On Reciprocity Formulas for Apostol’s Dedekind Sums and Their Analogues

M. Cihat Dağlı and Mümün Can Department of Mathematics Akdeniz University TR-07058 Antalya Turkey

                  mcihatdagli@akdeniz.edu.tr
                      mcan@akdeniz.edu.tr

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.

1 Introduction

Let

((x)) = \begin{cases}x − [x] − 1/2, & if x ∈ ℝ\ℤ; \\ 0, & if x ∈ ℤ, \end{cases}

with [x] being the largest integer ≤ x. For positive integers c and integers b the classical Dedekind sum s(b, c), arising in the theory of Dedekind η-function, was introduced by R. Dedekind in 1892 as

s(b, c) = ∑_{m(mod c)} ((\frac{m}{c})) ((\frac{bm}{c})) .

The most important property of Dedekind sums is the reciprocity theorem

s(b, c) + s(c, b) = −\frac{1}{4} + \frac{1}{12} (\frac{b}{c} + \frac{c}{b} + \frac{1}{bc})

when gcd (b, c) = 1. The standard reference for Dedekind sums is Rademacher and Grosswald [7]. Several generalizations of Dedekind sums have been defined and the corresponding reciprocity formulas have been obtained. One of these generalizations, due to Apostol [1], is

s_{p}(b, c) = ∑_{m=1}^{c−1} \frac{m}{c}\overline{B}_{p} (\frac{bm}{c}) ,

where \overline{B}_{p}(x) is the pth Bernoulli function defined by

\overline{B}_{1} (x) = ((x)) , and \overline{B}_{p}(x + m) = B_{p}(x) for 0 ≤ x < 1, m ∈ ℤ and p > 1.

Here B_{p}(x) is the pth Bernoulli polynomial. Apostol’s reciprocity formula is as follows.

Theorem 1. Let b and c be coprime positive integers. For odd p ≥ 1, we have

(p + 1) (bc^{p}s_{p}(b, c) + cb^{p}s_{p}(c, b)) = ∑_{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}.

Here B_{p} = B_{p}(0) is the pth 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 [2, 6]

s_{3}(b, c) = ∑_{m=1}^{c−1} (−1)^{m} \overline{B}_{1} (\frac{bm}{c}) , s_{4}(b, c) = −4 ∑_{m=1}^{c−1} \overline{B}_{1} (\frac{bm}{2c}) .

Goldberg [6] 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 [2, 6]

2s_{3}(b, c) − s_{4}(c, b) = 1 − \frac{b}{c}

when c is odd and gcd (b, c) = 1. The generalizations of these sums in the sense of Apostol have been given in [3] by

s_{3,p}(b, c) = ∑_{m=1}^{c−1} (−1)^{m} \overline{B}_{p} (\frac{bm}{c}) , s_{4,p}(b, c) = −4 ∑_{m=1}^{c−1} \overline{B}_{p} (\frac{bm}{2c})

which satisfy the following reciprocity formula.

Theorem 2. Let b and c be coprime positive integers with c odd. For odd p ≥ 1, we have

(p + 1) (2bc^{p}s_{3,p} (b, c) − 2^{−1}c (2b)^{p} s_{4,p} (c, b))

= 4 ∑_{j=1}^{p+1} \binom{p + 1}{j} (−1)^{j} b^{j}c^{p+1−j} (1 − 2^{j}) B_{j}B_{p+1−j}.

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 1 and 2 when p > 1 by applying the Euler-MacLaurin summation formula to Bernoulli function.

The method presented in the sequel is motivated by [4].

2 Proofs of the reciprocity theorems

Let us state the Euler-MacLaurin summation formula, which can be found in various books, for example, [5, p. 22].

