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\def\frheadline{\its S\'eminaire Lotharingien de
Combinatoire \bfs 53 \rms (2005), Article~B53c\hfill} 

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\define\Li{\operatorname{Li}}
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\topmatter
\title
Computing powers \\
of two generalizations of the logarithm
\endtitle
\author
Wadim Zudilin\footnotemark"$^\ddag$"\ {\rm(Moscow)}
\endauthor
\abstract
We prove multiple-series representations for
positive integer powers of the series
$$
L(z;\alpha)=\sum_{n=1}^\infty\frac{z^n}{n+\alpha},
\;\; |z|<1, \; \alpha\ge0,
\quad\text{and}\quad
\ell_q(z)=\sum_{n=1}^\infty\frac{z^nq^n}{1-q^n},
\;\; |z|\le1, \; |q|<1.
$$
The results generalize a known formula for powers
of the series for the ordinary logarithm $-\log(1-z)=L(z;0)$.
\endabstract
\address
\hbox to70mm{\vbox{\hsize=70mm%
\leftline{Department of Mechanics and Mathematics}
\leftline{Moscow Lomonosov State University}
\leftline{Vorobiovy Gory, GSP-2}
\leftline{119992 Moscow, RUSSIA}
\leftline{{\it URL\/}: \tt http://wain.mi.ras.ru/index.html}
}}
\endaddress
\email
{\tt wadim\@ips.ras.ru}
\endemail
\endtopmatter
\leftheadtext{W.~Zudilin}
%\rightheadtext{ }
\footnotetext"$^\ddag$"{The work is partially supported
by grant no.~03-01-00359 of the Russian Foundation for Basic Research.}
%==================================================
\document

The series for the ordinary logarithm,
$$
\Li_1(z)=-\log(1-z)=\sum_{n=1}^\infty\frac{z^n}n,
\qquad |z|<1,
\tag1
$$
admits the following formula
$$
\Li_1(z)^l=l!\sum_{n_l>n_{l-1}>\dots>n_1\ge1}
\frac{z^{n_l}}{n_1n_2\dotsb n_l}.
\tag2
$$
It plays an important r\^ole in evaluating
irrationality and transcendence measures
for values of the logarithm at rational points
(cf.~\cite{Ne}). On the other hand,
information on the arithmetic nature of values
of the following two generalizations of the
series~\thetag{1} is available only in particular cases at this moment.
The first generalization, the $\alpha$-shift
of the logarithm for $\alpha\ge0$, is defined
by the series
$$
L(z)=L(z;\alpha)=\sum_{n=1}^\infty\frac{z^n}{n+\alpha},
\qquad |z|<1.
$$
The second generalization, known as a $q$-extension
of the logarithm, is given by the formula
$$
\ell_q(z)
=\sum_{n=1}^\infty\frac{z^nq^n}{1-q^n}
=\sum_{n=1}^\infty\frac{z^n}{p^n-1},
\qquad |z|\le1, \quad |q|<1, \quad p=q^{-1}.
$$
Then one has
$$
\lim\Sb q\to1\\|q|<1\endSb(1-q)\ell_q(z)
=\Li_1(z)=L(z;0)
\qquad\text{for}\quad |z|<1.
$$

The aim of this note is to provide generalizations
of the formula \thetag{2} for both $L(z;\alpha)$
(Theorem~1) and $\ell_q(z)$ (Theorem~2), which
might become useful in the arithmetic study of values
of the series.

