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\begin{center}
\vskip 1cm{\LARGE\bf
On Reciprocity Formulas for Apostol's \\
\vskip .1in
Dedekind Sums and Their Analogues
}
\vskip 1cm
\large
M. Cihat Da\u{g}l\i\ and M\"{u}m\"{u}n Can\\
Department of Mathematics\\ Akdeniz University\\ TR-07058 Antalya\\ Turkey\\
\href{mailto:mcihatdagli@akdeniz.edu.tr}{\tt mcihatdagli@akdeniz.edu.tr} \\
\href{mailto:mcan@akdeniz.edu.tr}{\tt mcan@akdeniz.edu.tr}
\end{center}

\vskip .2 in

\begin{abstract}
Using the Euler-MacLaurin summation formula, we give alternative proofs for the
reciprocity formulas of Apostol's Dedekind sums and generalized Hardy-Berndt
sums $s_{3,p}(b,c)$ and $s_{4,p}(b,c)$. We also obtain an integral
representation for each sum.
\end{abstract}

\section{Introduction}

Let%
\[
\left(  \left(  x\right)  \right)  =%
\begin{cases}
x-\left[  x\right]  -1/2, &amp; \text{if }x\in\mathbb{R}\backslash\mathbb{Z}%
\text{;}\\
0, &amp; \text{if }x\in\mathbb{Z}\text{,}%
\end{cases}
\]
with $[x]$\ being the largest integer $\leq x$. For positive integers $c$ and
integers $b$ the classical Dedekind sum $s(b,c)$, arising in the theory of
Dedekind $\eta$-function, was introduced by R. Dedekind in 1892 as
\[
s(b,c)=\sum\limits_{m(\text{mod }c)}\left(  \left(  \frac{m}{c}\right)
\right)  \left(  \left(  \frac{bm}{c}\right)  \right)  .
\]
The most important property of Dedekind sums is the reciprocity
theorem\textbf{\ }%
\[
s(b,c)+s(c,b)=-\frac{1}{4}+\frac{1}{12}\left(  \frac{b}{c}+\frac{c}{b}%
+\frac{1}{bc}\right)
\]
when $\gcd\left(  b,c\right)  =1$. The standard reference for Dedekind sums is
Rademacher and Grosswald \cite{32}. Several generalizations of Dedekind sums
have been defined and the corresponding reciprocity formulas have been
obtained. One of these generalizations, due to Apostol \cite{1}, is
\[
s_{p}(b,c)=\sum\limits_{m=1}^{c-1}\frac{m}{c}\overline{B}_{p}\left(  \frac
{bm}{c}\right)  ,
\]
where $\overline{B}_{p}(x)$ is the $p$th Bernoulli function defined by
\[
\overline{B}_{1}\left(  x\right)  =\left(  \left(  x\right)  \right)  ,\text{
and }\overline{B}_{p}(x+m)=B_{p}(x)\text{ for }0\leq x&lt;1,\text{ }%
m\in\mathbb{Z}\text{ and }p&gt;1.
\]
Here $B_{p}(x)$ is the $p$th Bernoulli polynomial. Apostol's reciprocity
formula is as follows.

\begin{theorem}
\label{dedekind}Let $b$ and $c$ be coprime positive integers. For odd
$p\geq1,$ we have%
\[
\left(  p+1\right)  \left(  bc^{p}s_{p}(b,c)+cb^{p}s_{p}(c,b)\right)
=\sum\limits_{j=0}^{p+1}\binom{p+1}{j}(-1)^{j}b^{p+1-j}c^{j}B_{p+1-j}%
B_{j}+pB_{p+1}.
\]

\end{theorem}

\noindent Here $B_{p}=B_{p}(0)$ is the $p$th Bernoulli number.

