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\begin{center}
\vskip 1cm{\LARGE\bf 
Some Formulae for Products of Geometric \\
\vskip .1in
Polynomials with Applications
}
\vskip 1cm
\large
Levent Karg{\i}n\\
Akseki Vocational School\\
Alanya Alaaddin Keykubat University\\
TR-07630 Antalya\\
Turkey\\
\href{mailto:leventkargin48@gmail.com}{\texttt{leventkargin48@gmail.com}}
\end{center}

\vskip .2 in


\begin{abstract}
In this paper, we evaluate sums and integrals of products of two
geometric polynomials and obtain new explicit formulas for geometric
polynomials and numbers. As a consequence of these results, we give new
explicit formulas for $p$-Bernoulli numbers, Apostol-Bernoulli
functions, and integrals of products of two Apostol-Bernoulli
functions.
\end{abstract}

\vskip .2 in

\section{Introduction}

Let $
\genfrac{\{}{\}}{0pt}{}{n}{k}
$ be the Stirling numbers of the second kind \cite{Graham}. Geometric
polynomials are defined by \cite{T}
\begin{equation}
w_{n}\left(  y\right)  =\sum_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
k!y^{k}.\label{2}
\end{equation}
They have the exponential generating function
\begin{equation}
\frac{1}{1-y\left(  e^{t}-1\right)  }=\sum_{n=0}^{\infty}w_{n}\left(
y\right)  \frac{t^{n}}{n!}\text{,}\label{10}
\end{equation}
and are related to the geometric series by \cite{B}
\[
\left(  y\frac{d}{dy}\right)  ^{m}\frac{1}{1-y}=\sum_{k=0}^{\infty}k^{m}
y^{k}=\frac{1}{1-y}w_{m}\left(  \frac{y}{1-y}\right)  ,\text{ \ }\left\vert
y\right\vert &lt;1.
\]
In addition, the following recurrence relation holds for the
geometric polynomials \cite{Diletal}:
\begin{equation}
w_{n+1}\left(  y\right)  =y\frac{d}{dy}\bigl(  w_{n}\left(  y\right)
+yw_{n}\left(  y\right)  \bigr)  .\label{11}
\end{equation}


The $n$th geometric number (ordered Bell number, or Fubini number, or preferred arrangement number) \cite{Da,
Gr, T}, $w_{n}$, is defined by
\begin{equation}
w_{n}\left(  1\right)  =w_{n}=\sum_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
k!\text{,}\label{14}
\end{equation}
and counts all the possible set partitions of an $n$ element set such that the
order of the blocks matters. Besides with this combinatorial property, these
numbers are seen in the evaluation of the following series
\begin{equation}
\sum_{k=0}^{\infty}\frac{k^{n}}{2^{k}}=2w_{n}.\label{1}
\end{equation}
In the literature, numerous identities concerned with these polynomials and
numbers were obtained \cite{B, B2, B3, B4, Dil1, MEZO} and some
generalizations were given \cite{B, Dil2, Kargin, Mezo2}.

The sums of products of various polynomials and numbers with or without
binomial coefficients have been studied \cite{Agoh, Chu, Kamano,
Komatsu, Komatsu2, Singh, Zhao}. One of the famous results is \cite[Eq.\ 50.11.2]{Hansel}
\begin{equation}
\sum_{k=0}^{n}\binom{n}{k}B_{k}\left(  x\right)  B_{n-k}\left(  y\right)
=\left(  1-n\right)  B_{n}\left(  x+y-1\right)  +\left(  x+y-1\right)
nB_{n-1}\left(  x+y-1\right)  ,\label{b1}
\end{equation}
where $B_{n}\left(  x\right)  $ is the $n$th Bernoulli polynomial. In
addition, the integrals of products of various polynomials and functions have
been studied \cite{Agoh2, B2, Dagli1, Kim, Liu}. For example, the
following interesting integral for a product of two Bernoulli polynomials
appears in the book by N\"{o}rlund \cite[p.\ 31]{Norlund}: For all $k+m\geq2$,
\begin{equation}
\int\limits_{0}^{1}B_{k}\left(  x\right)  B_{m}\left(  x\right)
dx=\frac{k!m!}{\left(  k+m\right)  !}B_{k+m}.\label{b2}
\end{equation}
Here, $B_{n}$ is the $n$th Bernoulli number defined by $B_{n}=B_{n}\left(
0\right)  $.