Theorem 3. (Euler-MacLaurin) Let α and β be real numbers such that α ≤ β and assume that f ∈ C^{(l)} [α, β] for some l ≥ 1. Then

∑_{α<m≤β} f(m) =∫_{α}^{β}f(u)du + ∑_{j=1}^{l} \frac{(−1)^{j}}{j!} (\overline{B}_{j} (β) f^{(j−1)}(β) − \overline{B}_{j} (α) f^{(j−1)}(α))

+ \frac{(−1)^{l−1}}{l!} ∫_{α}^{β}f^{(l)}(u)\overline{B}_{l} (u) du.

We will also need the facts \overline{B}_{2r+1}(0) = \overline{B}_{2r+1}(1/2) = 0 for all r ≥ 0 and Raabe’s formula

∑_{m=0}^{n−1} \overline{B}_{r} (x + \frac{m}{n}) = n^{1−r}\overline{B}_{r} (nx) .  (1)

Firstly, we consider the function f (x) = \overline{B}_{p} (xy), y ∈ ℝ. The property

\frac{d}{dx}\overline{B}_{p} (x) = p\overline{B}_{p−1} (x) , p > 2

entails that

\frac{d^{j}}{dx^{j}} f (x) = \frac{d^{j}}{dx^{j}} \overline{B}_{p} (xy) = y^{j} \frac{p!}{(p − j)!}\overline{B}_{p−j} (xy)  (2)

for 1 ≤ j ≤ p − 2 and f ∈ C^{(p−2)} [α, β] . For α = 0 and 1 ≤ l ≤ p − 2, Theorem 3 can be written as

∑_{1≤m≤β} \overline{B}_{p} (my)

= \frac{1}{p + 1} ∑_{j=1}^{l} (−1)^{j} \binom{p + 1}{j}y^{j−1} (\overline{B}_{j} (β) \overline{B}_{p−j+1} (βy) − \overline{B}_{p−j+1}(0)\overline{B}_{j} (0))

+ \frac{1}{y} \frac{\overline{B}_{p+1} (βy) − \overline{B}_{p+1} (0)}{p + 1} − (−y)^{l} \binom{p}{l} ∫_{0}^{β} \overline{B}_{l} (u) \overline{B}_{p−l} (yu) du.  (3)

Let y = b/c and β = c ∈ ℕ. Then (3) becomes

∑_{m=1}^{c} \overline{B}_{p} (\frac{bm}{c}) = −\binom{p}{l} (−\frac{b}{c})^{l} ∫_{0}^{c} \overline{B}_{l} (u) \overline{B}_{p−l} (\frac{b}{c}u) du.  (4)

• For b = c in (4) we have the well-known relation [5, p. 120]

∫_{0}^{1} \overline{B}_{l} (u) \overline{B}_{r} (u) du = (−1)^{l−1} \frac{l!r!}{(r + l)!}B_{l+r}, for l ≥ 1 and p − l = r ≥ 2.

• Let gcd (b, c) = 1. Then, it follows from (1) and (4) that

∫_{0}^{1} \overline{B}_{l} (cu) \overline{B}_{p−l} (bu) du = \begin{matrix}(−1) & l−1 \\ ( & {p \atop l} & ) \end{matrix} \frac{c^{l−p}}{b^{l}} B_{p}.  (5)

• Now assume that gcd (b, c) = q and put c = qc_{1}, b = qb_{1}. From (5), we have

∫_{0}^{1} \overline{B}_{l} (cu) \overline{B}_{p−l} (bu) du = c ∫_{0}^{1} \overline{B}_{l} (c_{1}u) \overline{B}_{p−l} (b_{1}u) du = cq^{p} \begin{matrix}(−1) & l−1 \\ ( & {p \atop l} & ) \end{matrix} \frac{c^{l−p}}{b^{l}} B_{p}.

2.1 Proof of Theorem 1

Let f(x) = x\overline{B}_{p} (xy) , y ∈ ℝ. Then f ∈ C^{(p−2)} [α, β]. From (2) and the Leibniz rule for the derivative we have, for 1 ≤ j ≤ p − 2,

\frac{d^{j}}{dx^{j}} [x\overline{B}_{p} (xy)] = y^{j} \frac{p!}{(p − j)!}x\overline{B}_{p−j} (xy) + y^{j−1} \frac{p!}{(p + 1 − j)!}j\overline{B}_{p+1−j} (xy) .