\proclaim{Theorem 1}
For each $l=0,1,2,\dots$,
the following identity holds:
$$
L(z)^l=l!L_l(z),
\qquad l=0,1,2,\dots,
$$
where
$$
L_l(z)
=\sum_{n_l>\dots>n_2>n_1\ge1}
\frac{z^{n_l}}{(n_1+\alpha)\dotsb(n_l+l\alpha)},
\quad l=1,2,\dots,
\qquad L_0(z)=1.
$$
\endproclaim

\demo{Proof}
For given $l$, we have
$$
\align
z^{1-l\alpha}\frac{\d}{\d z}\bigl(z^{l\alpha}L_l(z)\bigr)
&=z^{1-l\alpha}\frac{\d}{\d z}
\sum_{n_l>\dots>n_1\ge1}
\frac{z^{n_l+l\alpha}}{(n_1+\alpha)\dotsb(n_{l-1}+(l-1)\alpha)(n_l+l\alpha)}
\\
&=\sum_{n_{l-1}>\dots>n_1\ge1}
\frac1{(n_1+\alpha)\dotsb(n_{l-1}+(l-1)\alpha)}
\sum_{n_l=n_{l-1}+1}^\infty z^{n_l}
\\
&=\frac z{1-z}
\sum_{n_{l-1}>\dots>n_1\ge1}
\frac{z^{n_{l-1}}}{(n_1+\alpha)\dotsb(n_{l-1}+(l-1)\alpha)}
\\
&=\frac z{1-z}L_{l-1}(z).
\endalign
$$
On the other hand,
$$
\align
z^{1-l\alpha}\frac{\d}{\d z}\bigl(z^{l\alpha}L_l(z)\bigr)
&=z^{1-l\alpha}\biggl(l\alpha z^{l\alpha-1}L_l(z)
+z^{l\alpha}\frac{\d}{\d z}L_l(z)\biggr)
\\
&=l\alpha L_l(z)+z\frac{\d}{\d z}L_l(z).
\endalign
$$
Therefore,
$$
z\frac{\d}{\d z}L_l(z)
=-l\alpha L_l(z)+\frac z{1-z}L_{l-1}(z),
\qquad l=1,2,\dotsc.
$$

If we define
$$
\tilde L_l(z)=\frac1{l!}L(z)^l
=\frac1{l!}\biggl(\sum_{n=1}^\infty\frac{z^n}{n+\alpha}\biggr)^l,
$$
then
$$
\align
z\frac{\d}{\d z}\tilde L_l(z)
&=\frac1{l!}z\frac{\d}{\d z}L(z)^l
=\frac1{(l-1)!}L(z)^{l-1}\cdot z\frac{\d}{\d z}L(z)
\\
&=\frac1{(l-1)!}L(z)^{l-1}
\biggl(-\alpha L(z)+\frac z{1-z}\biggr)
=-\alpha l\tilde L_l(z)+\frac z{1-z}\tilde L_{l-1}(z).
\endalign
$$
Since $L_0(z)=\tilde L_0(z)=1$ and
$L_l(z)=(z/(1+\alpha))^l/l!+O(z^{l+1})$,
$\tilde L_l(z)=(z/(1+\alpha))^l/l!+\allowmathbreak O(z^{l+1})$,
we obtain the desired result.
\qed
\enddemo

\proclaim{Theorem 2}
For each $l=1,2,\dots$,
$$
\align
\ell_q(z)^l
&=\sum_{n_l>n_{l-1}>\dots>n_1\ge1}
\frac{z^{n_l}q^{n_l}
\Phi_l(q^{n_1},q^{n_2-n_1},\dots,q^{n_l-n_{l-1}})}
{(1-q^{n_1})(1-q^{n_2})\dotsb(1-q^{n_l})}
\\
&=\sum_{n_l>n_{l-1}>\dots>n_1\ge1}
\frac{z^{n_l}\Phi_l(p^{n_1},p^{n_2-n_1},\dots,p^{n_l-n_{l-1}})}
{(p^{n_1}-1)(p^{n_2}-1)\dotsb(p^{n_l}-1)},
\tag3
\endalign
$$
where $\Phi_l(x_1,\dots,x_l)$ is the polynomial
$$
\align
\Phi_l(x_1,\dots,x_l)
&=(x_1^{l-1}+x_1^{l-2}+\dots+x_1+1)
(x_2^{l-2}+\dots+x_2+1)\dotsb(x_{l-1}+1)
\\
&=\prod_{j=1}^l\frac{x_j^{l+1-j}-1}{x_j-1}.
\tag4
\endalign
$$
In particular, one has
$$
\gathered
\ell_{1/p}(z)^2
=\sum_{n_2>n_1\ge1}\frac{z^{n_2}(p^{n_1}+1)}
{(p^{n_1}-1)(p^{n_2}-1)},
\\
\ell_{1/p}(z)^3
=\frac12\sum_{n_3>n_2>n_1\ge1}
\frac{z^{n_3}(p^{2n_1}+p^{n_1}+1)(p^{n_2-n_1}+1)}
{(p^{n_1}-1)(p^{n_2}-1)(p^{n_3}-1)}.
\endgathered
$$
\endproclaim

