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RECIPROCITY THEOREMS FOR BETTIN–CONREY SUMS 6 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK 1 0 DedicatedtothememoryofTomM.Apostol 2 g ABSTRACT. RecentworkofBettinandConreyontheperiodfunctionsofEisensteinseriesnaturallygaverise u totheDedekind-likesum A ca h = kak(cid:229)−1cot p mh z −a,m , k k k 3 (cid:18) (cid:19) m=1 (cid:18) (cid:19) (cid:16) (cid:17) wherea∈C,handkarepositivecoprimeintegers,andz (a,x)denotestheHurwitzzetafunction. Wederive ] anewreciprocitytheoremfortheseBettin–Conreysums,whichinthecaseofanoddnegativeintegeracanbe T explicitlygivenintermsofBernoullinumbers. This,inturn,impliesexplicitformulasfortheperiodfunctions N appearinginBettin–Conrey’swork.WestudygeneralizationsofBettin–Conreysumsinvolvingzetaderivatives . andmultiplecotangentfactorsandrelatethesetospecialvaluesoftheEstermannzetafunction. h t a m [ 1. INTRODUCTION AND STATEMENT OF RESULTS 2 OurpointofdepartureisrecentworkofBettinandConrey[7,8]ontheperiodfunctionsofEisensteinse- v ries. TheirinitialmotivationwasthederivationofanexactformulaforthesecondmomentsoftheRiemann 9 3 zetafunction, buttheirworknaturally gaverisetoafamilyoffinitearithmeticsumsoftheform 8 6 c h = kak(cid:229)−1cot p mh z −a,m , 0 a k k k 1. (cid:18) (cid:19) m=1 (cid:18) (cid:19) (cid:16) (cid:17) wherea∈C,handkarepositivecoprimeintegers, andz (a,x)denotes theHurwitzzetafunction 0 16 z (a,x) = (cid:229)¥ 1 , : (n+x)a v n=0 i initially defined for ´ (a)>1and meromorphically continued to the a-plane. Wecall c (h)and its natural X a k generalizations appearing belowBettin–Conrey sums. r a Therearetwomajormotivationstostudythesesums. Thefirstisthatc (h)isessentiallyaVasyuninsum, 0 k which in turn makes a critical appearance in the Nyman–Beurling–Ba´ez-Duarte approach to the Riemann hypothesis through the twisted mean-square of the Riemann zeta function on the critical line (see, e.g., [5,16]). Bettin–Conrey’s work,fora=0,impliesthatthereisahiddensymmetryofthismean-square. The second motivation, and the central theme of our paper, is that the Bettin–Conrey sums satisfy a reciprocity theorem: h k 1+a −k kaaz (1−a) c − c + a k h a h p h (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) Date:2August2016. 2010MathematicsSubjectClassification. Primary11F20;Secondary11L03,11M35. Keywordsandphrases. Dedekindsum,cotangentsum,Bettin–Conreysum,reciprocitytheorem,Hurwitzzetafunction,period function,quantummodularform,Estermannzetafunction. We thank Sandro Bettin for valuable comments, particularly for suggesting Theorem 3. Abdelmejid Bayad was partially supportedbytheFDIRoftheUniversite´d’EvryVald’Essonne;MatthiasBeckwaspartiallysupportedbytheUSNationalScience Foundation(DMS-1162638). 