For f(x) = x\overline{B}_{p} (xy) , α = 0, β = c and y = b/c with gcd (b, c) = 1 in Theorem 3, we have

∑_{m=1}^{c} m\overline{B}_{p} (m\frac{b}{c})

= c^{2} ∫_{0}^{1} x\overline{B}_{p} (bx) dx + \frac{c}{p + 1} ∑_{j=1}^{l}(−1)^{j}\binom{p + 1}{j} (\frac{b}{c})^{j−1} \overline{B}_{j} (0) \overline{B}_{p+1−j} (0)

+ \binom{p}{l} (−\frac{b}{c})^{l−1} c ∫_{0}^{1} \overline{B}_{l} (cx) (bx\overline{B}_{p−l} (bx) + \frac{l}{p + 1 − l}\overline{B}_{p+1−l}(bx)) dx,  (6)

where 1 ≤ l ≤ p − 2. It follows from integration by parts that

∫_{0}^{1} x\overline{B}_{p} (bx) dx = \frac{1}{b(p + 1)}\overline{B}_{p+1} (0)  (7)

and from (5)

(−1)^{l−1}\binom{p + 1}{l} (\frac{b}{c})^{l} c ∫_{0}^{1} \overline{B}_{l} (cx) \overline{B}_{p+1−l} (bx) dx = c^{−p}\overline{B}_{p+1} (0) .  (8)

Combining (6), (7) and (8), we have

∑_{m=1}^{c} m\overline{B}_{p} (\frac{bm}{c})

= \frac{c}{p + 1} ∑_{j=0}^{l}(−1)^{j}\binom{p + 1}{j} (\frac{b}{c})^{j−1} \overline{B}_{p+1−j} (0) \overline{B}_{j} (0) + \frac{l}{p + 1} \frac{c^{1−p}}{b} B_{p+1}

+ (−1)^{l−1}\binom{p}{l} (\frac{b}{c})^{l−1} bc ∫_{0}^{1} x\overline{B}_{l} (cx) \overline{B}_{p−l} (bx) dx.  (9)

It can be easily seen from (1) and the fact \overline{B}_{p} (−x) = (−1)^{p} \overline{B}_{p} (x) that 2s_{p}(b, c) = (c^{1−p} − 1) B_{p} when p is even. We then have

∑_{m=1}^{c} m\overline{B}_{p} (\frac{bm}{c}) = ∑_{m=1}^{c−1} m\overline{B}_{p} (\frac{bm}{c}) + c\overline{B}_{p} (0) = \begin{cases}cs_{p}(b, c), & if p is odd; \\ (c^{1−p} + 1) \frac{c}{2}B_{p}, & if p is even. \end{cases} (10)

• Let p > 1 be odd. Putting l = 2 in (9) and using (10), we get

cs_{p}(b, c) =\frac{c}{p + 1} (\frac{c}{b}\overline{B}_{p+1} (0) + \binom{p + 1}{2}\frac{b}{c}\overline{B}_{p−1} (0) \overline{B}_{2} (0))
+ \frac{2}{p + 1} \frac{c^{1−p}}{b} B_{p+1} − \binom{p}{2}b^{2} ∫_{0}^{1} x\overline{B}_{p−2} (bx) \overline{B}_{2} (cx) dx  (11)

Putting l = p − 2 and interchanging b and c in (9)

bs_{p}(c, b) =\frac{b}{p + 1} ∑_{j=0}^{p−2}(−1)^{j}\binom{p + 1}{j} (\frac{c}{b})^{j−1} \overline{B}_{p+1−j} (0) \overline{B}_{j} (0) + \frac{p − 2}{p + 1} \frac{b^{1−p}}{c} B_{p+1}
+ (−1)^{p−1}\binom{p}{2} (\frac{c}{b})^{p−3} bc ∫_{0}^{1} x\overline{B}_{p−2} (bx) \overline{B}_{2} (cx) dx.  (12)

Then, from (11) and (12), we arrive at the reciprocity formula

(p + 1) (bc^{p}s_{p}(b, c) + cb^{p}s_{p}(c, b)) = ∑_{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 (9) we have the following integral representation

s_{p}(b, c) = (c + c^{−p}) \frac{1}{b} \frac{B_{p+1}}{p + 1} + pb ∫_{0}^{1} x\overline{B}_{1} (cx) \overline{B}_{p−1} (bx) dx, for odd p > 1.

