Hindawi Publishing Corporation Journal of Mathematics Volume 2013, Article ID 181724, 17 pages http://dx.doi.org/10.1155/2013/181724

Research Article and Series Representations of Riemann’s Zeta Function and Dirichlet’s Eta Function and a Medley of Related Results

Michael S. Milgram

Consulting Physicist, Geometrics Unlimited, Ltd., P.O. Box 1484, Deep River, ON, Canada K0J 1P0

Correspondence should be addressed to Michael S. Milgram; [email protected]

Received 22 September 2012; Accepted 20 December 2012

Academic Editor: Alfredo Peris

Copyright © 2013 Michael S. Milgram. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Contour integral representations of Riemann’s Zeta function and Dirichlet’s Eta (alternating Zeta) function are presented and investigated. These representations flow naturally from methods developed in the 1800s, but somehow they do not appear in the standard reference summaries, textbooks, or literature. Using these representations as a basis, alternate derivations of known series and integral representations for the Zeta and Eta function are obtained on a unified basis that differs from the textbook approach, and results are developed that appear to be new.

Dedicated to the memory of Leslie M. Saunders, died September 8, 1972, age 29 (Physics Today obituary, issue of December 1972, page 63)

1. Introduction It is the purpose of this work to disinter these represen- tations and revisit and explore some of the consequences. 𝜁(𝑠) Riemann’s Zeta function and its sibling Dirichlet’s In particular, it is possible to obtain series and integral 𝜂(𝑠) (alternating zeta) function play an important role in representations of 𝜁(𝑠) and 𝜂(𝑠) that unify and generalize physics, complex analysis, and number theory and have been well-known results in various limits and combinations and studied extensively for several centuries. In the same vein, reproduceresultsthatareonlynowbeingdiscoveredand the general importance of a contour integral representation investigated by alternate means. Additionally, it is possible to of any function has been known for almost two centuries, explore the properties of 𝜁(𝑠) in the complex plane and on the so it is surprising that contour integral representations for critical line and obtain results that appear to be new. Many of 𝜁(𝑠) 𝜂(𝑠) both and exist that cannot be found in any of the the results obtained are disparate and difficult to categorize 25.5( ) modern handbooks (NIST, [1,Section iii ]; Abramowitz in a unified manner but share the common theme that they and Stegun, [2, Chapter 23]), textbooks (Apostol, [3,Chapter are all somehow obtained from a study of the revived integral 12]; Olver, [4, Chapter 8.2]; Titchmarsh, [5,Chapter4]; representations. That is the unifying theme of this work. To (13.13) Whittaker and Watson, [6,Section ]), summaries maintain a semblance of brevity, most of the derivations are (Edwards, [7], Ivic,´ [8,Chapter4];Patterson,[9]; Srivastava only sketched, with citations sufficient enough to allow the and Choi, [10, 11]), compendia (Erdelyi´ et al., [12,Section1.12] reader to reproduce any result for her/himself. and Chapter 17), tables (Gradshteyn and Ryzhik, [13], 9.512; Prudnikov et al, [14, Appendix II.7]), and websites [1, 15– 19] that summarize what is known about these functions. 2. Contour Integral Representations of One can only conclude that such representations, generally 𝜁(𝑠) and 𝜂(𝑠) discussed in the literature of the late 1800s, have been long- buried and their significance has been overlooked by modern Throughout, I will use 𝑠=𝜎+𝑖𝜌,where𝑠∈C (complex), scholars. 𝜎, 𝜌 ∈ R (reals), and 𝑛 and 𝑁 are positive integers. 2 Journal of Mathematics

The 𝜁(𝑠) and the Dirichlet alter- context, so that the specific results (7)and(9)neverappear, nating zeta function 𝜂(𝑠) are well known and defined by particularly in that section of the work devoted to 𝜁(𝑠).In (convergent) series representations: modern notation, for a meromorphic function 𝑓(𝑧), Lindelof

∞ writes ([24, Equation III(4)]) in general 𝜁 (𝑠) ≡ ∑𝑘−𝑠,𝜎>1, (1) ∞ 𝑘=1 1 ∑ 𝑓 (]) =− ∮ 𝜋 cot (𝜋𝑧) 𝑓 (𝑧) 𝑑𝑧, (10) ∞ ]=−∞ 2𝜋𝑖 𝑘 −𝑠 𝜂 (𝑠) ≡−∑(−1) 𝑘 ,𝜎>0, (2) 𝑘=1 where the contour of integration encloses all the singularities of the integrand but never specifically identifies 𝑓(𝑧) = with −𝑠 𝑧 .Muchlater,aftertheintegralhasbeensplitintotwo 1−𝑠 𝜂 (𝑠) =(1−2 )𝜁(𝑠) . (3) by the invocation of identities for cot(𝜋𝑧),hedoesmake this identification ([24, Equation 4.III (6)]), arriving (with It is easily found by an elementary application of the residue (3)) at ((26)—see below) a now-well-known result, thereby theorem that the following reproduces (1): bypassing the contour integral representation (7). In a similar vein, Olver ([4,page290])reproduces(10), employing it ∞ −𝑠 𝑖𝜋𝑡 󵄨 𝑡 𝑒 󵄨 to evaluate remainder terms, the Abel-Plana formula and a 𝜁 (𝑠) =𝜋∑ 󵄨 ,𝜎>1 residue (𝜋𝑡)󵄨 (4) 𝑘=1 sin 󵄨𝑡=𝑘 Jensen result but again never writes (7) explicitly. Perhaps the omission of an explicit statement of (7)and(9)fromChapter and the following reproduces (2): IV of Lindelof [24] is why these forms seem to have vanished from the historical record, although citations to Jensen and ∞ 󵄨 𝜂 (𝑠) =𝜋∑ (𝑡−𝑠 (𝜋𝑡) 𝑒𝑖𝜋𝑡)󵄨 ,𝜎>0. Lindelof abound in modern summaries. (Recently, Srivastava residue cot 󵄨𝑡=𝑘 (5) 𝑘=1 and Choi [11, pages 169–172], have reproduced Lindelof’s analysis, also failing to arrive explicitly at (7); although they Convert the result (4) into a contour integral enclosing the do write a general form of (8)([11, Section 2.3, Equation positive integers on the 𝑡-axis in a clockwise direction, and (38)]), it is studied only for the case 𝑐=1/2.)(Noteaddedin provided that 𝜎>1so that contributions from infinity proof: after a draft copy of this paper was posted, I was made vanish, the contour may then be opened such that it stretches aware that Ruijsenaars has previously obtained a generalized vertically in the complex 𝑡-plane, giving form of (7) for the ([25,Equation (4.22)]).Healsonotes(privatecommunication)thathecould 𝑐+𝑖∞ 𝑖 −𝑠 not find prior references to his result in the literature). 𝜁 (𝑠) = ∫ 𝑡 cot (𝜋𝑡) 𝑑𝑡, 0 < 𝑐 < 1, 𝑐∈ R, (6) 2 𝑐−𝑖∞ At this point, it is worthwhile to quote several related ∞ integral representations that define some of the relevant 1 −𝑠 =− ∫ (𝑐+𝑖𝑡) cot (𝜋 (𝑐+𝑖𝑡)) 𝑑𝑡, 𝜎 >1, (7) and important functions of complex analysis, starting with 2 −∞ Riemann’s original 1859 contour integral representation of the zeta function [26]: a result that cannot be found in any of the modern reference works cited (in particular [20]). It should be noted that Γ (1−𝑠) 𝑡𝑠−1 this procedure is hardly novel—it is often employed in an 𝜁 (𝑠) = ∮ 𝑑𝑡, (11) educational context (e.g., in the special case 𝑠=2)to 2𝜋𝑖 𝑒−𝑡 −1 demonstrate the utility of complex integration to evaluate special sums ([10, Section 4.1]), ([21,Lemma2.1])andforms where the contour of integration encloses the negative 𝑡-axis, the basis for an entire branch of physics [22]. looping from 𝑡=−∞−𝑖0to 𝑡=−∞+𝑖0enclosing the point A similar application to (5), with some trivial trigonomet- 𝑡=0. ric simplification, gives The equivalence of (11)and(1) is well described in texts (e.g., [11]) and is obtained by reducing three components of −𝑖 𝑐+𝑖∞ 𝑡−𝑠 𝜂 (𝑠) = ∫ 𝑑𝑡 (8) the contour in (11). A different analysis is possible however 2 𝑐−𝑖∞ sin (𝜋𝑡) (e.g., [7, Section 1.6])—open and translate the contour in (11) ∞ −𝑠 such that it lies vertically to the right of the origin in the 1 (𝑐+𝑖𝑡) 𝑡 𝜎<0 = ∫ 𝑑𝑡. (9) complex -plane (and thereby vanishes) with and 2 −∞ sin (𝜋 (𝑐+𝑖𝑡)) evaluatetheresiduesofeachofthepolesoftheintegrand lying at 𝑡=±2𝑛𝜋𝑖to find Equation (9)reduces,with𝑐=1/2,toaJensenresult [23], presented (without proof) in 1895, and, to the best of −𝑖 my knowledge, reproduced, again, but only in the special 𝜁 (𝑠) = Γ (1−𝑠)(2𝜋)𝑠 2𝜋 case 𝑐=1/2, only once in the modern reference literature [17]. In particular, although Lindelof’s 1905 work [24] arrives 𝑖𝜋𝑠 𝑖𝜋𝑠 ∞ (12) ×( ( )− (− )) ∑𝑘𝑠−1. at a result equivalent to that obtained here, attributing the exp 2 exp 2 technique to Cauchy (1827), it is performed in a general 𝑘=1 Journal of Mathematics 3