1 2 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK extends from its initiation domain Q to an (explicit) analytic function on C\R , making c nearly an ≤0 a example of a quantum modular form in the sense of Zagier [18]. In fact, Zagier’s “Example 0” is the Dedekindsum 1 k(cid:229)−1 p mh p m s(h,k) = cot cot , 4k k k m=1 (cid:18) (cid:19) (cid:16) (cid:17) whichis,uptoatrivialfactor, c (h). Dedekind sumsfirstappeared inthetransformation properties ofthe −1 k Dedekindetafunction andsatisfythereciprocity theorem [10,11] 1 1 h 1 k s(h,k)+s(k,h) = − + + + . 4 12 k hk h (cid:18) (cid:19) WenowrecallthepreciseformofBettin–Conrey’s reciprocity theorem. Theorem1(Bettin–Conrey [7]). Ifhandkarepositivecoprimeintegers then h k 1+a −k kaaz (1−a) h c − c + = −iz (−a)y a k h a h p h a k (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) where i z (1−a) i p a g (z) y (z) = − cot +i a a p z z (−a) z1+a 2 z (−a) and g (z) = −2 (cid:229) (−1)n B2n z (1−2n−a)(2p z)2n−1 a (2n)! 1≤n≤M 1 cosp a + z (s)z (s−a)G (s) 2 (2p z)−sds. p iZ(−21−2M) sinp s−2a Here B denotes the kth Bernoulli number, M is any integer ≥−1min(0,´ (a)), and the integral notation k 2 indicates thatourintegration pathisovertheverticalline´ (s)=−1−2M. 2 WenotethatBettinandConreyinitiallydefinedy (z)through a 1 1 y (z) = E (z)− E − , a a+1 za+1 a+1 z (cid:18) (cid:19) inotherwords,y (z)istheperiodfunction oftheEisenstein seriesofweight2k+2, a E (z) = 1+ 2 (cid:229) s (n)e2p inz, a+1 z (−a) a n≥1 wheres (n)=(cid:229) da,andthenshowedthaty (z)satisfiestheproperties ofTheorem1. a d|n a We have several goals. We start by showing that the right-hand side of Theorem 1 can be simplified by employinganintegration techniqueforDedekind-like sumsthatgoesbacktoRademacher[11]. Thisyields ourfirstmainresult: Theorem 2. Let ´ (a) > 1 and suppose h and k are positive coprime integers. Then for any 0 < e < min 1,1 , h k (cid:8) (cid:9) h k az (a+1) (hk)1−a cot(p hz)cot(p kz) h1−ac +k1−ac = − dz. −a k −a h p (hk)a 2i (e) za (cid:18) (cid:19) (cid:18) (cid:19) Z Theorem 2 implies the following analytic continuation result; in particular, in this sense Theorem 2 can beextendedtoallcomplexa. RECIPROCITYTHEOREMSFORBETTIN–CONREYSUMS 3 Theorem3. Lethandkarepositivecoprimeintegers. Thenforany0<e <min 1,1 ,thefunction h k cot(p hz)cot(p kz) (cid:8) (cid:9) F(a) = dz (e) za Z hasaholomorphic continuation tothewholecomplexplane. Second, we employ Theorem 2 to show that in the case that a is an odd negative integer, the right-hand sideofthereciprocity theorem canbeexplicitly givenintermsofBernoullinumbers. Theorem4. Letn>1beanoddinteger andsuppose handkarepositive coprimeintegers. Then h1−nc h +k1−nc k = 2p i n 1 nB +n(cid:229)+1 n+1 B B hmkn+1−m . −n −n n+1 m n+1−m k h hk i(n+1)! m (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) m=0(cid:18) (cid:19) ! Wenotethat,asfaraswecantell,thisidentitycannotbeeasilyderivedfromTheorem1. Ourthirdmain result is, in turn, a consequence of Theorem 4: in