a lattice sum involving the cosine function · 2018-10-12 · arzu boysal, fatih ecevit, cem...

J. Math. Anal. Appl. 463 (2018) 134–160

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A lattice sum involving the cosine function

Arzu Boysal, Fatih Ecevit, Cem Yalçın Yıldırım ∗

Department of Mathematics, Bog̃aziçi University, İstanbul 34342, Turkey

a r t i c l e i n f o a b s t r a c t

Article history:Received 19 May 2017Available online 13 March 2018Submitted by B.C. Berndt

Keywords:Lattice sumElementary classical functionsAsymptotic approximation

In this article we prove that, as n → ∞,

n−1∑j,k=1

13 − cos 2πj

n− cos 2πk

n− cos 2π(j+k)

n

∼ n2 logn√3π

.

We also obtain the secondary term of size � n2 to be followed by an error term of size O(logn). In this work, results and techniques from classical analysis involving roles by several special functions, from number theory, and from numerical analysis are needed.

© 2018 Elsevier Inc. All rights reserved.

1. Introduction

The sum

Sn :=n−1∑j,k=1

13 − cos 2πj

n − cos 2πkn − cos 2π(j+k)

n

(n ≥ 2) (1.1)

= 12

n−1∑j,k=1

1sin2 πj

n + sin2 πkn + sin2 π(j+k)

n

(1.2)

and other sums and integrals of similar types have been significant in studies of lattice Green’s functions which arise in many problems in condensed matter physics, random walks on lattices and the calculation of the resistance of resistor networks (see e.g. [7] and [16]), and in studies of metrized graphs [6]. Since even the order of magnitude of Sn as n → ∞ wasn’t known, in this paper we study this sum.

* Corresponding author.E-mail addresses: arzu .boysal @boun .edu .tr (A. Boysal), fatih .ecevit @boun .edu .tr (F. Ecevit), yalciny @boun .edu .tr

(C.Y. Yıldırım).

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A. Boysal et al. / J. Math. Anal. Appl. 463 (2018) 134–160 135

It is not difficult to obtain the non-trivial lower bound

Sn >n2 logn

4π2 + O(n2) (1.3)

by using

sin2 θ ≤ θ2, ∀θ ∈ R (1.4)

in (1.2). To get an upper bound one can employ a result due to Montgomery [11]:

(sin πθ)−2 ≤ (π||θ||)−2 + c, ||θ|| := min(θ − �θ, θ� − θ), c := 1 − 4π2 , ∀θ ∈ R. (1.5)

Upon plugging (1.5) in (1.2), in all of its occurrences the constant c may be neglected if we are just trying to determine the order of magnitude of Sn. This leads to

Sn � n2 logn, (1.6)

which in conjunction with (1.3) gives the order of magnitude of Sn:

Sn n2 logn. (1.7)

Our main result is

Theorem. As n → ∞,

n−1∑j,k=1

13 − cos 2πj

n − cos 2πkn − cos 2π(j+k)

n

= n2 logn√3π

+ n2√

3π

(γ −

√3π6 + log(4π 4

√3) − 3 log Γ

(13))

+ O(log n

), (1.8)

where γ is Euler’s constant.

The following two sections contain two proofs for the asymptotic value of Sn. The rest of the paper is for obtaining the secondary term and the error term. We achieve this by first relating Sn to an integral by the methods of numerical analysis (which is the main cause of the error term O(logn) in (1.8)), and then carrying out the evaluations of the terms which appear in the relation. It is desirable to obtain a more precise version of (1.8). In fact, since the summands are algebraic numbers in the cyclotomic field obtained by adjoining a primitive n-th root of unity to Q, due to the symmetry involved in the sum the precise value of Sn must be a rational number (S2 = 1

4 , S3 = 109 , S4 = 44

16 , S5 = 5811 , S6 = 1577

180 , . . .) We wonder whether a method for determining this rational number for any given n can be developed.

2. Proof of the asymptotic value

We first observe that by applying changes of variables the summation variables in (1.2) may be brought into the domain 0 < |j|, |k| ≤ n

2 without changing the summands. (For example, for the portion 0 < j ≤ n2 ,

n2 < k < n, n2 < j + k ≤ n we take k − n as our new k and leave j unchanged.) This will allow us to use Taylor expansion around the origin. First we deal with those j, k pairs which are not close to the origin. If j2 + k2 ≥ n2

, say, then max( |j| , |k| ) ≥ 1 . Hence each such summand will satisfy
72 n n 12

136 A. Boysal et al. / J. Math. Anal. Appl. 463 (2018) 134–160

1sin2 πj


n

≤ 1sin2 π

12= 4

2 −√

3. (2.1)

The number of such summands is < n2, so that the contribution of those j, k pairs which are far from the origin is

∑−n

2 ≤j,k≤n2

j2+k2≥n272

1sin2 πj


n

= O(n2). (2.2)

For the j, k pairs satisfying 0 < |j| + |k| ≤ n6 (which are all in the disk j2 + k2 ≤ n2

36 ) we have by Taylor’s theorem for functions of two variables ([3], p. 152)

sin2 πx + sin2 πy + sin2 π(x + y) = 2π2(x2 + xy + y2) + R3(x, y), (2.3)

with

R3(x, y) = −8π4{x4

4![cos 2πx + cos 2π(x + y)

]p∗ + x3y

3![cos 2π(x + y)

]p∗

+ x2y2

2!2![cos 2π(x + y)

]p∗ + xy3

3![cos 2π(x + y)

]p∗ + y4

4![cos 2πy + cos 2π(x + y)

]p∗

}(2.4)

where p∗ is a point on the line segment joining (0, 0) and (x, y). We see that

R3(j

n,k

n) = δj,k · 2π4

3n4 (j2 + |jk| + k2)2 for some δj,k ∈ [−1, 1]. (2.5)

Now, by formulas (2.2)–(2.5), we have

Sn = 12

∑j,k

0<|j|+|k|≤n6

1sin2 πj


n

+ O(n2)

= 12

∑j,k

0<|j|+|k|≤n6

12π2

n2 (j2 + jk + k2) + R3( jn ,

kn )

+ O(n2)

= n2

4π2

∑j,k

0<|j|+|k|≤n6

1(j2 + jk + k2) · 1

1 + δj,kπ2

3(j2+|jk|+k2)2n2(j2+jk+k2)

+ O(n2). (2.6)

Now we notice that

1 ≤ (j2 + |jk| + k2)(j2 + jk + k2) ≤ 3, (2.7)

and

j2 + |jk| + k2

n2 ≤ (|j| + |k|)2n2 ≤ 1

36 for |j| + |k| ≤ n

6 ,

so that

π2

3(j2 + |jk| + k2)2

n2(j2 + jk + k2) ≤ π2

36 < 1,


and the expansion

11 + δj,k · π2

3(j2+|jk|+k2)2n2(j2+jk+k2)

= 1 +∞∑

m=1

(δj,k

π2

3(j2 + |jk| + k2)2

n2(j2 + jk + k2)

)m

is valid and we plug it in (2.6). Then we have

Sn = n2

4π2

∑j,k

0<|j|+|k|≤n6

1j2 + jk + k2

+ n2

4π2

∑j,k

0<|j|+|k|≤n6

1j2 + jk + k2

∞∑m=1

(δj,k

π2

3(j2 + |jk| + k2)2

n2(j2 + jk + k2)

)m

+ O(n2). (2.8)

Now by (2.7) we see that

∞∑m=1

(δj,k

π2

3(j2 + |jk| + k2)2

n2(j2 + jk + k2)

)m

≤∞∑

m=1

(3π2 j

2 + jk + k2

n2

)m

= 3π2

n2 (j2 + jk + k2)∞∑

m=0

(3π2 j

2 + jk + k2

n2

)m

.

