4 \over 1 \cdot 3^6 \cdot 5}
\right)^{1/16} \cdots.
$$
\end{ejem}

\begin{ejem} \rm
Setting $m=1$, $z=-1$ and $u=1$ and multiplying by $4$, we get
$$
4 {\partial \Phi \over \partial s} (-1,-1,1)=\sum_{n=0}^{\infty}
{n+1 \over 2^{n}} \sum_{k=0}^{n} (-1)^{k+1} {n \choose k} \ln
(k+1).
$$
Using (\ref{valor1}), we obtain the product
$$
{A^{12} \over 2^{4/3} e }=\left( {2 \over 1} \right)^{2/2} \left(
{2^2 \over 1 \cdot 3} \right) ^{3/4} \left( {2^3 \cdot  4 \over 1
\cdot 3^3} \right)^{4/8} \left( {2^4 \cdot 4^4 \over 1 \cdot 3^6
\cdot 5} \right)^{5/16} \cdots,
$$
which converges faster than the products involving $A$ in Examples
\ref{prodgla2} and \ref{prodgla}.
\end{ejem}

\begin{ejem} \rm
Setting $m=2$, $z=-1$ and $u=1$, we get
$$
{\partial \Phi \over \partial s} (-1,-2,1)=\sum_{n=0}^{\infty}
{n^2+n \over 2^{n+3}} \sum_{k=0}^{n} (-1)^{k+1} {n \choose k} \ln
(k+1).
$$
Then (\ref{valor3}) gives
$$
e^{7 \zeta(3)/4 \pi^2}=\left( {2 \over 1} \right)^{1/8} \left(
{2^2 \over 1 \cdot 3} \right) ^{3/16} \left( {2^3 \cdot  4 \over 1
\cdot 3^3} \right)^{6/32} \left( {2^4 \cdot 4^4 \over 1 \cdot 3^6
\cdot 5} \right)^{10/64} \cdots.
$$
\end{ejem}

\begin{ejem} \rm \label{prodGammas}
Take $m=0$, $z=-1$ and $u=1/2$. Writing $\ln \left(k+{1 \over 2}
\right)$ as $\ln(2k+1)-\ln 2$ and using (\ref{eva0}), we see that
$$
{\partial \Phi \over \partial s} \left( -1 ,0,  {1 \over 2}
\right)-{\ln 2 \over 2}=\sum_{n=0}^{\infty} {1 \over 2^{n+1}}
\sum_{k=0}^{n} (-1)^{k+1} {n \choose k} \ln (2k+1).
$$
Using (\ref{valor0}), we get the product
$$
{\Gamma \left({1/4} \right) \over 2 \Gamma \left({3/4} \right)}=
\left( {3 \over 1} \right)^ {1/4} \left( {3^2 \over 1 \cdot 5}
\right) ^{1/8} \left( {3^3 \cdot  7 \over 1 \cdot 5^3}
\right)^{1/16} \left( {3^4 \cdot 7^4 \over 1 \cdot 5^6 \cdot 9}
\right)^{1/32} \cdots.
$$
\end{ejem}

\begin{ejem} \rm \label{prodcat}
Setting $m=1$, $z=-1$ and $u=1/2$, we get
$$
{\partial \Phi \over \partial s} \left( -1,-1,{1 \over 2}
\right)=\sum_{n=0}^{\infty} {n \over 2^{n+2}} \sum_{k=0}^{n}
(-1)^{k+1} {n \choose k} \ln (2k+1).
$$
Use (\ref{valor2}) and exponentiate to obtain
$$
e^{G/ \pi}=\left( {3 \over 1} \right)^ {1/8} \left( {3^2 \over 1
\cdot 5} \right) ^{2/16} \left( {3^3 \cdot  7 \over 1 \cdot 5^3}
\right)^{3/32} \left( {3^4 \cdot 7^4 \over 1 \cdot 5^6 \cdot 9}
\right)^{4/64} \cdots.
$$
\end{ejem}

\begin{coro} \label{coroinf2}
For all complex $s \neq 1$,
\begin{equation} \nonumber
\Phi(1,s,u)+(s-1) {\partial \Phi \over \partial s} (1,s,u)
=\sum_{n=0}^{\infty} {1 \over n+1} \sum_{k=0}^{n}(-1)^{k+1} {n
\choose k}  {\ln (u+k) \over (u+k)^{s-1}}.
\end{equation}
\end{coro}
\begin{proof} Multiply (\ref{hur1}) by $s-1$ and differentiate with
respect to $s$.
\end{proof}