conjunction with Theorem 1, it implies the following explicitformulasfory (z)andg (z)whenaisanoddnegativeinteger. a a Theorem5. Ifn>1isanoddinteger thenforallz∈C\R ≤0 y (z) = (2p i)n n(cid:229)+1 n+1 B B zm−1 −n z (n)(n+1)! m m n+1−m m=0(cid:18) (cid:19) and g (z) = (2p i)n (cid:229) n n+1 B B zm. −n m+1 n−m i(n+1)! m+1 m=0(cid:18) (cid:19) In [7, Theorem 2], Bettin and Conrey computed the Taylor series of g (z) and remarked that, if a is a a negativeinteger,p g(m)(1)isarationalpolynomialinp 2. Theorem5generalizesthisremark. Wewillprove a Theorems2–5inSection2. Our next goal is to study natural generalizations of c (h). Taking a leaf from Zagier’s generalization of a k s(h,k) to higher-dimensional Dedekind sums [17] and its variation involving cotangent derivatives [9], let k ,k ,...,k be positive integers such that (k ,k )=1 for j=1,2,...,n, let m ,m ,...,m be nonnegative 0 1 n 0 j 0 1 n integers, a6=−1acomplexnumber,anddefinethegeneralized Bettin–Conrey sum c k0 k1 ··· kn = kak(cid:229)0−1z (m0) −a, l (cid:213) n cot(mj) p kjl . a(cid:18) m0 m1 ··· mn (cid:19) 0 l=1 (cid:18) k0(cid:19)j=1 (cid:18) k0 (cid:19) Herez (m0)(a,z)denotesthemth derivativeoftheHurwitzzetafunction withrespect toz. 0 Thisnotation mimicsthatofDedekindcotangent sums;notethat h k h c = c . s k s 0 0 (cid:18) (cid:19) (cid:18) (cid:19) In Section 3, we will prove reciprocity theorems for generalized Bettin–Conrey sums, paralleling Theo- rems2and4,aswellasmorespecialcasesthatgive,wethink, interesting identities. Ourfinalgoalistorelatetheparticular generalized Bettin–Conrey sum q−1 (cid:229) cot(k)(p mx)z −a,m q m=1 (cid:16) (cid:17) withevaluations oftheEstermannzetafunction (cid:229) s (n)e2pinx atintegers s;seeSection4. n≥1 a ns 4 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK 2. PROOFS OF MAIN RESULTS InordertoproveTheorem2,weneedtwolemmas. Lemma6. Letmbeanonnegative integer. Then ∓i ifm=0, limcot(m)p (x±iy) = y→¥ (0 ifm>0. Furthermore, thisconvergence isuniform withrespecttoxinafixedboundedinterval. Proof. Sincecotz= i(eiz+e−iz),wemayestimate eiz−e−iz 2 2 2 |i+cotp (x+iy)| = ≤ = . ei(2p x)−e2p y ei(2p x) −|e2p y| |1−e2p y| Giventhattherightmostterminthisin(cid:12)equality vanis(cid:12)hesa(cid:12)s(cid:12)y→¥ (cid:12),wesee(cid:12)that (cid:12) (cid:12) (cid:12)(cid:12) (cid:12) (cid:12) limcotp (x+iy) = −i. y→¥ Similarly,theinequality 2 2 2 |−i+cotp (x−iy)| = ≤ = ei(2p x)e2p y−1 ei(2p x)e2p y −1 |e2p y−1| impliesthatlimy→¥ cotp (x−iy)=i. Si(cid:12)nce (cid:12) (cid:12)(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)(cid:12) (cid:12) (cid:12) 2ep y 2ep y 2ep y |cscp (x+iy)| = ≤ = , |eip x−e−ip xe2p y| ||eip x|−|e−ip xe2p y|| |1−e2p y| itfollowsthatlimy→¥ cscp (x+iy)=0. Similarly, 2ep y 2ep y 2ep y |cscp (x−iy)| = ≤ = |eip xe2p y−e−ip x| ||eip xe2p y|−|e−ip x|| |e2p y−1| impliesthatlimy→¥ cscp (x−iy)=0. Weremarkthat