Hence the second term on the right hand side of (2.8) is majorized by

34

∑j,k

0<|j|+|k|≤n6

∞∑m=0

(3π2 · j

2 + jk + k2

n2

)m

<34

∑j,k

0<|j|+|k|≤n6

∞∑m=0

(π2

12)m = 9

12 − π2

∑j,k

0<|j|+|k|≤n6

1 = O(n2).

Thus (2.8) simplifies into

Sn = n2

4π2

∑j,k

0<|j|+|k|≤n6

1j2 + jk + k2 + O(n2)

= n2

2π2

( ∑0≤j,k≤n

60<j+k≤n

6

1j2 + jk + k2 +

∑0≤j,k≤n

60<j+k≤n

6

1j2 − jk + k2

)+ O(n2)

= n2

2π2

( ∑1≤j,k≤n

6

1j2 + jk + k2 +

∑1≤j,k≤n

6

1j2 − jk + k2

)+ O(n2). (2.9)

Observing that

∑1≤j,k≤n

6

1j2 ± jk + k2 =

n6∑

k=1

( ∞∑j=1

1j2 ± jk + k2 −

∑j>n

6

1j2 ± jk + k2

)

=n6∑

k=1

( ∞∑j=1

1j2 ± jk + k2 −O( 1

n)), (2.10)


we can cast (2.9) into

Sn = n2

2π2

n6∑

k=1

∞∑j=1

( 1j2 + jk + k2 + 1

j2 − jk + k2

)+ O(n2). (2.11)

Expanding into partial fractions we have

Sn = n2

2π2

n6∑

k=1

∞∑j=1

i√3j

( 1j + ke

π3 i

− 1j + ke−

π3 i

+ 1j − ke−

π3 i

− 1j − ke

π3 i

)+ O(n2). (2.12)

The sum over j is evaluated in terms of the digamma function (see §6.8 of [1]) and we have

Sn = n2

2√

3π2

n6∑

k=1

i

k

(− ψ(1 + ke

π3 i) + ψ(1 + ke−

π3 i) − ψ(1 − ke−

π3 i) + ψ(1 − ke

π3 i)

)+ O(n2)

= n2

2√

3π2

n6∑

k=1

i

k

(− 2i�ψ(1 + ke

π3 i) + 2i�ψ(1 − ke

π3 i)

)+ O(n2)

= n2√

3π2

n6∑

k=1

1k�[ψ(1 + ke

π3 i) − ψ(1 − ke

π3 i)

]+ O(n2), (2.13)

since ψ(z) = ψ(z) (for this and other formulas involving the digamma function see §6.3 of [1]). In the last summand there is a quantity of the form ψ(z) −ψ(2 −z). The recurrence formula ψ(z+1) = ψ(z) + 1

z , turns this into ψ(z) − ψ(1 − z) − 1

1−z . Then by the reflection formula ψ(1 − z) = ψ(z) + π cotπz this quantity is expressed as −π cotπz − 1

1−z . Hence (2.13) becomes

Sn = n2√

3π2

n6∑

k=1

1k�[− π cot(π(1 + ke

π3 i)) + 1

keπ3 i

]+ O(n2)

= − n2√

3π

n6∑

k=1

1k� cot(π(1 + ke

π3 i)) + O(n2). (2.14)

Now

cot(π(1 + keπ3 i)) =

{cot

(√3πk2 i

)= −i coth

(√3πk2

)if k is even,

− tan(√

3πk2 i

)= −i tanh

(√3πk2

)if k is odd,

so that

Sn = n2√

3π

( ∑1≤k≤n

6k: odd

1k

tanh(√3πk

2)

+∑

1≤k≤n6

k: even

1k

coth(√3πk

2))

+ O(n2)

= n2√

3π

( ∑1≤k≤n

6k: odd

1k· 1 − e−

√3πk

1 + e−√

3πk+

∑1≤k≤n

6k: even

1k· 1 + e−

√3πk

1 − e−√

3πk

)+ O(n2)

= n2√

3π

n6∑ 1

k·(1 + O(e−

√3πk)

)+ O(n2)

k=1


= n2√

3π

( n6∑

k=1

1k

+ O( n

6∑k=1

e−√

3πk

k

))+ O(n2)

= n2√

3π

([log n

6+ γ + O( 1

n)] + O

( n6∑

k=1

e−√

3πk))

+ O(n2)

∼ n2 logn√3π

. (2.15)

We remark that the calculation of the infinite series in (2.11) can also be done by using the residue theorem. For fixed non-zero integer k, let

fk(z) := 1(z + ke

π3 i)(z − ke

2π3 i)

11 − e2πzi .

Then fk(z) has simple poles at −keπ3 i, ke

2π3 i, and at all integers, and it has no pole at infinity. Then, by

residue theorem on the extended complex plane (see, for example, §5.3 of Chapter III of [4]), the total residue is zero, that is,

Resz=−ke

π3 ifk(z) + Res

z=ke2π3 i

fk(z) +∑m∈Z

Resz=m

fk(z) = 0.

Evaluating the residues, we have

1k(eπ

3 i + e2π3 i)

( 11 − e2πke

2π3 ii

− 11 − e−2πke

π3 ii

)+

∑m∈Z

1m2 + mk + k2

−12πie2πmi

= 0.

Then∑m∈Z

1m2 + mk + k2 = 2πi

k(eπ3 i + e

2π3 i)

( 11 − e2πke

2π3 ii

− 11 − e−2πke

π3 ii

),

so that∑

m∈Z\{0}

1m2 + mk + k2 = 2π√

3k( 11 − e2πke

2π3 ii

− 11 − e−2πke

π3 ii

)− 1

k2 ,

and plugging this in (2.11) we obtain (2.14).

3. A number theoretic approach for the asymptotic value

For basic facts used in this section we refer the reader to [2]. The denominators in (2.9) are norms of certain algebraic integers in Q(

√−3). The set {1, eπ

3 i} is an integral basis for the ring of integers of Q(√−3)

seen as a vector space over Z. We have

j2 + jk + k2 = NQ(√−3)/Q(j + ke

π3 i). (3.1)

We will relate the sums in (2.9) to the Dedekind zeta-function of the field

ζQ(√−3)(s) :=

∑a

1N(a)s , (� s > 1), (3.2)


where the norm is relative to the field extension Q(√−3)/Q and a runs through all non-zero integral ideals

of Q(√−3). This Dedekind zeta-function factors as

ζQ(√−3)(s) = ζ(s)L(s, χ), (3.3)

where ζ(s) is the Riemann zeta-function and

L(s, χ) =∞∑

m=1

χ(m)ms

(3.4)

is the Dirichlet L-function associated with the character χ ( mod 3) defined on the integers as

χ(m) =

⎧⎪⎨⎪⎩

0 if m ≡ 0 (mod 3)1 if m ≡ 1 (mod 3)

−1 if m ≡ 2 (mod 3).(3.5)

Since the field Q(√−3) has class number 1, all of its ideals are principal ideals (and the unique factorization

property holds for its ring of integers). Take a principal ideal (j + ke

π3 i). Multiplying this ideal by the

numbers (eπ3 i)�, = 0, 1, . . . , 5, doesn’t change the ideal since e

π3 i is a unit, and therefore doesn’t change

the norm either. Then for = 0, 1, . . . , 5 for the same value of the norm we obtain the representations j2 + jk+k2, (−k)2 +(−k)(j +k) +(j +k)2, (−(j +k))2 +(−(j+k))j + j2, (−j)2 +(−j)(−k) +(−j)2, k2 +k(−(j + k)) + (−(j + k))2, (j + k)2 + (j + k)(−j) + (−j)2 respectively. So there is a cyclic process in which the ordered pair (j, k) goes to (k, −(j + k)), which in turn goes to (−(j + k), j) and this is mapped back to (j, k). Hence each such principal ideal