This can be further reduced, giving the well-known func- 3. Simple Reductions tional equation for 𝜁(𝑠): 3.1. The Case 0 ≤ 𝑐 ≤ 1. The simplest reductions of (7) 2 (𝜋𝑠/2) Γ (𝑠) and (9) are obtained by uniting the two halves of their range 𝜁 (1−𝑠) = cos 𝜁 (𝑠) , corresponding to 𝑡<0and 𝑡>0.Withtheuseof (2𝜋)𝑠 (13) (𝑠/2) (−(𝑐+𝑖𝑡)𝑠 + (𝑐−𝑖𝑡)𝑠)𝑖= 2(𝑐2 +𝑡2) 𝑠 𝜎< which is valid for all by analytic continuation from 𝑡 (20) 0. This demonstrates that (11)and(6)arefundamentally × (𝑠 ( )) , sin arctan 𝑐 different representations of 𝜁(𝑠), since further deformations (𝑠/2) 𝑡 of (11) do not lead to (7). (𝑐+𝑖𝑡)𝑠 + (𝑐−𝑖𝑡)𝑠 =2(𝑐2 +𝑡2) (𝑠 ( )) , Ephemerally, at one time it was noted in [17]thatthe cos arctan 𝑐 following complex integral representation (21) one finds (with 0<𝑐<1and 0≤𝑐≤1,resp.) 𝜋 ∞ (1/2 + 𝑖𝑡)1−𝑠 𝜁 (𝑠) = ∫ 𝑑𝑡, 𝜎 >0 ∞ 𝑡 2 (14) 2 (𝑠−1) −∞ (𝜋𝑡) 𝜁 (𝑠) =−∫ (cos (𝑠 arctan ( )) cos (𝜋𝑐) sin (𝜋𝑐) cosh 0 𝑐 𝑡 can be obtained by the invocation of the Cauchy-Schlomilch¨ − (𝜋𝑡) (𝜋𝑡) (𝑠 ( ))) sinh cosh sin arctan 𝑐 transformation [27]. In fact, the representation (14)followsby −1 the simple expedient of integrating (7) by parts in the special 2 2 𝑠/2 2 2 case 𝑐=1/2(see (35)), as also noted in [17]. Another result ×((𝑐 +𝑡 ) (cosh (𝜋𝑡) − cos (𝜋𝑐))) 𝑑𝑡, 𝜎>1, 2𝑠−1 ∞ 𝑡𝑠 𝜁 (𝑠) = ∫ 𝑑𝑡, 𝜎 >−1 1−𝑠 2 (15) (22) (1 − 2 )Γ(𝑠+1) 0 cosh (𝑡) ∞ 𝑡 𝜂 (𝑠) = ∫ (− sin (𝑠 arctan ( )) cos (𝜋𝑐) sinh (𝜋𝑡) obtained in [27] on the basis of the Cauchy-Schlomilch¨ 0 𝑐 transformation does not follow from (14)aserroneously 𝑡 + (𝑠 ( )) (𝜋𝑐) (𝜋𝑡)) claimed at one time, but later corrected, in [17]. cos arctan 𝑐 sin cosh Based on (11), a number of other representations of 𝜁(𝑠) 𝑠/2 −1 arewellknown.Fromthetablescited,thefollowingareworth ×((𝑐2 +𝑡2) ( 2 (𝜋𝑡) − 2 (𝜋𝑐))) 𝑑𝑡, noting here: cosh cos ([12,Equation1.12(6)]) (23) the latter valid for all 𝑠.Inthecasethat𝑐=1/2,(22)reduces 1 ∞ 𝑡𝑠−1 𝜁 (𝑠) = ∫ 𝑑𝑡, 𝜎 >1, to −𝑠 (16) ∞ 2Γ (𝑠)(1−2 ) 0 sinh (𝑡) (𝜋𝑡) (𝑠 (2𝑡)) 𝜁 (𝑠) =2𝑠 ∫ sinh sin arctan 𝑑𝑡, 𝜎 >1, 𝑠/2 (24) 0 (1 + 4𝑡2) cosh (𝜋𝑡) ([12,Equation1.12(4)]) equivalent to the well-known result (attributed to Jensen by Lindelof, and a special case of Hermite’s result for the 1 ∞ 𝑡𝑠−1 𝜁 (𝑠) = ∫ 𝑑𝑡, 𝜎 >1, (17) Hurwitzzetafunction([6, Section 13.2)]) (which also appears Γ (𝑠) 0 exp (𝑡) −1 as an exercise in many textbooks—for example, [11,Equation 2.3(41)]): and ([13,Equation3.527(1)]) 2(𝑠−1) ∞ (𝑠 (2𝑡)) 𝑒−𝜋𝑡 𝜁 (𝑠) = −2𝑠 ∫ sin arctan 𝑑𝑡. 𝑠−1 2 𝑠/2 (25) 0 (1 + 4𝑡 ) cosh (𝜋𝑡) 1 (2𝑎)(𝑠+1) ∞ 𝑡𝑠 𝜁 (𝑠) = ∫ 𝑑𝑡, 𝜎 >1 2 (18) Similarly, (23)reducestotheknown([24, IV.III(6)]) result: 4 Γ (𝑠+1) 0 sinh (𝑎𝑡) ∞ (𝑠 (𝑡)) 𝜂 (𝑠) =2𝑠−1 ∫ cos arctan 𝑑𝑡. 𝑠/2 (26) By way of comparison, Laplace’srepresentation ([13,Equation 0 (1 + 𝑡2) cosh (𝜋𝑡/2) 8.315(2)]) Comparison of (24)and(25)immediatelyyieldstheknown result ([13,Equation3.975(2)]): 1 1 ∞ 𝑒(𝑐+𝑖𝑡) = ∫ 𝑑𝑡 (𝑠−1) ∞ 𝜋𝑡 𝑠 (19) 2 𝑠 sin (𝑠 arctan (2𝑡)) 𝑒 Γ (𝑠) 2𝜋𝑖 −∞ (𝑐+𝑖𝑡) 𝜁 (𝑠) = +2 ∫ 𝑑𝑡, 𝜎 >1. 1−𝑠 2 𝑠/2 0 (1 + 4𝑡 ) cosh (𝜋𝑡) defines the inverse . (27) 4 Journal of Mathematics

Notice that (25)and(26)arevalidforall𝑠. Simple substitution nonnegative integers 𝑐=𝑁, when it becomes indefinite. of 𝑐→0in (23)gives Therefore, it is of interest to investigate this operation in the neighbourhood of 𝑐=𝑁/2as well as the limit 𝑐→𝑁. 𝑠𝜋 ∞ 𝑡−𝑠 𝜂 (𝑠) =−sin ( ) ∫ 𝑑𝑡, 𝜎 <0, (28) Applied to (9) (and equivalent to integration by parts), the 2 0 sinh (𝜋𝑡) requirement that which, after taking3 ( ) into account, leads to 𝜕𝜂 (𝑠) =0, 𝑐=𝑁̸ (33) 𝜕𝑐 (𝑠𝜋/2) ∞ 𝑡−𝑠 𝜁 (𝑠) = sin ∫ 𝑑𝑡, 𝜎 <0, 1−𝑠 (29) (2 −1) 0 sinh (𝜋𝑡) and identification of one of the resulting terms yields 𝜋 ∞ (𝑐+𝑖𝑡)−𝑠 (𝜋 (𝑐+𝑖𝑡)) displaying the existence of the trivial zeros of 𝜁(𝑠) at 𝑠=−2𝑁, 𝜂 (𝑠+1) =− ∫ cos 𝑑𝑡 2𝑠 2 𝜋 𝑐+𝑖𝑡 since the integral in (29)isclearlyfiniteateachofthesevalues −∞ sin ( ( )) (34) 𝑠 of .Theresult(29)isequivalenttotheknownresult(16) 0≤𝑐<1, after taking (13) into account. Other possibilities abound. For 𝑐=1/3 example, set in (22)toobtain a new result with improved convergence at infinity. Similarly ∞ 𝜕/𝜕𝑐 𝑠−1 2𝜋𝑡 operating with on (7)yields 𝜁 (𝑠) =3 ∫ (2 sin (𝑠 arctan (𝑡)) sinh ( ) 0 3 𝜋 ∞ (𝑐+𝑖𝑡)−𝑠 𝜁 (𝑠+1) = ∫ 𝑑𝑡 0 < 𝑐 <1. 2 (35) 2𝑠 −∞ sin (𝜋 (𝑐+𝑖𝑡)) −√3 cos (𝑠 arctan (𝑡))) (It should be noted that Srivastava et al. [20]studyan 𝑠/2 2𝜋𝑡 −1 ×((1+𝑡2) (2 ( )+1)) 𝑑𝑡, incomplete form of (35) in order to obtain representations cosh 3 for sums both involving 𝜁(𝑛),and𝜁(𝑛) by itself, without commenting on the significance of the contour integral with 𝜎>1. infinite bounds (35).) Using (20)and(21)toreducethe (30) integration range of (34)givesafairlylengthyresultfor general values of 𝑐.If𝑐=1/2,(34)becomes Further, a novel representation can be found by noting that (9) is invariant under the transformation of variables 𝑡→𝑡+ 2𝑠−1𝜋 𝜂 (𝑠+1) = 𝜌 together with the choice 𝑐=𝜎along with the requirement 𝑠 0≤𝜎≤1 𝑐=𝑠 that .Thiseffectivelyallowsonetochoose , ∞ 𝜂(𝑠) 2 −𝑠/2 𝜋𝑡 yielding the following complex representation for ,aswell × ∫ (1 + 𝑡 ) sinh ( ) sin (𝑠 arctan (𝑡)) as the the corresponding representation for 𝜁(𝑠) because of 0 2 (3), which is valid in the critical strip 0≤𝜎≤1: −2 𝜋𝑡 × cosh ( )𝑑𝑡, −𝑖 𝑠+𝑖∞ 𝑡−𝑠 2 𝜂 (𝑠) = ∫ 𝑑𝑡 (31) (36) 2 𝑠−𝑖∞ sin (𝜋𝑡) 𝑠→𝑠−1 1 ∞ (𝑠+𝑖𝑡)−𝑠 which, by letting ,canberewrittenas = ∫ 𝑑𝑡. (32) 𝑠−1 2 −∞ sin (𝜋 (𝑠+𝑖𝑡)) 2 (𝑠−1) 𝜂 (𝑠) 2 𝜋 𝜂 (𝑠+1) = + 𝑠 𝑠 Because (20)and(21)arevalidfor𝑐∈C,itfollowsthat ∞ (23)with𝑐=𝑠is also a valid representation for 𝜂(𝑠),as 2 −𝑠/2 𝜋𝑡 𝜁(𝑠) × ∫ cos (𝑠 arctan (𝑡)) (1 + 𝑡 ) sinh ( ) well as the the corresponding representation for with 0 2 recourse to (3), again only valid in the critical strip 0≤𝜎≤ 1 𝑐=1/𝜌 −2 𝜋𝑡 . Another interesting variation is the choice in × cosh ( )𝑡𝑑𝑡, (23), valid for 𝜌>1, again applicable to the corresponding 2 representation for 𝜁(𝑠) because of (3). Such a representation (37) may find application in the analysis of 𝜁(𝜎 + 𝑖𝜌) in the and, with regard to (3), asymptotic limit 𝜌→∞.Finally,thechoice𝑐=𝜎,0≤ 𝜎≤1 1−𝑠 leads to an interesting variation that is discussed in 2 (𝑠−1) (1 − 2 )𝜁(𝑠) 2𝑠−1𝜋 Section 9. 𝜁 (𝑠+1) = + 𝑠 (1−2−𝑠) 𝑠 (1−2−𝑠) ∞ 3.2. Simple Recursion via Partial Differentiation. As a func- 2 −𝑠/2 𝜋𝑡 𝑐 𝜁(𝑠) 𝜂(𝑠) × ∫ cos (𝑠 arctan (𝑡)) (1 + 𝑡 ) sinh ( ) tion of , and are constant, as are their integral 0 2 representations over a range corresponding to 0<𝑁− 𝑐<1 −2 𝜋𝑡 ; however the integrals change discontinuously at each × cosh ( )𝑡𝑑𝑡. new value of 𝑐 that corresponds to a new value of 𝑁.Thus, 2 the operator 𝜕/𝜕𝑐 acting on (7)and(9) vanishes, except at (38) Journal of Mathematics 5