Similar arithmetic sums arise in the theory of logarithms of the classical
theta functions. They were studied by Hardy and Berndt, and for this reason
they are called Hardy or Hardy-Berndt sums. There are six such sums, two of
which are \cite{7,19}\textbf{\ }%
\[
s_{3}(b,c)=\sum\limits_{m=1}^{c-1}\left(  -1\right)  ^{m}\overline{B}%
_{1}\left(  \frac{bm}{c}\right)  ,\text{ }s_{4}(b,c)=-4\sum\limits_{m=1}%
^{c-1}\overline{B}_{1}\left(  \frac{bm}{2c}\right)  .
\]
Goldberg \cite{19} showed that these sums also arise in the theory of
$r_{m}(n)$, the number of representations of $n$ as a sum of $m$ integral
squares and in the study of the Fourier coefficients of the reciprocals of the
classical theta functions. Like Dedekind sums, Hardy-Berndt sums also satisfy
a reciprocity (or reciprocity-like) formula \cite{7,19}%
\[
2s_{3}(b,c)-s_{4}(c,b)=1-\frac{b}{c}%
\]
when $c$ is odd and $\gcd\left(  b,c\right)  =1$. The generalizations of these
sums in the sense of Apostol have been given in \cite{12} by%
\[
s_{3,p}(b,c)=\sum\limits_{m=1}^{c-1}\left(  -1\right)  ^{m}\overline{B}%
_{p}\left(  \frac{bm}{c}\right)  ,\text{ \ }s_{4,p}(b,c)=-4\sum\limits_{m=1}%
^{c-1}\overline{B}_{p}\left(  \frac{bm}{2c}\right)
\]
which satisfy the following reciprocity formula.

\begin{theorem}
\label{hb}Let $b$ and $c$ be coprime positive integers with $c$ odd. For odd
$p\geq1,$ we have%
\begin{align*}
&amp;  \left(  p+1\right)  \left(  2bc^{p}s_{3,p}\left(  b,c\right)
-2^{-1}c\left(  2b\right)  ^{p}s_{4,p}\left(  c,b\right)  \right) \\
&amp;  \ =4\sum\limits_{j=1}^{p+1}\binom{p+1}{j}\left(  -1\right)  ^{j}%
b^{j}c^{p+1-j}\left(  1-2^{j}\right)  B_{j}B_{p+1-j}.
\end{align*}

\end{theorem}

The reciprocity formulas in this concept are proved by employing various
techniques and theories such as transformation formulas, residue theory,
Franel integral and arithmetic methods.

In this study we give rather elementary but new proofs for Theorems
\ref{dedekind} and \ref{hb} when $p&gt;1$\ by applying the Euler-MacLaurin
summation formula to Bernoulli function.

The method presented in the sequel is motivated by \cite{29}.

\section{Proofs of the reciprocity theorems}

Let us state the Euler-MacLaurin summation formula, which can be found in
various books, for example, \cite[p.\ 22]{HC}.

\begin{theorem}
\label{E-M}\textbf{(Euler-MacLaurin)} Let $\alpha$ and $\beta$ be real numbers
such that $\alpha\leq\beta$ and assume that $f\in C^{\left(  l\right)
}\left[  \alpha,\beta\right]  $ for some $l\geq1.$ Then%
\begin{align*}
\sum_{\alpha&lt;m\leq\beta}f(m)=  &amp;
%TCIMACRO{\dint \limits_{\alpha}^{\beta}}%
%BeginExpansion
{\displaystyle\int\limits_{\alpha}^{\beta}}
%EndExpansion
f(u)du+\sum\limits_{j=1}^{l}\frac{(-1)^{j}}{j!}\left(  \overline{B}_{j}\left(
\beta\right)  f^{(j-1)}(\beta)-\overline{B}_{j}\left(  \alpha\right)
f^{(j-1)}(\alpha)\right) \\
&amp;  +\frac{(-1)^{l-1}}{l!}%
%TCIMACRO{\dint \limits_{\alpha}^{\beta}}%
%BeginExpansion
{\displaystyle\int\limits_{\alpha}^{\beta}}
%EndExpansion
f^{(l)}(u)\overline{B}_{l}\left(  u\right)  du.
\end{align*}

\end{theorem}

We will also need the facts\textbf{\ }$\overline{B}_{2r+1}(0)=\overline
{B}_{2r+1}(1/2)=0$ for all $r\geq0$ and Raabe's formula
\begin{equation}
\sum\limits_{m=0}^{n-1}\overline{B}_{r}\left(  x+\frac{m}{n}\right)
=n^{1-r}\overline{B}_{r}\left(  nx\right)  . \label{raabe}%
\end{equation}


