Finite Trigonometric Sums And Class Numbers

Math. Ann. 330, 551–575 (2004)

Mathematische Annalen

DOI: 10.1007/s00208-004-0559-5

Finite trigonometric sums and class numbers Bruce C. Berndt · Alexandru Zaharescu Received: 17 December 2003 / Revised version: 19 April 2004 / Published online: 13 July 2004 – © Springer-Verlag 2004 Abstract. Explicit evaluations of finite trigonometric sums arose in proving certain theta function identities of Ramanujan. In this paper, without any appeal to theta functions, several classes of finite trigonometric sums, including the aforementioned sums, are evaluated in closed form in terms of class numbers of imaginary quadratic fields. Mathematics Subject Classification (2000): Primary, 11L03; Secondary, 11R29, 11L10

1. Introduction As corollaries of two difficult theta function identities from Ramanujan’s notebooks [16], the first author and L.–C. Zhang [6] discovered two interesting identities for trigonometric sums, namely, √ sin(2π/7) sin(π/7) sin(3π/7) 7 − + = 2 sin2 (3π/7) sin2 (2π/7) sin2 (π/7)

(1.1)

sin2 (3π/7) sin2 (2π/7) sin2 (π/7) − + = 0. sin(2π/7) sin(π/7) sin(3π/7)

(1.2)

and

(Much of the material from [6] can also be found in [4]; in particular, see pages 183–184.) Using similar ideas, Z.–G. Liu [15, pp. 107–108] established both (1.1) and (1.2) and eight further new identities involving sin(nπ/7), three of which are given by sin(π/7) sin(3π/7) 64 √ sin(2π/7) 7, − + = 4 4 4 7 sin (π/7) sin (3π/7) sin (2π/7)

(1.3)

5√ sin4 (3π/7) sin4 (π/7) sin4 (2π/7) − − = 7, sin(π/7) sin(2π/7) sin(3π/7) 8

(1.4)

B. C. Berndt , A. Zaharescu Department of Mathematics, University of Illinois, 1409 West Green Street, Urbana, IL 61801, USA (e-mail: {berndt, zaharesc}@math.uiuc.edu)  Research partially supported by grant MDA904-00-1-0015 from the National Security Agency.

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and sin3 (3π/7) sin3 (π/7) sin3 (2π/7) 368 − + = √ . 6 6 6 sin (π/7) sin (2π/7) sin (3π/7) 7

(1.5)

It is natural to ask if the identities (1.1)–(1.5) are unique in their relationship to 7, or if they are special cases of infinite classes of such identities. The purpose of this paper is to show that (1.1)–(1.5) are each special cases of infinite classes of identities which we shall establish. Moreover, we generalize in several directions. For example, we find a closed form expression when the powers of sin in the numerator and denominator of (1.1) are replaced by a and a + 1, respectively, where a is an odd positive integer. Some of our results, as well as Liu’s other identities, can be further generalized. Our evaluations are effected in terms of class numbers of imaginary quadratic number fields. Thus, our theorems may be regarded as representations of class numbers by trigonometric sums. The earliest theorems in this direction were first established by G. L. Dirichlet [12, p. 152], V. A. Lebesgue [13], and V. Schemmel [17]; see also L. E. Dickson’s History [11, pp. 117–118]. Despite the origins of our original identities, our proofs are devoid of any connections these trigonometric identities may have with theta function identities. Our methods depend on contour integration and properties of Gauss sums. In Sections 2 and 3, we generalize (1.1) and (1.2), respectively. The former generalization is broader than the latter and contains several elegant special cases and classical results. In Sections 4 and 5, we generalize (1.3)–(1.5), with the first and third of these identities being special cases of one general theorem in Section 4, and (1.4) being a special case of Theorem 5.1 in Section 5. We use our techniques in Section 6 to establish Schemmel’s results [17]. Lastly, in Section 7, we evaluate a trigonometric sum first examined in a letter from R. G. Stoneham to R. J. Evans in 1982 [18]. Throughout this paper, χ denotes a nonprincipal, real, primitive, odd character modulo k, where k is an odd positive integer at least equal to 7. Define the Gauss sum G(z, χ) for any complex number z by G(z, χ ) :=

k−1


χ(j )e2π ij z/k .

(1.6)

j =0

Then, for each integer n, we have the factorization theorem [5, p. 9, Thm. 1.1.3] G(n, χ ) = χ (n)G(1, χ) =: χ(n)G(χ).

(1.7)

In fact, (1.7) characterizes real primitive characters, i.e., (1.7) holds if and only if χ is real and primitive [1, p. 482, Thm. 1], [10, pp. 65–66]. We need Gauss’s famous evaluation [7, p. 349, Thm. 7], [5, p. 22, Thm. 1.3.4] √ G(χ ) = i k. (1.8)

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√ We denote the class number of the imaginary quadratic field Q( −k) by h(−k). For k ≥ 7 and odd, recall the classical formulas for the class number [2, p. 299], [7, p. 344, eq. (4.3)] 1
j χ(j ) h(−k) = − k j =1 k−1

(1.9)

and [7, p. 346, eq. (4.7)] h(−k) =


1 χ (j ). 2 − χ(2) 0<j
(1.10)

In the sequel, Rα (f ) = Rα denotes the residue of a meromorphic function f at a pole α. 2. Generalization of (1.1) Theorem 2.1. Let χ denote an odd, real, nonprincipal, primitive character of modulus k, where k is odd and k ≥ 7. Let


S1 (k, χ, a, b) :=

0
χ (n)

sina (bπn/k) , sina+1 (πn/k)

(2.1)

where a and b are positive integers with a odd, b even, and ab − a − 3 < 2k. Define   
a a+r m (−1) χ(n) . (2.2) Ca,b := m r n,m,r≥0 n+bm+r=(ab−a−1)/2

Then S1 (k, χ, a, b) =

√  a  k b h(−k) − 2Ca,b .