(j + ke

π3 i)

gives rise to its norm written in three ways as denominators in (2.9). In the particular case of an ideal with jk = 0, say the ideal (j) where j �= 0, the set of ordered pairs arising in this process consists of (j, 0), (0, −j) and (−j, j). Of these the first two won’t lead to representations which appear as denominators in (2.9) since the cases with jk = 0 are not included there, so these have to be subtracted. Thus we obtain, for � s > 1,

∞∑j,k=1

( 1(j2 + jk + k2)s + 1

(j2 − jk + k2)s)

= 3ζQ(√−3)(s) − 2ζ(2s)

= 3ζ(s)L(s, χ) − 2ζ(2s). (3.6)

Now note that instead of the sums in the last line of (2.9) we may consider

n212∑

m=1

f(m)m

, (3.7)

where f(m) is the number of ways m can be expressed as j2 ± jk + k2, j, k ≥ 1, within an error of O(1)because passing to (3.7) involves a difference of O(n2) terms each of which is of size O

( 1n2

). This error is

absorbed by the error term O(n2) in the last line of (2.9). We evaluate (3.7) as

3L(1, χ) log(n2

12 ) + O(1) = 3 π

3 32

log(n2

12 ) + O(1) = 2π√3

log n + O(1), (3.8)

using a Tauberian theorem for Dirichlet series from [12]. Thus, returning to (2.9), we find

Sn = n2 log n√3π

+ O(n2). (3.9)


4. Relating the sum Sn to an integral

With the calculations of the preceding sections we could only determine the asymptotic value of Sn. We wish to calculate the secondary main term of the value of Sn as n → ∞. For this purpose we begin by relating the sum Sn to a certain double integral.

When one encounters a sum such as (1.1), one would perhaps be tempted to express it as a Riemann integral. In particular, for (1.1) with a large n the first impulse might be to think of it as a Riemann sum with the square (0, 2π) × (0, 2π) chopped into little squares of side length 2π

n , and express the sum as

( n

2π)2 2πˆ

0

2πˆ

0

dθ dφ

3 − cos θ − cosφ− cos(θ + φ) . (4.1)

However, this integral is divergent. This is in accordance with what we already know from (1.3), for if the integral in (4.1) were convergent then that would imply Sn = O(n2). The divergence is caused by the behaviour of the integrand at the corners of the square.

The integral

In :=( n

2π)2 π(2n−1)

nˆπn

π(2n−1)nˆ

πn

dθ dφ

3 − cos θ − cosφ− cos(θ + φ) (4.2)

is appropriate for our purpose. The endpoints in this integral have been determined so as to make the midpoints of the subintervals to be constructed below coincide with the points in the summands of (1.1)(see (4.9) and (4.5)).

In order to establish a relation between Sn and In we begin with a lemma.

Lemma. For a twice continuously differentiable function h(x, y) on the domain R = [a − Δx

2 , a + Δx

2 ] ×[b − Δy

2 , b + Δy

2 ] ⊂ R2, we have∣∣∣∣∣∣¨

R

h(x, y) dx dy − Δx Δy h(a, b)

∣∣∣∣∣∣ ≤Δ3

x Δy maxR

|hxx| + Δx Δ3y max

R|hyy|

24 . (4.3)

Proof. We first write

¨

R

h(x, y) dx dy − Δx Δy h(a, b) =b+Δy

2ˆ

b−Δy2

( a+Δx2ˆ

a−Δx2

h(x, y) dx− Δxh(a, y))dy

+ Δx

( b+Δy2ˆ

b−Δy2

h(a, y)dy − Δyh(a, b)).

To the quantities inside parentheses on the right-hand side we can apply the midpoint rule (see, e.g., (9.5)–(9.6) of [14])

bˆ

a

H(x)dx = (b− a)H(a + b

2 ) + (b− a)3

24 H ′′(ξ) for some ξ ∈ (a, b) (4.4)


valid for a twice continuously differentiable H. Hence we have

¨

R

h(x, y) dx dy − Δx Δy h(a, b) = Δ3x

24

b+Δy2ˆ

b−Δy2

hxx(a′(y), y) dy +Δx Δ3

y

24 hyy(a, b′)

for some a′(y) ∈ [a − Δx

2 , a + Δx

2 ] and b′ ∈ [b − Δy

2 , b + Δy

2 ], and (4.3) follows immediately. �Let f(x, y) = 1

3 − cosx− cos y − cos(x + y) , and tj = 2πjn , (j ∈ Z), so that

Sn =∑

0<tj ,tk<2πf(tj , tk) =

∑−π<tj,tk≤π

tj �=0,tk �=0

f(tj , tk) = Fn,1 + Fn,2, say, (4.5)

with

Fn,1 :=∑

0<|tj |,|tk|≤ 12

f(tj , tk) and Fn,2 := Sn −Fn,1. (4.6)

As x2 +xy+ y2 is the first non-zero term in the Taylor expansion of the denominator of f about the origin, it is relevant to consider also the sum

Gn,1 :=∑

0<|tj |,|tk|≤ 12

g(tj , tk) where g(x, y) := 1x2 + xy + y2 . (4.7)

We define

h := f − g and Hn,1 := Fn,1 − Gn,1 (4.8)

and set

xj = 2πjn

− π

n, Xj = [xj , xj+1] , Δ = xj+1 − xj = 2π

n, (j ∈ Z). (4.9)

Note that tj = xj+1+xj

2 . Also set

Dn,1 =⋃

0<|tj |,|tk|≤ 12

Xj ×Xk, Dn,2 =( ⋃

−π<tj,tk≤π

tj �=0,tk �=0

Xj ×Xk

)\Dn,1. (4.10)

We now write¨

Dn,1

h(x, y) dxdy − Δ2Hn,1 =∑

0<|tj |,|tk|≤ 12

( ¨

Xj×Xk

h(x, y) dxdy − Δ2h(tj , tk))

and appeal to the Lemma to deduce that

∣∣∣ ¨Dn,1

h(x, y) dxdy − Δ2Hn,1

∣∣∣ ≤ Δ4

24∑

0<|tj |,|tk|≤ 12

(M1j,k + M2

j,k) (4.11)


where M1j,k := max

Xj×Xk

|hxx|, M2j,k := max

Xj×Xk

|hyy|. In order to estimate hxx = fxx − gxx, we calculate

fxx = fN

fD, gxx = gN

gD, (4.12)

where

fN = 2 (sin x + sin(x + y))2 − (cosx + cos(x + y)) (3 − cosx− cos y − cos(x + y)) ;

fD =u3, u(x, y) = 3 − cosx− cos y − cos(x + y);

gN =6(x2 + xy

); gD = v3, v(x, y) = x2 + xy + y2. (4.13)

By trigonometric identities fN can be re-expressed as

fN = 12

(− 5 cos(x + y) + cos(x− y) + cos(x + 2y) − 2 cos(2x + y)

− cos 2(x + y) + 6 − 5 cosx− cos(2x) + 6 cos y). (4.14)

Let p = (x, y) and 0 = (0, 0). We now apply Taylor’s theorem as in (2.3), but with the fourth order remainder, to fN and to u, and find after pedestrian calculations

fN (p) = gN (p) + RN (p,p′) and u(p) = v(p) + RD(p,p′′) (4.15)

for some p′ = (x′, y′) and p′′ = (x′′, y′′) between 0 and p, where

RN (p,p′) = RN4 (p,p′) = 1

48

(− 5(x + y)4 cos(x′ + y′) + (x− y)4 cos(x′ − y′)

+ (x + 2y)4 cos(x′ + 2y′) − 2(2x + y)4 cos(2x′ + y′) (4.16)

− (2x + 2y)4 cos(2x′ + 2y′) − x4(5 cosx′ + 16 cos 2x′) + 6y4 cos y′),

and

RD(p,p′′) = RD4 (p,p′′) = − 1

24(x4 cosx′′ + (x′′ + y′′)4 cos(x′′ + y′′) + y4 cos y′′

). (4.17)

Note that

fD(p) =(gD(p) + RD(p,p′′)

)3. (4.18)