ddz(cotz)=−csc2zand d (cscz) = −csczcotz, dz soallthederivativesofcotzhaveacsczfactor, andtherefore, ∓i ifm=0, limcot(m)p (x±iy) = y→¥ (0 ifm>0. Since the convergence above is independent of x, the limit is uniform with respect to x in a fixed bounded interval. (cid:3) Lemma6impliesthat ∓i ifm=0, limcot(m)p h(x±iy) = limcot(m)p k(x±iy) = y→¥ y→¥ (0 ifm>0, uniformlywithrespecttoxinafixedbounded interval. TheproofofthefollowinglemmaishintedatbyApostol[3]. Lemma7. If´ (a)>1andR>0,thenz (a,x+iy)vanishesuniformlywithrespecttox∈[0,R]asy→±¥ . RECIPROCITYTHEOREMSFORBETTIN–CONREYSUMS 5 Proof. Webeginbyshowingthatz (a,z)vanishesas` (z)→±¥ if´ (z)>0. Since´ (a)>1and´ (z)>0, wehavetheintegral representation [14,eq.25.11.25] 1 ¥ ta−1e−zt z (a,z) = dt, G (a) 1−e−t 0 Z whichmaybewrittenas 1 ¥ ta−1e−t´ (z) (1) z (a,z) = e−it` (z)dt. G (a) 1−e−t 0 Z Notethatforfixed´ (z), ¥ ta−1e−t´ (z) dt = z (a,´ (z))G (a) 1−e−t 0 Z and ¥ ta−1e−t´ (z) ¥ t´ (a)−1e−t´ (z) dt = dt = z (´ (a),´ (z))G (´ (a)), 1−e−t 1−e−t 0 (cid:12) (cid:12) 0 Z Z (cid:12) (cid:12) sotheRiemann–Lebe(cid:12)sgue lemma(cid:12)(see,forexample,[13,Theorem16])impliesthat (cid:12) (cid:12) (cid:12) (cid:12) ¥ ta−1e−t´ (z) e−it` (z)dt 1−e−t 0 Z vanishes as ` (z)→±¥ . By (1), this means that for ´ (z) fixed, z (a,z) vanishes as ` (z)→±¥ . In other words,z (a,x+iy)→0pointwisewithrespect tox>0asy→±¥ . Moreover, the vanishing of z (a,x+iy) as y→±¥ is uniform with respect to x∈[0,R]. Indeed, denote g(t)= ta−1e−tR,then(1)impliesthat 1−e−t ¥ g(t)dt = G (a)z (a,R) 0 Z and ¥ |g(t)|dt = G (´ (a))z (´ (a),R). 0 It then follows from the Riemann–ZLebesgue lemma that lim|z|→¥ 0¥ g(t)e−itzdt =0. If x∈(0,R], we may write ¥ ta−1e−tx ¥R G (a)z (a,x±iy) = e∓itydt = g(t)e−it(±y−i(x−R))dt. 1−e−t 0 0 Z Z Sinceg(t)doesnotdependonx,thespeedatwhichz (a,x±iy)vanishesdependsonRandy2+(x−R)2. However,weknowthat0≤|x−R|<R,sothespeedofthevanishing dependsonlyonR. Finally,notethat ¥ ¥ z (a,iy) = (cid:229) 1 = (cid:229) 1 + 1 = z (a,1+iy)+ 1 , (iy+n)a (1+iy+n)a (iy)a (iy)a n=0 n=0 so z (a,iy)→0 as y→±¥ , and the speed at which z (s,iy) vanishes depends on that of z (s,1+iy). Thus, z (s,x+iy)→0uniformly asy→±¥ ,aslongasx∈[0,R]. (cid:3) ProofofTheorem2. TheideaistouseCauchy’sresiduetheorem tointegrate thefunction f(z) = cot(p hz)cot(p kz)z (a,z) alongC(M,e )asM→¥ ,whereC(M,e )denotes thepositively oriented rectangle withvertices 1+e +iM, e +iM,e −iM and1+e −iM,forM>0and0<e <min 1,1 (seeFigure1). h k Henceforth, a ∈ C is such that ´ (a) > 1, (h,k) is a pair of coprime positive intergers, and f(z) and C(M,e )areasabove,unlessotherwisestated. Sincez (a,z)(cid:8)isana(cid:9)lyticinsideC(M,e ),theonlypolesof f(z) 6 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK FIGURE 1. TheclosedcontourC(M,e ). arethose ofthecotangent factors. Thus,thefactthathandk arecoprimeimpliesthatacompletelistofthe possible polesof