(2.3)

Observe that Ca,b is an integer. For fixed a and b and k large, the sum in (2.1) is a “long" sum, while the sum in (2.2) has a bounded number of terms and can be computed explicitly. The requirement ab − a − 3 < 2k is technical; we could dispense with this condition, but then terms arising from the left side of (2.19) below would appear, and Ca,b would have to be replaced by a more complicated sum. Proof. For N > 0, let CN denote the positively oriented indented rectangle with horizontal sides through ±iN and vertical sides through 0 and k. On the left side of CN , there is a semicircular indentation I0 of radius less than 1 centered at 0

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and to its left. On the right side of CN , the semicircular indentation comprises the points I0 + k. Consider the meromorphic function f (z) :=

G(z, χ ) sina (bπ z/k) 1 , a+1 2πiz G(χ ) sin (π z/k) e −1

(2.4)

where G(z, χ) and G(χ ) are defined in (1.6) and (1.7), respectively. We shall integrate f (z) over the contour CN , on the interior of which the function f (z) has simple poles (at most) at z = 1, 2, . . . , k − 1. Observe that on the interior of CN , the pole of f (z) at z = 0 is also simple, because G(0, χ ) = 0. For 1 ≤ n ≤ (k − 1)/2, a simple calculation with the use of (1.7) shows that Rn =

χ (n) sina (bπ n/k) 2π i sina+1 (π n/k)

(2.5)

and, since χ is odd, a is odd, and b is even, Rk−n = −

χ(n) sina (bπ(k − n)/k) χ (n) sina (bπn/k) = . 2πi sina+1 (π(k − n)/k) 2πi sina+1 (πn/k)

(2.6)

Next, by the definition (1.6), (1.8), and (1.9), sina (bπ z/k) z G(z, χ ) R0 = lim a 2πiz z→0 G(χ )(e − 1) sin (π z/k) sin(πz/k) k−1 a 1 j =1 χ (j )(2π ij/k) (bπ/k) = a (π/k) π/k G(χ )2π i √ k−1 ba ba
ba k h(−k). χ(j )j = − √ k h(−k) = − = √ πi πi k j =1 πi k Hence, by the residue theorem, (2.5), (2.6), and (2.7),  √ f (z)dz = 2S1 (k, χ, a, b) − 2ba h(−k) k.

(2.7)

(2.8)

CN

Next, we let N → ∞ in order to calculate the integral in (2.8) directly. First observe from (2.4) that, because a is odd and b is even, f (z) has period k, and so the integrals over the indented vertical sides of CN cancel. Now examine the integral over the top horizontal side CNT of CN . Set z = x + iN, 0 ≤ x ≤ k. We regard f (z) as a rational function of eπN/k and eπix/k and expand f (z) as a power series in eπ N/k with coefficients depending only on x, say f (z) = f (x + iN) = A0 (x)eDπN/k + A1 (x)e(D−1)πN/k + · · · .

(2.9)

Observe that we can ignore the terms of the form An (x)e(D−n)πN/k with n > D, since their contributions to the integral of f (z) over CNT tend to 0 as N tends

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555

to ∞. Thus, we truncate (2.9), keeping only the polynomial part in eπN/k . Hence, the integral of f (z) over CNT has the shape  f (z)dz = B0 eDπN/k + B1 e(D−1)πN/k + · · · + BD + O(e−πN/k ), (2.10) CN T

as N → ∞, where B0 , B1 , . . . , BD are constants. Observe that the degree D is given by D = ab − a − 3. Similarly, the integral of f (z) over the bottom side CNB of CN has the form     f (z)dz = B0 eDπN/k + B1 e(D−1)πN/k + · · · + BD + O(e−πN/k ), (2.11) CN B







as N → ∞, where B0 , B1 , . . . , BD are constants. Thus, by (2.8), (2.10), (2.11), and the aforementioned cancellation over the vertical sides of CN , 





(B0 + B0 )eDπ N/k + (B1 + B1 )e(D−1)πN/k + · · · + (BD + BD ) + O(e−πN/k ) √ (2.12) = 2S1 (k, χ, a, b) − 2ba k h(−k). The right side of (2.12) is independent of N . Therefore, by letting N → ∞, we find successively that 

B0 + B0 = 0,





B1 + B1 = 0, . . . , BD−1 + BD−1 = 0.

(2.13)

Thus, letting N → ∞ in (2.12), we conclude that

√  BD + BD = 2S1 (k, χ, a, b) − 2ba k h(−k).

(2.14)



It remains to compute BD and BD . We first calculate BD , which arises from the constant term in the expansion of f (z) in powers of eπN/k on CNT . Set z = x +iN and µ := e−πiz/k = eπN/k e−πix/k .

(2.15)

Then, by their definitions, G(z, χ ) =

k−1
χ (n) n=1

µ2n

,

a µ−b − µb (−1)a ab = µ (1 − µ−2b )a sin (bπ z/k) = 2i (2i)a   a (−1)a ab
m a µ (−1) µ−2bm , = (2i)a m m=0

(2.16)



a

(2.17)

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sin

−a−1

−a−1 µ−1 − µ (πz/k) = 2i a+1 = (−1) (2i)a+1 µ−a−1 (1 − µ−2 )−a−1   ∞
a+1 a+1 −a−1 r −a − 1 = (−1) (2i) µ (−1) µ−2r r r=0 

= (−1)

a+1

a+1

(2i)

−a−1

µ

 ∞ 
a+r r=0

r

µ−2r ,

(2.18)

and 1 = −1 + O(µ−2k ). e2πiz − 1 Thus, by (2.16)–(2.19) and the definition (2.4) of f , we find that

(2.19)

k−1
χ(n) 2i µab−a−1 G(χ ) µ2n n=1

f (z) = f (x + iN ) =

   ∞ 
a a + r −2r −2bm × (−1) µ µ m r m=0 r=0 a


m

+O(µab−a−3−2k ).