Firstly, we consider the function $f\left(  x\right)  =\overline{B}_{p}\left(
xy\right)  $, $y\in\mathbb{R}$. The property
\[
\frac{d}{dx}\overline{B}_{p}\left(  x\right)  =p\overline{B}_{p-1}\left(
x\right)  ,\text{ }p&gt;2
\]
entails that
\begin{equation}
\frac{d^{j}}{dx^{j}}f\left(  x\right)  =\frac{d^{j}}{dx^{j}}\overline{B}%
_{p}\left(  xy\right)  =y^{j}\frac{p!}{\left(  p-j\right)  !}\overline
{B}_{p-j}\left(  xy\right)  \label{3}%
\end{equation}
for $1\leq j\leq p-2$ and $f\in C^{\left(  p-2\right)  }\left[  \alpha
,\beta\right]  .$ For $\alpha=0$ and $1\leq l\leq p-2,$ Theorem \ref{E-M} can
be written as%
\begin{align}
&amp;  \sum_{1\leq m\leq\beta}\overline{B}_{p}\left(  my\right) \nonumber\\
&amp;  \ \quad=\frac{1}{p+1}\sum\limits_{j=1}^{l}\left(  -1\right)  ^{j}%
\binom{p+1}{j}y^{j-1}\left(  \overline{B}_{j}\left(  \beta\right)
\overline{B}_{p-j+1}\left(  \beta y\right)  -\overline{B}_{p-j+1}%
(0)\overline{B}_{j}\left(  0\right)  \right) \nonumber\\
&amp;  \qquad+\frac{1}{y}\frac{\overline{B}_{p+1}\left(  \beta y\right)
-\overline{B}_{p+1}\left(  0\right)  }{p+1}-\left(  -y\right)  ^{l}\binom
{p}{l}\int\limits_{0}^{\beta}\overline{B}_{l}\left(  u\right)  \overline
{B}_{p-l}\left(  yu\right)  du. \label{23}%
\end{align}


Let $y=b/c$ and $\beta=c\in\mathbb{N}.$ Then (\ref{23}) becomes
\begin{equation}
\sum_{m=1}^{c}\overline{B}_{p}\left(  \frac{bm}{c}\right)  =-\binom{p}%
{l}\left(  -\frac{b}{c}\right)  ^{l}\int\limits_{0}^{c}\overline{B}_{l}\left(
u\right)  \overline{B}_{p-l}\left(  \frac{b}{c}u\right)  du. \label{24}%
\end{equation}


$\bullet$ For $b=c$ in (\ref{24}) we have the well-known relation \cite[p.\ 120]{HC}
\begin{equation}
\int\limits_{0}^{1}\overline{B}_{l}\left(  u\right)  \overline{B}_{r}\left(
u\right)  du=(-1)^{l-1}\frac{l!r!}{\left(  r+l\right)  !}B_{l+r},\text{ for
}l\geq1\text{ and }p-l=r\geq2.\nonumber
\end{equation}


$\bullet$ Let $\gcd\left(  b,c\right)  =1$.\ Then, it follows from
(\ref{raabe}) and (\ref{24}) that
\begin{equation}
\int\limits_{0}^{1}\overline{B}_{l}\left(  cu\right)  \overline{B}%
_{p-l}\left(  bu\right)  du=\frac{\left(  -1\right)  ^{l-1}}{\binom{p}{l}%
}\frac{c^{l-p}}{b^{l}}B_{p}. \label{8}%
\end{equation}


$\bullet$ Now assume that $\gcd\left(  b,c\right)  =q$ and put $c=qc_{1}%
,$\ $b=qb_{1}.$\ From (\ref{8}), we have
\[
\int\limits_{0}^{1}\overline{B}_{l}\left(  cu\right)  \overline{B}%
_{p-l}\left(  bu\right)  du=c\int\limits_{0}^{1}\overline{B}_{l}\left(
c_{1}u\right)  \overline{B}_{p-l}\left(  b_{1}u\right)  du=cq^{p}\frac{\left(
-1\right)  ^{l-1}}{\binom{p}{l}}\frac{c^{l-p}}{b^{l}}B_{p}.
\]