It follows that

hxx = fxx − gxx = gN + RN (p,p′)(v + RD(p,p′′))3 − gN

v3 = hN

hD(4.19)

where

hN = v3RN (p,p′) − gNRD(p,p′′)(3v2 + 3vRD(p,p′′) + RD(p,p′′)2) (4.20)

and

hD = (v + RD(p,p′′))3v3. (4.21)


Writing x = |p| cos θ and y = |p| sin θ, we have

|RN (p,p′)| ≤ 5(x + y)4 + (x− y)4 + (x + 2y)4 + 2(2x + y)4 + (2x + 2y)4 + 21x4 + 6y4

48 ;

and from this we deduce

|RN (p,p′)| ≤ 4|p|4. (4.22)

The estimates

|RD(p,p′′)| ≤ |p|44 , |v| ≤ 3

2 |p|2, |gN | ≤ 3(1 +

√2)|p|2 (4.23)

are found in a similar way. Using (4.22) and (4.23) in (4.20), we see

|hN (p)| ≤ 28|p|10, (|p| ≤ 1). (4.24)

We have from

v ≥ |p|22 , (4.25)

v + RD(p,p′′) ≥ |p|22 − |p|4

4 ≥ |p|24 , (|x| ≤ 1). (4.26)

Use of (4.25) and (4.26) in (4.21) yields

|hD(p)| ≥ |p|1283 , (|x| ≤ 1). (4.27)

Combining (4.24) and (4.27), and setting C = 7 · 47, we obtain

|hxx(p)| ≤ C

|p|2 , (0 < |p| ≤ 1). (4.28)

It follows that, for all sufficiently large n,

∑0<|tj |,|tk|t2j+t2

k≤ 1

2

maxXj×Xk

|hxx| ≤ C∑

0<|tj |,|tk|t2j+t2

k≤ 1

2

maxXj×Xk

1|p|2 = 4C

∑0<tj,tk

t2j+t2k≤ 1

2

1x2j + x2

k

= 4C{ ∑

0<tkt21+t2

k≤ 1

2

2x2

1 + x2k

+ 1Δ2

∑t1<tj,tk

t2j+t2k≤ 1

2

1x2j + x2

k

Δ2}

≤ 4C{ ∞∑

k=1

2x2

1 + x2k

+ 1Δ2

π2ˆ

0

1ˆπn

1r2 rdrdθ

}

= 4C{2n2

π2

∞∑k=1

11 + (2k − 1)2 + 1

Δ2π

2 (logn− log π)}


and therefore ∑0<|tj |,|tk|t2j+t2

k≤ 1

2

maxXj×Xk

|hxx| = O(n2 log n). (4.29)

Clearly the same estimate holds with hyy in place of hxx, so that using (4.29) in (4.11) gives¨

Dn,1

h(x, y) dxdy − Δ2Hn,1 = O( log nn2 ), (4.30)

which we rewrite as

Δ2Fn,1 =¨

Dn,1

f(x, y) dxdy −¨

Dn,1

g(x, y) dxdy + Δ2Gn,1 + O(logn/n2). (4.31)

Considering next the remainder Fn,2 in (4.6), we note that the same argument leading to (4.11) delivers

∣∣∣ ¨Dn,2

f(x, y) dxdy − Δ2Fn,2

∣∣∣ ≤ Δ4

24∑

(tj ,tk)∈Dn,2

( maxXj×Xk

|fxx| + maxXj×Xk

|fyy|). (4.32)

These maxima are O(1) in Dn,2 and there are O(n2) summands, so that¨

Dn,2

f(x, y) dxdy − Δ2Fn,2 = O(Δ4n2) = O( 1n2 ). (4.33)

Using (4.31) and (4.33) in (4.6), we obtain

Δ2Sn =¨

Dn,1∪Dn,2

f(x, y) dxdy −¨

Dn,1

g(x, y) dxdy + Δ2Gn,1 + O( log nn2 )

=2π−π

nˆπn

2π−πnˆ

πn

f(x, y) dxdy −¨

Dn,1

g(x, y) dxdy + Δ2Gn,1 + O( log nn2 ) (4.34)

by the periodicity of f in x and y. Let

G′n,1 =

∑0<|tj |,|tk|<2π

g(tj , tk) and D′n,1 =

⋃0<|tj |,|tk|<2π

Xj ×Xk. (4.35)

The inequality (4.32) continues to hold with Dn,2, f , and Fn,2 replaced by D′n,1\Dn,1, g, and G′

n,1 − Gn,1, so that just as (4.32) implies (4.33), we have

¨

D′n,1\Dn,1

g(x, y) dxdy − Δ2(G′n,1 − Gn,1) = O( 1

n2 ). (4.36)

Plugging this into (4.34), we obtain

Δ2Sn =2π−π

nˆπn

2π−πnˆ

πn

f(x, y) dxdy −¨

D′n,1

g(x, y) dxdy + Δ2G′n,1 + O( log n

n2 ). (4.37)


Writing

G(x, y) = g(x, y) + g(−x, y), (4.38)

we have

¨

D′n,1

g(x, y) dxdy = 22π−π

nˆπn

2π−πnˆ

πn

G(x, y) dxdy, (4.39)

and

G′n,1 =

∑0<|tj |,|tk|<2π

g(tj , tk) = 2∑

0<tj ,tk<2πG(tj , tk) = 2

n−1∑j,k=1

G(tj , tk). (4.40)

Using (4.39) and (4.40) in (4.37), we reach

Sn = In + 2( n

2π)2[ n−1∑

j,k=1

( 1j2 + jk + k2 + 1

j2 − jk + k2

)

−

π(2n−1)nˆ

πn

π(2n−1)nˆ

πn

( 1x2 + xy + y2 + 1

x2 − xy + y2

)dxdy

]+ O(logn)

= In + 2( n

2π)2[

Gn − IG

]+ O(logn), say. (4.41)

There are now two ways to complete the evaluation of the right-hand side of (4.41): Either evaluate the double sum and the double integral separately or take them together. For the latter alternative there is an Euler–Maclaurin summation formula for double series due to Riesel which is very handy except that the error terms were not specified in [15]. It is possible to develop a version of this formula which includes estimates for the errors, but we won’t take this way here since in our particular example including more terms (involving higher order derivatives) leads to worse results as a comparison with the direct numerical calculations for Sn reveals. This stems from the fact that the higher order derivatives of the particular function in the summand assume very large values near the corner of the region which is close to the origin. Therefore in the following sections we will calculate the three terms appearing on the right hand-side of (4.41) separately.

5. Evaluation of In =( n

2π)2 π(2n−1)

n̂

πn

π(2n−1)n̂

πn

dθ dφ

3 − cos θ − cosφ − cos(θ + φ)

In the domain of integration of In for the portion θ, φ ∈ [π, (2n−1)πn ] we use 2π − θ, 2π − φ and see

that this portion is the same as the portion θ, φ ∈ [πn , π]. What remains of the square is twice the portion

θ ∈ [πn , π], φ ∈ [π, (2n−1)πn ] (because of symmetry of the integrand and of the regions with respect to θ

and φ). Changing φ to 2π−φ brings the domain of integration to θ, φ ∈ [πn , π], but also alters the integrand by making the denominator 3 − cos θ − cosφ − cos(θ − φ). Thus we have


In = 2 · n2

4π2

π̂

πn

π̂

πn

( 13 − cos θ − cosφ− cos(θ + φ)

+ 13 − cos θ − cosφ− cos(θ − φ)

)dθ dφ. (5.1)

Putting u = tan θ2 , v = tan φ

2 transforms (5.1) to

In = n2

π2

∞̂

tan π2n

∞̂

tan π2n

u2 + v2 + u2v2

(u2 + v2 + u2v2)2 − u2v2 du dv. (5.2)