(2.20)

The constant term in (2.20) is equal to   
2i a a+r 2 m (−1) χ (n) = √ Ca,b , AD := m r G(χ) k n,m,r≥0 n+bm+r=(ab−a−1)/2

by (1.8) and (2.2). Hence,



0

BD =

√ AD dx = −kAD = −2Ca,b k.

(2.21)

k

The calculation of the integral of f (z) over the bottom side CNB is similar. Set z = x − iN and ν := eπiz/k = eπN/k eπix/k .

(2.22)

Then, in analogy with (2.20), we find that f (z) = f (x − iN) =

k−1 2iχ(−1) ab−a−1
χ(n) ν G(χ ) ν 2n n=1

   ∞  a −2bm
a + r −2r × (−1) ν ν m r m=0 r=0 a


m

+O(ν ab−a−3−2k ).

(2.23)

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The constant term is equal to 

AD :=

2iχ(−1) 2 Ca,b = − √ Ca,b , G(χ ) k

by (1.8) and (2.2). Hence, 

BD =



k 0

√   AD dx = kAD = −2Ca,b k.

(2.24)

Lastly, from (2.14), (2.21), and (2.24), we conclude that √ √ −4Ca,b k = 2S1 (k, χ, a, b) − 2ba h(−k) k,  

which completes the proof of Theorem 2.1.

Corollary 2.2. Let χ be a character as in Theorem 2.1, and let a be a positive odd integer such that a − 3 < 2k. Then k−1


√   χ(n) cot(π n/k) cosa−1 (π n/k) = 2 k h(−k) − 21−a Ca,2 .

(2.25)

n=1

Proof. Set b = 2 in Theorem 2.1. Using the elementary identity sin(2θ ) = 2 sin θ cos θ, we find that
0
χ(n)


sina (2π n/k) χ (n) cot(πn/k) cosa−1 (πn/k) =2a a+1 sin (π n/k) 0
√   = k 2a h(−k) − 2Ca,2 . The desired result now easily follows.

 

Corollary 2.3. For χ as in Theorem 2.1, k−1


√ χ (n) cot(π n/k) = 2 k h(−k).

(2.26)

n=1

Proof. Set a = 1 in Corollary 2.2. Observe that C1,2 = 0. Corollary 2.3 then easily follows.   Corollary 2.3 is due to Dirichlet [12]. See also papers by Lebesgue [13], V. Schemmel [17], M. Lerch [14], S. Chowla [8], [9, pp. 192–193], A. L. Whiteman [19], and Berndt [3].

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Corollary 2.4. If χ satisfies the conditions of Theorem 2.1, and if b is an even positive integer such that b − 4 < 2k, then S1 (k, χ, 1, b) =




χ (n)

0
sin(bπ n/k) sin2 (π n/k)

 √ = k b h(−k) − 2






χ(n)(r + 1) .

(2.27)

n,r≥0 n+r= 21 b−1

Proof. Let a = 1 in Theorem 2.1. The equality n + bm + r = 21 (ab − a − 1) = 21 b − 1 forces m = 0. The desired result now follows.

 

Corollary 2.5. For χ as above, let S1 (k, χ) := S1 (k, χ , 1, 4) :=


0
χ(n)

sin(4πn/k) . sin2 (πn/k)

Then S1 (k, χ ) =

√ k (4h(−k) − 2) .

(2.28)

Proof. Set b = 4 in Corollary 2.4. Observe that C1,4 = 1, since the only nonzero term in the sum on the right side of (2.27) arises from the case when n = 1 and r = 0. The result then follows from Corollary 2.4.   Corollary 2.6. We have √ S1 (7, χ) = 2 7, √ S1 (23, χ) = 10 23,

√ S1 (11, χ ) = 2 11, √ S1 (35, χ ) = 6 35.

(2.29) (2.30)

These values follow readily from Corollary 2.5, since [7, p. 425] h(−7) = h(−11) = 1,

h(−23) = 3,

The first of these evaluations is (1.1).

h(−35) = 2.

(2.31)

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3. Generalization of (1.2) We next prove a generalization of (1.2). Theorem 3.1. Let χ denote an odd, real, primitive, nonprincipal character of modulus k, where k is odd and k ≥ 7. Define


S2 (k, χ ) :=

0
χ (n)

sin2 (πn/k) . sin(4πn/k)

(3.1)

Then √ 3 k (χ (2) − 1)h(−k). S2 (k, χ ) = 4

(3.2)

Proof. For N > 0, let CN denote the same positively oriented indented rectangle that we employed in the proof of Theorem 2.1. We shall integrate the function f (z) :=

1 G(z, χ ) sin2 (π z/k) G(χ ) sin(4π z/k) e2πiz − 1

(3.3)

over the contour CN , where G(z, χ ) and G(χ) are defined in (1.6) and (1.7), respectively. On the interior of CN , the function f (z) has simple poles (at most) at z = 1, 2, . . . , k − 1. For 1 ≤ n ≤ (k − 1)/2, simple calculations with the use of (1.7) show that Rn =

χ (n) sin2 (π n/k) 2π i sin(4π n/k)

(3.4)

and, since χ is odd, Rk−n =

χ (n) sin2 (πn/k) . 2π i sin(4πn/k)

(3.5)

In contrast to the proof of Theorem 2.1, f (z) has a removable singularity at z = 0. On the other hand, there are additional simple poles at z = k/4, z = k/2, and z = 3k/4, with residues, respectively, kG(k/4, χ) kG(k/2, χ ) , , Rk/2 = − πik/2 8π G(χ )(e 8πG(χ) − 1) kG(3k/4, χ) =− . 8π G(χ )(e−πik/2 − 1)

Rk/4 = − R3k/4

(3.6)

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We examine each of the Gauss sums appearing above. To evaluate G(k/2, χ ), we write