\begin{ejem} \rm \label{pipare}
Set $u=1$ and $s=0$ and multiply by $2$. The relation (\ref{fzeta})
and the values $\zeta(0)=-1/2$ and $\zeta'(0)=-\frac{1}{2} \ln 2
\pi$ give
$$
 \ln 2 \pi -1=\sum_{n=0}^{\infty} {2 \over n+1} \sum_{k=0}^{n}
(-1)^{k+1} {n \choose k} (k+1) \ln (k+1).
$$
Exponentiating, we obtain the product
$$
{2 \pi \over e}= \left( {2^2 \over 1} \right)^{2/2} \left( {2^4
\over 1 \cdot 3^3} \right)^{2/3} \left( {2^6 \cdot 4^4 \over 1
\cdot 3^9} \right)^{2/4} \left( {2^8 \cdot 4^{16} \over 1 \cdot
3^{18} \cdot 5^5} \right)^{2/5} \cdots.
$$
\end{ejem}

\begin{ejem} \rm \label{prodgla2}
Set $s=-1$ and $u=1$ and multiply by $1/2$. Using relations
(\ref{fzeta}) and (\ref{gla}), and the value $\zeta(-1)=-1/12$, we
get the formula
\begin{equation}\label{progla}
\ln A -{1 \over 8} =\sum_{n=0}^{\infty} {1 \over 2n+2}
\sum_{k=0}^{n} (-1)^{k+1} {n \choose k} (k+1)^2 \ln (k+1),
\end{equation}
which yields
$$
{A \over e^{1/8}}= \left( {2^4 \over 1} \right)^{1/4} \left( {2^8
\over 1 \cdot 3^9} \right)^{1/6} \left( {2^{12} \cdot 4^{16} \over
1 \cdot 3^{27}} \right)^{1/8} \cdots.
$$
\end{ejem}

\begin{teor} \label{teordigamma}
For $u>0$,
\begin{equation}\label{coropro3}
\psi(u)=\sum_{n=0}^{\infty} {1 \over n+1} \sum_{k=0}^{n}(-1)^{k}
{n \choose k} \ln (u+k).
\end{equation}
\end{teor}

\begin{proof}
Multiply (\ref{hur1}) by $s-1$ and compute the derivative
with respect to $s$ at $s=1$, using (\ref{ww}).
\end{proof}

\begin{ejem} \rm \label{prodeuler}
Take $u=1$, multiply by $-1$, and exponentiate. Using (\ref{psi1}),
we recover Ser's product \cite{ser} (rediscovered in \cite{sondow3}
and \cite{sondow2})
$$
e^{\gamma} =\left( {2 \over 1} \right)^ {1/2} \left( {2^2 \over 1
\cdot 3} \right) ^{1/3} \left( {2^3 \cdot  4 \over 1 \cdot 3^3}
\right)^{1/4} \left( {2^4 \cdot 4^4 \over 1 \cdot 3^6 \cdot 5}
\right)^{1/5} \cdots \; .
$$
Compare the remarkably similar product for $e$ in Example
\ref{raize1}.
\end{ejem}

\begin{ejem} \rm \label{prodexppi}
Take first $u=1/4$ and then $u=3/4$. Using the relation
$\psi(3/4)-\psi(1/4)=\pi$ (see \cite[p. 427]{spanier}), we obtain
the series
$$
\pi=\sum_{n=0}^{\infty} {1 \over n+1} \sum_{k=0}^{n} (-1)^k {n
\choose k} \ln {4k+3 \over 4k+1},
$$
which gives the product
$$
e^{\pi} \! = \! \left( {3 \over 1} \right)^{1 \over 1} \! \!
\left( {3 \cdot 5 \over 1 \cdot 7} \right)^{1 \over 2} \! \!
\left( {3 \cdot 5^2 \cdot 11 \over 1 \cdot 7^2 \cdot 9} \right)^{1
\over 3} \! \! \left( {3 \cdot 5^3 \cdot 11^3 \cdot 13 \over 1
\cdot 7^3 \cdot 9^3 \cdot 15} \right)^{1 \over 4} \! \! \left( {3
\cdot 5^4 \cdot 11^6 \cdot 13^4 \cdot 19 \over 1 \cdot 7^4 \cdot
9^6 \cdot 15^4 \cdot 17} \right)^{1 \over 5} \cdots.
$$
\end{ejem}