\subsection{Proof of Theorem \ref{dedekind}}

Let $f(x)=x\overline{B}_{p}\left(  xy\right)  ,$ $y\in\mathbb{R}.$ Then $f\in
C^{\left(  p-2\right)  }\left[  \alpha,\beta\right]$. From (\ref{3}) and the
Leibniz rule for the derivative we have, for $1\leq j\leq p-2,$%
\[
\frac{d^{j}}{dx^{j}}\left[  x\overline{B}_{p}\left(  xy\right)  \right]
=y^{j}\frac{p!}{\left(  p-j\right)  !}x\overline{B}_{p-j}\left(  xy\right)
+y^{j-1}\frac{p!}{\left(  p+1-j\right)  !}j\overline{B}_{p+1-j}\left(
xy\right)  .
\]


For $f(x)=x\overline{B}_{p}\left(  xy\right)  ,$ $\alpha=0,$ $\beta=c$ and
$y=b/c$ with $\gcd\left(  b,c\right)  =1$ in Theorem \ref{E-M}, we have%
\begin{align}
&amp;  \sum_{m=1}^{c}m\overline{B}_{p}\left(  m\frac{b}{c}\right) \nonumber\\
&amp;  =c^{2}\int\limits_{0}^{1}x\overline{B}_{p}\left(  bx\right)  dx+\frac
{c}{p+1}\sum\limits_{j=1}^{l}(-1)^{j}\binom{p+1}{j}\left(  \frac{b}{c}\right)
^{j-1}\overline{B}_{j}\left(  0\right)  \overline{B}_{p+1-j}\left(  0\right)
\nonumber\\
&amp;  +\binom{p}{l}\left(  -\frac{b}{c}\right)  ^{l-1}c\int\limits_{0}%
^{1}\overline{B}_{l}\left(  cx\right)  \left(  bx\overline{B}_{p-l}\left(
bx\right)  +\frac{l}{p+1-l}\overline{B}_{p+1-l}(bx)\right)  dx, \label{1}%
\end{align}
where $1\leq l\leq p-2.$ It follows from integration by parts that%
\begin{equation}
\int\limits_{0}^{1}x\overline{B}_{p}\left(  bx\right)  dx=\frac{1}%
{b(p+1)}\overline{B}_{p+1}\left(  0\right)  \label{2}%
\end{equation}
and from (\ref{8})
\begin{equation}
(-1)^{l-1}\binom{p+1}{l}\left(  \frac{b}{c}\right)  ^{l}c\int\limits_{0}%
^{1}\overline{B}_{l}\left(  cx\right)  \overline{B}_{p+1-l}\left(  bx\right)
dx=c^{-p}\overline{B}_{p+1}\left(  0\right)  . \label{4}%
\end{equation}
Combining (\ref{1}), (\ref{2}) and (\ref{4}), we have%
\begin{align}
&amp;  \sum_{m=1}^{c}m\overline{B}_{p}\left(  \frac{bm}{c}\right) \nonumber\\
&amp;  =\frac{c}{p+1}\sum\limits_{j=0}^{l}(-1)^{j}\binom{p+1}{j}\left(  \frac
{b}{c}\right)  ^{j-1}\overline{B}_{p+1-j}\left(  0\right)  \overline{B}%
_{j}\left(  0\right)  +\frac{l}{p+1}\frac{c^{1-p}}{b}B_{p+1}\nonumber\\
&amp;  +(-1)^{l-1}\binom{p}{l}\left(  \frac{b}{c}\right)  ^{l-1}bc\int%
\limits_{0}^{1}x\overline{B}_{l}\left(  cx\right)  \overline{B}_{p-l}\left(
bx\right)  dx. \label{5}%
\end{align}


It can be easily seen from (\ref{raabe}) and the fact $\overline{B}_{p}\left(
-x\right)  =\left(  -1\right)  ^{p}\overline{B}_{p}\left(  x\right)  $ that
2$s_{p}(b,c)=\left(  c^{1-p}-1\right)  B_{p}$ when $p$ is even. We then have
\begin{equation}
\sum_{m=1}^{c}m\overline{B}_{p}\left(  \frac{bm}{c}\right)  =\sum_{m=1}%
^{c-1}m\overline{B}_{p}\left(  \frac{bm}{c}\right)  +c\overline{B}_{p}\left(
0\right)  =%
\begin{cases}
cs_{p}(b,c), &amp; \text{if }p\text{ is odd;}\\
\left(  c^{1-p}+1\right)  \dfrac{c}{2}B_{p}, &amp; \text{if }p\text{ is even.}%
\end{cases}
\label{9}%
\end{equation}