The proof requires two auxiliary identities.

\proclaim{Lemma 1}
For $l=2,3,\dots$, the equality
$$
\align
&
\frac1{(x_1-1)(x_1x_2-1)\dotsb(x_1x_2\dotsb x_{l-1}-1)\cdot(x-1)}
\\ &\qquad
=\sum_{j=1}^l\frac{x_1\dots x_{j-1}}
{\prod_{k=1}^{j-1}(x_1\dotsb x_k-1)
\cdot\prod_{k=j-1}^{l-1}(x_1\dotsb x_kx-1)}
\endalign
$$
holds identically in the variables $x_1,\dots,x_{l-1}$
and~$x$. \rm(Empty products should be replaced by~$1$.)
\endproclaim

The proof exploits a simple inductive argument,
and therefore is omitted.

\proclaim{Lemma 2}
For each $l=1,2,\dots$,
the following identity holds:
$$
\align
&
\frac1{(x_1-1)(x_2-1)\dotsb(x_l-1)}
\\ &\qquad
=\frac1{l!}\sum_{\sigma\in\frak S_l}
\frac{\Phi_l(x_{\sigma(1)},\dots,x_{\sigma(l)})}
{(x_{\sigma(1)}-1)(x_{\sigma(1)}x_{\sigma(2)}-1)
\dotsb(x_{\sigma(1)}x_{\sigma(2)}\dotsb x_{\sigma(l)}-1)},
\tag5
\endalign
$$
where
$$
\Phi_l(x_1,\dots,x_l)
=\sum\Sb k_i=0\\i=1,\dots,l\endSb^{l-i}
x_1^{k_1}x_2^{k_2}\dotsb x_l^{k_l}
\tag6
$$
is the polynomial defined in~\thetag{4} and $\frak S_l$ denotes
the set of all permutations of $\{1,2,\dots,\allowmathbreak l\}$.
\endproclaim