For the next result, recall the formulas for the \textit{Euler beta
function} \cite[p. 254]{whittaker},
$$
B(u,v)=\int_0^1 x^{u-1} (1-x)^{v-1} dx={\Gamma(u) \Gamma(v) \over
\Gamma(u+v)}.
$$

\begin{teor} \label{teorbeta}
For $j=1,2, \dots,$
\begin{equation}\label{gsejem14}
B(u,j)= \sum_{n=j}^{\infty} {1 \over n-j+1} \sum_{k=0}^{n}
(-1)^{k+1} { n \choose k} \ln(u+k).
\end{equation}
\end{teor}

\begin{proof}
We first establish the formula
\begin{equation}\label{lemasimp}
\int_{0}^{1} {x^{u-1} (1-x)^n \over -\ln x} dx = \sum_{k=0}^{n}
(-1)^{k+1} {n \choose k} \ln (u+k)
\end{equation}
for $n=1,2, \dots \; .$ Fix $n$ and define $f(u)$ to be the
difference between the integral and the sum in (\ref{lemasimp}).
Then $f'(u)=0$. (To see this, differentiate $f(u)$ under the
integral sign, expand $(1-x)^n$, and evaluate the integral.) Thus
$f(u)=C$ is a constant. By (\ref{eva0}) with $m=0$, we may replace
$\ln(u+k)$ with $\ln \left(1+{k \over u} \right)$ in $f(u)$. It
follows that $f(u) \to 0$ as $u \to \infty$. Hence $C=0$ and
(\ref{lemasimp}) follows. Now to prove (\ref{gsejem14}), multiply
(\ref{lemasimp}) by $(n-j+1)^{-1}$ and sum from $n=j$ to $\infty$.
\end{proof}

\begin{ejem} \rm \label{prodsigma}
Set $u=1$ in (\ref{lemasimp}), multiply the equation by
$(-1)^{n+1}$, and sum from $n=1$ to $\infty$. Using Example
\ref{intsomos} (let $x=X/Y$, $y=Y$ and integrate with respect to
$Y$), we obtain
$$
\ln \sigma = \int_0^1 {1-x \over (2-x) (-\ln x)} dx=
\sum_{n=1}^{\infty} (-1)^{n} \sum_{k=0}^{n} (-1)^{k} {n \choose k}
\ln (k+1),
$$
where $\sigma$ is Somos's constant (\ref{somos}). Exponentiation
gives a product for $\sigma$ with rational factors:
\begin{equation}\label{somos2}
\sigma= {2 \over 1} \, \cdot \, {1 \cdot 3 \over 2^2} \, \cdot \,
{2^3 \cdot 4 \over 1 \cdot 3^3} \, \cdot \, {1 \cdot 3^6 \cdot 5
\over 2^4 \cdot 4^4} \cdots.
\end{equation}
\rm Since the partial products are alternately above and below
$\sigma$, and converge slowly, the product is a good candidate for
acceleration by Euler's transformation \cite{sondowzeta},
\cite{sondow2}. Applying it and simplifying (as in \cite[section
3]{sondow2}), we arrive at the product
\begin{equation}\label{somosfast}
\sigma = \left( {2 \over 1} \right)^{1/2} \left( {3 \over 2}
\right)^{1/4} \left( {4 \over 3} \right)^{1/8} \left( {5 \over 4}
\right)^{1/16} \cdots,
\end{equation}
which converges faster than the products for $\sigma$ in both
(\ref{somos}) and (\ref{somos2}). (Note that (\ref{somosfast}) also
follows by writing $\sigma$ as $\sigma^2/\sigma$ and substituting
the product in (\ref{somos}).)
\end{ejem}

\begin{ejem} \rm \label{prodgla}
Set $u=1$ and $j=4$ in (\ref{gsejem14}), and replace $n$ with
$n+4$. Multiplying the equation by $1/2$, so that the left side
becomes $B(1,4)/2=1/8$, we  add the result to (\ref{progla}).
Replacing $k$ with $k-1$, and $n$ with $n-1$, we obtain
\begin{equation}
\ln A=\sum_{n=1}^{\infty} {1 \over 2n} \left[ \sum_{k=1}^{n}
(-1)^{k} {n-1 \choose k-1} k^2 \ln k + \sum_{k=1}^{n+4} (-1)^{k}
{n+3 \choose k-1} \ln k \right]. \nonumber
\end{equation}
This yields a product for the Glaisher-Kinkelin constant:
$$
A=\left( {2^4 \cdot 4^4 \over 1 \cdot 3^6 \cdot 5} \right) ^{1/2}
\left( {2^9 \cdot 4^{10} \cdot 6 \over 1 \cdot 3^{10} \cdot 5^5}
\right)^{1/4} \left( {2^{14} \cdot 4^{20} \cdot 6^6 \over 1 \cdot
3^{24} \cdot 5^{15} \cdot 7} \right)^{1/6} \cdots.
$$
\end{ejem}