(k−1)/2

G(k/2, χ) =




(k−1)/2

χ (2j ) −

j =1

χ(2j + 1)

j =0




(k−1)/2

=2χ(2)

χ (j ) −

j =1




k−1


χ(j )

j =1

(k−1)/2

=2χ (2)

χ (j ) = 2χ (2) (2 − χ (2)) h(−k),

(3.7)

j =1

by (1.10). We evaluate the remaining two Gauss sums in (3.6) together. Using the definition (1.6), putting both terms under a common denominator, replacing j by k − j in two of the resulting four sums, and using several times that χ is odd, we find that G(k/4, χ) G(3k/4, χ) + −πik/2 eπik/2 − 1 e −1 k−1
  1 χ (j ) eπi(j −k)/2 − eπij/2 + eπi(3j +k)/2 − e3πij/2 = 2 j =1   1
χ(j ) e−πij/2 + eπij/2 + eπij/2 + e−πij/2 2 j =1 k−1

=−

= −2

k−1


χ(j ) cos(πj/2)

j =1

= −2χ (2)




χ(j )(−1)j

0<j
= −χ(2)

k−1


χ(j )(−1)j

j =1

= −χ(2)G(k/2, χ) = −2 (2 − χ (2)) h(−k),

(3.8)

by (3.7). Summing the residues in (3.6) and using the evaluations in (3.7) and (3.8), we deduce that k (χ (2) − 1) (2 − χ (2)) h(−k) 4π G(χ ) √ 3 k (χ (2) − 1)h(−k), =− 4π i

Rk/4 + Rk/2 + R3k/4 = −

(3.9)

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by (1.8). Hence, by the residue theorem, (3.4), (3.5), and (3.9), √  3 k f (z)dz = 2S2 (k, χ ) − (χ(2) − 1)h(−k). 2 CN

(3.10)

We now calculate the integral in (3.10) directly by letting N → ∞. The integrals over the indented vertical sides cancel, because of the periodicity of the integrand. Inverting the order of integration and summation for the integrals on the horizontal sides, we see that for each j, 1 ≤ j ≤ k − 1, the integrals over the top and bottom sides of CN approach 0 as N tends to ∞. Hence,  f (z) = 0. (3.11) CN

Hence, combining (3.10) and (3.11), we deduce (3.2) to complete the proof.

 

We now state an immediate corollary of Theorem 3.1. Corollary 3.2. We have  √  3 k h(−k), if k ≡ 3 (mod 8), − S2 (k, χ ) = 2  0, if k ≡ 7 (mod 8).

(3.12)

Corollary 3.3. We have 3√ 11, 2√ S2 (35, χ ) = −3 35.

S2 (7, χ) = 0,

S2 (11, χ ) = −

S2 (23, χ) = 0,

(3.13) (3.14)

These values follow readily from Corollary 3.2 and the class numbers given in (2.31). The first of these evaluations is (1.2).

4. Generalization of (1.3) and (1.5) Our objective in this section is to prove a general theorem from which (1.3) and (1.5) follow as special cases. Theorem 4.1. Let χ denote a nonprincipal, real, primitive, odd character of modulus k, k ≥ 7. Let S3 (k, χ, a, b) :=


0
χ (n)

sina (bπn/k) , sina+3 (πn/k)

(4.1)

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where a and b are positive integers with a odd, b even, and ab − a − 5 < 2k. Define   
a a+2+r m (−1) χ (n) , (4.2) Ea,b := m r n,m,r≥0 n+bm+r=(ab−a−3)/2

and let M3 (χ ) =

k−1


χ (j )j 3 .

(4.3)

j =1

Then √  ba  S3 (k, χ, a, b) = √ k(a + 3 − ab2 + 4k 2 )h(−k) + 4M3 (χ) + 8Ea,b k. 6 k (4.4) Proof. We integrate 1 G(z, χ ) sina (bπ z/k) a+3 2πiz G(χ ) sin (π z/k) e −1

f (z) :=

(4.5)

over the same contour CN that we utilized in the proof of Theorem 2.1, where G(z, χ) and G(χ) are defined in (1.6) and (1.7), respectively. The function f (z) has simple poles (at most) at z = 1, 2, . . . , k − 1 and a triple pole at z = 0 on the interior of CN . By straightforward calculations with the use of (1.7), for 0 < n < k/2, Rn =

χ (n) sina (bπ n/k) = Rk−n . 2π i sina+3 (π n/k)

(4.6)

To calculate the residue at z = 0, we first observe that, since χ is odd, k−1


χ(j )j = −

j =1

2

k−1


χ (j )(k − j )2

j =1

= − k2

k−1


χ (j ) + 2k

j =1

= − 2k 2 h(−k) −

k−1


χ (j )j −

j =1 k−1


k−1


χ (j )j 2

j =1

χ(j )j 2 ,

j =1

by (1.9). Hence, k−1
j =1

χ(j )j 2 = −k 2 h(−k).

(4.7)

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Thus, using (1.9), (4.3), and (4.7), we find that the Laurent expansion of f (z) about the origin is f (z) =

 ba k 3 h(−k)i ba i  k(a + 3 − ab2 + 4k 2 )h(−k) + 4M3 (χ ) + · · · , + √ √ π 3 k z3 6π kz

and hence R0 =

 ba i  √ k(a + 3 − ab2 + 4k 2 )h(−k) + 4M3 (k) . 6π k

(4.8)

Hence, by the residue theorem, (4.1), (4.6), and (4.8), we find that   ba  f (z)dz = 2S3 (k, χ, a, b)− √ k(a + 3 − ab2 + 4k 2 )h(−k) + 4M3 (k) . 3 k CN (4.9) Next, we calculate the integral in (4.9) directly. From the definition of f (z) in (4.5), we see that the integrals over the indented vertical sides of CN cancel by periodicity. Now we examine the integral over the top horizontal side CNT of CN . Set z = x + iN, 0 ≤ x ≤ k. As before, we expand f (z) as a power series in eπN/k with coefficients depending only on x, say f (z) = f (x + iN ) = A0 (x)eDπN/k + A1 (x)e(D−1)πN/k + · · · . Then 