Similarly, ((37)and(38)arethefirsttwoentriesinafamily such that the contour in (44) bypasses the origin to the right. of recursive relationships that will be discussed elsewhere It is easy to show that the sum of the two in (45) (inpreparation)).Inthecasethat𝑐=1/2,(35)reducesto reduces to the well-known result ([24, Equation 4, III(5)] (attributed to ∞ 𝜋𝑠 −𝑠 Jensen [23])) 𝜁1 (𝑠) = sin ( ) ∫ 𝑡 coth (𝜋𝑡) 𝑑𝑡, 𝜎 >1, (46) 2 1 𝑠−1 ∞ 𝜋2 cos (𝑠 arctan (𝑡)) (𝜋𝑠/2) 𝜋𝑠 ∞ 𝑡−𝑠𝑒(−𝜋𝑡) 𝜁 (𝑠+1) = ∫ 𝑑𝑡. = sin + ( ) ∫ 𝑑𝑡, 𝑠 2 𝑠/2 2 (39) sin (47) 0 (1 + 𝑡 ) cosh (𝜋𝑡/2) 𝑠−1 2 1 sinh (𝜋𝑡) the latter being valid for all 𝑠.One,ofseveralcommonly If 𝑐→0in (34)(forthecase𝑐=0in (35)seethe used possibilities of reducing (44), is to apply ([28,Equation following section), one finds a finite result (also equivalent (54.3.2)]) to integrating (28)byparts): ∞ 2 2𝑘 𝜋 (𝑠𝜋/2) ∞ 𝑡1−𝑠 (𝜋𝑡) 𝑡 cot (𝜋𝑡) =− ∑𝑡 𝜁 (2𝑘) , |𝑡| <1, (48) 𝜂 (𝑠) = sin ∫ cosh 𝑑𝑡, 𝜎 <0, 𝜋 2 (40) 𝑘=0 𝑠−1 0 sinh (𝜋𝑡) giving and, when 𝑐=1, one finds the unusual forms: 2 𝜋𝑠 ∞ 𝜁 (2𝑘) 𝜁 (𝑠) =− ( ) ∑(−1)𝑘 , 𝑖𝜋𝑟 0 𝜋 sin 2 2𝑘 − 𝑠 (49) 𝜂 (𝑠+1) = {0,1} + 𝑘=0 2𝑠 being convergent for all 𝑠. Such sums appear frequently in the 𝑖𝜃 −𝑠 𝑖𝜃 𝑖𝜃 𝜋 (1 ± 𝑟𝑖𝑒 ) cosh (𝜋𝑟𝑒 )𝑒 literature (e.g., [10, 29]Section(3.2))andhavebeenstudied × ∫ 𝑑𝜃 2 𝑖𝜃 extensively when 𝑠=−𝑛[10, 30]. In the case 𝑠=2𝑛,(49) 0 sinh (𝜋𝑟𝑒 ) (41) reduces to 𝜁0(𝑠) = 𝜁(𝑠) and (47)vanishes. ∞ 𝑟𝜋 2 2 −𝑠/2 Along with (42) the integral in (35)canbesimilarlysplit, − ∫ (1 + 𝑟 𝑡 ) cos (𝑠 arctan (𝑟𝑡)) 𝑠 1 using the convergent series representation: × (𝜋𝑟𝑡) −2 (𝜋𝑟𝑡) 𝑑𝑡, 1 2 ∞ cosh sinh = ∑ (2𝑘 − 1) 𝜁 (2𝑘) V2𝑘−2, |V| <1, 2 𝜋2 (50) sin (𝜋V) 𝑘=0 which is valid for any 0<𝑟<1and all 𝑠 (particularly 𝑟=1/𝜋 or 𝑟=1/2), after deformation of the contour of integration where the Bernoulli numbers 𝐵2𝑘 appearing in the original in (34) either to the left or to the right of the singularity at 𝑡= ([28, Equation (50.6.10)]) are related to 𝜁(2𝑘) by 1, respectively, belonging to a choice of ± where the symbol {0, 1} 1 (−1)𝑘(2𝜋)2𝑘𝐵 means zero or one as the case may be. 𝜁 (2𝑘) =− 2𝑘 . (51) 2 Γ (2𝑘 + 1)

4. Special Cases Thus the representation 4.1. The Case 𝑐=0. With respect to (35), setting 𝑐=0 1 𝑖𝜋 𝑖 𝑡1−𝑠 𝜁 (𝑠) =− ∫ 𝑑𝑡 together with some simplification gives 2 2 (𝑠−1) −𝑖 sin (𝜋𝑡) (52) (𝜋𝑠/2) ∞ 𝑡1−𝑠 𝜋 (𝜋𝑠/2) ∞ 𝑡1−𝑠 𝜁 (𝑠) =−𝜋𝑠−1 sin ∫ 𝑑𝑡, 𝜎 <0, − sin ∫ 𝑑𝑡 2 (42) 2 𝑠−1 0 sinh (𝑡) 𝑠−1 1 sinh (𝜋𝑡) reproducing (18), with 𝑎=1,aftertaking(13) into account. becomes (formally) 𝑐=0 This same case ( ) naturally suggests that the integral in (𝜋𝑠/2) 𝜁 (𝑠) =−sin (6)bebrokenintotwoparts: 𝑠−1

∞ 𝑘 𝜁 (𝑠) =𝜁0 (𝑠) +𝜁1 (𝑠) , (43) 2 (−1) (2𝑘 − 1) 𝜁 (2𝑘) ×( ∑ 𝜋 2𝑘 − 𝑠 (53) 𝑘=0 where ∞ 𝑡1−𝑠 𝑖 +𝜋∫ 𝑑𝑡) 𝑖 2 −𝑠 1 (𝜋𝑡) 𝜁0 (𝑠) = ∮ 𝑡 cot (𝜋𝑡) 𝑑𝑡, (44) sinh 2 −𝑖 𝜎<0 𝑖 𝑖∞ after straightforward integration of50 ( )with .In 𝑖 −𝑠 𝑖 −𝑠 𝜁1 (𝑠) = ∫ 𝑡 cot (𝜋𝑡) 𝑑𝑡 + ∫ 𝑡 cot (𝜋𝑡) 𝑑𝑡 (45) (53), the integral converges, and the formal sum conditionally 2 −𝑖∞ 2 𝑖 converges or diverges according to several tests. Howver, 6 Journal of Mathematics furtheranalysisisstillpossible.Followingthemethodof 4.2. The Case 𝑐 ≥ 1. Interesting results can be obtained by [10, Section 3.3] applied to (49), together with [10,Equation translating the contour in (6)and(9) and adding the residues 3.4(15)], we have the following general result for |V|<1: so-omitted. From (1)and(2) define the partial sums ∞ 𝜁 (2𝑘) V2𝑘 1 V ∞ 𝑁 ∑ = + ∫ 𝑡−𝑠 ∑𝜁 (2𝑘) 𝑡2𝑘−1𝑑𝑡 𝜁 (𝑠) ≡ ∑𝑘−𝑠 2𝑘 − 𝑠 2𝑠 𝑁 𝑘=0 0 𝑘=1 𝑘=1 (60) (54) 𝑁 𝜂 (𝑠) ≡−∑(−1)𝑘𝑘−𝑠, 1 1 V 𝑁 = + V𝑠 ∫ 𝑡−𝑠 (Ψ (1+𝑡) −Ψ(1−𝑡)) 𝑑𝑡, 𝑘=1 2𝑠 2 0 let 𝑐→𝑐𝑁, 𝑁 = 1, 2, 3, . . with 𝜎<2 𝑁≤𝑐𝑁 <𝑁+1, (61) (55) and adjust all the above results accordingly by redefining so that (formally), after premultiplying (55)with1/V,then 𝑐→𝑐𝑁.(If𝑐𝑁 =𝑁, an extra half-residue is removed and V2(𝜕/𝜕V) operating with , the contour deformed to pass to the right of the point 𝑡=𝑁.) ∞ 𝜁 (2𝑘)(2𝑘 − 1) V2𝑘 This gives, for example, the generalization of (6): ∑ 𝑐 +𝑖∞ 2𝑘 − 𝑠 1 𝑖 𝑁 𝑘=0 𝜁 (𝑠) =𝜁 (𝑠) − 𝛿 + ∫ 𝑡−𝑠 (𝜋 ) 𝑑𝑡, 𝜎 >1, 𝑁 𝑠 𝑁,𝑐 cot t 2𝑁 𝑁 2 𝑐 −𝑖∞ 1 1 𝑁 =− + V𝑠 (56) (62) 2𝑠 2 V and (9): × ∫ 𝑡1−𝑠 (Ψ󸀠 (1+𝑡) +Ψ󸀠 (1−𝑡))𝑑𝑡 𝑁 𝑐𝑁+𝑖∞ −𝑠 0 (−1) 𝑖 𝑡 𝜂 (𝑠) =𝜂𝑁 (𝑠) + 𝛿𝑁,𝑐 − ∫ 𝑑𝑡, (63) 2𝑁𝑠 𝑁 2 (𝜋𝑡) 1 1 𝑐𝑁−𝑖∞ sin =− + V𝑠 2𝑠 2 so (22)and(23)with𝑐→𝑐𝑁 become integral representa- 𝜁(𝑠) −𝜁 (𝑠) V 1 tions of the remainder of the partial sums 𝑁 and × ∫ 𝑡1−𝑠 (𝜋2 (1 + 2 (𝜋𝑡))− )𝑑𝑡, (57) 𝜂(𝑠)−𝜂 (𝑠) cot 𝑡2 𝑁 , the former corresponding to a special case of the 0 Hurwitz zeta function 𝜁(𝑠,. 𝑁) 𝜎<2. In the case that 𝑐𝑁 =𝑁=1,onefinds 1 ∞ (𝜋𝑡) (𝑠 (𝑡)) Notice that the right-hand side of (55) provides an analytic 𝜁 (𝑠) = + ∫ cosh sin arctan 𝑑𝑡, 𝜎 >1, 2 2 𝑠/2 (64) continuation of the left-hand side for |V|>1and the right- 0 (1 + 𝑡 ) sinh (𝜋𝑡) hand side of (57)givesaregularizationoftheleft-hand 𝑠+1 ∞ 𝑒−𝜋𝑡 (𝑠 (𝑡)) side. Putting together (53)and(57), incorporating the first = + ∫ sin arctan 𝑑𝑡, V =𝑖 2 (𝑠−1) 2 𝑠/2 (65) term of (57)intothesecondtermof(53), setting , 0 (1 + 𝑡 ) sinh (𝜋𝑡) and constraining 𝜎>0eventually yields a new integral (2.8) 1 ∞ (𝑠 (𝑡)) representation (compare with (121), [31,Equation ]and 𝜂 (𝑠) = + ∫ sin arctan 𝑑𝑡, [25,Equation(4.80)]): 2 2 𝑠/2 (66) 0 sinh (𝜋𝑡) (1 + 𝑡 ) 𝜋𝑠−1 (𝜋𝑠/2) ∞ 1 1 𝜁 (𝑠) =− sin ∫ 𝑡1−𝑠 ( − )𝑑𝑡, with the latter two results being valid for all 𝑠. Neither (64)nor 2 2 (58) 𝑠−1 0 sinh 𝑡 𝑡 (66) appear in references; (65) corresponds to a limiting case 0<𝜎<2 of Hermite’s representation of the Hurwitz zeta function ([12, which is valid for , a nonoverlapping analytic 1.10(7) continuation of (42) that spans the critical strip. Additionally, Equation ]). In analogy to the developments leading to 𝑠→1−𝑠 (27), equations (64), (66), and (3), together with the following using (13)followedbyreplacing gives a second (9.51.5) representation: modified form of [13,Equation ] 𝑠−1 𝑠−1 𝑠−1 ∞ 2 𝑠 2 2 𝑠 1 1 𝜁 (𝑠) = + 𝜁 (𝑠) = ∫ 𝑡 ( − )𝑑𝑡 𝑠 𝑠 2 2 (59) (2 −1)(𝑠−1) 2 −1 Γ (1+𝑠) 0 sinh 𝑡 𝑡 (67) ∞ (𝑠 (𝑡)) (1 + 𝑒−𝜋𝑡) valid for −1<𝜎<1,anon-overlappinganalytic × ∫ sin arctan 𝑑𝑡, 0<𝜎<1 𝑠/2 continuation of (18). Over the critical strip 0 (1 + 𝑡2) sinh (𝜋𝑡) representations (58)and(59) are simultaneously true and augment the rather short list of other representations that canbeusedtoobtainthepossiblynewresult: share this property ([32], as cited in [12](unnumbered 𝑠−3 ∞ (𝑠 (𝑡)) 𝑒𝜋𝑡 equations immediately following 1.12(1),alsoreproduced 𝜁 (𝑠) = + ∫ sin arctan 𝑑𝑡, 𝜎 >1. 2 (𝑠−1) 2 𝑠/2 in [11,Equations2.3(44)–(46)].)). See Section 9 for further 0 (1 + 𝑡 ) sinh (𝜋𝑡) application of this result. (68) Journal of Mathematics 7