The main purpose of this paper is to obtain analogues of (\ref{b1}) and
(\ref{b2}) for geometric polynomials, which generalize the binomial formulas
\cite{Diletal}
\begin{align}
\sum_{k=0}^{n}\binom{n}{k}w_{k} &amp;  =2w_{n},\text{ \ }n&gt;0,\label{5}\\
2\sum_{k=0}^{n}\binom{n}{k}\left(  -1\right)  ^{k}w_{k} &amp;  =\left(  -1\right)
^{n}w_{n}+1,\text{ \ }n\geq0,\label{6}
\end{align}
and the integral \cite{KELLER}
\begin{equation}
\int_{-1}^{0}w_{n}\left(  y\right)  dy=B_{n},\text{ \ }n&gt;0.\label{26}
\end{equation}


The paper is outlined as follows. We derive sums of products of two geometric
polynomials and explicit formulas for these polynomials and numbers in Section
\ref{sec2}. In Section \ref{sec3}, we give integrals of products of two geometric polynomials. As
an application of this result, an explicit formula for $p$-Bernoulli numbers is
obtained. Finally, in Section \ref{sec4}, we give
a new explicit formula and integrals of
products of Apostol-Bernoulli functions.

\section{Sums of products of geometric polynomials}\label{sec2}

In this section, we define two variable geometric polynomials and obtain some
basic properties which give us new formulas for $w_{n}\left(  y\right)$.
Moreover, we consider the sums of products of two geometric polynomials.

Two variable geometric polynomials are defined by means of the following
generating function:
\begin{equation}
\sum_{n=0}^{\infty}w_{n}\left(  x;y\right)  \frac{t^{n}}{n!}=\frac{e^{xt}
}{1-y\left(  e^{t}-1\right)  }.\label{8}
\end{equation}
As some special cases of (\ref{8}), we have
\begin{equation}
w_{n}\left(  0;y\right)  =w_{n}\left(  y\right)  \text{ and }w_{n}\left(
0;1\right)  =w_{n}.\label{29}
\end{equation}


We can rewrite (\ref{8}) as
\begin{align*}
\sum_{n=0}^{\infty}w_{n}\left(  x;y\right)  \frac{t^{n}}{n!}  &amp;  =\frac
{1}{1-y\left(  e^{t}-1\right)  }e^{xt}\\
&amp;  =\sum_{n=0}^{\infty}w_{n}\left(  y\right)  \frac{t^{n}}{n!}\sum
_{n=0}^{\infty}x^{n}\frac{t^{n}}{n!}\\
&amp;  =\sum_{n=0}^{\infty}\bigl(  \sum_{k=0}^{n}\binom{n}{k}w_{k}\left(
y\right)  x^{n-k}\bigr)  \frac{t^{n}}{n!}.
\end{align*}
Comparing the coefficients of $\frac{t^{n}}{n!}$ yields
\begin{equation}
w_{n}\left(  x;y\right)  =\sum_{k=0}^{n}\binom{n}{k}w_{k}\left(  y\right)
x^{n-k}.\label{3}
\end{equation}
Besides, from (\ref{8}) we have
\begin{align*}
\sum_{n=0}^{\infty}\bigl(  w_{n}\left(  x+1;y\right)  -w_{n}\left(
x;y\right)  \bigr)  \frac{t^{n}}{n!}  &amp;  =\frac{e^{xt}\left(  e^{t}-1\right)
}{1-y\left(  e^{t}-1\right)  }\\
&amp;  =\frac{1}{y}\bigl(  \frac{e^{xt}}{1-y\left(  e^{t}-1\right)  }
-e^{xt}\bigr) \\
&amp;  =\frac{1}{y}\sum_{n=0}^{\infty}\bigl(  w_{n}\left(  x;y\right)
-x^{n}\bigr)  \frac{t^{n}}{n!}.
\end{align*}
Comparing the coefficients of $\frac{t^{n}}{n!}$ gives
\begin{equation}
yw_{n}\left(  x+1;y\right)  =\left(  1+y\right)  w_{n}\left(  x;y\right)
-x^{n}.\label{4}
\end{equation}
Thus, setting $x=0$ and $x=-1$ in (\ref{4}), we find
\begin{align}
yw_{n}\left(  1;y\right)   &amp;  =\left(  1+y\right)  w_{n}\left(  y\right)
,\text{ \ }n&gt;0,\label{7}\\
\left(  1+y\right)  w_{n}\left(  -1;y\right)   &amp;  =yw_{n}\left(  y\right)
+\left(  -1\right)  ^{n},\text{ \ }n\geq0,\label{9}
\end{align}
respectively. Combining these relations with (\ref{3}) gives equations
(\ref{5}) and (\ref{6}) which were obtained by using Euler-Seidel matrix
method in \cite{Diletal}.