We switch to polar coordinates by putting u = r cosα, v = r sinα and have

In = 2n2

π2

π4ˆ

α=0

∞̂

r=tan π

2nsin α

1 + r2 cos2 α sin2 α

r((1 + r2 cos2 α sin2 α)2 − (cosα sinα)2

) dr dα. (5.3)

Letting β = 2α we re-write (5.3) as

In = n2

π2

π2ˆ

β=0

∞̂

r=√

2 tan π2n√

1−cos β

1 + r2 sin2 β4

r((1 + r2 sin2 β

4 )2 − sin2 β4

) dr dβ. (5.4)

Now we define a new variable t := r sin β√2 for fixed β. Then

In = n2

π2

π2ˆ

β=0

∞̂

t=tan π

2n sin β√

1−cos β

1 + t2

2

t((1 + t2

2 )2 − sin2 β4

) dt dβ. (5.5)

Reversing the order of integration gives

In = n2

π2

( ∞̂

t=√

2 tan π2n

π2ˆ

β=0

+

√2 tan π

2nˆ

t=tan π2n

π2ˆ

β=arccos(( ttan π

2n)2−1

)) 1 + t2

2

t((1 + t2

2 )2 − sin2 β4

) dβ dt. (5.6)

The integral with respect to β has the evaluation

ˆ 11 − sin2 β

a2

dβ =a arctan

(√a2−1 tan β

a

)√a2 − 1

+ C, (a > 1), (5.7)

where C is a constant of integration. We will employ (5.7) with a = 2(1 + t2

2 ) in (5.6), so that

π2ˆ

dβ

1 − sin2 β(t2 )2 =

(1 + t2

2 )π√(1 + t2)(3 + t2)

,

0 2(1+ 2 )


andπ2ˆ

arccos(( ttan π

2n)2−1

)dβ

1 − sin2 β(2(1+ t2

2 ))2 =

(1 + t2

2 )π√(1 + t2)(3 + t2)

− 2 + t2√(1 + t2)(3 + t2)

arctan(√

(1 + t2)(3 + t2)2 + t2

tan(

arccos(( t

tan π2n

)2 − 1)))

.

Since tan(arccosu) =√

1−u2

u , we have

tan(

arccos(( t

tan π2n

)2 − 1))

= t

tan π2n

√2 − ( t

tan π2n

)2

( ttan π

2n)2 − 1

.

Using these (5.6) becomes

In = n2

π2

∞̂

tan π2n

π

t√

(1 + t2)(3 + t2)dt (5.8)

− n2

π2

√2 tan π

2nˆ

tan π2n

2t√

(1 + t2)(3 + t2)arctan

⎛⎝ t

tan π2n

√(1 + t2)(3 + t2)

2 + t2

√2 − ( t

tan π2n

)2

( ttan π

2n)2 − 1

⎞⎠ dt.

We can easily estimate the very last integral for n ≥ 3. The arctan takes values in [0, π2 ], the factor before arctan is of size 1

t 1tan π

2n, and the interval of integration is of length tan π

2n . Hence this integral is bounded, and (5.8) simplifies to

In = n2

π2

∞̂

tan π2n

π

t√

(1 + t2)(3 + t2)dt + O(n2). (5.9)

The substitution tan ν =√

1+t2

3+t2 transforms the last integral to

n2

π

π4ˆ

arctan√

1+tan2 π2n

3+tan2 π2n

14 sin2 ν − 1

dν.

Now letting tan ν2 = x the last integral becomes

− 2n2

π

tan π8ˆ

tan( 12 arctan

√1+tan2 π

2n3+tan2 π

2n

)1 + x2

x4 − 14x2 + 1 dx

= − 2n2

π

√2−1ˆ

√4+2 tan2 π

2n−√

3+tan2 π2n√

1+tan2 π

1 + x2

x4 − 14x2 + 1 dx (5.10)

2n


= − 2n2

π

√2−1ˆ

n

1 + x2

x4 − 14x2 + 1 dx, say.

Note that, using √

1 + y = 1 + y2 − y2

8 + O(y3) for small y and the Taylor–Maclaurin series tanw =w + w3

3 + O(w5) for small |w|, for n large enough we see (with t := tan π2n for convenience)

�n :=

√4 + 2 tan2 π

2n −√

3 + tan2 π2n√

1 + tan2 π2n

=2√

1 + t2

2 −√

3√

1 + t2

3√1 + t2

=2(1 + t2

4 − t4

32 + O(t6))−√

3(1 + t2

6 − t4

72 + O(t6))

1 + t2

2 − t4

8 + O(t6)

=((2 −

√3) +

(3 −√

36

)t2 +

(−9 + 2√

3144

)t4 + O(t6)

)·(1 − t2

2 + 3t4

8 + O(t6))

= (2 −√

3) +(−3 + 2

√3

6)t2 +

(63 − 40√

3144

)t4 + O(t6)

= (2 −√

3) ·(1 + 1

8√

3π2

n2 + 6 −√

32304

π4

n4 + O( 1n6 )

)> 2 −

√3 > 0 (5.11)

Returning to (5.10), we expand the integrand into partial fractions to have

− n2

2√

3π

√2−1ˆ

n

( 1x− (2 +

√3)

− 1x + (2 +

√3)

− 1x− (2 −

√3)

+ 1x + (2 −

√3)

)dx

= − n2

2√

3π(log(�n − (2 −

√3)) + O(1)

)= − n2

2√

3πlog

(π2

n22 −

√3

8√

3+ O( 1

n4 ))

+ O(n2) (5.12)

= n2 log n√3π

+ O(n2),

where we have made use of (5.11) twice in the penultimate line. Inserting this in (5.9), we obtain

In = n2 log n√3π

+ O(n2). (5.13)

We are interested in the estimation of the O(n2) term in (5.12) (in order to get an explicit secondary main term instead of leaving it as just O(n2) as in (5.13)) although there still remains the problem of calculating the second integral in (5.8) more precisely. The precise evaluation of the first line of (5.12) is

= − n2

2√

3π

(log(2 +

√3) + log

( �n + 2 +√

3(�n + 2 −

√3)(2 +

√3 −�n)

)+ log(�n − (2 −

√3))

), (5.14)

where the very last term gives the main contribution of size n2 logn and the first two terms give O(n2). To calculate the expression in (5.14) we write from (5.11)

�n = 2 −√

3 + εn, εn = (2 −√

3) ·( 1

8√

3π2

n2 + 6 −√

32304

π4

n4 + O( 1n6 )

). (5.15)


Then

log( �n + 2 +

√3

(�n + 2 −√

3)(2 +√

3 −�n))

= log(2 +

√3√

3)−

(9 + 4√

312

)εn + O(ε2n)

= log(2 +

√3√

3)− 6 −

√3

96√

3π2

n2 + O( 1n4 ),

and

log εn = −2 logn + log( (2 −

√3)π2

8√

3)

+ 2√

3 − 196

π2

n2 + O( 1n4 ),

so that (5.14) becomes

n2√

3π

(log n− 1

2 log(2 +

√3

24 π2) + O( 1n4 )

). (5.16)

Thus, denoting the second term on the right-hand side of (5.8) by

Rn := −n2

π2

√2 tan π

2nˆ

tan π2n

2t√

(1 + t2)(3 + t2)arctan

⎛⎝ t

tan π2n

√(1 + t2)(3 + t2)

2 + t2

√2 − ( t

tan π2n

)2

( ttan π

2n)2 − 1

⎞⎠ dt, (5.17)

(it was seen between (5.8) and (5.9) that Rn = O(n2)), we have obtained

In = n2 logn√3π

− n2

2√

3πlog

(2 +√

324 π2) + Rn + O( 1

n2 ). (5.18)

To understand Rn we begin by letting w = ttan π

2n, so that (5.17) becomes

Rn = −n2

π2

√2ˆ

1

2w√

(1 + (w tan π2n )2)(3 + (w tan π

2n )2)

× arctan(w√

2 − w2

w2 − 1

√(1 + (w tan π

2n )2)(3 + (w tan π2n )2)