f(z)insideC(M,e )is 1 h−1 1 k−1 E = ,..., , ,..., ,1 h h k k (cid:26) (cid:27) and each of these poles is (at most) simple, with the exception of 1, which is (at most) double. For m∈ {1,2,...,h−1}, p km m 1 1 p km m Resf(z) = cot cos(p m)z a, Res = cot z a, . z=m h h z=m sin(p hz) p h h h h (cid:18) (cid:19) h (cid:18) (cid:19) (cid:16) (cid:17) (cid:16) (cid:17) Ofcourse,ananalogous resultistrueforResz=m f(z)forallm∈{1,2,...,k−1},andtherefore k (cid:229) Resf(z) = Resf(z)+ 1 h(cid:229)−1cot p km z a,m + 1 k(cid:229)−1cot p hm z a,m z0∈Ez=z0 z=1 p hm=1 (cid:18) h (cid:19) (cid:16) h(cid:17) p km=1 (cid:18) k (cid:19) (cid:16) k(cid:17) or,equivalently, (2) h1−ac h +k1−ac k = p (hk)1−a (cid:229) Resf(z) −Resf(z) . −a −a (cid:18)k(cid:19) (cid:18)h(cid:19) z0∈Ez=z0 ! z=1 ! RECIPROCITYTHEOREMSFORBETTIN–CONREYSUMS 7 Wenowdetermine Res f(z). TheLaurentseriesofthecotangent function about0isgivenby z=1 1 1 1 2 cotz = − z− z3− z5+···, z 3 45 945 so,bytheperiodicity ofcotz,forz6=1inasmallneighborhood ofz=1, 1 1 p k (p k)3 2(p k)5 cot(p kz) = − (z−1)− (z−1)3− (z−1)5+··· p k z−1 3 45 945 (cid:18) (cid:19) and,similarly, 1 1 p h (p h)3 2(p h)5 cot(p hz) = − (z−1)− (z−1)3− (z−1)5+···. p h z−1 3 45 945 (cid:18) (cid:19) Sincez (a,z)isanalytic inasmallneighborhood of1,Taylor’stheoremimpliesthat ¥ z (a,z)= (cid:229) b (z−1)n, n n=0 where b = z (n)(a,1) for n=0,1,2,... (derivatives relative to z). Thus, the expansion of f(z) about 1 is of n n! theform 2 b 1 b 1 0 1 + +(analyticpart). p 2hk z−1 p 2hk z−1 (cid:18) (cid:19) (cid:18) (cid:19) Giventhata6=0,1,weknowthat ¶ z (a,z)=−az (a+1,z)[14,eq.25.11.17], sob =−az (a+1,1)= ¶ z 1 −az (a+1). Weconclude thatRes f(z)=−az (a+1) anditthenfollowsfrom(2)that z=1 p 2hk (3) h1−ac h +k1−ac k = az (a+1)+ p (cid:229) Resf(z). −a(cid:18)k(cid:19) −a(cid:18)h(cid:19) p (hk)a (hk)a−1z0∈Ez=z0 (cid:229) We now turn to the computation of Resf(z) via Cauchy’s residue theorem, which together with (3) z0∈Ez=z0 willprovide thereciprocity weare after. Notethat thefunction f(z) isanalytic onanytwoclosed contours C(M ,e )andC(M ,e )and since the poles inside these twocontours arethe same, wemayapply Cauchy’s 1 2 residuetheorem tobothcontours anddeducethat f(z)dz = f(z)dz. ZC(M1,e) ZC(M2,e) Inparticular, thisimpliesthat (4) lim f(z)dz = 2p i (cid:229) Resf(z). M→¥ ZC(M,e) z0∈Ez=z0 Letg bethepathalongC(M,e )from1+e +iMtoe +iM. Similarly,defineg frome −iMto1+e −iM, 1 2 g frome +iM toe −iM,andg from1+e −iM to1+e +iM (seeFigure1). Since´ (a)>1, 3 4 ¥ ¥ z (a,z+1) = (cid:229) 1 = (cid:229) 1 = z (a,z)− 1 , (n+z+1)a (n+z)a za n=0 n=1 andsotheperiodicity ofcotzimpliesthat cot(p hz)cot(p kz) f(z)dz = − f(z)dz+ dz. g g g za Z 4 Z 3 Z 3 8 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK Lemmas6and7implythat f(z)vanishesuniformlyasM→¥ (uniformitywithrespectto´ (z)∈[e ,1+e ]) so lim f(z)dz = 0 = lim f(z)dz. M→¥ g