Now, we want to give the generalization of the binomial formula (\ref{5}).
Derivative of (\ref{8}) can be written as
\[
\frac{\partial}{\partial t}\left(  \frac{e^{xt}}{1-y\left(  e^{t}-1\right)
}\right)  =\frac{xe^{xt}}{1-y\left(  e^{t}-1\right)  }+\frac{ye^{t}
}{1-y\left(  e^{t}-1\right)  }\frac{e^{xt}}{1-y\left(  e^{t}-1\right)  }.
\]
Taking $x=x_{1}+x_{2}-1$ leads to
\begin{align*}
\frac{\partial}{\partial t}\left(  \frac{e^{xt}}{1-y\left(  e^{t}-1\right)
}\right)   &amp;  =\sum_{n=0}^{\infty}w_{n+1}\left(  x_{1}+x_{2}-1;y\right)
\frac{t^{n}}{n!},\\
\frac{xe^{xt}}{1-y\left(  e^{t}-1\right)  }  &amp;  =\left(  x_{1}+x_{2}-1\right)
\sum_{n=0}^{\infty}w_{n}\left(  x_{1}+x_{2}-1;y\right)  \frac{t^{n}}{n!},
\end{align*}
and
\begin{align*}
\frac{ye^{t}}{1-y\left(  e^{t}-1\right)  }\frac{e^{xt}}{1-y\left(
e^{t}-1\right)  }  &amp;  =y\left(  \sum_{n=0}^{\infty}w_{n}\left(  x_{1}
;y\right)  \frac{t^{n}}{n!}\right)  \left(  \sum_{n=0}^{\infty}w_{n}\left(
x_{2};y\right)  \frac{t^{n}}{n!}\right) \\
&amp;  =y\sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}w_{k}\left(  x_{1};y\right)
w_{n-k}\left(  x_{2};y\right)  \frac{t^{n}}{n!}.
\end{align*}
By equating the coefficients of $\frac{t^{n}}{n!}$, we get
\[
y\sum_{k=0}^{n}\binom{n}{k}w_{k}\left(  x_{1};y\right)  w_{n-k}\left(
x_{2};y\right)  =w_{n+1}\left(  x_{1}+x_{2}-1;y\right)  -\left(  x_{1}
+x_{2}-1\right)  w_{n}\left(  x_{1}+x_{2}-1;y\right)  .
\]
For $x_{1}=x_{2}=0$ in the above equation, by using (\ref{9}) gives the sums of
products of the geometric polynomials.

\begin{theorem}
\label{teo1}For $n\geq0$,
\begin{equation}
\left(  y+1\right)  \sum_{k=0}^{n}\binom{n}{k}w_{k}\left(  y\right)
w_{n-k}\left(  y\right)  =w_{n+1}\left(  y\right)  +w_{n}\left(  y\right)  .
\label{13}
\end{equation}
When $y=1$ this becomes
\begin{equation}
2\sum_{k=0}^{n}\binom{n}{k}w_{k}w_{n-k}=w_{n+1}+w_{n}. \label{12}
\end{equation}
\end{theorem}

We note that Boyadzhiev and Dil \cite[Proposition\ 9]{BoyadzhievandDil} proved
a more general form of Theorem \ref{teo1} by different way.

Now, we investigate the sums of products of the geometric polynomials for
different $y$ values in the following theorem.

\begin{theorem}
For $n\geq0$ and $y_{1}\neq y_{2}$,
\begin{equation}
\sum_{k=0}^{n}\binom{n}{k}w_{k}\left(  y_{1}\right)  w_{n-k}\left(
y_{2}\right)  =\frac{y_{2}w_{n}\left(  y_{2}\right)  -y_{1}w_{n}\left(
y_{1}\right)  }{y_{2}-y_{1}}. \label{23}
\end{equation}
\bigskip
\end{theorem}


\begin{proof}
We write
\begin{align}
&amp;  \frac{e^{x_{1}t}}{\left(  1-y_{1}\left(  e^{t}-1\right)  \right)  }
\frac{e^{x_{2}t}}{\left(  1-y_{2}\left(  e^{t}-1\right)  \right)  }
\label{69}\\
&amp;  \qquad\qquad=\frac{y_{2}}{y_{2}-y_{1}}\frac{e^{\left(  x_{1}+x_{2}\right)
t}}{1-y_{2}\left(  e^{t}-1\right)  }-\frac{y_{1}}{y_{2}-y_{1}}\frac{e^{\left(
x_{1}+x_{2}\right)  t}}{1-y_{1}\left(  e^{t}-1\right)  }.\nonumber
\end{align}
Using the same method as in the proof of Theorem \ref{teo1}, we have{}
\[
\sum_{k=0}^{n}\binom{n}{k}w_{k}\left(  x_{1};y_{1}\right)  w_{n-k}\left(
x_{2};y_{2}\right)  =\frac{y_{2}w_{n}\left(  x_{1}+x_{2};y_{2}\right)
-y_{1}w_{n}\left(  x_{1}+x_{2};y_{1}\right)  }{y_{2}-y_{1}}.
\]
Setting $x_{1}=x_{2}=0$ in the above equation gives the desired equation.
\end{proof}