2 + (w tan π2n )2

)dw. (5.19)

As in (5.11) we calculate√

(1 + (w tan π

2n )2)(3 + (w tan π

2n )2) =√

3(1 + w2

6π2

n2 + O( 1n4 )

),

√(1 + (w tan π

2n )2)(3 + (w tan π2n )2)

2 + (w tan π2n )2 =

√3

2

(1 + w2

24π2

n2 + O( 1n4 )

),

and we can re-write (5.19) as

Rn = −n2

π2

√2ˆ

1

2√3w

(1 − w2

6π2

n2 + O( 1n4 )

)

× arctan(w√

2 − w2

w2 − 1

√3

2

(1 + w2

24π2

n2 + O( 1n4 )

))dw. (5.20)


Now since

arctan(a + aε) = arctan(a) + aε

1 + a2 + O(ε2(1 + a3)

(1 + a2)2)

with

a =√

32

w√

2 − w2

w2 − 1 , ε = w2

24π2

n2 + O( 1n4 ),

valid for a ∈ [0, ∞) and formally valid even for w = 1+, we express (5.20) as

Rn = −n2

π2

√2ˆ

1

2√3w

(1 − w2

6π2

n2 + O( 1n4 )

)

×[arctan

(√32

w√

2 − w2

w2 − 1

)+ w3(w2 − 1)

√2 − w2

4√

3((w2 − 1)2 + 3

) π2

n2 + O( 1n4 )

]

= −n2

π2

√2ˆ

1

2√3w

arctan(√3

2w√

2 − w2

w2 − 1

)dw

+

√2ˆ

1

[ w

3√

3arctan

(√32

w√

2 − w2

w2 − 1

)− 1

6w2(w2 − 1)

√2 − w2

(w2 − 1)2 + 3

]dw + O( 1

n2 ). (5.21)

By the last sentence of §4 we need to evaluate only the term of size n2 in Rn. We let w2 = 21 + u2 and

then integrate by parts ( 2u1 + u2 du = d(log(1 + u2)) to have

√2ˆ

1

2√3w

arctan(√3

2w√

2 − w2

w2 − 1

)dw =

1ˆ

0

2u√3(1 + u2)

arctan( √

3u1 − u2

)du

= π log 22√

3−

1ˆ

0

(1 + u2) log(1 + u2)u4 + u2 + 1 du. (5.22)

By expanding into partial fractions we have

−1ˆ

0

(1 + u2) log(1 + u2)u4 + u2 + 1 du =

i

2√

3

1ˆ

0

[log(1 + ui) + log(1 − ui)][ −1u + e

2π3 i

+ 1u + e−

π3 i

+ 1u + e−

2π3 i

+ −1u + e

π3 i

] du

= 1√3�

1ˆ

0

log(1 + ui)u + e

2π3 i

− log(1 + ui)u + e−

π3 i

− log(1 + ui)u + e−

2π3 i

+ log(1 + ui)u + e

π3 i

du

= 1√3�

iˆ

0

log(1 + v)v + e−

5π6 i

− log(1 + v)v + e

π6 i

− log(1 + v)v + e−

π6 i

+ log(1 + v)v + e

5π6 i

dv. (5.23)


Let L be the line segment from 0 to i. In (5.23) we have four integrals of the type ˆ

L

log(1 + v)v + b

dv. (There

are formulas in [10] for such integrals but the conditions for the applicability of those formulas were not stated clearly, and in fact they are not applicable in some cases relevant to us.) With the values of b in (5.23) it is valid to separate the logarithm into two parts and write

log(1 + v) ={

log(v + b) + log(1 + 1−bv+b ) if |v + b| > |1 − b| ∀v ∈ L

log(1 − b) + log(1 + v+b1−b ) if |v + b| < |1 − b| ∀v ∈ L

=

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

log(v + b) +∞∑

m=1

(−1)m+1

m

(1 − b

v + b

)m

if |v + b| > |1 − b| ∀v ∈ L

log(1 − b) +∞∑

m=1

(−1)m+1

m

(v + b

1 − b

)m

if |v + b| < |1 − b| ∀v ∈ L.

(5.24)

Now we integrate term by term and find

ˆ

L

log(1 + v)v + b

dv = (5.25)

{12(log2(b + i) − log2 b

)+ Li2

(b−1b+i

)− Li2

(b−1b

)if |v + b| > |1 − b| ∀v ∈ L

log(1 − b) log(b+ib

)+ Li2

(b

b−1)− Li2

(b+ib−1

)if |v + b| < |1 − b| ∀v ∈ L,

where (see [17])

Li2(z) := −zˆ

0

log(1 − u)u

du,(z ∈ C \ [1,∞), arg(1 − z) ∈ (−π, π)

)(5.26)

=∞∑k=1

zk

k2 , (|z| ≤ 1). (5.27)

Using (5.25) in (5.23) (for the first and fourth terms in (5.23) the second case of (5.25) applies, for the second and third terms the first case of (5.25) applies) the imaginary part in (5.23) is evaluated as

−π

2 log(1 +

√3√

2)− 5π

24 log 3

−�[Li2

( 1 − i

1 +√

3)

+ Li2( 1 + i

3 +√

3)

+ Li2(−(1 + i)

1 +√

3)

+ Li2(√3(1 − i)

1 +√

3)]. (5.28)

Now we use, with ω = arg(1 − re−θi),

�Li2(reiθ) = ω log r + 12[Cl2(2θ) + Cl2(2ω) − Cl2(2θ + 2ω)], (0 < r < 1), (5.29)

a formula due to Kummer (see (5.2)–(5.5) in [10]), where

Cl2(θ) := �Li2(eiθ) =∞∑k=1

sin kθ

k2 (5.30)


is Clausen’s function. The result of these calculations, in which we use

Cl2(−θ) = −Cl2(θ), 2Cl2(π3)

= 3Cl2(2π

3), (5.31)

is√

2ˆ

1

2√3w

arctan(√3

2w√

2 − w2

w2 − 1

)dw = π√

3log(

√3 − 1) + 5

3√

3Cl2(

π

3 ). (5.32)

To complete this section we find the value of the integral in the last line of (5.21). Letting w2 − 1 = x

turns this integral into

16√

3

1ˆ

0

arctan(√3

2

√1 − x2

x

)dx− 1

12

1ˆ

0

x√

1 − x2(x2 + 3

) dx,

which upon letting x = cos θ becomes

16√

3

π/2ˆ

0

sin θ arctan(√3

2 tan θ)dθ − 1

12

π/2ˆ

0

cos θ sin2 θ

4 − sin2 θdθ.

Using integration by parts in the first integral here we have

16√

3

[− cos θ arctan

(√32 tan θ

)∣∣∣π/20

+π/2ˆ

0

cos θ 2√

34 − sin2 θ

dθ]− 1

12

π/2ˆ

0

cos θ sin2 θ

4 − sin2 θdθ,

which simplifies to

13

π/2ˆ

0

cos θ4 − sin2 θ

dθ − 112

π/2ˆ

0

cos θ sin2 θ

4 − sin2 θdθ = 1

12

π/2ˆ

0

cos θdθ = 112 .

Hence√

2ˆ

1

[ w

3√

3arctan

(√32

w√

2 − w2

w2 − 1

)− 1

6w2(w2 − 1)

√2 − w2

(w2 − 1)2 + 3

]dw = 1

12 . (5.33)

Combining (5.18), (5.21), (5.32), and (5.33) we reach

In = n2 log n√3π

+ n2√

3π

(log

(2√

3π

)− 5

3πCl2(π

3 ))

+ 112 + O( 1

n2 ). (5.34)

6. Evaluation of Gn :=n−1∑j,k=1

( 1j2 + jk + k2 +

1j2 − jk + k2

)

In this section we consider the sum Gn which occurs in (4.41). Expanding the summand into partial fractions as in (2.12) we have


n−1∑j=1

( 1j2 + jk + k2 + 1

j2 − jk + k2

)

= i√3k

( ∞∑j=1

−∞∑j=n

)( 1j + ke

π3 i

− 1j + ke−

π3 i

+ 1j − ke−

π3 i

− 1j − ke

π3 i

).