$\bullet$ Let $p&gt;1$ be odd. Putting $l=2$ in (\ref{5}) and using (\ref{9}), we
get%
\begin{align}
cs_{p}(b,c)=  &amp;  \frac{c}{p+1}\left(  \frac{c}{b}\overline{B}_{p+1}\left(
0\right)  +\binom{p+1}{2}\frac{b}{c}\overline{B}_{p-1}\left(  0\right)
\overline{B}_{2}\left(  0\right)  \right) \nonumber\\
&amp;  +\frac{2}{p+1}\frac{c^{1-p}}{b}B_{p+1}-\binom{p}{2}b^{2}\int\limits_{0}%
^{1}x\overline{B}_{p-2}\left(  bx\right)  \overline{B}_{2}\left(  cx\right)
dx \label{6}%
\end{align}
Putting $l=p-2$ and interchanging $b$ and $c$\ in (\ref{5})
\begin{align}
bs_{p}(c,b)=  &amp;  \frac{b}{p+1}\sum\limits_{j=0}^{p-2}(-1)^{j}\binom{p+1}%
{j}\left(  \frac{c}{b}\right)  ^{j-1}\overline{B}_{p+1-j}\left(  0\right)
\overline{B}_{j}\left(  0\right)  +\frac{p-2}{p+1}\frac{b^{1-p}}{c}%
B_{p+1}\nonumber\\
&amp;  +(-1)^{p-1}\binom{p}{2}\left(  \frac{c}{b}\right)  ^{p-3}bc\int%
\limits_{0}^{1}x\overline{B}_{p-2}\left(  bx\right)  \overline{B}_{2}\left(
cx\right)  dx. \label{6a}%
\end{align}
Then, from (\ref{6}) and (\ref{6a}), we arrive at the reciprocity formula%
\[
\left(  p+1\right)  \left(  bc^{p}s_{p}(b,c)+cb^{p}s_{p}(c,b)\right)
=\sum\limits_{j=0}^{p+1}\binom{p+1}{j}(-1)^{j}b^{p+1-j}c^{j}B_{p+1-j}%
B_{j}+pB_{p+1}.
\]


Note that for $l=1$\ in (\ref{5}) we have the following integral
representation%
\begin{equation}
s_{p}(b,c)=\left(  c+c^{-p}\right)  \frac{1}{b}\frac{B_{p+1}}{p+1}%
+pb\int\limits_{0}^{1}x\overline{B}_{1}\left(  cx\right)  \overline{B}%
_{p-1}\left(  bx\right)  dx,\text{ for odd }p&gt;1.\nonumber
\end{equation}


$\bullet$ Let $p&gt;2$ be even. From (\ref{5}) and (\ref{9}) we have
\begin{equation}
b\int\limits_{0}^{1}x\overline{B}_{l}\left(  cx\right)  \overline{B}%
_{p-l}\left(  bx\right)  dx=\frac{1}{2}\left(  1+c^{1-p}\right)  \left(
-\frac{c}{b}\right)  ^{l-1}\binom{p}{l}^{-1}B_{p},\text{ for }1\leq l\leq
p-2.\nonumber
\end{equation}


\subsection{Proof of Theorem \ref{hb}}

We will use (\ref{23}) for the following cases;

\qquad\textbf{I)} $y=b/\left(  2c\right)  $ and $\beta=c,$

\qquad\textbf{II)} $y=2b/c$ and $\beta=c/2.$

\bigskip

\textbf{I)} Let $b$\ be odd and consider $y=b/2c,$ $\beta=c$ with $\gcd\left(
b,c\right)  =1$ in (\ref{23}). From (\ref{raabe}), we have%
\begin{equation}
\overline{B}_{p-j+1}\left(  \frac{b}{2}\right)  =\overline{B}_{p-j+1}\left(
\frac{1}{2}\right)  =\left(  2^{j-p}-1\right)  \overline{B}_{p-j+1}\left(
0\right)  . \label{25}%
\end{equation}
Therefore, (\ref{23}) becomes%
\begin{align}
\sum_{n=1}^{c}\overline{B}_{p}\left(  \frac{b}{2c}n\right)   &amp;  =-\frac{1}%
{4}s_{4,p}\left(  b,c\right)  +\overline{B}_{p}\left(  \frac{1}{2}\right)
\nonumber\\
&amp;  =\frac{2^{1-p}}{p+1}\sum\limits_{m=p-l+1}^{p+1}(-1)^{p+1-m}\binom{p+1}%
{m}\left(  \frac{b}{c}\right)  ^{p-m}(1-2^{m})\overline{B}_{m}(0)\overline
{B}_{p+1-m}(0)\nonumber\\
&amp;  \quad-c\binom{p}{l}\left(  -\frac{b}{2c}\right)  ^{l}\int\limits_{0}%
^{1}\overline{B}_{l}\left(  cu\right)  \overline{B}_{p-l}\left(  \frac{b}%
{2}u\right)  du \label{26}%
\end{align}
by setting $j=p+1-m$.