As we know, for $y=1$, geometric polynomials reduce to geometric numbers. We now
point out (see \eqref{15}) that geometric numbers also arise for other values of $y$. If we take $y-1$ in place of $y$ in (\ref{8}), we have
\begin{equation}
w_{n}\left(  x;y-1\right)  =\left(  -1\right)  ^{n}w_{n}\left(  1-x;-y\right)
.\label{18}
\end{equation}
Setting $x=0$ in the above equation and using the relation (\ref{7}), we have
a reflection formula

\begin{equation}
w_{n}\left(  y\right)  =\left(  -1\right)  ^{n}\frac{y}{y+1}w_{n}\left(
-y-1\right)  ,\text{ \ }n&gt;0.\label{19}
\end{equation}
Therefore, employing (\ref{2}) gives a new explicit formula for geometric polynomials as
in the following theorem.

\begin{theorem}
\label{teo4}For $n&gt;0$, we obtain
\begin{equation}
w_{n}\left(  y\right)  =y\sum_{k=1}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
\left(  -1\right)  ^{n+k}k!\left(  y+1\right)  ^{k-1}. \label{21}
\end{equation}
\end{theorem}


Note that, when $y=1$ (\ref{21}) reduces to \cite[Theorem\ 4.2]{GuoandQi}.
Moreover, from (\ref{19}), we get two conclusions
\begin{equation}
w_{2k}\left(  \frac{-1}{2}\right)  =0\text{ and }w_{n}\left(  -2\right)
=\left(  -1\right)  ^{n}2w_{n}.\label{15}
\end{equation}
The first part of (\ref{15}) was given in \cite[Corollary\
17]{BoyadzhievandDil}. If we take $y_{1}=-2$ and $y_{2}=1$ in (\ref{23}) and use the second
part of (\ref{15}), we obtain the alternating sums of products of geometric numbers.

\begin{corollary}
For $n&gt;0$, we have
\begin{displaymath}
\sum_{k=0}^{n}\binom{n}{k}\left(  -1\right)  ^{k}w_{k}w_{n-k}=
\begin{cases}
0, &amp; \text{if $n$ is odd;}\\
\frac{4}{3}w_{n}, &amp; \text{if $n$ is even.}
\end{cases}
\label{17}
\end{displaymath}

\end{corollary}


Finally, we obtain a new explicit formula for geometric polynomials and numbers
in the following theorem.

\begin{theorem}
\label{teo5}For $y\neq\frac{-1}{2}$ and $n\geq0$,
\begin{equation}
w_{n}\left(  y\right)  =\sum_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
k!y^{k}\frac{\bigl(  2^{n+1}\left(  y+1\right)  y^{k}+\left(  -1\right)
^{k+1}\bigr)  }{\left(  2y+1\right)  ^{k+1}}. \label{84}
\end{equation}
When $y=1$ this becomes
\begin{equation}
w_{n}=\sum_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
k!\frac{\bigl(  2^{n+2}+\left(  -1\right)  ^{k+1}\bigr)  }{3^{k+1}},\text{
\ }n\geq0. \label{85}
\end{equation}
When $y=-2$ this becomes
\begin{equation}
w_{n}=\sum_{k=0}^{n}\left(  -1\right)  ^{n-k}
\genfrac{\{}{\}}{0pt}{}{n}{k}
k!\frac{2^{k-1}\bigl(  2^{n+k+1}+1\bigr)  }{3^{k+1}},\text{ \ } n&gt;0. \label{86}
\end{equation}
\end{theorem}


\begin{proof}
If we take $\frac{1}{y^{2}-1}$ in place of $y$ in (\ref{10}), we arrive at
\begin{equation}
\frac{1}{1-\frac{1}{y^{2}-1}\left(  e^{2t}-1\right)  }=\frac{y^{2}-1}{2y^{2}
}\bigl(  \frac{1}{y-e^{t}}+\frac{1}{y+e^{t}}\bigr)  . \label{79}
\end{equation}
Each of the functions in the above equation can be written as
\begin{align}
\frac{1}{1-\frac{1}{y^{2}-1}\left(  e^{2t}-1\right)  }  &amp;  =\sum_{n=0}
^{\infty}2^{n}w_{n}\left(  \frac{1}{y^{2}-1}\right)  \frac{t^{n}}
{n!},\label{80}\\
\frac{1}{y-e^{t}}  &amp;  =\frac{y}{y-1}\sum_{n=0}^{\infty}w_{n}\left(  \frac
{1}{y-1}\right)  \frac{t^{n}}{n!},\label{81}\\
\frac{1}{y+e^{t}}  &amp;  =\frac{y}{y+1}\sum_{n=0}^{\infty}w_{n}\left(  \frac
{-1}{y+1}\right)  \frac{t^{n}}{n!}. \label{82}
\end{align}
By equating the coefficients of $\frac{t^{n}}{n!}$, we have
\begin{equation}
w_{n}\left(  y\right)  =2^{n+1}\left(  1+y\right)  w_{n}\left(  \frac{y^{2}
}{1+2y}\right)  -\left(  1+2y\right)  w_{n}\left(  -y\right)  . \label{24}
\end{equation}
Finally, using (\ref{2}) in the right hand side of the above equation yields
(\ref{84}).
\end{proof}