In the sum ∞∑j=n

we replace j by j + n − 1 so that we shift the sum to start from 1. Then as in (2.13) we

have the expression in terms of the digamma function

n−1∑j=1

( 1j2 + jk + k2 + 1

j2 − jk + k2

)

= i√3k

[− ψ(1 + ke

π3 i) + ψ(1 + ke−

π3 i) − ψ(1 − ke−

π3 i) + ψ(1 − ke

π3 i)

+ ψ(n + keπ3 i) − ψ(n + ke−

π3 i) + ψ(n− ke−

π3 i) − ψ(n− ke

π3 i)

]= 2√

3k�[ψ(1 + ke

π3 i) + ψ(1 − ke−

π3 i) − ψ(n + ke

π3 i) − ψ(n− ke−

π3 i)

].

Therefore

Gn = 2√3

n−1∑k=1

�(ψ(1 + ke

π3 i) + ψ(1 − ke−

π3 i)

)k

−�(ψ(n + ke

π3 i) + ψ(n− ke−

π3 i)

)k

= 2√3(Gn,1 −Gn,2), say. (6.1)

We use the recurrence and reflection formulas for the digamma function, as was done in passing to (2.14), to have

Gn,1 = −√

32

n−1∑k=1

1k2 + π

n−1∑k=1

1k� cot

( πk

eiπ/3).

Since cot z = ie2iz + 1e2iz − 1 the last term simplifies into

n−1∑k=1

1k� cot

( πk

eiπ/3)

=n−1∑k=1

1k

+ 2n−1∑k=1

1k

qk

1 − qk, (q = −e−

√3π),

so that

Gn,1 = πn−1∑k=1

1k−

√3

2

n−1∑k=1

1k2 + 2π

n−1∑k=1

1k

qk

1 − qk. (6.2)

For the first two terms of the right-hand side of (6.2) we employ

n−1∑k=1

1k

= − 1n

+n∑

k=1

1k

= − 1n

+(

log n + γ + 12n − 1

12n2 + O( 1n4

))(6.3)


from p. 560 of [8], and

n−1∑k=1

1k2 = π2

6 − 1n− 1

2n2 − 16n3 + O

( 1n5

)(6.4)

which is written by combining (6.4.3) and (6.4.12) of [1]. As for the last member of (6.2), by the alternating series test we have

n−1∑k=1

1k

qk

1 − qk=

∞∑k=1

1k

qk

1 − qk+ O

( 1ne

√3πn

), (6.5)

(the last infinite series is an instance of the quantum dilogarithm Li2(x; q) =∞∑k=1

1k

xk

1 − qk, see [17]), and

then

∞∑k=1

1k

qk

1 − qk=

∞∑k=1

∞∑m=0

1k

(qm+1)k = −∞∑

m=0

∞∑k=1

−1k

(qm+1)k = −∞∑

m=0log(1 − qm+1)

= − log( ∞∏

m=0(1 − qm+1)

)=: − log ((q; q)∞) (6.6)

where (q; q)∞ is the q-Pochhammer symbol. Writing q = e2πτi, for �τ > 0, (q; q)∞ is related to the Dedekind eta-function as (q; q)∞ = q−

124 η(τ). To q = −e−

√3π there corresponds τ = e

π3 i. We can evaluate η

(e

π3 i)

by the result given in [13] on the Chowla–Selberg formula [5] employing the facts that eπ

3 i is in the field Q(√−3)

which has class number 1 and x2 + xy + y2 is the only reduced binary quadratic form of discriminant −3. Hence we obtain

∞∏m=0

(1 −

(− e−

√3π)m+1

)=

3 18 e

√3π24

(Γ(1

3)) 3

2

2π . (6.7)

Plugging the results of (6.3)–(6.7) in (6.2), we obtain

Gn,1 = π logn +(πγ −

√3π2

6 + 2π log 2π3 1

8− 3π log Γ

(13))

+√

3 − π

2n + 3√

3 − π

12n2 +√

312n3 + O

( 1n4

). (6.8)

In Gn,2 of (6.1) we use (6.3.18) of [1],

ψ(z) = log z − 12z − 1

12z2 + O( 1z4

)(see [8] for a discussion about this asymptotic) with z = n ± ke

π3 i. This leads to

Gn,2 =n−1∑k=1

1k

arctan( √

3knn2 − k2

)+

√3

4

n−1∑k=1

( 1n2 + nk + k2 + 1

n2 − nk + k2

)

+√

312

n−1∑k=1

( n + k2

(n2 + nk + k2)2 +n− k

2(n2 − nk + k2)2

)+ O

( lognn4

). (6.9)


We are going to evaluate these sums by applying the following version of the Euler–Maclaurin summation formula due to Lampret [9]: For any integers n, p ≥ 1 and any function f ∈ Cp[a, b],

h

n−1∑k=0

f(a + kh) =bˆ

a

f(x) dx +p∑

j=1hjBj

j![f (j−1)(x)

]ba

+ rp(a, b, n), (6.10)

where Bj are the Bernoulli numbers, h = b− a

n, and

rp(a, b, n) = −hp

p!

bˆ

a

Bp

(a− x

h

)f (p)(x) dx (6.11)

with Bp(x), the p-th Bernoulli polynomial in [0, 1) extended to all x ∈ R via Bp(x +1) = Bp(x). Expressing the first sum in (6.9) as

−√

3n

+n−1∑k=0

1n

( 1kn

arctan( √

3 kn

1 − ( kn )2

))(6.12)

for this application we take a = 0, b = 1, h = 1n, p = 4, f(x) = 1

xarctan

( √3x

1 − x2

)(we can take f(0) =

√3,

f(1) = π

2 , f′(0) = 0, f ′(1) = 2√

3− π

2 by considering limits as x → 0+ and x → 1−). Then

r4(0, 1, n) = − 124n4

1ˆ

0

B4(−nx)f (iv)(x) dx, (6.13)

and since f (iv)(x) remains bounded on [0, 1] (the values at the endpoints determined by limits), we have

n−1∑k=1

1k

arctan( √

3knn2 − k2

)=

1ˆ

0

1x

arctan( √

3x1 − x2

)dx−

√3

2 + π4

n+

16√

3 − π24

n2 + O( 1n4

). (6.14)

Now, integration by parts gives1ˆ

0

1x

arctan( √

3x1 − x2

)dx = −

√3

1ˆ

0

(1 + x2) log xx4 + x2 + 1 dx, (6.15)

an integral similar to the one in (5.23). Expanding into partial fractions as in (5.23), we arrive at

1ˆ

0

1x

arctan( √

3x1 − x2

)dx = −�

1ˆ

0

log(x)( 1x + e−

π3 i

+ 1x + e−

2π3 i

)dx. (6.16)

Now, integration by parts gives1ˆ

0

log xx + b

dx = 1b

1ˆ

0

log x1 + x

b

dx = (log x) log(1 + x

b)∣∣∣10−

1ˆ

0

log(1 + xb )

xdx

= −1ˆ

0

log(1 + xb )

xdx = −

− 1bˆ

0

log(1 − u)u

du. (6.17)


For −1b ∈ C \ [1, ∞), the last integral is evaluated by (5.26). Thus we find that

1ˆ

0

1x

arctan( √

3x1 − x2

)dx = −�

(Li2(−e

π3 i) + Li2(−e

2π3 i)

)= 5

3Cl2(π

3 ), (6.18)

and by (6.14) that

n−1∑k=1

1k

arctan( √

3knn2 − k2

)= 5

3Cl2(π

3 ) −√

32 + π

4n

+1

6√

3 − π24

n2 + O( 1n4

). (6.19)

We rewrite the second sum of (6.9) as

− 2n2 + 1

n

n−1∑k=0

1n

( 11 + k

n + ( kn )2

+ 11 − k

n + ( kn )2

). (6.20)

To apply (6.10), we take a = 0, b = 1, h = 1n, p = 4, f(x) = 1

1 + x + x2 + 11 − x + x2 . Then f(0) = 2,

f(1) = 43 , f

′(0) = 0, f ′(1) = −43 , f ′′′(x) and f (iv)(x) are bounded on [0, 1], and the expression in (6.20) is

equal to

1n

1ˆ

0

( 11 + x + x2 + 1

1 − x + x2

)dx− 5

3n2 − 19n3 + O

( 1n5

).