(4.10)

f (z)dz = B0 eDπN/k + B1 e(D−1)πN/k + · · · + BD + O(e−πN/k ), (4.11)

CN T

as N → ∞, where B0 , B1 , . . . , BD are constants. Similarly, the integral of f (z) over the bottom side CNB of CN has the form     f (z)dz = B0 eDπN/k + B1 e(D−1)πN/k + · · · + BD + O(e−πN/k ), (4.12) CN B







as N → ∞, where B0 , B1 , . . . , BD are constants. By (4.9), (4.11), (4.12), and the cancellation over the vertical sides of CN , 





(B0 + B0 )eDπN/k + (B1 + B1 )e(D−1)πN/k + · · · + (BD + BD ) + O(e−πN/k )  ba  = 2S3 (k, χ, a, b) − √ k(a + 3 − ab2 + 4k 2 )h(−k) + 4M3 (k) . 3 k (4.13) Letting N → ∞, we find successively that 

B0 + B0 = 0,





B1 + B1 = 0, . . . , BD−1 + BD−1 = 0,

(4.14)

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B. C. Berndt, A. Zaharescu

and further, that  ba   BD + BD = 2S3 (k, χ, a, b) − √ k(a + 3 − ab2 + 4k 2 )h(−k) + 4M3 (k) . 3 k (4.15) In order to calculate BD , which arises from the constant term in the expansion of f (z) in powers of eπN/k on CNT , set z = x + iN and define µ by (2.15). Then, using (2.16), (2.17), (2.18) with a replaced by a + 2, and (2.19), we find that k−1
8i χ(n) ab−a−3 f (z) = f (x + iN) = − µ G(χ ) µ2n n=1

×

a


(−1)m

m=0

   ∞ 
a a + 2 + r −2r µ−2bm µ m r r=0

+O(µab−a−5−2k ).

(4.16)

The constant term in (4.16) is equal to 8i AD := − G(χ)




   a a+2+r 8 (−1) χ (n) = − √ Ea,b , m r k m

n,m,r≥0 n+bm+r=(ab−a−3)/2

by (1.8) and (4.2), and so 

0

BD =

√ AD dx = −kAD = 8Ea,b k.

(4.17)

k

To compute the integral of f (z) over the bottom side CNB , set z = x − iN and define µ by (2.22). Then we find that k−1 8iχ(−1) ab−a−3
χ(n) f (z) = f (x − iN) = − ν G(χ ) ν 2n n=1

   ∞  a −2bm
a + 2 + r −2r × (−1) ν ν m r m=0 r=0 a


m

+O(ν ab−a−5−2k ). The constant term is equal to 

AD := −

8iχ(−1) 8 Ea,b = √ Ea,b , G(χ ) k

(4.18)

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565

and so 



BD =

k 0

√   AD dx = kAD = 8Ea,b k.

(4.19)

By (4.15), (4.17), and (4.19), we derive that √  ba  16Ea,b k = 2S3 (k, χ, a, b) − √ k(a + 3 − ab2 + 4k 2 )h(−k) + 4M3 (k) , 3 k which completes the proof of Theorem 4.1.

 

For a = 1 and b = 2, we find that E1,2 = 0, and so Theorem 4.1 gives the following result. Corollary 4.2. Let χ denote a nonprincipal, real, primitive, odd character of modulus k, k ≥ 7. Then
 sin(2π n/k) 4  χ (n) 4 = √ k 3 h(−k) + M3 (χ) . S3 (k, χ, 1, 2) = sin (π n/k) 3 k 0
Corollary 4.4. Let χ denote a nonprincipal, real, primitive, odd character of modulus k, k ≥ 7. Then
sin3 (4πn/k) χ (n) S3 (k, χ , 3, 4) = sin6 (πn/k) 0
sin3 (3π/7) sin3 (π/7) sin3 (2π/7) 368 − + = √ . 6 6 6 sin (π/7) sin (2π/7) sin (3π/7) 7

566

B. C. Berndt, A. Zaharescu

5. Generalization of (1.4) Next, we establish a general theorem for which (1.4) is a special case. Theorem 5.1. Let χ be as in the previous theorems, and let a and b be positive integers with a even, b odd, and ab − 4 < 2k. Define


S4 (k, χ, a, b) :=

0<j
Then

χ (j )

sina (bπj/k) . sin(2πj/k)

  1√ 1 k Da,b − {2χ(2) − 1} h(−k) , S4 (k, χ, a, b) = 2 (2i)a−2

where Da,b :=


n,m,r≥0 2n+2bm+4r=ab−2

  a (−1)m χ(n) . m

(5.1)

(5.2)

(5.3)

The condition ab − 4 < 2k is required for technical reasons. If this restriction were lifted, Da,b would have to be replaced by a more complicated sum arising from the contributions of the left side of (2.19). Proof. Define f (z) :=

G(z, χ ) sina (bπ z/k) 1 . G(χ ) sin(2π z/k) e2πiz − 1

(5.4)

As in previous proofs, we integrate f (z) over the contour CN of Theorem 2.1. On the interior of CN , f (z) has simple poles (at most) at z = 1, 2, . . . , k − 1 with residues Rn =

χ (n) sina (bπ n/k) = Rk−n , 2π i sin(2π n/k)

(5.5)

where 0 < n < k/2 and where we used (1.7) and the fact that a is even. However, there is also a simple pole of f (z) at z = k/2 with residue √ 1 1 G(k/2, χ ) k G(k/2, χ) = , (5.6) Rk/2 = G(χ ) (−2π/k) (−2) 4πi by (1.8). Using (5.6) and (3.7), we deduce that √ k χ(2) {2 − χ (2)} h(−k). Rk/2 = 2π i

(5.7)

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567

Hence, by the residue theorem, (5.5), and (5.7),  √ f (z)dz = 2S4 (k, χ, a, b) + kχ(2) {2 − χ (2)} h(−k).