\section{Integrals of products of geometric polynomials}\label{sec3}

In this section, we deal with an integral for a product of two geometric
polynomials. First we need following lemmas.

\begin{lemma}
\label{lem1}For all $k\geq0$ and $n\geq1$, we have
\begin{equation}
{\displaystyle\int\limits_{-1}^{0}}
y^{k}w_{n}\left(  y\right)  dy=\frac{\left(  -1\right)  ^{k}}{k!}\sum
_{j=0}^{k}
\genfrac{[}{]}{0pt}{}{k+1}{j+1}
B_{n+j}, \label{25}
\end{equation}
where $\genfrac{[}{]}{0pt}{}{n}{k}$ is the Stirling number of the fist kind \cite{Graham}.
\end{lemma}


\begin{proof}
We prove (\ref{25}) by induction on $k$. The case $k=0$ of (\ref{25}) is known
from (\ref{26}). If we integrate both sides of (\ref{11}) with respect to $y$
from $-1$ to $0$ and apply integration by parts, we have
\begin{align*}
{\displaystyle\int\limits_{-1}^{0}}
w_{n+1}\left(  y\right)  dy  &amp;  =
{\displaystyle\int\limits_{-1}^{0}}
y\frac{d}{dy}\bigl(  w_{n}\left(  y\right)  +yw_{n}\left(  y\right)  \bigr)dy
\\
&amp;  =\left[  y\left(  w_{n}\left(  y\right)  +yw_{n}\left(  y\right)  \right)
\right]  _{-1}^{0}-
{\displaystyle\int\limits_{-1}^{0}}
\bigl(  w_{n}\left(  y\right)  +yw_{n}\left(  y\right)  \bigr)  dy.
\end{align*}
So, using (\ref{26}) yields the case $k=1$ of (\ref{25}) as
\[
{\displaystyle\int\limits_{-1}^{0}}
yw_{n}\left(  y\right)  dy=-\left(  B_{n+1}+B_{n}\right)  .
\]
Multiplying both sides of (\ref{11}) by $y$ and integrating it with respect
to $y$ from $-1$ to $0$, we obtain
\[
{\displaystyle\int\limits_{-1}^{0}}
yw_{n+1}\left(  y\right)  dy=
{\displaystyle\int\limits_{-1}^{0}}
y^{2}\frac{d}{dy}\bigl(  w_{n}\left(  y\right)  +yw_{n}\left(  y\right)
\bigr)dy  .
\]
Applying integration by parts and using (\ref{26}) yield the case $k=2$ of
(\ref{25}) as
\[
2{\displaystyle\int\limits_{-1}^{0}}
y^{2}w_{n}\left(  y\right)  dy=B_{n+2}+3B_{n+1}+2B_{n}.
\]
If we multiply both sides of (\ref{11}) by $y^{k}$ and integrating it with
respect to $y$ from $-1$ to $0$, we obtain
\[
{\displaystyle\int\limits_{-1}^{0}}
y^{k}w_{n+1}\left(  y\right)  dy=
{\displaystyle\int\limits_{-1}^{0}}
y^{k+1}\frac{d}{dy}\bigl(  w_{n}\left(  y\right)  +yw_{n}\left(  y\right)
\bigr)dy  .
\]
Applying integration by parts to the right hand side of the above equation and
considering
\[
{\displaystyle\int\limits_{-1}^{0}}
y^{k}w_{n}\left(  y\right)  dy=\frac{\left(  -1\right)  ^{k}}{k!}\sum
_{j=0}^{k}
\genfrac{[}{]}{0pt}{}{k+1}{j+1}
B_{n+j},
\]
we have
\[
{\displaystyle\int\limits_{-1}^{0}}
y^{k+1}F_{n+1}\left(  y\right)  dy=\frac{\left(  -1\right)  ^{k+1}}{\left(
k+1\right)  !}\sum_{j=0}^{k}
\genfrac{[}{]}{0pt}{}{k+1}{j+1}
B_{n+j+1}+\frac{\left(  -1\right)  ^{k+1}}{\left(  k+1\right)  !}\sum
_{j=0}^{k}\left(  k+1\right)
\genfrac{[}{]}{0pt}{}{k+1}{j+1}
B_{n+j}.
\]
Finally, the well-known relations
\[
\genfrac{[}{]}{0pt}{}{n+1}{k}
=n
\genfrac{[}{]}{0pt}{}{n}{k}
+
\genfrac{[}{]}{0pt}{}{n}{k-1}
\text{ and }
\genfrac{[}{]}{0pt}{}{n}{1}
=\left(  n-1\right)  !,
\]
give that the statement is true for $k+1$.
\end{proof}