The last integral is evaluated by first expanding the integrand into partial fractions as π√3, hence we find

n−1∑k=1

( 1n2 + nk + k2 + 1

n2 − nk + k2

)= π√

3n− 5

3n2 − 19n3 + O

( 1n5

). (6.21)

Similarly, the last sum of (6.9) is rewritten as

− 2n3 + 1

n2

n−1∑k=0

1n

( 1 + k2n

(1 + kn + ( k

n )2)2+

1 − k2n

(1 − kn + ( k

n )2)2)

(6.22)

To apply (6.10), we take a = 0, b = 1, h = 1n, p = 2, f(x) =

1 + x2

(1 + x + x2)2 +1 − x

2(1 − x + x2)2 . Then f(0) = 2,

f(1) = 23 , f

′(0) = 0, f ′(1) = −169 , f ′′(x) is bounded on [0, 1], and the expression in (6.22) is equal to

1n2

1ˆ

0

( 1 + x2

(1 + x + x2)2 +1 − x

2(1 − x + x2)2

)dx− 4

3n3 + O( 1n4

).

The last integral is easily evaluated as 4 +√

3π6 , so that

n−1∑k=1

( n + k2

(n2 + nk + k2)2 +n− k

2(n2 − nk + k2)2

)= 4 +

√3π

6n2 − 43n3 + O

( 1n4

). (6.23)


Putting (6.19), (6.21) and (6.23) together in (6.9), we have

Gn,2 = 53Cl2(

π

3 ) −√

32n − 11

√3

36n2 − 5√

336n3 + O

( lognn4

), (6.24)

and (6.24) and (6.8) plugged in (6.1) gives

Gn = 2π√3

log n +(2πγ√

3− π2

3 + 4π√3

log 2π3 1

8− 2

√3π log Γ

(13)− 10

3√

3Cl2(

π

3 ))

+2 − π√

3n

+109 − π

6√

3n2 +

49n3 + O

( log nn4

). (6.25)

7. Evaluation of IG := −

π(2n−1)n̂

πn

π(2n−1)n̂

πn

( 1x2 + xy + y2 +

1x2 − xy + y2

)dxdy

In this section upon evaluating IG of (4.41) we will finish the proof of the theorem. In IG the region of integration is symmetric with respect to the line x = y, the integrand is invariant under the interchange of x and y, and as in (2.12) the integrand can be expanded into partial fractions, so that we have

IG = 2i√3

2π−πnˆ

πn

yˆπn

1y

( 1x + ye

π3 i

− 1x + ye−

π3 i

+ 1x− ye−

π3 i

− 1x− ye

π3 i

)dx dy

= 2i√3

2π−πnˆ

πn

1y

[log(x + ye

π3 i) − log(x + ye−

π3 i)

+ log(x− ye−π3 i) − log(x− ye

π3 i)

]x=y

x=πn

dy

= 2i√3

2π−πnˆ

πn

1y

[2iArg(1 + e

π3 i) + 2iArg(1 − e−

π3 i)

−(log(1 + n

πye

π3 i) − log(1 + n

πye−

π3 i) + log(1 − n

πye−

π3 i) − log(1 − n

πye

π3 i)

)]dy

= − 2π√3

log(2n− 1) + 2i√3

[Li2(−

neπ3 i

πy) − Li2(−

ne−π3 i

πy)

+ Li2(ne−

π3 i

πy) − Li2(

neπ3 i

πy)]2π−π

n

πn

(7.1)

= − 2π√3

log(2n− 1) + 2i√3

[L(2π − π

n) − L(π

n)], say,

where to pass to the penultimate member (5.26) is employed. We see that

L(πn

) = Li2(−eπ3 i) − Li2(−e−

π3 i) + Li2(e−

π3 i) − Li2(e

π3 i) (7.2)

= −2i�Li2(−e−π3 i) − 2i�Li2(e

π3 i) = −2i

(Cl2(

2π3 ) + Cl2(

π

3 ))

= −10i3 Cl2(

π

3 )


by (5.31). Next, in

L(2π − π

n) = Li2(−(2n− 1)eπ

3 i) − Li2(−(2n− 1)e−π3 i)

+ Li2((2n− 1)e−π3 i) − Li2((2n− 1)eπ

3 i)

the arguments of the dilogarithm are not on the ray [0, ∞) allowing us to apply the inversion formula

Li2(z) = −Li2(1z

)− π2

6 − 12 log2(−z), (7.3)

so that these values of the dilogarithm are expressed in terms of the values of the dilogarithm at the reciprocal points which are inside the unit disk. Hence we have

L(2π − π

n) = −Li2

( −1(2n− 1)eπ

3 i

)+ Li2

( −1(2n− 1)e−π

3 i

)− Li2

( 1(2n− 1)e−π

3 i

)+ Li2

( 1(2n− 1)eπ

3 i

)− 1

2 log2((2n− 1)eπ3 i) + 1

2 log2((2n− 1)e−π3 i)

− 12 log2(−(2n− 1)e−π

3 i) + 12 log2(−(2n− 1)eπ

3 i).

The logarithm terms here add up to −2π(log(2n − 1))i using the principal branch of the logarithm. For the values of the dilogarithm we use (5.31). Hence we have

L(2π − π

n) = − 2π(log(2n− 1))i

− 2√

3i∞∑

m=0

( 1(6m + 1)2(2n− 1)6m+1 − 1

(6m + 5)2(2n− 1)6m+5

)

= − 2π(log(2n− 1))i− 2√

32n− 1 i + O

( 1(2n− 1)5 ). (7.4)

Combining (7.1), (7.3) and (7.4), we obtain

IG = 2π√3

log(2n− 1) − 203√

3Cl2(

π

3 ) + 42n− 1 + O

( 1n5

)

= 2π√3

logn +(2π log 2√

3− 20

3√

3Cl2(

π

3 ))

+2 − π√

3n

+1 − π

4√

3n2 +

12 − π

12√

3n3 + O

( 1n4

). (7.5)

8. Conclusion

Putting the results (5.34), (6.25) and (7.5) together gives

In + 2( n

2π)2[

Gn − IG

]= n2 log n√

3π+ n2

√3π

(γ −

√3π6 + log(4π 4

√3) − 3 log Γ

(13))

+( 1

12 + 118π2 + 1

24√

3π

)+

124

√3π − 1

36π2

n+ O

( lognn2

), (8.1)

and from (4.41) we obtain the result announced in our Theorem.The referee informed us that the total of the terms of magnitude smaller than n2 seems to converge

numerically to 1 as n → ∞. So in the final form of our paper we gave the terms of size o(logn) in §5–7
12


even though there is an error of O(logn) term from §4. Now that the constant term is seen to be different from

112 , it becomes clear that the difference between Sn and the quantity in (8.1) is � 1. This points to

the limitation of the methods of this paper.

Acknowledgments

We thank the referee for carefully reading the paper and for the remark about the convergence of the smaller terms. To include the smaller terms we made some changes in the arguments of §6 which in fact led to a simpler presentation. Although it is not manifest in the final version of this paper, we have made use of MAPLE, WolframAlpha and Wikipedia while studying for this article.

References

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a lattice sum involving the cosine function · 2018-10-12 · arzu boysal, fatih ecevit, cem...

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