For 𝑐𝑁 =1/2+𝑁corresponding to general partial sums, (23) (for comparison, see also [4,Section8.3],and[7,Section two possibilities are apparent. Based on (20), (21), (22), and (6.4)] and Chapter 7), we find

2 −𝑠/2 𝑁 1 1 1 1−𝑠 ∞ sinh (𝜋𝑡 (1+2𝑁)) (1 + 𝑡 ) sin (𝑠 arctan (𝑡)) 𝜁 (𝑠) − ∑ = ( +𝑁) ∫ 𝑑𝑡 𝜎 >1 (69) 𝑘𝑠 2 2 2 𝑘=1 0 cosh (𝜋𝑡 (1/2 + 𝑁))

2 −𝑠/2 (1/2 + 𝑁)1−𝑠 1 1−𝑠 ∞ (1 + 𝑡 ) sin (𝑠 arctan (𝑡)) = −2( +𝑁) ∫ 𝑑𝑡, (70) 𝜋𝑡(1+2𝑁) 𝑠−1 2 0 1+𝑒

2 −𝑠/2 𝑁 (−1)𝑘 1 1−𝑠 ∞ (1 + 𝑡 ) cos (𝑠 arctan (𝑡)) 𝜂 (𝑠) + ∑ = (−1)𝑁( +𝑁) ∫ 𝑑𝑡 (71) 𝑘𝑠 2 (𝜋𝑡 (1/2 + 𝑁)) 𝑘=1 0 cosh

which together extend (24)and(26), respectively. Application of (34)and(35)with(20)and(21), also with 𝑐𝑁 =1/2+𝑁, gives the remainders

2 −𝑠/2 𝑁 1 𝜋 (1/2 + 𝑁)1−𝑠 ∞ (1 + 𝑡 ) cos (𝑠 arctan (𝑡)) 𝜁 (𝑠+1) − ∑ = ∫ 𝑑𝑡, 𝑘𝑠+1 𝑠 2 𝑘=1 0 cosh (𝜋𝑡 (1/2 + 𝑁)) (72) 2 −𝑠/2 𝑁 (−1)𝑘 𝜋(−1)𝑁(1/2 + 𝑁)1−𝑠 ∞ sinh (𝜋𝑡 (1/2 + 𝑁)) (1 + 𝑡 ) sin (𝑠 arctan (𝑡)) 𝜂 (𝑠+1) + ∑ = ∫ 𝑑𝑡 𝑘𝑠+1 𝑠 2 𝑘=1 0 cosh (𝜋𝑡 (1/2 + 𝑁))

which are valid for all 𝑠.Forthecase𝑐𝑁 =𝑁applying (62) and (63) gives the remainders

2 −𝑠/2 𝑁 1 1 ∞ cosh (𝜋𝑁𝑡) (1 + 𝑡 ) sin (𝑠 arctan (𝑡)) 𝜁 (𝑠) − ∑ =− +𝑁1−𝑠 ∫ 𝑑𝑡, 𝜎 >1 (73) 𝑘𝑠 2𝑁𝑠 (𝜋𝑁𝑡) 𝑘=1 0 sinh

2 −𝑠/2 1 𝑁1−𝑠 ∞ (1 + 𝑡 ) sin (𝑠 arctan (𝑡)) =− + +2𝑁1−𝑠 ∫ 𝑑𝑡, (74) 𝑠 2𝑁𝜋𝑡 2𝑁 𝑠−1 0 𝑒 −1

2 −𝑠/2 𝑁 (−1)𝑘 (−1)𝑁 ∞ (1 + 𝑡 ) sin (𝑠 arctan (𝑡)) 𝜂 (𝑠) + ∑ = + (−1)1+𝑁𝑁1−𝑠 ∫ 𝑑𝑡, (75) 𝑘𝑠 2𝑁𝑠 (𝜋𝑁𝑡) 𝑘=1 0 sinh

which extend (64)and(66), the latter being valid for all 𝑠. remainder, provided that 𝑠∈R in both cases. In addition, It should be noted that (74), valid for 𝑠∈C, generalizes many of the above results, in particular (70)and(74), appear well-established results (e.g., [13,Equation3.411(12)]) that to be new and should be compared to classical results such as are only valid in the limiting case 𝑠=𝑛.(Although[13, [8,Theorem1.8,page23andpages99-100],andasymptotic Equation 3.411(21)] gives (with a missing minus sign) an approximations such as [33,Equation(8.2.1)]. integral representation for finite sums such as those appearing 𝑛 in (74), specified to be valid only for integer exponents𝑘 ( ), that particular result in [13] also appears to apply after the 5. Series Representations generalization 𝑛→𝑠.) Omitting the integral term from (70) gives an upper bound for the remainder; similarly, 5.1. By Parts. By simple manipulation of the integral omitting the integral term in (74)givesalowerboundforthe representations given, it is possible to obtain new series 8 Journal of Mathematics representations. Starting from the integral representation With reference to (7), express the numerator 𝑐𝑜𝑠𝑖𝑛𝑒 factor (29), integrate by parts, giving as a difference of exponentials each of which in turn is written (𝜋𝑠/2) 𝜋2 as a convergent power series in 𝜋(𝑐 + 𝑖𝑡),andrecognizethat 𝜁 (𝑠) = sin (2(1−𝑠) −1)(−𝑠/2 +) 1 the resulting series can be interchanged with the integral and each term identified using (9)tofind ∞ 2 2 2−𝑠 𝑠 3 𝑠 𝑡 𝜋 (76) × ∫ 𝑡 (− +1; ,2− ; ) 𝑘 1F2 ∞ 2 0 2 2 2 4 (−𝜋 /4) 𝜂 (𝑠−2𝑘) 𝜁 (𝑠) =−√𝜋∑ ,𝜎>1. (82) −3 Γ (𝑘+1/2) Γ (𝑘+1) × cosh (𝜋𝑡) sinh (𝜋𝑡) 𝑑𝑡, 𝜎 <0, 𝑘=0 where the hypergeometric function arises because of [34, A more interesting variation of (82)canbeobtainedby Equation 6.2.11(6)]. Write the as a applying (3) and explicitly identifying the 𝑘=0term of the (uniformly convergent) series, thereby permitting the sum sum giving and integral to be interchanged, apply a second integration by parts, and find 𝜁 (𝑠) ∞ 2𝑘 𝜋 𝜋𝑠 𝜋 𝑘 𝜁 (𝑠) =( ) ( ) ∑ ∑∞ (−𝜋2) (1 − 21−𝑠+2𝑘)/Γ(2𝑘 + 1) 𝜁 (𝑠−2𝑘) (83) 21−𝑠 −1 sin 2 Γ (2𝑘 + 2) 1 𝑘=1 𝑘=0 =− , (77) 2 1−2−𝑠 ∞ 𝑡1−𝑠+2𝑘 × ∫ 𝑑𝑡, 𝜎 <0. 2 a convergent representation if 𝜎>1(by Gauss’ test). An 0 sinh (𝜋𝑡) immediate consequence is the identification Identify the integral in (77)using(18)giving 2 2𝑠𝜋𝑠−1 (𝜋𝑠/2) 𝜋 𝜁 (𝑠) = sin 𝜁 (2) =− 𝜁 (0) , (84) 21−𝑠 −1 3 ∞ 2−2𝑘𝜁 (−𝑠+1+2𝑘) Γ (−𝑠+2+2𝑘) (78) 𝑘=1 × ∑ , because only the term in the sum contributes when Γ (2𝑘 + 2) 𝑠=2. Additionally, when 𝑠=2𝑛,(83)terminatesat𝑛 𝑘=0 terms, becoming a recursion for 𝜁(2𝑛) in terms of 𝜁(2𝑛 − which is valid for all values of 𝜎 bytheratiotestandthe 2𝑘), 𝑘 = 1,...,𝑛—see Section 8 for further discussion. If principle of . This result, which appears the functional equation (13) is used to extend its region of to be new, can be rewritten by extracting the 𝑘=0term, and convergence, (83)becomes applying (13)toyield ∞ 𝜁 (1−𝑠+2𝑘) (2𝑠−2𝑘−1 −1) 2𝑠𝜋𝑠−1 (𝜋𝑠/2) 1 𝑠 𝜁 (𝑠) = sin 𝜁 (𝑠) = (𝜋 ∑ 1−𝑠 2 Γ (2𝑘 + 1) Γ (𝑠−2𝑘) 2 −2+𝑠 𝑘=1 ∞ 2−2𝑘 𝜁 (−𝑠+1+2𝑘) Γ (−𝑠 + 2 + 2𝑘) (79) (85) × ∑ , 𝜋𝑠 −1 Γ (2 +2) ×((2−𝑠 −1) ( )) ), 𝜎>1, 𝑘=1 k cos 2 a representation, valid for all values of 𝑠,thatisreminiscent of the following similar sum recently obtained by Tyagi and which augments similar series appearing in [10,Sections4.2 Holm ([31,Equation(3.5)]): and 4.3], and corresponds to a combination of entries in [28, (54.6.3) (54.6.4) 𝜋𝑠−1 𝜋𝑠 Equations and ]. 𝜁 (𝑠) = ( ) 1−21−𝑠 sin 2 5.2. By Splitting ∞ (2 − 2𝑠−2𝑘)𝜁(−𝑠+1+2𝑘) Γ (−𝑠+1+2𝑘) × ∑ , Γ (2𝑘 + 2) 5.2.1. The Lower Range. After applying (13), the integral in 𝑘=1 (42)isfrequently(e.g.,[11]) and conveniently split in two, 𝜎>0 defining (80) 𝜔 𝑡𝑠 𝜁 (𝑠) = ∫ 𝑑𝑡, 𝜎 >1, − 2 (86) (convergent by Gauss’ test). Equation (79)canalsoberewrit- 0 sinh (𝑡) tenintheform ∞ 𝑡𝑠 1 𝜁 (𝑠) = ∫ 𝑑𝑡 𝜁 (𝑠) = + 2 (87) Γ (𝑠)(2𝑠 −1−𝑠) 𝜔 sinh (𝑡) ∞ 2−2𝑘𝜁 (𝑠+2𝑘) Γ (1+𝑠+2𝑘) (81) × ∑ , with Γ (2𝑘 + 2) 𝑘=1 Γ (𝑠+1) 𝜁 (𝑠) =𝜁− (𝑠) +𝜁+ (𝑠) (88) by replacing 𝑠→1−𝑠and applying (13). 2𝑠−1 Journal of Mathematics 9