\demo{Proof}
We apply induction on~$l$. If $l=1$, then both sides
of \thetag{5} are simply $1/(x_1-1)$. Suppose that $l>1$
and that we have proved
identity \thetag{5} for $l$ replaced by~$l-1$.
Then applying Lemma~1 (with $x=x_l$) we obtain
$$
\align
&
\frac1{(x_1-1)(x_2-1)\dotsb(x_{l-1}-1)\cdot(x_l-1)}
\\ &\qquad
=\frac1{(l-1)!}\sum_{\sigma\in\frak S_{l-1}}
\frac{\Phi_{l-1}(x_{\sigma(1)},\dots,x_{\sigma(l-1)})}
{(x_{\sigma(1)}-1)(x_{\sigma(1)}x_{\sigma(2)}-1)
\dotsb(x_{\sigma(1)}\dotsb x_{\sigma(l-1)}-1)
\cdot(x_l-1)}
\\ &\qquad
=\frac1{(l-1)!}\sum_{\sigma\in\frak S_{l-1}}\sum_{j=1}^l
\frac{\Phi_{l-1}(x_{\sigma(1)},\dots,x_{\sigma(l-1)})}
{\prod_{k=1}^{j-1}(x_{\sigma(1)}\dotsb x_{\sigma(k)}-1)}
\\ &\qquad\qquad\qquad\qquad\times
\frac{x_{\sigma(1)}\dotsb x_{\sigma(j-1)}}
{\prod_{k=j-1}^{l-1}(x_{\sigma(1)}\dotsb x_{\sigma(j-1)}
x_lx_{\sigma(j)}\dotsb x_{\sigma(k)}-1)}
\\ &\qquad
=\frac1{(l-1)!}\sum_{\tau\in\frak S_l}
\frac{\Phi_{l-1}(x_{\tau(1)},\dots,\widehat{x_{\tau(j)}},\dots,x_{\tau(l)})
\cdot x_{\tau(1)}\dotsb x_{\tau(j-1)}}
{\prod_{k=1}^l(x_{\tau(1)}\dotsb x_{\tau(k)}-1)},
\tag7
\endalign
$$
where in the latter sum $j$ abbreviates $\tau^{-1}(l)$, that is
$$
\tau=\tau(j,\sigma)\in\frak S_l\:
(1,2,\dots,l-1,l)\mapsto
\bigl(\sigma(1),\dots,\sigma(j-1),l,
\sigma(j),\dots,\sigma(l-1)\bigr),
$$
and the notation $\widehat x$ means omitting
the corresponding parameter.
Since the product on the left-hand side of~\thetag{7}
is symmetric in $x_1,x_2,\dots,x_l$, we may replace
$x_l$ by $x_i$ with $i\ne l$, to deduce an identity
similar to but different from~\thetag{7}.
If we now average the results over all $i=1,\dots,l$,
then we obtain
$$
\align
&
\frac1{(x_1-1)(x_2-1)\dotsb(x_{l-1}-1)(x_l-1)}
\\ &\qquad
=\frac1{(l-1)!}\cdot\frac1l\sum_{j=1}^l\sum_{\tau\in\frak S_l}
\frac{\Phi_{l-1}(x_{\tau(1)},\dots,\widehat{x_{\tau(j)}},\dots,x_{\tau(l)})
\cdot x_{\tau(1)}\dotsb x_{\tau(j-1)}}
{\prod_{k=1}^l(x_{\tau(1)}\dotsb x_{\tau(k)}-1)}
\\ &\qquad
=\frac1{l!}\sum_{\tau\in\frak S_l}
\frac{\wt\Phi_l(x_{\tau(1)},\dots,x_{\tau(l)})}
{\prod_{k=1}^l(x_{\tau(1)}\dotsb x_{\tau(k)}-1)},
\endalign
$$
where we set
$$
\wt\Phi_l(x_1,\dots,x_l)
=\sum_{j=1}^l\Phi_{l-1}(x_1,\dots,\widehat{x_j},\dots,x_l)
\cdot x_1x_2\dotsb x_{j-1}.
\tag8
$$
It remains to verify that the polynomials
$\wt\Phi_l$ in~\thetag{8} and $\Phi_l$ in~\thetag{6}
coincide:
$$
\align
\wt\Phi_l(x_1,\dots,x_l)
&=\sum_{j=1}^l\sum\Sb k_i=0\\i=1,\dots,l-1\endSb^{l-1-i}
x_1^{k_1+1}\dotsb x_{j-1}^{k_{j-1}+1}
x_{j+1}^{k_j}\dotsb x_l^{k_{l-1}}
\allowdisplaybreak
&=\sum_{j=1}^l\sum\Sb k_i'=1\\i=1,\dots,j-1\endSb^{l-i}
x_1^{k_1'}\dotsb x_{j-1}^{k_{j-1}'}
\sum\Sb k_i'=0\\i=j+1,\dots,l\endSb^{l-i}
x_{j+1}^{k_{j+1}'}\dotsb x_l^{k_l'}
\\
&=\sum\Sb k_i'=0\\i=1,\dots,l\endSb^{l-i}
x_1^{k_1'}x_2^{k_2'}\dotsb x_l^{k_l'}
=\Phi_l(x_1,\dots,x_l),
\endalign
$$
where we use the bijection
$$
(j,k_1,\dots,k_{l-1})
\mapsto(k_1',k_2',\dots,k_l')
=(k_1+1,\dots,k_{j-1}+1,0,k_j,\dots,k_{l-1}).
$$
This completes our proof of the lemma.
\qed
\enddemo