\begin{lemma}
\label{lem2}For any non-negative integer $m$ and $j$,
\begin{equation}
\sum_{k=j}^{m}
\genfrac{\{}{\}}{0pt}{}{m}{k}
\genfrac{[}{]}{0pt}{}{k+1}{j+1}
(-1)^{k}=(-1)^{m}\binom{m}{j}. \label{33}
\end{equation}
\end{lemma}


\begin{proof}
We rewrite this equation into matrix form by using the matrices
\[
(\mathcal{S}_{1})_{i,j}=(-1)^{i+j}
\genfrac{[}{]}{0pt}{}{i+1}{j+1}
,\quad(\mathcal{S}_{2})_{i,j}=
\genfrac{\{}{\}}{0pt}{}{i}{j}
,\quad(\mathcal{B})_{i,j}=\binom{i}{j}.
\]
These can be considered as infinite matrices so that the statement we are
going to prove takes the form
\[
\mathcal{S}_{2}\mathcal{S}_{1}=\mathcal{B}^{-1},
\]
where the element-wise inverse of the matrix $\mathcal{B}$ is $(\mathcal{B}
)_{i,k}^{-1}=(-1)^{i+k}\binom{i}{k}$. The above equation is equivalent to
\[
\mathcal{S}_{1}=\mathcal{B}^{-1}\mathcal{S}_{2}^{-1}=\left(  \mathcal{S}
_{2}\mathcal{B}\right)  ^{-1}.
\]
The matrix on the right hand side is easily decipherable. Element-wise, it is
\[
(\left(  \mathcal{S}_{2}\mathcal{B}\right)  ^{-1})_{i,j}=\sum_{k=0}^{i}
\genfrac{\{}{\}}{0pt}{}{i}{k}
\binom{k}{j}.
\]
The latter sum simply equals to
\[
\sum_{k=0}^{i}
\genfrac{\{}{\}}{0pt}{}{i}{k}
\binom{k}{j}=
\genfrac{\{}{\}}{0pt}{}{i+1}{j+1}
\]
\cite[Eq.\ 6.15]{Graham}. Hence, our original
statement equals to the matrix equation
\[
(\mathcal{S}_{1})_{i,j}^{-1}=
\genfrac{\{}{\}}{0pt}{}{i+1}{j+1}
.
\]
This is nothing but the reformulation of the fact that the second and
signed first kind Stirling matrices are inverses of each other.
\end{proof}


Now, we are ready to give the integrals of products of geometric polynomials.
Using (\ref{2}), we have
\[
{\displaystyle\int\limits_{-1}^{0}}
w_{m}\left(  y\right)  w_{n}\left(  y\right)  dy=
{\displaystyle\int\limits_{-1}^{0}}
\sum_{k=0}^{m}
\genfrac{\{}{\}}{0pt}{}{m}{k}
k!y^{k}w_{n}\left(  y\right)  dy.
\]
Then, interchanging the sum and integral in the above equation and using
(\ref{25}) yield
\[
{\displaystyle\int\limits_{-1}^{0}}
w_{m}\left(  y\right)  w_{n}\left(  y\right)  dy=\sum_{j=0}^{m}\sum_{k=j}^{m}
\genfrac{\{}{\}}{0pt}{}{m}{k}
\genfrac{[}{]}{0pt}{}{k+1}{j+1}
\left(  -1\right)  ^{k}B_{n+j}.
\]
Finally, using Lemma \ref{lem2} gives the following theorem.

\begin{theorem}
\label{teo2}For all $m\geq0$ and $n\geq1$, we have
\begin{equation}
{\displaystyle\int\limits_{-1}^{0}}
w_{m}\left(  y\right)  w_{n}\left(  y\right)  dy=(-1)^{m}\sum_{j=0}^{m}
\binom{m}{j}B_{n+j}. \label{30}
\end{equation}
\end{theorem}