(5.8)

CN

We next evaluate the integral above by a direct calculation. Noting from (5.4) that f (z) has period k, we see that the integrals over the indented vertical sides of CN cancel. We now examine the individual terms on the top and bottom of CN , i.e., on CNT and CNB , respectively. As before, on CNT , we write f (z) as a power series in eπN/k with the leading term having degree D := ab − 4. We now proceed as in the proof of Theorem 2.1. Using analogous notation and the same sort of reasoning, we deduce that √  (5.9) BD + BD = 2S4 (k, χ, a, b) + kχ(2) {2 − χ (2)} h(−k). 

It remains to compute BD and BD . We first calculate BD , which arises from the constant term in the expansion of f (z) in powers of eπN/k on CNT . As in (2.15), set z = x + iN and µ := e−πiz/k = eπN/k e−πix/k .

(5.10)

We use the expansions (2.16), (2.17), (2.19), and −1  −2 ∞
1 µ − µ2 2i −1 −2 =− 2 = −2iµ µ−4r sin (2πz/k) = 2i µ 1 − µ−4 r=0 (5.11) in the definition (5.4) of f (z) to find that f (z) = f (x + iN) =

1 (2i)a−1 G(χ )

µab−2

k−1
χ(n) n=1

µ2n

  ∞
a −2bm × (−1) µ−4r + O(µab−4−2k ). µ m m=0 r=0 a


m

(5.12) The constant term in (5.12) is equal to  
1 a 1 m (−1) χ(n) = AD := √ Da,b , a−1 m (2i) G(χ ) (2i)a−1 i k n,m,r≥0 2n+2bm+4r=ab−2

by (1.8) and (5.3). Hence,  0 BD = AD dx = −kAD = − k

√ k Da,b . a−1 (2i) i

(5.13)

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B. C. Berndt, A. Zaharescu

The calculation of the integral of f (z) over CNB is similar, and we find that √  k k    AD dx = kAD = − (5.14) Da,b . BD = a−1 (2i) i 0 Lastly, from (5.9), (5.13), and (5.14), we conclude that √ √ k Da,b = 2S4 (k, χ, a, b) + kχ(2) {2 − χ (2)} h(−k), a−2 (2i)  

which completes the proof of Theorem 5.1. Corollary 5.2. If χ is as in previous theorems, then
√ χ (n) tan(π n/k) = −2 k {2χ(2) − 1} h(−k). 0
Proof. Set a = 2 and b = 1 in Theorem 5.1. Observe that D2,1 = 0. Hence, by Theorem 5.1 and the elementary identity sin(2θ) = 2 sin θ cos θ,


χ (n)

0
1
sin2 (π n/k) = χ(n) tan(πn/k) sin(2π n/k) 2 0
 

from which the desired corollary follows. Corollary 5.2 is due to Lebesgue [13]. Corollary 5.3. Let χ be as in Theorems 2.1 and 5.1 and set S4 (k, χ ) :=


0<j
Then

χ(j )

sin4 (πj/k) . sin(2πj/k)

√ k S4 (k, χ ) = − (1 + 4 {2χ (2) − 1} h(−k)) . 8

(5.15)

(5.16)

Proof. Set a = 4 and b = 1 in Theorem 5.1. We easily see that D4,1 = 1. The corollary then immediately follows.   Corollary 5.4. We have S4 (7, χ) = −

5√ 7, 8

S4 (11, χ ) =

11 √ 11. 8

(5.17)

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569

First observe that in the case k = 7 the sum in (5.15) has the opposite sign as the sum in (1.4). If we set k = 7 and k = 11, respectively, in (5.16), recall that h(−7) = h(−11) = 1, and note that χ(2) = 1 for k = 7 and χ (2) = −1 for k = 11, both evaluations in (5.17) follow. By the same kind of argument, we can prove an analogue of Theorem 5.1 in which sina (bπj/k) is replaced by cosa (bπj/k). Because the proof is similar, we provide only the most salient details. Theorem 5.5. Let χ be as in Theorem 5.1, let a be a nonnegative integer, and let b be an even positive integer. Assume, as in the previous theorem, that ab − 4 < 2k. Define S5 (k, χ, a, b) :=




χ(j )

0<j
Then S5 (k, χ, a, b) =

cosa (bπj/k) . sin(2πj/k)

(5.18)

    1√ 1 k − a−2 Ea,b + 1 − (−1)ab/2 {2χ (2) − 1} h(−k) , 2 2 (5.19)

where Ea,b :=


n,m,r≥0 2n+2bm+4r=ab−2

  a χ(n) . m

(5.20)

As with Theorem 5.1, if the condition ab − 4 < 2k were lifted, Ea,b would have to be replaced by a more complicated sum arising from the contributions of the left side of (2.19). Proof. Define f (z) :=

1 G(z, χ ) cosa (bπ z/k) . G(χ ) sin(2π z/k) e2πiz − 1

(5.21)

We integrate f (z) over the contour CN of Theorem 2.1. On the interior of CN , f (z) has simple poles (at most) at z = 1, 2, . . . , k − 1 with residues Rn =

χ (n) cosa (bπ n/k) = Rk−n , 2π i sin(2π n/k)

1 ≤ n < k/2,

(5.22)

since b is even and χ is odd. In contrast to the proof of Theorem 5.1, our new function f (z) has a simple pole at 0 with residue √ k−1 k h(−k) n=1 χ (n)n , (5.23) R0 = =− √ 2πi 2π i k

570

B. C. Berndt, A. Zaharescu

by (1.8) and (1.9). As in the proof of Theorem 5.1, f (z) also has a simple pole at z = k/2, here with residue √ k ab/2 {2χ(2) − 1} h(−k). (5.24) Rk/2 = (−1) 2π i Hence, invoking the residue theorem and using (5.22)–(5.24), we find that  √ √ f (z) = 2S5 (k, χ, a, b) − k h(−k) + (−1)ab/2 {2χ (2) − 1} k h(−k). CN