M→¥ g Z 1 Z 2 Thismeansthat lim f(z)dz = lim f(z)dz+ f(z)dz M→¥ ZC(M,e) M→¥ (cid:18)Zg3 Zg4 (cid:19) anditfollowsfrom(3)and(4)that h k az (a+1) (hk)1−a e−i¥ cot(p hz)cot(p kz) h1−ac +k1−ac = + dz. −a k −a h p (hk)a 2i e+i¥ za (cid:18) (cid:19) (cid:18) (cid:19) Z ThiscompletestheproofofTheorem2. (cid:3) ProofofTheorem3. FromTheorem2,for´ (a)>1, (hk)1−a az (a+1) h k F(a) = − h1−ac +k1−ac , 2i p (hk)a −a k −a h (cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19) sonowwestudythefunction a7→ az (a+1)− h1−ac h +k1−ac k . p (hk)a −a k −a h From the definition of c (h) (resp., c ((cid:0)k)) the fu(cid:0)nc(cid:1)tion h1−ac (cid:0) (cid:1)h(cid:1) +k1−ac k is a linear com- −a k −a h −a k −a h bination (with holomorphic coefficients) of the Hurwitz function z (s,x) with s = −a and x = m with m=1,...,k−1(resp., s=−aandx= m withm=1,...,k−1). Thisg(cid:0)ive(cid:1)saholomor(cid:0)ph(cid:1)ic continuaktion to h C\{1}. However,ata=1, h k h1−ac +k1−ac = 2p (s(h,k)+s(k,h)) −a −a k h (cid:18) (cid:19) (cid:18) (cid:19) iswelldefined. Thereforethefunction h1−ac h +k1−ac k hasaholomorphic continuation toC. −a k −a h Next,itisclearthatthefunction s(cid:0)7→(cid:1)sz (s+1) (cid:0) (cid:1) hasaholomorphiccontinuation toC\{0}. Theanalyticityofthisfunctionats=0followsfromtheanalytic continuation oftheRiemannz -function. Thus,ourdesiredresultfollowsfromthemonodromytheorem (see,e.g.,[15]). (cid:3) Toprove Theorem 4, we now turn to the particular case in which a=n>1 is an odd integer and study Bettin–Conrey sumsoftheformc . −n LetY (n)(z)denotethe(n+2)-thpolygammafunction(see,forexample,[14,Sec.5.15]). Itiswellknown thatfornapositiveinteger, (−1)n+1Y (n)(z) z (n+1,z) = n! whenever´ (z)>0(see,forinstance, [14,eq.25.11.12]), soforn>1,wemaywrite c h = (−1)n k(cid:229)−1cot p mh Y (n−1) m . −n k kn(n−1)! k k (cid:18) (cid:19) m=1 (cid:18) (cid:19) (cid:16) (cid:17) Bythereflectionformulaforthepolygammafunctions [14,eq.5.15.6], Y (n)(1−z)+(−1)n+1Y (n)(z) = (−1)np cot(n)(p z), weknowthatifnisodd,then m m p m Y (n−1) 1− −Y (n−1) = p cot(n−1) k k k (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) RECIPROCITYTHEOREMSFORBETTIN–CONREYSUMS 9 foreachm∈{1,2,...,k−1}. Therefore, 2k(cid:229)−1cot p mh Y (n−1) m = k(cid:229)−1cot p mh Y (n−1) m k k k k m=1 (cid:18) (cid:19) m=1 (cid:18) (cid:19) (cid:16) (cid:17) (cid:16) (cid:17) +k(cid:229)−1cot p (k−m)h Y (n−1) 1−m , k k m=1 (cid:18) (cid:19) (cid:16) (cid:17) whichimpliesthat 2k(cid:229)−1cot p mh Y (n−1) m = k(cid:229)−1cot p mh Y (n−1) m −Y (n−1) 1−m k k k k k m=1 (cid:18) (cid:19) m=1 (cid:18) (cid:19) (cid:16) (cid:17) (cid:16) (cid:16) (cid:17) (cid:16) (cid:17)(cid:17) = −p k(cid:229)−1cot p mh cot(n−1) p m . k k m=1 (cid:18) (cid:19) (cid:16) (cid:17) This means that for n> 1 odd, c is essentially a Dedekind cotangent sum. Indeed, using the notation −n in[9], c h = p k(cid:229)−1cot p mh cot(n−1) p m −n k 2kn(n−1)! k k (cid:18) (cid:19) m=1 (cid:18) (cid:19) (cid:16) (cid:17) k h 1 p = c n−1 0 n−1 . 