\bigskip

$\bullet$ Let $p&gt;1$ be odd and put $l=2$ in (\ref{26}). Then,%
\begin{align}
-\frac{2^{p}}{4}s_{4,p}\left(  b,c\right)  =  &amp;  \frac{2}{p+1}\sum
\limits_{m=p-1}^{p+1}\binom{p+1}{m}\left(  -1\right)  ^{m}\left(  \frac{b}%
{c}\right)  ^{p-m}\left(  1-2^{m}\right)  \overline{B}_{m}(0)\overline
{B}_{p+1-m}(0)\nonumber\\
&amp;  -2^{p-3}p(p-1)\frac{b^{2}}{c}\int\limits_{0}^{1}\overline{B}_{2}\left(
cu\right)  \overline{B}_{p-2}\left(  \frac{b}{2}u\right)  du. \label{29}%
\end{align}


For $l=1$\ in (\ref{26}) we have the following integral representation
\begin{align*}
-\frac{1}{4}s_{4,p}\left(  b,c\right)  =  &amp;  \frac{2^{1-p}}{p+1}\frac{c}%
{b}\left(  1-2^{p+1}\right)  B_{p+1}\\
&amp;  +p\frac{b}{2}\int\limits_{0}^{1}\overline{B}_{1}\left(  cu\right)
\overline{B}_{p-1}\left(  \frac{b}{2}u\right)  du,\text{ for odd }p&gt;1.
\end{align*}


$\bullet$ If $p&gt;2$ is even, it is seen from (\ref{26}), (\ref{25}) and
\[
s_{4,p}(b,c)=2^{2-p}\left(  1-c^{1-p}\right)  B_{p}\text{, for even }p\text{
and odd }b
\]
\cite[Proposition 2.5]{12} that
\[
c\int\limits_{0}^{1}\overline{B}_{l}\left(  cu\right)  \overline{B}%
_{p-l}\left(  \frac{b}{2}u\right)  du=2^{l-p}(2^{p}-c^{1-p}-1)\left(
-\frac{c}{b}\right)  ^{l}\binom{p}{l}^{-1}B_{p}.
\]


\textbf{II)} Let $c$ be odd and consider $y=2b/c,$ $\beta=c/2$ with
$\gcd\left(  b,c\right)  =1$. Then, from (\ref{23}) and (\ref{25}) we have
\begin{align}
&amp;  \sum_{0&lt;n\leq c/2}\overline{B}_{p}\left(  \frac{2b}{c}n\right)  =\sum
_{n=1}^{\left(  c-1\right)  /2}\overline{B}_{p}\left(  \frac{2b}{c}n\right)
\nonumber\\
&amp;  \ =\frac{1}{p+1}\sum\limits_{j=1}^{l}\left(  -1\right)  ^{j}\binom{p+1}%
{j}\left(  \frac{2b}{c}\right)  ^{j-1}\left(  2^{1-j}-2\right)  \overline
{B}_{j}(0)\overline{B}_{p+1-j}(0)\nonumber\\
&amp;  \quad-\left(  -\frac{2b}{c}\right)  ^{l}\binom{p}{l}\frac{c}{2}%
\int\limits_{0}^{1}\overline{B}_{l}\left(  \frac{c}{2}u\right)  \overline
{B}_{p-l}\left(  bu\right)  du. \label{27}%
\end{align}
By the definition of the sum $s_{3,p}\left(  b,c\right)  $ we have%
\begin{equation}
s_{3,p}\left(  b,c\right)  =\sum\limits_{n=1}^{c-1}\left(  -1\right)
^{n}\overline{B}_{p}\left(  \frac{bn}{c}\right)  =2\sum\limits_{n=1}^{\left(
c-1\right)  /2}\overline{B}_{p}\left(  \frac{2bn}{c}\right)  -\sum
\limits_{n=1}^{c-1}\overline{B}_{p}\left(  \frac{bn}{c}\right)  . \label{28}%
\end{equation}