• Let p > 2 be even. From (9) and (10) we have

b ∫_{0}^{1} x\overline{B}_{l} (cx) \overline{B}_{p−l} (bx) dx = \frac{1}{2} (1 + c^{1−p}) (−\frac{c}{b})^{l−1} \binom{p}{l}^{−1}B_{p}, for 1 ≤ l ≤ p − 2.

2.2 Proof of Theorem 2

We will use (3) for the following cases; I) y = b/ (2c) and β = c, II) y = 2b/c and β = c/2.

I) Let b be odd and consider y = b/2c, β = c with gcd (b, c) = 1 in (3). From (1), we have

\overline{B}_{p−j+1} (\frac{b}{2}) = \overline{B}_{p−j+1} (\frac{1}{2}) = (2^{j−p} − 1) \overline{B}_{p−j+1} (0) .  (13)

Therefore, (3) becomes

∑_{n=1}^{c} \overline{B}_{p} (\frac{b}{2c}n) = −\frac{1}{4}s_{4,p} (b, c) + \overline{B}_{p} (\frac{1}{2})

= \frac{2^{1−p}}{p + 1} ∑_{m=p−l+1}^{p+1}(−1)^{p+1−m}\binom{p + 1}{m} (\frac{b}{c})^{p−m} (1 − 2^{m})\overline{B}_{m}(0)\overline{B}_{p+1−m}(0)

− c\binom{p}{l} (−\frac{b}{2c})^{l} ∫_{0}^{1} \overline{B}_{l} (cu) \overline{B}_{p−l} (\frac{b}{2}u) du  (14)

by setting j = p + 1 − m. • Let p > 1 be odd and put l = 2 in (14). Then,

−\frac{2^{p}}{4} s_{4,p} (b, c) =\frac{2}{p + 1} ∑_{m=p−1}^{p+1} \binom{p + 1}{m} (−1)^{m} (\frac{b}{c})^{p−m} (1 − 2^{m}) \overline{B}_{m}(0)\overline{B}_{p+1−m}(0)
− 2^{p−3}p(p − 1)\frac{b^{2}}{c} ∫_{0}^{1} \overline{B}_{2} (cu) \overline{B}_{p−2} (\frac{b}{2}u) du.  (15)

For l = 1 in (14) we have the following integral representation

−\frac{1}{4}s_{4,p} (b, c) =\frac{2^{1−p}}{p + 1} \frac{c}{b} (1 − 2^{p+1}) B_{p+1}

+ p\frac{b}{2} ∫_{0}^{1} \overline{B}_{1} (cu) \overline{B}_{p−1} (\frac{b}{2}u) du, for odd p > 1.

• If p > 2 is even, it is seen from (14), (13) and

s_{4,p}(b, c) = 2^{2−p} (1 − c^{1−p}) B_{p}, for even p and odd b

[3, Proposition 2.5] that

c ∫_{0}^{1} \overline{B}_{l} (cu) \overline{B}_{p−l} (\frac{b}{2}u) du = 2^{l−p}(2^{p} − c^{1−p} − 1) (−\frac{c}{b})^{l} \binom{p}{l}^{−1}B_{p}.