Using the representation (\ref{2}) in (\ref{30}) and integrating termwise, one
obtains
\[
\sum_{k=0}^{n}\sum_{j=0}^{m}
\genfrac{\{}{\}}{0pt}{}{n}{k}
\genfrac{\{}{\}}{0pt}{}{m}{j}
\frac{\left(  -1\right)  ^{k+j}k!j!}{k+j+1}=(-1)^{m}\sum_{j=0}^{m}\binom{m}
{j}B_{n+j}.
\]
This double sum identity extends to the explicit formula \cite[p.\ 560]{Graham}
\[
B_{n}=\sum_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
\left(  -1\right)  ^{k}\frac{k!}{k+1}.
\]


In order to give an application of Lemma \ref{lem1}, we now emphasize the
summation in the right hand of (\ref{25}). Rahmani \cite{Rahmani} defined
$p$-Bernoulli numbers as
\[
\sum_{n=0}^{\infty}B_{n,p}\frac{t^{n}}{n!}=\text{ }_{2}F_{1}\left(
1,1;p+2;1-e^{t}\right)  ,
\]
where $_{2}F_{1}\left(  a,b;c;z\right)  $ denotes the Gaussian hypergeometric
function \cite{Andrews}. These numbers can be written in terms of Stirling
numbers of the first kind as follows:
\[
\sum_{j=0}^{p}\left(  -1\right)  ^{j}
\genfrac{[}{]}{0pt}{}{p}{j}
B_{n+j}=\frac{p!}{p+1}B_{n,p},\text{ \ }n,p\geq0.
\]
From the above equation, we have
\begin{equation}
\sum_{j=0}^{p}\left(  -1\right)  ^{j+1}
\genfrac{[}{]}{0pt}{}{p+1}{j+1}
B_{n+j}=\frac{\left(  p+1\right)  !}{p+2}B_{n-1,p+1},\text{ \ }n\geq
1,p\geq0.\label{27}
\end{equation}
Moreover, when $n$\ is odd or even, we have
\[
\left(  -1\right)  ^{j+1}B_{n+j}=B_{n+j}\ \text{or }\left(  -1\right)
^{j+1}B_{n+j}=-B_{n+j},\text{ \ }n&gt;1,
\]
\ respectively. Therefore, we obtain
\begin{displaymath}
\sum_{j=0}^{p}
\genfrac{[}{]}{0pt}{}{p+1}{j+1}
B_{n+j}=
\begin{cases}
\frac{\left(  p+1\right)  !}{p+2}B_{n-1,p+1}, &amp; \text{if $n$ is odd;}\\
-\frac{\left(  p+1\right)  !}{p+2}B_{n-1,p+1}, &amp; \text{if $n$ is even.}
\end{cases}
\end{displaymath}
Using the above equation, (\ref{25}) can be written as
\begin{displaymath}
{\displaystyle\int\limits_{-1}^{0}}
y^{p}w_{n}\left(  y\right)  dy=
\begin{cases}
\left(  -1\right)  ^{p}\frac{p+1}{p+2}B_{n-1,p+1}, &amp; \text{if $n$ is
odd;}\\
\left(  -1\right)  ^{p+1}\frac{p+1}{p+2}B_{n-1,p+1}, &amp; \text{if $n$ is
even,}
\end{cases}
\label{28}
\end{displaymath}
where $n&gt;1$, $p\geq0$. On the other hand, using (\ref{2}) in the left part of
(\ref{28}), a new explicit formula for $p$-Bernoulli numbers is obtained.

\begin{theorem}
For $n&gt;1$ and $p&gt;0$,
\[
B_{2n-1,p}=\frac{p+1}{p}\sum_{k=0}^{2n-1}
\genfrac{\{}{\}}{0pt}{}{2n-1}{k+1}
\frac{\left(  -1\right)  ^{k+1}\left(  k+1\right)  !}{k+p+1}
\]
and
\[
B_{2n,p}=\frac{p+1}{p}\sum_{k=0}^{2n}
\genfrac{\{}{\}}{0pt}{}{2n+1}{k+1}
\frac{\left(  -1\right)  ^{k}\left(  k+1\right)  !}{k+p+1}.
\]
\end{theorem}