2 for arbitrary 𝜔. Similar to the derivation of (53), after oftheright-handsidewithrespecttothevariable𝑦 at 𝑦=1, application of (13)wehave leading to a rational polynomial in coth(𝜔) and sinh(𝜔).The first few terms are given in Table 1 for two choices of 𝜔. ∞ 𝐵 𝜔2𝑘+𝑠−1 𝜁 (𝑠) =−2√𝜋 ∑ 2𝑘 , Application of (93)to(92) leads to a less rapidly converg- − Γ (𝑘−1/2) Γ (𝑘+1)(2𝑘−1+𝑠) (89) 𝑘=0 ing representation that can be expressed in terms of more familiar functions: which can be rewritten using (51)as 𝑠−1 1 𝑠 𝜁− (𝑠) =𝑠𝜔 Γ( + ) 𝑘 2 2 ∞ (−𝜔2/𝜋2) Γ (𝑘+1/2) 𝜁 (2𝑘) 𝜁 (𝑠) =4𝜔𝑠−1 ∑ ,𝜔≤𝜋. ∞ ∞ (−1)𝑚Γ (𝑚+𝑘+1) 𝜁 (2𝑚 + 2𝑘 +2) − Γ (𝑘−1/2)(2𝑘−1+𝑠) × ∑ ∑ 𝑘=0 𝑚=0 Γ (𝑘 + 3/2 + 𝑠/2) Γ (𝑚+1) (90) 𝑘=0 2 (𝑚+𝑘+1) 𝑠−1 𝜔 𝑠 𝑠𝜔 Substitute (1)in(90), interchange the order of summation, ×( ) −𝜔coth (𝜔) + . and identify the resulting series to obtain 𝜋2 𝑠−1 (95) 2𝑠𝜔𝑠+1 𝜁 (𝑠) = − 𝜋2 (𝑠+1) A second transform of the intermediate hypergeometric function yields another useful result: ∞ 2 1 1 𝑠 3 𝑠 𝜔 (𝑘+1) × ∑ 𝐹 (1, + ; + ;− ) (91) 2𝑠𝜔𝑠−1 ∞ 𝜔2 𝑚2 2 1 2 2 2 2 𝜋2𝑚2 𝜁 (𝑠) = ∑ ( ) 𝑚=1 − Γ (1/2 + 𝑠/2) 𝜋2 𝑘=0 𝑠−1 𝑠 𝑠𝜔 −𝜔 coth (𝜔) + , ∞ 𝜔2 𝑚 1 𝑠 𝑠−1 × ∑ ( ) (−1)𝑚Γ(𝑚+𝑘+ + ) 𝜋2 2 2 𝑠 𝑚=0 a convergent representation that is valid for all . (96) Other representations are easily obtained by applying any ×𝜁(2𝑚 + 2𝑘 +2) of the standard hypergeometric linear transforms as in (118)— −1 see Section 6. One such leads to × (Γ (𝑚+1) Γ (𝑘+1)(2𝑘+1+𝑠))

𝑠−1 1 𝑠 𝑠−1 𝜁 (𝑠) =𝑠𝜔 Γ( + ) 𝑠 𝑠𝜔 − 2 2 −𝜔 (𝜔) + . coth 𝑠−1 ∞ Γ (𝑘+1) ∞ 1 𝑘+1 × ∑ ∑ ( ) Both (95)and(96) can be reordered along a diagonal of Γ (𝑘 + 3/2 +) 𝑠/2 𝜋2𝑚2/𝜔2 +1 (92) ∞ ∞ ∞ 𝑘 𝑘=0 𝑚=1 thedoublesum(∑𝑘=0 ∑𝑚=0 𝑓(𝑚, 𝑘) =∑𝑘=0 ∑𝑚=0 𝑓(𝑚, 𝑘− 𝑚)), allowing one of the resulting (terminating) series to be 𝑠𝜔𝑠−1 −𝜔𝑠 (𝜔) + , evaluatedinclosedform.Ineithercase,botheventuallygive coth 𝑠−1 ∞ −𝜔2 𝑘 𝜁 (2𝑘) 𝜔 𝜁 (𝑠) =−2𝑠𝜔𝑠−1 ∑( ) −𝜔𝑠 (𝜔) , a rapidly converging representation for small values of , − 𝜋2 (2𝑘−1+𝑠) coth valid for all 𝑠.Theinnersumof(92)canbewritteninseveral 𝑘=0 2 2 ways. By expanding the denominator as a series in 1/(𝑡 𝑚 ) (97) 𝑟<2𝑝−1 with and interchanging the two series, we find the reducing to (49)inthecase𝜔=𝜋taking (13)into general form consideration. Series of the form (97) have been studied 𝑠=𝑛 𝜔 ∞ 𝑚𝑟 extensively [10, 11]when for special values of .These 𝜅 (𝑟, 𝑝, 𝑡)≡ ∑ representations will find application in Section 8. 2 2 𝑝 𝑚=1 (1 + 𝑡 𝑚 ) ∞ Γ(𝑝+𝑚)(−1)𝑚𝑡−2𝑚−2𝑝𝜁(2𝑚+2𝑝−𝑟) 5.2.2. The Upper Range. Rewrite (87) in exponential form and = ∑ , (1 − −2𝑡)−2 Γ(𝑝)Γ(𝑚+1) expand the factor e in a convergent power series, 𝑚=0 interchange the sum and integral, and recognize that the (93) resulting integral is a generalized exponential integral 𝐸𝑠(𝑧) usually defined in terms of incomplete Gamma functions where the right-hand side continues the left if |𝑡| > 1 and ([1, 35], Section 8.19) as follows: 𝑟−2𝑝≥−1.Alternatively,[28,Equation(6.1.32)] identifies 𝑠−1 𝐸𝑠 (𝑧) =𝑧 Γ (1−𝑠,𝑧) , (98) ∞ 1 1 𝜋 𝜋𝑦 ∑ = + ( ) , 2 2 2 2 coth (94) giving a rapidly converging series representation: 𝑚=0 (𝑚 𝑡 +𝑦 ) 2𝑦 2𝑡𝑦 𝑡 ∞ 𝑠+1 so for 𝑝=𝑘+1and 𝑟=0in (93), each term of the inner series 𝜁+ (𝑠) =4𝜔 ∑𝑘𝐸−𝑠 (2𝜔𝑘) . (99) (92) can be obtained analytically by taking the 𝑘th derivative 𝑘=1 10 Journal of Mathematics

2 2 Table 1: The first few values of 𝜅(0, 𝜅 + 1,𝜋 /𝜔 )in(93).

𝑘 Analytic 𝜔=1 𝜔=𝜋 1 𝜔 (𝜔) − + coth 0 2 2 0.156518 1.076674 1 1 𝜔2 1 − + + 𝜔 (𝜔) × −2 1 2 coth 0.927424 10 0.306837 2 4 sinh 𝜔 4 2 3𝜔 1 𝜔 coth 𝜔(−3+2𝜔 ) 8+3𝜔2 3𝜔− 2𝜔+ + × −3 2 coth coth 2 2 0.795491 10 0.134296 16 2 16sinh 𝜔 16sinh 𝜔