II) Let c be odd and consider y = 2b/c, β = c/2 with gcd (b, c) = 1. Then, from (3) and (13) we have

∑_{0<n≤c/2} \overline{B}_{p} (\frac{2b}{c}n) = ∑_{n=1}^{(c−1)/2} \overline{B}_{p} (\frac{2b}{c}n)

= \frac{1}{p + 1} ∑_{j=1}^{l} (−1)^{j} \binom{p + 1}{j} (\frac{2b}{c})^{j−1} (2^{1−j} − 2) \overline{B}_{j}(0)\overline{B}_{p+1−j}(0)

− (−\frac{2b}{c})^{l} \binom{p}{l}\frac{c}{2} ∫_{0}^{1} \overline{B}_{l} (\frac{c}{2}u) \overline{B}_{p−l} (bu) du.  (16)

By the definition of the sum s_{3,p} (b, c) we have

s_{3,p} (b, c) = ∑_{n=1}^{c−1} (−1)^{n} \overline{B}_{p} (\frac{bn}{c}) = 2 ∑_{n=1}^{(c−1)/2} \overline{B}_{p} (\frac{2bn}{c}) − ∑_{n=1}^{c−1} \overline{B}_{p} (\frac{bn}{c}) .  (17)

• Let p > 1 be odd. Put l = p − 2 in (16). Then, (1), (16) and (17) yield

(p + 1) bc^{p}s_{3,p} (b, c) =2 ∑_{j=1}^{p−2} \binom{p + 1}{j} (−1)^{j} b^{j}c^{p+1−j} (1 − 2^{j}) \overline{B}_{p+1−j}(0)\overline{B}_{j}(0)
+ 2^{p−3}b^{p−1}c^{3}p (p − 1) (p + 1) ∫_{0}^{1} \overline{B}_{p−2} (\frac{c}{2}u) \overline{B}_{2} (bu) du.  (18)

Combining (15) and (18) we obtain the reciprocity formula

(p + 1) (bc^{p}s_{3,p} (b, c) − 2^{−2}c (2b)^{p} s_{4,p} (c, b))

= 2 ∑_{j=1}^{p+1} \binom{p + 1}{j} (−1)^{j} b^{j}c^{p+1−j} (1 − 2^{j}) B_{j}B_{p+1−j},

when c and p > 1 are odd.

• If p > 2 is even, then from (1), (16) and s_{3,p} (b, c) = 0 for odd (p + c) we get

c\binom{p}{l} ∫_{0}^{1} \overline{B}_{l} (\frac{c}{2}u) \overline{B}_{p−l} (bu) du = (−\frac{c}{2b})^{l} (1 − c^{1−p}) B_{p}.

Putting l = 1 in (16) gives an integral representation for s_{3,p} (b, c) as

s_{3,p} (b, c) = 2bp ∫_{0}^{1} \overline{B}_{1} (\frac{c}{2}u) \overline{B}_{p−1} (bu) du, for odd p > 1.

References

[1] T. M. Apostol, Generalized Dedekind sums and transformation formulae of certain Lambert series. Duke Math. J. 17 (1950) 147–157.

[2] B. C. Berndt, Analytic Eisenstein series, theta functions and series relations in the spirit of Ramanujan, J. Reine Angew. Math. 303/304 (1978) 332–365.

[3] M. Can, M. Cenkci, and V. Kurt, Generalized Hardy-Berndt sums, Proc. Jangjeon Math. Soc. 9 (2006) 19–38.

[4] M. Can and V. Kurt, Character analogues of certain Hardy-Berndt sums, Int. J. Number Theory, 10 (2014), 737–762.

[5] H. Cohen, Number Theory Volume II: Analytic and Modern Tools, Springer, 2007.

[6] L. A. Goldberg, Transformations of theta-functions and analogues of Dedekind sums, thesis, University of Illinois, 1981.

[7] H. Rademacher and E. Grosswald, Dedekind Sums, Carus Math. Monographs, Vol. 16, Math. Assoc. Amer., 1972.

2010 Mathematics Subject Classification: Primary 11F20; Secondary 11B68, 65B15. Keywords: Dedekind sum, Hardy-Berndt sum, Bernoulli polynomial, Euler-MacLaurin formula.

Received January 10 2014; revised version received February 28 2014. Published in Journal of Integer Sequences, March 23 2014.

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