(5.25) We next evaluate directly the integral above. Noting from (5.21) that f (z) has period k, we see that the integrals over the indented vertical sides of CN cancel. We now examine the individual terms on the top and bottom of CN , i.e., on CNT and CN B , respectively. As before, on CNT , we write f (z) as a power series in eπN/k with the leading term having degree D = ab − 4. As in the proof of Theorem 5.1, we need only to determine the constant term in this power series. Using the expansion a   1 ab 1 ab
a a −2b a (5.26) µ−2bm , cos (bπz/k) = a µ (1 + µ ) = a µ 2 2 m m=0 where µ = e−πiz/k , (2.16), (2.19), and (5.11) in the definition (5.21) of f (z), we find that the constant term in this expansion is equal to AD :=

1 √ Ea,b , k

2a−1

where Ea,b is defined in (5.20). Hence, √   0 k f (z)dz = AD dx = −kAD = − a−1 Ea,b . 2 CN T k

(5.27)

By a similar calculation, 

√ k f (z)dz = − a−1 Ea,b . 2 CN B

(5.28)

Hence, combining our two evaluations of the integral of f (z) over CN arising from (5.25) and then (5.27) and (5.28), we conclude that √ √ √ k − a−2 Ea,b = 2S5 (k, χ, a, b) − kh(−k) + (−1)ab/2 {2χ (2) − 1} kh(−k), 2 which upon a slight rearrangement yields (5.19).

 

Trigonometric sums

571

Corollary 5.6. For χ as above, k−1


√ χ (n) cot(2π n/k) = χ (2) k h(−k).

(5.29)

n=1

Proof. Set a = 1 and b = 2 in Theorem 5.5. Noting that E1,2 = 0, we complete the proof.   6. Schemmel’s Theorems The method used in the proofs of the results in this paper is very flexible, and one can prove in the same way various variants of the results above. For instance, one can use this method to prove the following identity obtained by Schemmel [17]. Theorem 6.1. For any real parameter α, and χ as in the statement of Theorem 2.1, k−1


k−1 1
sin kα sin(2πn/k) . χ (n) sin nα = √ χ (n) cos(2πn/k) − cos α 2 k n=1 n=1

(6.1)

As was shown by Schemmel, one can use this identity to prove certain formulas involving class numbers. By taking α = π/2 in (6.1), he derived the equality k−1 1
χ (n) tan(2πn/k) . h(−k) = − √ 2 k n=1

(6.2)

Also, after differentiating (6.1) with respect to α and then taking α = 0, Schemmel deduced Lebesgue’s class-number formula [13] 2 − χ (2)
χ (n) cot(πn/k) . √ 2 k n=1 k−1

h(−k) =

(6.3)

Proof. In order to prove (6.1), we integrate the function f (z) :=

1 G(z, χ ) sin kα sin(2πz/k) G(χ ) cos(2π z/k) − cos α e2πiz − 1

(6.4)

over the contour CN from the proof of Theorem 2.1. Taking into account the periodicity of both sides of (6.1) as functions of α, we may assume in what follows that 0 ≤ α ≤ 2π. We may also assume that α = π and that α is not equal to one of the numbers 2π n/k, with n ∈ {0, 2, . . . , k}, since, if we prove (6.1) for any 0 ≤ α ≤ 2π except for these finitely many values, then (6.1) will also hold for these values by the continuity of both sides of (6.1) as functions of α. Assume then in what follows that 0 ≤ α ≤ 2π, α = π, and α = 2πn/k, 0 ≤ n ≤ k. Then, on the interior of CN , the function f (z) has simple poles (at

572

B. C. Berndt, A. Zaharescu

kα most) at z = 2π , at z = k − that, for 1 ≤ n ≤ k − 1,

Rn = The residue of f (z) at z =

kα , 2π

and at z = 1, 2, . . . , k − 1. Using (1.7), we see

χ (n) sin kα sin(2π n/k) . 2π i cos(2π n/k) − cos α

kα 2π

(6.5)

is given by

G(kα/2π, χ ) sin kα sin α 1   2π √ ikα e − 1 − k sin α i k √ k−1 − k sin kα
= χ (n)einα . 2π i eikα − 1 n=1

Rkα/2π =

Similarly, the residue of f (z) at z = k −

kα 2π

(6.6)

equals

G(k − kα/2π, χ ) (− sin kα sin α) √ e−ikα − 1 i k √ k−1 − k sin kα
= χ (n)e−inα . 2π i e−ikα − 1 n=1

Rk−kα/2π =

2π k

1 sin α (6.7)

Note that 2πiRk−kα/2π is the complex conjugate of 2πiRkα/2π , and therefore

2πi(Rk−kα/2π

√ + Rkα/2π ) = −2 k sin kα Re



k−1
1 inα χ(n)e . (6.8) eikα − 1 n=1

Here  k−1 k−1

1 1 χ (n)einα = Re ikα/2 χ(n)ei(n−k/2)α Re ikα e − 1 n=1 e − e−ikα/2 n=1  k−1
1 = Re χ (n)(cos(n − k/2)α + i sin(n − k/2)α 2i sin(kα/2) n=1 


1 χ (n) sin(n − k/2)α. 2 sin(kα/2) n=1 k−1

=

(6.9)

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573

We thus deduce from (6.8) and (6.9) that k−1
√ 2πi(Rk−kα/2π + Rkα/2π ) = − 2 k cos(kα/2) χ(n) sin(n − k/2)α n=1 k−1 √
k χ (n)(sin nα + sin(n − k)α)

=−

n=1 k−1 √
=−2 k χ (n) sin nα,

(6.10)

n=1

where we used the substitution n → k − n and the assumption that χ is odd in order to conclude that the contribution of the terms χ(n) sin(n − k)α equals the contribution of the terms χ (n) sin nα in the sum above. By (6.5), (6.10), and the residue theorem,  k−1 k−1
√
sin kα sin(2π n/k) f (z)dz = χ (n) χ(n) sin nα. (6.11) −2 k cos(2π n/k) − cos α CN n=1 n=1 Next, we let N → ∞ in order to calculate the integral in (6.11) directly. Note first that f (z) has period k, and so the integrals over the indented vertical sides of CN cancel. Inverting the order of integration and summation for the integrals on the horizontal sides, we see that for each j, 1 ≤ j ≤ k − 1, the integrals over the top and the bottom sides of CN tend to zero as N tends to infinity. Hence,  f (z)dz = 0. (6.12) CN

The identity (6.1) now follows by combining (6.11) with (6.12).