$\bullet$ Let $p&gt;1$ be odd. Put $l=p-2$ in (\ref{27}). Then, (\ref{raabe}),
(\ref{27}) and (\ref{28}) yield
\begin{align}
\ \left(  p+1\right)  bc^{p}s_{3,p}\left(  b,c\right)  =  &amp;  2\sum
\limits_{j=1}^{p-2}\binom{p+1}{j}\left(  -1\right)  ^{j}b^{j}c^{p+1-j}\left(
1-2^{j}\right)  \overline{B}_{p+1-j}(0)\overline{B}_{j}(0)\nonumber\\
&amp;  +2^{p-3}b^{p-1}c^{3}p\left(  p-1\right)  \left(  p+1\right)  \int%
\limits_{0}^{1}\overline{B}_{p-2}\left(  \frac{c}{2}u\right)  \overline{B}%
_{2}\left(  bu\right)  du. \label{30}%
\end{align}
Combining (\ref{29}) and (\ref{30}) we obtain the reciprocity formula
\begin{align*}
&amp;  \left(  p+1\right)  \left(  bc^{p}s_{3,p}\left(  b,c\right)  -2^{-2}%
c\left(  2b\right)  ^{p}s_{4,p}\left(  c,b\right)  \right) \\
&amp;  \ \qquad=2\sum\limits_{j=1}^{p+1}\binom{p+1}{j}\left(  -1\right)  ^{j}%
b^{j}c^{p+1-j}\left(  1-2^{j}\right)  B_{j}B_{p+1-j},
\end{align*}
when $c$ and $p&gt;1$ are odd.

\bigskip

$\bullet$ If $p&gt;2$ is even, then from (\ref{raabe}), (\ref{27}) and
$s_{3,p}\left(  b,c\right)  =0$ for odd $\left(  p+c\right)  $ we get%
\[
c\binom{p}{l}\int\limits_{0}^{1}\overline{B}_{l}\left(  \frac{c}{2}u\right)
\overline{B}_{p-l}\left(  bu\right)  du=\left(  -\frac{c}{2b}\right)
^{l}\left(  1-c^{1-p}\right)  B_{p}.
\]


Putting $l=1$\ in (\ref{27}) gives an integral representation for
$s_{3,p}\left(  b,c\right)  $ as\
\[
s_{3,p}\left(  b,c\right)  =2bp\int\limits_{0}^{1}\overline{B}_{1}\left(
\frac{c}{2}u\right)  \overline{B}_{p-1}\left(  bu\right)  du,\text{ for odd
}p&gt;1.
\]


\begin{thebibliography}{9}                                                                                                %

\bibitem {1}T. M. Apostol, Generalized Dedekind sums and transformation
formulae of certain Lambert series. {\it Duke Math. J.} {\bf 17} (1950)
147--157.

\bibitem {7}B. C. Berndt, Analytic Eisenstein series, theta functions
and series relations in the spirit of Ramanujan, {\it J. Reine Angew.
Math.} {\bf 303/304} (1978) 332--365.

\bibitem {12}M. Can, M. Cenkci, and V. Kurt, Generalized Hardy-Berndt sums,
{\it Proc. Jangjeon Math. Soc.} {\bf 9} (2006) 19--38.

\bibitem {29}M. Can and V. Kurt, Character analogues of certain Hardy-Berndt
sums, {\it Int. J. Number Theory}, 
{\bf 10} (2014), 737--762.

\bibitem {HC}H. Cohen, {\it Number Theory Volume II: Analytic and Modern Tools},
Springer, 2007.

\bibitem {19}L. A. Goldberg, Transformations of theta-functions and analogues
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\bibitem {32}H. Rademacher and E. Grosswald, {\it Dedekind Sums}, Carus Math.
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11F20; Secondary 11B68, 65B15.

\noindent \emph{Keywords: }
Dedekind sum, Hardy-Berndt sum, Bernoulli polynomial, Euler-MacLaurin formula.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received January 10 2014;
revised version received February 28 2014.
Published in {\it Journal of Integer Sequences}, March 23 2014.

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\noindent
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