\remark{Remark}
The identity of Lemma~2 belongs to a family of identities
in the style of Littlewood~\cite{Li}, p.~85:
$$
\prod_{j=1}^l\frac1{x_j}
=\sum_{\sigma\in\frak S_l}
\frac1{x_{\sigma(1)}(x_{\sigma(1)}+x_{\sigma(2)})
\dotsb(x_{\sigma(1)}+x_{\sigma(2)}+\dots+x_{\sigma(l)})}.
$$
Identities of similar type and their applications
may be found in~\cite{Me}, Section~10.9, Appendix~A.23.
Lassalle in~\cite{La} gives formulae, where
monomial symmetric functions are specialized;
taking as variables the parts of the partition,
which indexes the function (i.e., replacing
$x_j$ by~$q^{j-1}$ for $j=1,\dots,l$), one
obtains curious generalizations of Littlewood-type
identities.
\endremark

\demo{Proof of Theorem\/ \rom2}
From Lemma~2 we deduce that
$$
\align
\ell_q(z)^l
&=\biggl(\sum_{m=1}^\infty\frac{z^m}{p^m-1}\biggr)^l
=\sum\Sb m_i=1\\i=1,\dots,l\endSb^\infty
\frac{z^{m_1+\dots+m_l}}{(p^{m_1}-1)\dotsb(p^{m_l}-1)}
\\
&=\frac1{l!}\sum\Sb m_i=1\\i=1,\dots,l\endSb^\infty
\sum_{\sigma\in\frak S_l}
\frac{\Phi_l(x_{\sigma(1)},\dots,x_{\sigma(l)})}
{\prod_{k=1}^l(x_{\sigma(1)}\dotsb x_{\sigma(k)}-1)}
\bigg|_{x_i=p^{m_i},\ i=1,\dots,l}
\\
&=\sum\Sb m_i=1\\i=1,\dots,l\endSb^\infty
\frac{\Phi_l(x_1,\dots,x_l)}
{\prod_{k=1}^l(x_1\dotsb x_k-1)}
\bigg|_{x_i=p^{m_i},\ i=1,\dots,l},
\endalign
$$
and after the change $n_i=m_1+\dots+m_i$, $i=1,\dots,l$,
we arrive at the claimed formula~\thetag{3}.
\qed
\enddemo

\subhead
Acknowledgements
\endsubhead
I thank C.~Krattenthaler,
J.~Sondow and the two anonymous referees of the journal
for several suggestions.

%==================================================
\Refs
\widestnumber\key{W0}

\ref\key La
\by M.~Lassalle
\paper Une $q$-sp\'ecialisation pour les fonctions sym\'etriques monomiales
\jour Adv\. Math\.
\vol162
\yr2001
\issue2
\pages217--242
\endref

\ref\key Li
\by D.\,E.~Littlewood
\book The theory of group characters and matrix representations
of groups
\bookinfo 2nd ed.
\publ Oxford Univ\. Press
\publaddr Oxford
\yr1950
\endref

\ref\key Me
\by M.\,L.~Mehta
\book Random matrices
\bookinfo 2nd ed\.
\publ Academic Press
\publaddr Boston, MA
\yr1991
\endref

\ref\key Ne
\by Yu.\,V.~Nesterenko
\paper On an identity of Mahler
\jour preprint
\yr2003
\lang Russian
\endref

\endRefs
\enddocument