Ausefulvariantof(99) is obtained by manipulating the well- of digits between the two terms 𝜁−(𝑠) and 𝜁+(𝑠) if (88)isused known recurrence relation ([1,Equation8.19.12]) for 𝐸𝑠(𝑧) to to attempt a numerical evaluation of 𝜁(𝑠),ashasbeenreported obtain elsewhere [37]. Essentially, (88) is numerically useless (with 1 common choices of arithmetic precision) for any choices of 𝐸 (𝑧) = ( −𝑧 (𝑠+𝑧) +𝑠(𝑠−1) 𝐸 (𝑧)). 𝜌>10 𝜔 −𝑠 𝑧2 e 2−𝑠 (100) ,evenif is carefully chosen in an attempt to balance the cancellation of digits. However, with 20 digits of precision, Apply (100) repeatedly 𝑁 times to (99)andfind using 20 terms in the series, it is possible to find the first zero of 𝜁(1/2 + 𝑖𝜌) correct to 10 digits, deteriorating rapidly 𝑠 −1+𝑠 2𝜔 𝑠 2𝜔 thereafter as 𝜌 increases. For any choice of 𝜔,theconvergence 𝜁+ (𝑠) =−𝑠𝜔 log (𝑒 −1)+2𝑠𝜔 − 1−𝑒2𝜔 properties of the two sums representing 𝜁−(𝑠) and 𝜁+(𝑠) anti- correlate,ascanbeseenfromTable1.Thiscanalsobeseen 𝑁+1 −2𝜔 Li𝑘 (𝑒 ) +2𝜔𝑠Γ (𝑠+1) ∑ by rewriting (102)toyieldthecombinedrepresentation: 𝑘 (101) 1−𝑠 𝑘=2 Γ (𝑠−𝑘+1)(2𝜔) 2 Γ (𝑠+1) 𝜁 (𝑠) 𝑠−𝑁−1 ∞ 2𝜔 Γ (𝑠+1) 𝐸𝑁+2−𝑠 (2𝑘𝜔) ∞ + ∑ 𝑠+1 1𝐹1 (𝑠 + 1; 2 + 𝑠; −2𝑘𝜔) Γ (𝑠−𝑁−1) 2𝑁+1 𝑘𝑁+1 =4𝜔 (∑𝑘(𝐸−𝑠 (2𝑘𝜔) + )) . 𝑘=1 𝑠+1 𝑘=1 (104) in terms of and (Li𝑘(𝑧)), a result 𝑁 that loses numerical accuracy as increases, but which also, In (104), the first inner term (𝐸−𝑠(2𝑘𝜔))vanisheslike 𝑠=𝑛 𝑁≥𝑛 ≥0 usefully, terminates if , . This is discussed in exp(−2𝑘𝜔)/(2𝑘𝜔) whereas the second term (1𝐹1/(𝑠 + 1)) (1+𝑠) more detail in Section 8. vanishes like 1/𝑘 − exp(−2𝑘𝜔)/(2𝑘𝜔),soforlargevalues Another useful representation emerges from (99)by of 𝑘 cancellation of digits is bound to occur. (In fact, 𝐸 (2𝑘𝜔) writing the function 𝑠 in terms of confluent hyper- the coefficients of both asymptotic series prefaced by the (2.6 ) geometric functions. From [35,Equation b ]and[36, exponential terms cancel exactly term by term.) (9.10.3) Equation ], we find It should be noted that (102)isrepresentativeofafamilyof ∞ related transformations. For example, although setting 𝑠=0 𝑠+1 𝜕/𝜕𝑠 𝜁+ (𝑠) =4𝜔 ∑𝑘𝐸−𝑠 (2𝑘𝜔) in (102) does not lead to a new result, operating first with 𝑘=1 followed by setting 𝑠=0leads to a transformation of a related 𝑠+1 sum: 1−𝑠 4𝜔 =2 Γ (𝑠+1) 𝜁 (𝑠) − (102) ∞ 𝜔 arctanh (√−𝜔2/𝜋) 𝑠+1 ∑𝐸 (2𝑘𝜔) =− 1 √ 2 ∞ 𝑘=1 𝜋 −𝜔 −2𝑘𝜔 × ∑𝑘𝑒 𝐹1 (1; 2 + 𝑠; 2𝑘𝜔) , 1 ∞ √ 2 𝑘=1 𝜔 arctanh ( −𝜔 /𝜋𝑘) − ∑ (105) 𝜋𝑘√−𝜔2 wheretheseriesontheright-handsideconvergesabsolutelyif 𝑘=2 𝜎>1independent of 𝜔.Comparisonof(99), (88), and (102) 1 𝜔 1 𝛾 + ( )+ + , identifies 2 log 𝜋 2𝜔 2 4𝜔𝑠+1 ∞ 𝑘=1 𝜁 (𝑠) = ∑𝑘𝑒−2𝑘𝜔 𝐹 (1; 2 + 𝑠; 2𝑘𝜔) where the first term corresponding to has been isolated, − 𝑠+1 1 1 (103) 𝑘=1 since it contains the singularity that occurs due to divergence of the left-hand sum if 𝜔=𝑖𝜋.(In(86)and(87), 𝜔∈C which can be interpreted as a transformation of sums, or an is permitted by a fundamental theorem of integral calculus.) analytic continuation, for the various guises of 𝜁−(𝑠) given in Similarly, operating with 𝜕/𝜕𝑠 on (102)at𝑠=1leads to the Section 5.2.1, if both, or either, of the two sides converge for identification some choice of the variables 𝑠 and/or 𝜔, respectively. The fact ∞ 1 𝜁(𝑠) ∑ (1 + )=−1− (2) + (𝑒2 −1) that a term containing naturally appears in the expression log 𝜋2𝑘2 log log (106) 𝑘=1 (102)for𝜁+(𝑠) suggests that there will be a severe cancellation Journal of Mathematics 11

(which can alternatively be obtained by writing the infinite The contour may now be deformed to enclose the singularity sum as a product). A second application of 𝜕/𝜕𝑠 on (102), at 𝑡=−𝑠−1, translated to the right to pick up residues of again at 𝑠=1using (106), leads to the following transfor- Γ(−𝑡) at 𝑡 = 𝑛, 𝑛 = 0, 1,. and the residue of 𝜁(−1 − 𝑡) at 𝑡= mation of sums: −2. After evaluation, each of the residues so obtained cancels one of the terms in the series expression (90)for𝜁−(𝑠) leaving 1 ∞ 𝜁 (2𝑘)(−1)𝑘 1 𝛾2 𝜋2 ∑ = (2)2 − + (after a change of integration variables) 2 2𝑘 2 2 ln 2 12 𝑘=1 𝜋 𝑘 (107) Γ (𝑠+1) 𝜁 (𝑠) 1 Γ (𝑡+1) 𝜁 (𝑡) ∞ 𝐸 (2𝑘) =− ∮ 𝑑𝑡, (114) −1−𝛾(1) − ∑ 1 . 2𝑠−1𝜔𝑠+1 2𝜋𝑖 (𝑠−𝑡) 2𝑡−1𝜔𝑡+1 𝑘 𝑘=1 the standard Cauchy representation of the left-hand side, This transformation can be further reduced by applying an valid for all 𝑠, where the contour encloses the simple pole at integral representation given in [1,Section(6.2)], leading to 𝑡=𝑠. −2𝑡 ∞ 𝐸 (2𝑘) ∞ log (1 − 𝑒 ) ∑ 1 =−∫ 𝑑𝑡. (108) 𝑘 𝑡 𝑘=1 1 6. Integration by Parts

Correcting a misprinted expression given in ([38,Equation Integration by parts applied to previous results yields other (4.8)], [10, 11], Equation 3.3(43)) which should read representations. From (17), we obtain (known to the Maple computer code, ∞ 𝑡𝑘 𝑡 𝑑V butonlyintheformofaMellintransform) ∑ 𝜁 (2𝑘) = ∫ (𝜋√V (𝜋√V)) , 2 log csc (109) 𝑘 0 V 𝑘=1 ∞ 1 𝑠−2 𝜁 (𝑠) =− ∫ 𝑡 log (1−exp (−𝑡)) 𝑑𝑡, 𝜎 >1, followed by a change of variables, eventually yields the Γ (𝑠−1) 0 representation (115) ∞ 𝜁 (2𝑘)(−1)𝑘 1 1 1 1 ∑ =−2∫ ( ( 𝑒𝑡 − 𝑒−𝑡)− (𝑡))𝑑𝑡. and from (64), we find 2𝑘 2 𝑡 log 2 2 log 𝑘=1 𝜋 𝑘 0 1 𝑠 (110) 𝜁 (𝑠) = − 2 𝜋 Substituting (108)and(110)into(107)followedbysome ∞ ( (𝜋𝑡)) (( +1) (𝑡)) simplification finally gives × ∫ log sinh cos s arctan 𝑑𝑡, 2 (1+𝑠)/2 −2/V 2V 0 (1 + 𝑡 ) 1 (1 − 𝑒 )(𝑒 −1) 𝑑V 𝜋2 𝛾2 ∫ log ( ) =− + +𝛾(1) 0 2V V 12 2 𝜎>1 (116) 1 − 2 (2) +2, 2log as well as the following intriguing result (discovered by the (111) Maple computer code): a result that appears to be new. 1 1 √𝜋 (𝜋𝑠/2) Γ (𝑠/2 − 1/2) 𝜁 (𝑠) = + sin 5.3. Another Interpretation. These expressions for the 2 2 Γ (𝑠/2) split representation can be interpreted differently. For ∞ 𝑡 (𝑠 (𝑡)) 𝜎<−3 +𝜋∫ sin arctan the moment, let , and insert the contour integral 2 0 (𝜋𝑡) representation [35,Equation(2.6a)], for 𝐸−𝑠(𝑧) into sinh the converging series representation (99), noting that 1 𝑠 3 × 𝐹 ( , ; ;−𝑡2)𝑑𝑡 (117) contributions from infinity vanish, giving 2 1 2 2 2 ∞ 1 𝑐+𝑖∞ Γ (−𝑡)(2𝑘𝜔)𝑡 ∞ 𝑡 (𝜋𝑡) (𝑠 (𝑡)) 𝜁 (𝑠) =−4𝜔𝑠+1 ∑ ∫ 𝑑𝑡, −𝑠∫ cosh cos arctan + 2𝜋𝑖 𝑠+1+𝑡 (112) 2 𝑘=1 𝑐−𝑖∞ 0 sinh (𝜋𝑡) (1 + 𝑡 ) 1 𝑠 3 where the contour is defined by the real parameter 𝑐<−2. 2 × 2𝐹1 ( , ; ;−𝑡)𝑑𝑡, 𝜎>0. With this caveat, the series in (112)converges,allowingthe 2 2 2 integration and summation to be interchanged, leading to the identification This result has the interesting property that it is the only 𝑠+1 𝑐+𝑖∞ 𝑡 integral representation of which I am aware in which the 4𝜔 Γ (−𝑡)(2𝜔) 𝜁 (−1 − 𝑡) 𝑠 𝜁+ (𝑠) =− ∫ 𝑑𝑡. (113) independent variable does not appear as an exponent. Many 2𝜋𝑖 𝑐−𝑖∞ 𝑠+1+𝑡 variations of (117) can be obtained through the use of any 12 Journal of Mathematics of the transformations of the hypergeometric function, for 8.1. Finite Series Representations. From (79), we find, for even example, values of the argument,

𝑛 𝑘 2𝑘 1 𝑠 3 2 2 −𝑠/2 1 (−1) 𝜋 (2𝑛 − 2𝑘 −1) 𝜁 (2𝑛 − 2𝑘) 2𝐹1 ( , ; ;−𝑡)=(1+𝑡) 𝜁 (2𝑛) =− (∑ ( ) 2 2 2 2 Γ (2𝑘 + 2) (118) 𝑘=1 𝑠 3 𝑡2 × 𝐹 (1, ; ; ), −1 2 1 2 2 (1 + 𝑡2) ×(2−2 𝑛 +𝑛−1) ),

(122) which choice happens to restore the variable 𝑠 to its usual role as an exponent. from (83), we find