\section{Applications}\label{sec4}

Apostol-Bernoulli functions $\mathcal{B}_{n}\left(  \lambda\right)  $ have the
following explicit expression:
\begin{equation}
\mathcal{B}_{n}\left(  \lambda\right)  =\frac{n}{\lambda-1}\sum_{k=0}^{n-1}
\genfrac{\{}{\}}{0pt}{}{n-1}{k}
k!\left(  \frac{\lambda}{1-\lambda}\right)  ^{k},\,\ \ \lambda\in
\mathbb{C}
\backslash\{1\}.\label{53}
\end{equation}
Thus, for $\lambda\neq1$,
\[
\mathcal{B}_{0}\left(  \lambda\right)  =0,\text{ }\mathcal{B}_{1}\left(
\lambda\right)  =\frac{1}{\lambda-1},\text{ }\mathcal{B}_{2}\left(
\lambda\right)  =\frac{-2\lambda}{\left(  \lambda-1\right)  ^{2}
},\ \ldots\ \
\]
The functions $\mathcal{B}_{n}\left(  \lambda\right)  $ are rational functions
in the variable $\lambda$. Apostol \cite{APOSTOL} introduced these functions in order to evaluate the Lerch transcendent (also known as the Lerch zeta
function) for negative integer values of $s$. Also, these functions were studied and
generalized recently in a number of papers, under the name Apostol-Bernoulli numbers.

Comparing (\ref{53}) and (\ref{2}), Boyadzhiev \cite{B3} showed that Apostol-Bernoulli functions can be
expressed by geometric polynomials as
\begin{equation}
\mathcal{B}_{n+1}\left(  \lambda\right)  =\frac{n+1}{\lambda-1}w_{n}\left(
\frac{\lambda}{1-\lambda}\right)  ,\text{ \ \ \ }\lambda\in
\mathbb{C}
\backslash\{1\}.\label{22}
\end{equation}


We can use this relation to obtain some new properties of $\mathcal{B}
_{n}\left(  \lambda\right)$. For example, setting $y=\frac{-\lambda}
{\lambda-1}$ in (\ref{21}), we have \cite[Theorem\ 4.3]{GuoandQi}
\[
\frac{\mathcal{B}_{n+1}\left(  \lambda\right)  }{\left(  n+1\right)  }=\left(
-1\right)  ^{n}\lambda\sum_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
k!\left(  \frac{1}{\lambda-1}\right)  ^{k+1},\text{ \ }\lambda\neq1,\text{
}n\geq0.
\]
Similarly, from Theorem \ref{teo1}, we get the sums of products of Apostol-Bernoulli functions as given
in \cite[Corollary\ 1.3]{KimandHu}. Moreover, using equation (\ref{84}) of
Theorem \ref{teo5} gives a new explicit formula for Apostol-Bernoulli functions.

\begin{corollary}
For $\lambda\neq\pm1$ and $n\geq0$,
\[
\frac{\mathcal{B}_{n+1}\left(  \lambda\right)  }{\left(  n+1\right)  }
=\sum_{k=0}^{n}
\genfrac{\{}{\}}{0pt}{}{n}{k}
k!\frac{\left(  -\lambda\right)  ^{k}\bigl(  2^{n+1}\lambda^{k}+\left(
\lambda-1\right)  ^{k+1}\bigr)  }{\left(  \lambda^{2}-1\right)  ^{k+1}}.
\]
\end{corollary}


To give a different application of the relation (\ref{22}), we first deal with
Lemma \ref{lem1}. Replacing $y$ with $\frac{\lambda}{1-\lambda}$ in
(\ref{25}), we have
\[
{\displaystyle\int\limits_{-\infty}^{0}}
\frac{\lambda^{k}}{\left(  \lambda-1\right)  ^{k+1}}\mathcal{B}_{n+1}\left(
\lambda\right)  d\lambda=\frac{n+1}{k!}\sum_{j=0}^{k}
\genfrac{[}{]}{0pt}{}{k+1}{j+1}
B_{n+j},
\]
where $k\geq0$ and $n\geq1$. Then, from Theorem \ref{teo2}, we obtain the
integrals of products of Apostol-Bernoulli functions as given in the following corollary.

\begin{corollary}
For all $m\geq0$ and $n\geq1$, we have
\[
{\displaystyle\int\limits_{-\infty}^{0}}
\mathcal{B}_{m}\left(  \lambda\right)  \mathcal{B}_{n}\left(  \lambda\right)
d\lambda=(-1)^{m}\left(  m+1\right)  \left(  n+1\right)  \sum_{j=0}^{m}
\binom{m}{j}B_{n+j}.
\]
\end{corollary}


\section{Acknowledgments}

The author would like to thank to Professor Istvan Mez\H{o} for his valuable
suggestions regarding improvement of the paper.

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\bigskip\hrule
\bigskip

\noindent2010 \textit{Mathematics Subject Classification}: Primary 11B83;
Secondary 11B68, 11B75.

\noindent\emph{Keywords: } geometric number, geometric
polynomial, Apostol-Bernoulli
function, $p$-Bernoulli number.

\bigskip\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000670}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received September 21 2016;
revised version received  February 2 2017; February 6 2017.
Published in {\it Journal of Integer Sequences}, February 9 2017.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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