2(n−1)!   0 0 0   ThusTheorem 4isaninstance ofTheorem2. Itssignificance isareciprocity instance forBettin–Conrey sumsoftheformc intermsofBernoullinumbers. Forthisreasonwegivethedetailsofitsproof. −n ProofofTheorem4. We consider the closed contour C(M,e ) defined as the positively oriented rectangle with vertices 1+iM, iM, −iM and 1−iM, with indentations (to the right) of radius 0<e <min 1,1 h k around0and1(seeFigure2). e (cid:8) (cid:9) SinceC(M,e )contains thesamepoles of f(z)=cot(p hz)cot(p kz)z (n,z) astheclosed contourC(M,e ) inFigure1usedtoproveTheorem2,wemayapplyCauchy’sresiduetheorem,lettingM→¥ ,andweonly needtodeeterminelimM→¥ C(M,e) f(z)dzinordertodeduceareciprocity lawforthesumsc−n. Asin the case ofC(M,e ), the integrals along the horizontal paths vanish, so using the periodicity of the R cotangent toaddintegralsaloengparallelpaths, aswedidwhenconsideringC(M,e ),weobtain ie −iM (5) lim f(z)dz = lim g(z)dz+ g(z)dz + g(z)dz, M→¥ ZC(M,e) M→¥ (cid:18)ZiM Z−ie (cid:19) Zg3 whereg denotestheindented patharound 0and 3 e cot(p kz)cot(p hz) g(z) = . zn Giventhatg(z)isanoddfunction,theverticalintegralscancelandwemayapplyCauchy’sresiduetheorem tointegrate g(z)alongthepositivelyoriented circleofradiuse andcentered at0,todeduce that lim f(z)dz = −p iResg(z). M→¥ C(M,e) z=0 Z This is the main reason to use the contouer C(M,e ) instead ofC(M,e ). Indeed, integration along C(M,e ) exploitstheparityofthefunction g(z),allowingustocanceltheverticalintegrals in(5). e e 10 JUANS.AULI,ABDELMEJIDBAYAD,ANDMATTHIASBECK FIGURE 2. TheclosedcontourC(M,e ). Theexpansion ofthecotangent function is e p zcot(p z) = (cid:229)¥ (2p i)mBmzm, m! m=0 withtheconvention thatB mustberedefinedtobezero. Thus,wehavetheexpansion 1 cot(p kz) = (cid:229)¥ (2i)(2p ik)mBm+1zm (m+1)! m=−1 andofcourse, ananalogous resultholdsforh. Hence, (6) Resg(z) = (2i)(2p i)n n(cid:229)+1 n+1 B B hmkn+1−m, z=0 p hk(n+1)! m m n+1−m m=0(cid:18) (cid:19) andgiventhatz (n+1)=−(2p i)n+1B [14,eq.25.6.2],theCauchyresidue theoremand(3)yield 2(n+1)! n+1 (7) 2p i n 1 nB +n(cid:229)+1 n+1 B B hmkn+1−m = h1−nc h +k1−nc k . n+1 m n+1−m −n −n hk i(n+1)! m k h (cid:18) (cid:19) m=0(cid:18) (cid:19) ! (cid:18) (cid:19) (cid:18) (cid:19) Finally, notethat theconvention B :=0isirrelevant in(7),since B inthis sum isalways multiplied by 1 1 aBernoullinumberwithoddindexlargerthan1. (cid:3) NotethatTheorem4isessentiallythesameasthereciprocitydeducedbyApostolforDedekind–Apostol sums [2]. This is a consequence of the fact that for n > 1 an odd integer, c h is a multiple of the −n k Dedekind–Apostol sums (h,k). Indeed,forsuchn[3,Theorem1] n (cid:0) (cid:1) h s (h,k) = in!(2p i)−nc . n −n k (cid:18) (cid:19)

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