1 7. Conversion to Integral Representations 𝜁 (2𝑛) = 2(2−2𝑛 −1) From (79), identify the product 𝜁(⋅⋅⋅)Γ(⋅⋅⋅)in the form of an 𝑘 (123) integral representation from one of the primary definitions 𝑛 (−𝜋2) (1 − 21−2𝑛+2𝑘)𝜁(2𝑛 − 2𝑘) of 𝜁(𝑠) ([13,Equation(9.511)]) and interchange the series and × ∑ , Γ (2𝑘 + 1) sum. The sums can now be explicitly evaluated, leaving an 𝑘=1 integral representation valid over an important range of 𝑠: and, based on (80), we find [39] a similar, but independent, 𝜋𝑠−14𝑠 (𝜋𝑠/2) result: 𝜁 (𝑠) = sin 2−21+𝑠 +𝑠2𝑠 2−2𝑛 ∞ 𝑡−𝑠 (𝑠−(2𝑠 (𝑡/2) /𝑡) −1+ (𝑡/2)) 𝜁 (2𝑛) =2 × ∫ sinh cosh 𝑑𝑡, 1−21−2𝑛 𝑡 0 𝑒 −1 𝑛 (1 − 22𝑛−2𝑘−1) (2𝜋)2𝑘𝜁 (2𝑛 − 2𝑘)(−1)𝑘 𝜎<2. × ∑ ( ), Γ (2𝑘 + 2) (119) 𝑘=1 (124) Each of the terms in (119) corresponds to known integrals ([13,Equations(3.552.1) and (8.312.2)]), and if the identi- three of an infinite number of such recursive relationships fication is taken to completion, (119) reduces to the identity [40]. The latter two may be identified with known results by 𝜁(𝑠) = 𝜁(𝑠) with the help of (13). Other useful representations reversing the sums, giving are immediately available by evaluating the integrals in (119) 𝑛 corresponding to pairs of terms in the numerator, giving (−𝜋2) 𝜁 (2𝑛) =− 21−2𝑛 +2𝑛−2 𝜋𝑠2𝑠−2 (125) 𝜁 (𝑠) = 𝑛−1 (2𝑘 − 1) 𝜁 (2𝑘) (1−2𝑠) cos (𝜋𝑠/2) ×(∑ ), (−𝜋2)𝑘Γ (2𝑛 − 2𝑘 +2) 𝑠 −𝑠 𝑘=0 sin (𝜋𝑠) 2 ∞ 𝑡 (−1 + cosh (𝑡/2)) ×( ∫ 𝑑𝑡 (120) 𝜋 𝑒𝑡 −1 0 thoughttobenew, 1 + ), 𝜎<2, 𝑛 Γ (𝑠) (−𝜋2) 𝑛−1 (1 − 21−2𝑘)𝜁(2𝑘) 𝜁 (2𝑛) = (∑ ) (126) 𝜋𝑠 21−2𝑛 −2 (−𝜋2)𝑘Γ (2𝑛 − 2𝑘 +1) 𝜁 (𝑠) =𝜋𝑠−1 ( ) 𝑘=0 sin 2 ∞ 4.4(11) −𝑠−1 V −V equivalent to [11,Equation ], and ×(∫ V ( −1)𝑒 𝑑V (121) 0 sinh (V) 𝑛 2(−𝜋2) 𝑛−1 (1 − 22𝑘−1)𝜁(2𝑘)(−1)𝑘 𝜋 𝜁 (2𝑛) = (∑ ), (127) − ), 𝜎<2. 1−21−2𝑛 (2𝜋) 2𝑘Γ (2𝑛−2𝑘+2) sin (𝜋𝑠) Γ (1+𝑠) 𝑘=0

equivalent to [11,Equation4.4(9)]. 8. The Case 𝑠=𝑛 Several of the results obtained in previous sections reduce to 8.2. Novel Series Representations. By adding and subtract- helpful representations when 𝑠=𝑛. ing the first 𝑛−1terms in the series representation of Journal of Mathematics 13 the hypergeometric function in (91), we find ([36,Chapter Comparison of the previous infinite series with [10,Equation 9]), for the case 𝑠=2𝑛+2(𝑛≥0), 3.4(464)]using,𝑧=𝑖/𝜋identifies

2 2 2 2𝐹1 (1, 𝑛 + 1/2; 𝑛 + 3/2, −𝜔 /𝜋 𝑚 ) 𝐺 (1+𝑖/𝜋) 𝑖 −2 log ( )= (6 + 4 log (2𝜋) +2Li2 (𝑒 ) 𝑚2 𝐺 (1−𝑖/𝜋) 4𝜋 −𝑛 1 −𝜔2 𝜋2 =(𝑛+ )( ) −4 (𝑒2 −1)− ) (132) 2 𝜋2𝑚2 log 3

2 2 2 𝑘 = 0.232662175652953𝑖, 2𝜋 𝜔 1 𝑛−1 (−𝜔 /𝜋 𝑚 ) ×( ( )− ∑ ), 𝑚𝜔 arctan 𝜋𝑚 𝑚2 1/2 + 𝑘 𝑘=0 𝐺 (128) where is Barnes double Gamma function. From [10,Section 1.3], unnumbered equation following 26, we also find and for the case 𝑠=2𝑛+3 𝐺 (1+𝑖/𝜋) 𝑖 log (2𝜋) 𝑖 log ( )= − 2 2 2 𝐺 (1−𝑖/𝜋) 𝜋 𝜋 2𝐹1 (1, 𝑛 + 1; 𝑛 + 2; −𝜔 /𝜋 𝑚 ) ∞ 𝑚2 2𝑖 𝑘+𝑖/𝜋 +( ∑ (− +𝑘ln ( ))) −𝑛 𝜋 𝑘−𝑖/𝜋 = (𝑛+1) (−𝜔2/𝜋2𝑚2) 𝑘=1 (133) 2 2 2 𝑘 𝜋2 𝜔2 1 𝑛−1 (−𝜔 /𝜋 𝑚 ) ×( (1 + )− ∑ ). 𝜔2 ln 𝜋2𝑚2 𝑚2 𝑘+1 thereby recovering (130) after representing the arctan func- 𝑘=0 tion in (130) as a (with 𝜔=1). (129)

9. The Critical Strip Substituting (128)or(129)in(88), (91), and (101) gives series representations for 𝜁(2𝑛) and 𝜁(2𝑛+1),respectively,forwhich In (58)and(59)with0≤𝜎≤1let I find no record in the literature, with particular reference to works involving Barnes double Gamma function (e.g., [30]). 𝑠−1 𝑛=0 𝜋 sin (𝜋𝑠/2) For example, employing (128)inthecase ,weobtain 𝑓 (𝑠) =− , (134) what appears to be a new representation (or the evaluation of 𝑠−1 anewsum): 2𝑠−1 𝑔 (𝑠) = , (135) Γ (𝑠+1) ∞ 𝜋𝑚 𝜔 𝜁 (2) =−4𝜔∑ ( arctan ( )−1) 1 1 𝜔 𝜋𝑚 ℎ (𝑡) = √𝑡( − ). 𝑚=1 2 2 (136) sinh 𝑡 𝑡 −2𝜔 2𝜔 (130) + Li2 (𝑒 )−2𝜔log (−1 + 𝑒 )

+3𝜔2 +2𝜔. Then, after adding and subtracting, we find a composition of symmetric and antisymmetric integrals about the critical line 𝜎=1/2: The series (130)(comparewith([28, Equation (42.1.1)])) is absolutely convergent by Gauss’ test for arbitrary values of 𝜔 2𝑓 (𝑠) 𝑔 (𝑠) 𝜔→0 𝜁 (𝑠) = and reduces to an identity in the limiting case . 𝑔 (𝑠) −𝑓(𝑠) ∞ 1 8.3. Connection with Multiple Gamma Function. From Sec- × ∫ ℎ (𝑡) (𝜌 (𝑡)) (( −𝜎) (𝑡))𝑑𝑡 cos ln sinh 2 ln tion 5.2 with 𝜔=1, we have, after some simplification, 0 2𝑖𝑓 (𝑠) 𝑔 (𝑠) + ∞ (−1)𝑚𝜋(−2𝑚)𝜁 (2𝑚) 𝑓 (𝑠) −𝑔(𝑠) 𝜁 (2) =5−4(∑ ) 2𝑚 + 1 ∞ 1 𝑚=1 (131) × ∫ ℎ (𝑡) (𝜌 (𝑡)) (( −𝜎) (𝑡))𝑑𝑡, sin ln cosh 2 ln −2 2 0 + Li2 (𝑒 )−2log (𝑒 −1). (137) 14 Journal of Mathematics and simultaneously Then, by breaking 𝑓(1/2 + 𝑖𝜌) and 𝑔(1/2 + 𝑖𝜌) into their real and imaginary parts, (141)reducesto

2𝑓 (𝑠) 𝑔 (𝑠) 𝜁𝑅 (1/2 + 𝑖𝜌) 𝜁 (𝑠) = 𝑓 (𝑠) +𝑔(𝑠) 𝜁𝐼 (1/2 + 𝑖𝜌)

∞ 𝐶𝑚 (𝜌) cos (𝜌 ln (2𝜋))−𝐶𝑝 (𝜌) sin (𝜌 ln (2𝜋)) 1 = , × ∫ ℎ (𝑡) cos (𝜌 ln (𝑡)) cosh (( −𝜎)ln (𝑡))𝑑𝑡 𝐶 (𝜌) (𝜌 (2𝜋))+𝐶 (𝜌) (𝜌 (2𝜋))−√𝜋 0 2 𝑝 cos ln 𝑚 sin ln (143) 2𝑖𝑓 (𝑠) 𝑔 (𝑠) − 𝑓 (𝑠) +𝑔(𝑠) where 1 𝜋𝜌 1 𝜋𝜌 ∞ 1 𝐶 (𝜌) = Γ ( +𝑖𝜌) ( )+Γ ( +𝑖𝜌) ( ), × ∫ ℎ (𝑡) (𝜌 (𝑡)) (( −𝜎) (𝑡))𝑑𝑡. 𝑝 𝑅 cosh 𝐼 sinh sin ln sinh 2 ln 2 2 2 2 0 (144) (138) 1 𝜋𝜌 1 𝜋𝜌 𝐶 (𝜌) = Γ ( +𝑖𝜌) ( )−Γ ( +𝑖𝜌) ( ), 𝑚 𝐼 2 cosh 2 𝑅 2 sinh 2 Let (145)

in terms of the real and imaginary parts of Γ(1/2 + 𝑖𝜌).This ∞ is a known result [41] that depends only on the functional C (𝜌) = ∫ ℎ (𝑡) cos (𝜌 ln (𝑡))𝑑𝑡, equation (13) (See the Appendix for further discussion on 0 this point). In [41], it was shown that the numerator and (139) ∞ denominator on the right-hand side of (11) can never vanish 𝑆(𝜌)=∫ ℎ (𝑡) sin (𝜌 ln (𝑡))𝑑𝑡. simultaneously, so this result defines the half-zeros (only 0 one of 𝜁𝑅(1/2 + 𝑖𝜌) or 𝜁𝐼(1/2 + 𝑖𝜌) vanishes) (and usually brackets the full-zeros) of 𝜁(1/2 + 𝑖𝜌) according to whether the numerator or denominator vanishes. This was studied in 𝜁 (1/2 + 𝑖𝜌 )=𝜁(1/2 + 𝑖𝜌 )=0 For 𝜁(𝑠) =0 on the critical line 𝜎=1/2,both(137)and [41]. The full-zeros ( 𝑅 0 𝐼 0 )of (138) can be simultaneously satisfied when both 𝐶(𝜌) =0̸ and 𝜁(1/2 + 𝑖𝜌) occur when the following condition 𝑆(𝜌) =0̸ only if 𝑆(𝜌0)=0, (146) 𝐶(𝜌 )=0 󵄨 󵄨2 󵄨 󵄨2 0 󵄨 1 󵄨 󵄨 1 󵄨 󵄨𝑓 ( +𝑖𝜌)󵄨 = 󵄨𝑔 ( +𝑖𝜌)󵄨 , (140) 󵄨 2 󵄨 󵄨 2 󵄨 is simultaneously satisfied for some value(s) of 𝜌=𝜌0.Inthat case, the defining equation (143)becomes a true relationship that can be verified by direct computation. 𝜁 (1/2 + 𝑖𝜌)󸀠 𝜁󸀠(1/2 + 𝑖𝜌) 𝜎=1/2 𝑅 =− 𝐼 Set in (137)and(138)andfind 󸀠 𝜁󸀠(1/2 + 𝑖𝜌) 𝜁𝐼(1/2 + 𝑖𝜌) 𝑅