The remaining results use only the series for the Euler beta
function in Theorem \ref{teorbeta}. In particular, their
construction does not involve Lerch's function.

\begin{ejem} \rm \label{raize1}
As $B(1,j)=1/j$, exponentiating (\ref{gsejem14}) with $u=1$ yields a
product for $e^{1/j}$. When $j=1$, this is the first author's
product for $e$ in \cite{sondow2}:
$$
e =\left( {2 \over 1} \right)^ {1/1} \left( {2^2 \over 1 \cdot 3}
\right) ^{1/2} \left( {2^3 \cdot  4 \over 1 \cdot 3^3} \right)^{1/3}
\left( {2^4 \cdot 4^4 \over 1 \cdot 3^6 \cdot 5} \right)^{1/4}
\cdots \; .
$$
The extension to $e^{1/j}$ for $j=2,3, \dots$ was first found by C.
Goldschmidt and J. B. Martin \cite{goldschmidt} using probability
theory. With $j=2$, it gives
$$
\sqrt{e}=\left( {2^2 \over 1 \cdot 3} \right) ^{1/1} \left( {2^3
\cdot 4 \over 1 \cdot 3^3} \right)^{1/2} \left( {2^4 \cdot 4^4
\over 1 \cdot 3^6 \cdot 5} \right)^{1/3} \cdots.
$$
\end{ejem}

\begin{teor} \label{expx}
The infinite product representation of the exponential function
\begin{equation}\label{prodexpx}
e^x=\prod_{n=1}^{\infty} \left( \prod_{k=1}^{n}
(kx+1)^{{}^{(-1)^{k+1} {n \choose k}}} \right)^{1/n}
\end{equation}
is valid for any real number $x \geq 0$.
\end{teor}
\begin{proof}
The formula holds when $x=0$. If $x>0$, then since $u$, unlike $j$,
is not restricted to integer values in Theorem \ref{teorbeta}, we
may take $u=1/x$. By (\ref{eva0}) with $m=0$, we may then replace
$\ln \left( {1 \over x }+k \right)$ with $\ln(kx+1)$ in
(\ref{gsejem14}). Setting $j=1$ and using $B(1/x,1)=x$, we
exponentiate (\ref{gsejem14}) and obtain (\ref{prodexpx}).
\end{proof}

\begin{ejem} \rm \label{prode2}
Taking $x=2$, we get
$$
e^2=\left( {3 \over 1} \right)^ {1/1} \left( {3^2 \over 1 \cdot 5}
\right)^{1/2} \left( {3^3 \cdot  7 \over 1 \cdot 5^3}
\right)^{1/3} \left( {3^4 \cdot 7^4 \over 1 \cdot 5^6 \cdot 9}
\right)^{1/4} \cdots.
$$
When $x=p/q$ is a rational number with $p>0$ and $q>0$, we can
simplify (\ref{prodexpx}) by replacing $kx+1$ with $kp+q$, again
using (\ref{eva0}) with $m=0$. An example is
$$
e^{2/3}=\left( {5 \over 3} \right)^ {1/1} \left( {5^2 \over 3
\cdot 7} \right)^{1/2} \left( {5^3 \cdot  9 \over 3 \cdot 7^3}
\right)^{1/3} \left( {5^4 \cdot 9^4 \over 3 \cdot 7^6 \cdot 11}
\right)^{1/4} \cdots.
$$
In particular, with $x=1/q$ and $q=2,3, \dots,$ the roots
$e^{1/q}$ are represented as different products from those in
Example \ref{raize1}. For instance,
$$
\sqrt{e}=\left( {3 \over 2} \right)^ {1/1} \left( {3^2 \over 2
\cdot 4} \right)^{1/2} \left( {3^3 \cdot  5 \over 2 \cdot 4^3}
\right)^{1/3} \left( {3^4 \cdot 5^4 \over 2 \cdot 4^6 \cdot 6}
\right)^{1/4} \cdots.
$$
\end{ejem}

\subsection*{Acknowledgments} For bringing Ser's and Hasse's papers to our attention, we thank Donal
Connon and Alex Schuster, respectively.

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\end{thebibliography}


\end{document}
��
math0506319.extracted_bib
\begin{thebibliography}{99} \label{refer}

\bibitem{bailey} D. H. Bailey, P. Borwein, and S. Plouffe,
"On the rapid computation of various polylogarithmic constants,"
{\it Math. Comp.} \textbf{66} (1997), 903-913.