 

7. Stoneham’s Letter to Evans Theorem 7.1. Let χ be a character satisfying the hypotheses of Theorem 2.1, and let b be a positive even integer such that b < 2k + 2. Let S6 (k, χ, b) :=

k−1


χ (n) cos(bπn/k) cot(πn/k).

(7.1)

n=1

Then S6 (k, χ, b) = (2h(−k) − Fb )

√

k,

(7.2)

where Fb := 2

1 b−1 2
n=1

χ (n) + χ ( 21 b).

(7.3)

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B. C. Berndt, A. Zaharescu

The sum S6 (k, χ, b) is equivalent to a sum described by Stoneham in his letter of September 3, 1982, to Evans [18]. Stoneham calculated his sum for several values of b and k = 19 and asked if the sum had ever been previously studied. No conjectures were given. Proof. Integrate f (z) :=

1 G(z, χ ) cos(bπ z/k) cot(πz/k) 2πiz G(χ ) e −1

(7.4)

over the contour CN in the proof of Theorem 2.1. For 1 ≤ n ≤ k − 1, χ (n) cos(bπ n/k) cot(πn/k). 2π i A straightforward calculation with the use of (1.8) and (1.9) shows that √ k h(−k) . R0 = − πi Hence, by (7.5), (7.6), and the residue theorem,  √ f (z)dz = S6 (k, χ, b) − 2 k h(−k). Rn =

(7.5)

(7.6)

(7.7)

CN

Next, we provide a direct calculation of the integral above. Since, b is even, by the periodicity of f (z) in (7.4), we see that the integrals over the indented vertical sides of CN cancel. As before, we next expand f (z) in powers of µ on CN T . Using (2.18) (with a = 0), (2.19), and the expansion cos(bπ z/k) =

1 b µ (1 + µ−2b ) 2

twice, we find that on CNT , f (z) =

∞ k−1
i χ (n)
−2r µb (1 + µ−2b )(1 + µ−2 ) µ + O(µb−2−2k ). 2n 2G(χ) µ n=1 r=0

The only contribution to the integral of f (z) over CNT arises √ from the constant term in this expansion which is readily seen to equal Fb /(2 k), where Fb is defined in (7.3). We use a similar argument on CNB . Thus, we find that   1√ f (z)dz = − k Fb = f (z)dz. (7.8) 2 CN T CN B Combining (7.7) and (7.8), we conclude that √ √ S6 (k, χ, b) − 2 k h(−k) = − k Fb , which completes the proof of Theorem 7.1.

 

The authors are grateful to the referee for corrections and helpful suggestions.

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References 1. Apostol, T.M.: Euler’s φ-function and separable Gauss sums. Proc. Am. Math. Soc. 24, 482–485 (1970) 2. Ayoub, R.: An Introduction to the Analytic Theory of Numbers. American Mathematical Society, Providence, RI, 1963 3. Berndt, B.C.: Classical theorems on quadratic residues. Enseign. Math. 22, 261–304 (1976) 4. Berndt, B.C.: Ramanujan’s Notebooks, Part IV. Springer-Verlag, New York, 1994 5. Berndt, B.C., Evans, R.J., Williams, K.S.: Gauss and Jacobi Sums. Wiley, New York, 1998 6. Berndt, B.C., Zhang, L.-C.: Ramanujan’s identities for eta-functions. Math. Ann. 292, 561–573 (1992) 7. Borevich, Z.I., Shafarevich, I.R.: Number Theory. Academic Press, New York, 1966 8. Chowla, S.: The evaluation of a trigonometric sum. J. Indian Math. Soc. 18, 147–148 (1929) 9. Chowla, S.: The Collected Papers of Sarvadaman Chowla. Vol. I, Les Publications CRM, Montreal, 1999 10. Davenport, H.: Multiplicative Number Theory. 3rd ed., Springer-Verlag, New York, 2000 11. Dickson, L.E.: History of the Theory of Numbers. Vol. 3, Chelsea, New York, 1966 12. Dirichlet, G.L.: Recherches sur diverses applications de l’analyse infinit´esimale a` le th´eorie des nombres, seconde partie. J. Reine Angew. Math. 21, 134–155 (1840)   13. Lebesgue, V.A.: Suite du Memoire sur les applications du symbole ab . J. de Math. 15, 215–237 (1850) 14. Lerch, M.: Essais sur le calcul du nombre des classes de formes quadratiques binaires aux coefficients entiers. Acta Math. 30, 203–293 (1906) 15. Liu, Z.-G.: Some Eisenstein series identities related to modular equations of the seventh order. Pacific J. Math. 209, 103–130 (2003) 16. Ramanujan, S.: Notebooks. (2 volumes), Tata Institute of Fundamental Research, Bombay, 1957 17. Schemmel, V.: De Multitudine formarum secundi gradus disquisitiones. Dissertation, Breslau, 1863 18. Stoneham, R.G.: Letter to R. J. Evans. September 3, 1982 19. Whiteman, A.L.: Theorems on quadratic residues. Math. Magazine 23, 71–74 (1949)