\bibitem{bateman} H. Bateman and A. Erdelyi, {\it Higher Transcendental Functions}, Vol. 1, McGraw-Hill, New York, 1953.

\bibitem{ramanujan} B. C. Berndt, {\it Ramanujan's Notebooks,} Parts I and IV, Springer-Verlag,
New York, 1985 and 1994.

\bibitem{beukers} F. Beukers, ''A note on the irrationality of $\zeta(2)$ and $\zeta(3)$,''
{\it Bull. London Math. Soc.} \textbf{11} (1979), 268-272.

\bibitem{chapman} R. Chapman, ''A proof of Hadjicostas's
conjecture'' (2004, preprint); available at
\\ http://arXiv.org/abs/math/0405478.

\bibitem{emery} M. Emery, ''On a multiple harmonic power
series'' (2004, preprint); available at \\
http://arxiv.org/abs/math/0411267v2.

\bibitem{finch} S. Finch, {\it Mathematical Constants}, Cambridge
University Press, Cambridge, 2003.

\bibitem{goldschmidt} C. Goldschmidt and J. B. Martin,
''Random recursive trees and the Bolthausen-Sznitman coalescent,''
{\it Electron. J. Probab.} \textbf{10} (2005), 718-745.

\bibitem{hadjicostas0} P. Hadjicostas, ''Some generalizations of Beukers' integrals,''
{\it Kyungpook Math. J.} \textbf{42} (2002), 399-416.

\bibitem{hadjicostas} P. Hadjicostas, ''A conjecture-generalization of Sondow's
formula'' (2004, preprint); available at
http://arXiv.org/abs/math/0405423.

\bibitem{hadjicostas2} P. Hadjicostas, {\it personal communication}, 1 June 2004.

\bibitem{hasse} H. Hasse, ''Ein Summierungsverfahren f\"{u}r die Riemannsche  $\zeta$-Reihe,'' {\it Math.
Z.} \textbf{32} (1930), 458-464.

\bibitem{lewin2} L. Lewin, {\it Polylogarithms and Associated Functions,} Elsevier, New York, 1981.

\bibitem{lewin} L. Lewin, ed., {\it Structural Properties of
Polylogarithms,} Math. Surveys and Monographs Vol. 37, American
Mathematical Society, Providence, 1991.

\bibitem{milnor} J. Milnor, ''On polylogarithms, Hurwitz zeta
functions, and the Kubert identities,'' {\it L'Enseignement Math.}
\textbf{29} (1983), 281-322.

\bibitem{ramanujan2} S. Ramanujan, {\it Collected Papers of Srinivasa
Ramanujan,} G. H. Hardy et al., eds., AMS Chelsea, Providence, 2000.

\bibitem{ser} J. Ser, ''Sur une expression de la function $\zeta(s)$
de Riemann,'' {\it C. R. Acad. Sci. Paris S\'{e}r. I Math.} \textbf{182}
(1926), 1075-1077.

\bibitem{somos} M. Somos, ''Several constants related to quadratic recurrences,''
unpublished note, 1999.

\bibitem{sondowzeta} J. Sondow, ''Analytic continuation of Riemann's zeta function
and values at negative integers via Euler's transformation of
series,'' {\it Proc. Amer. Math. Soc.} \textbf{120} (1994),
421-424.

\bibitem{sondow0} J. Sondow, ''Criteria for irrationality of Euler's
constant,'' {\it Proc. Amer. Math. Soc.} \textbf{131} (2003),
3335-3344.

\bibitem{sondow3} J. Sondow, ''An infinite product for $e^ \gamma$ via hypergeometric formulas for
Euler's constant, $\gamma$''  (2003, preprint); available at
http://arXiv.org/abs/math/0306008.

\bibitem{sondow1} J. Sondow, ''Double integrals for Euler's constant and $\ln
(4/\pi)$ and an analog of Hadjicostas's formula,'' {\it Amer.
Math. Monthly} \textbf{112} (2005), 61-65.

\bibitem{sondow2} J. Sondow, ''A faster product for $\pi$ and a new integral for $\ln
(\pi/2)$,'' {\it Amer. Math. Monthly}  \textbf{112} (2005), 729-734.

\bibitem{spanier} J. Spanier and K. B. Oldham, {\it An Atlas of
Functions}, Hemisphere, New York, 1987.

\bibitem{whittaker}
E. T. Whittaker and G. N. Watson, {\it A Course of Modern
Analysis}, 4th ed., Cambridge University Press, Cambridge, 1996.

\end{thebibliography}