=(𝐶𝑚 (𝜌) cos (𝜌 ln (2𝜋))−𝐶𝑝 (𝜌) sin (𝜌 ln (2𝜋))) 1 2𝑓 (1/2 + 𝑖𝜌) 𝑔 (1/2 +𝑖𝜌) (147) 𝜁( +𝑖𝜌)= 𝐶(𝜌) 2 𝑓 (1/2 + 𝑖𝜌) + 𝑔 (1/2 +𝑖𝜌) ×(𝐶𝑝 (𝜌) cos (𝜌 ln (2𝜋))

(141) −1 2𝑖𝑓 (1/2 + 𝑖𝜌) 𝑔 (1/2 +𝑖𝜌) +𝐶𝑚 (𝜌) sin (𝜌 ln (2𝜋))−√𝜋) , = 𝑆(𝜌). 𝑓 (1/2 + 𝑖𝜌) − 𝑔 (1/2 +𝑖𝜌) where the derivatives are taken with respect to the variable 𝜌,and(147) is evaluated at 𝜌=𝜌0.Manysolutionsto(146) are known [42], so there is no doubt that these two functions Equation (141) can be used to obtain a relationship between can simultaneously vanish at certain values of 𝜌.(Aproofthat 𝜁(1/2 + 𝑖𝜌) the real and imaginary components of .Let this cannot occur in (137)and(138)when𝜎 =1/2̸ is left as an exercise for the reader.) A further interesting representation can be obtained 1 1 1 inside the critical strip by setting 𝑐=𝜎in (9)andseparating 𝜁( +𝑖𝜌)=𝜁 ( +𝑖𝜌) + 𝑖𝜁 ( +𝑖𝜌). 2 𝑅 2 𝐼 2 (142) 𝜂(𝑠) into its real and imaginary components: Journal of Mathematics 15

𝜂𝑅 (𝜎 + 𝜌𝑖)

2 −𝜎/2 𝜌 arctan(𝑡) 1 ∞ (1 + 𝑡 ) 𝑒 (cos (Ω (𝑡)) sin (𝜋𝜎) cosh (𝜋𝜎𝑡) − sin (Ω (𝑡)) cos (𝜋𝜎) sinh (𝜋𝜎𝑡)) (148) = 𝜎(1−𝜎) ∫ 𝑑𝑡, 2 2 −∞ cosh (𝜋𝜎𝑡) − cos2 (𝜋𝜎)

𝜂𝐼 (𝜎 + 𝜌𝑖)

2 −𝜎/2 𝜌 arctan(𝑡) 1 ∞ (1 + 𝑡 ) 𝑒 (sin (Ω (𝑡)) sin (𝜋𝜎) cosh (𝜋𝜎𝑡) + cos (Ω (𝑡)) cos (𝜋𝜎) sinh (𝜋𝜎𝑡)) (149) =− 𝜎(1−𝜎) ∫ 𝑑𝑡 2 2 −∞ cosh (𝜋𝜎𝑡) − cos2 (𝜋𝜎) where further ([43, Equation (13)]) identifying 1 𝐺 (1−𝑡) (𝜋𝑡) 1 Ω (𝑡) = 𝜌 (𝜎2 (1 + 𝑡2)) + 𝜎 (𝑡) . ( )=𝑡 (sin )+ (2𝜋𝑡) , 2 log arctan (150) log 𝐺 (1+𝑡) log 𝜋 2𝜋Cl2 (154) In analogy to (137)and(138), the representations (148)and 0<𝑡<1. (𝜋𝜎) (149) include a term cos that vanishes on the critical line 𝑡∈C 𝜎=1/2, both reducing to the equivalent of (26)inthatcase. To allow for the possibility that , this should instead read (I am speculating here that such forms contain information 𝐺 (1−𝑡) (𝜋𝑡) 𝑖𝜋 ( )=𝑡 (sin )+ 𝐵 (𝑡) about the distribution and/or existence of zeros of 𝜂(𝑠) and log 𝐺 (1+𝑡) log 𝜋 2 2 hence 𝜁(𝑠) inside the critical strip.) The integrands in both (155) (−𝜋𝜎𝑡) 𝑡→∞ −𝑖𝜋/2 representations vanish as exp as and are finite + e ( 2𝜋𝑖𝑡), at 𝑡=0; most of the numerical contribution to the integrals 2𝜋 Li2 e comes from the region 0<𝑡≅𝜌. Numerical evaluation in which is valid for all 𝑡 in terms of the range 0≤𝑡canbeaccleratedbywriting 𝐵𝑛(𝑡) and functions. However, similar results 𝜌 arctan(𝑡) 𝜌(arctan(𝑡)−𝜋/2) are given in the equivalent, general form in [20,Equation e 󳨀→ e , (151) (5.12)]. Interestingly, both the Maple and Mathematica com- 𝜌𝜋/2 puter codes somehow (neither [44] nor references contained and extracting a factor e outside the integral in that range. therein, both of which study such integrals, list the computer It remains to be shown how the remaining integral conspires (5.12) 𝜌→∞ results, which are likely based on [20,Equation ]) to cancel this factor in the asymptotic range . manage to evaluate the integral in (153)intermssimilarto Finally, (148) is invariant and (149) changes sign under the (155) for a variety of values of 𝑛 provided that the user restricts transformations 𝜌→ −𝜌and 𝑡→−𝑡, demonstrating that |𝑡| < 1 𝜂(𝑠) , although the right-hand side provides the analytic is self-conjugate. continuation of the left when that restriction is lifted. This has application in the evaluation of 𝜁(3) according to the method 10. Comments of Section 8.3, where, using the methods of [10,Chapter3], we identify Anumberofrelatedresultsthatarenoteasilycategorized ∞ 𝜁 (2𝑘) 𝑡(2𝑘) 1 𝜋 𝑡 and which either do not appear in the reference literature or ∑ = − ∫ V2 (𝜋V) 𝑑V, 𝑘+1 2 𝑡2 cot (156) appear incorrectly should be noted. 𝑘=1 0 Comparison of (49)and(54) suggests the identification V =𝑖. Such sums and integrals have been studied [30, 43]for a generalization of integrals studied in [44]. From the com- the case 𝑠=−𝑛in terms of Barnes double Gamma function puter evaluation of the integral, (156)eventuallyyields 𝐺(𝑧) and Clausen’s function defined by ∞ 2𝑘 𝜁 (2𝑘) 𝑡 1 1 2𝑖𝜋𝑡 ∑ = 𝑖𝜋𝑡 + − log (1 − 𝑒 ) ∞ (𝑧𝑘) 𝑘+1 3 2 (𝑧) = ∑ sin , 𝑘=1 Cl2 𝑘2 (152) 𝑘=1 2𝑖𝜋𝑡 Li2 (𝑒 ) +𝑖 (157) with the (tacit) restriction that 𝑧∈R.Thatisnotthe 𝜋𝑡 case here. Thus straightforward application of that analysis 2𝑖𝜋𝑡 Li3 (𝑒 )−𝜁(3) will sometimes lead to incorrect results. For example [30, − , Equation (4.3)], gives (with 𝑛=1) 2𝜋2𝑡2 (3.4) 𝑡 as a new addition to the lists found in [10,Section ], 𝑛 𝐺 (1−𝑡) (3.3) (3.4) 𝜋 ∫ V cot (𝜋V) 𝑑V =𝑡log (2𝜋) + log ( ), (153) and [11,Sections and ], and a generalization of (97) 0 𝐺 (1+𝑡) when 𝑠=3. 16 Journal of Mathematics

Appendix [2] M. Abramowitz and I. Stegun, Handbook of Mathematical Functions,Dover,NewYork,NY,USA,1964. In [41], an attempt was made to show that the functional [3] T. M. Apostol, Introduction to , equation (13) and a minimal number of analytic properties Springer,NewYork,NY,USA,2010. 𝜁(𝑠) 𝜁(𝑠) =0̸ 𝜎 =1/2̸ of might suffice to show that when .This [4]F.W.J.Olver,Asymptotics and Special Functions, Academic was motivated by a comment attributed to H. Hamburger Press, New York, NY,USA, 1974, Computer Science and Applied ([8, page 50]) to the effect that the functional equation for Mathematics. 𝜁(𝑠) perhaps characterizes it completely. (An unequivocal [5] E. C. Titchmarsh, The Theory of Functions, Oxford University equivalent claim can also be found in [15]. As demonstrated Press, New York, NY, USA, 2nd edition, 1949. by the counterexample given here, and in [45], this claim [6] E.T.WhittakerandG.N.Watson,ACourseofModernAnalysis, is false.) The purported demonstration contradicts a result Cambridge Mathematical Library, Cambridge University Press, of Gauthier and Zeron [45], who show that infinitely many Cambridge, UK, 1963. 𝜁(𝑠) functions exist, arbitrarily close to ,thatsatisfy(13)and [7] H. M. Edwards, Riemann’s Zeta Function, Academic Press, New the same analytic properties that are invoked in [41], but York, NY, USA, 1974. 𝜎 =1/2̸ which vanish when . (To quote Prof. Gauthier: “We [8] A. Ivic,´ The Riemann Zeta-Function: Theory and Applications, prove the existence of functions approximating the Riemann Dover, New York, NY, USA, 1985. zeta-function, satisfying the Riemann functional equation [9]S.J.Patterson,An introduction to the Theory of the Riemann andviolatingtheanalogueoftheRiemannhypothesis.” Zeta-Function,vol.14ofCambridge Studies in Advanced Math- (private communication).) ematics, Cambridge University Press, Cambridge, UK, 1988. In fact, an example of such a function is easily found. Let [10] H. M. Srivastava and J. Choi, Series Associated with the Zeta and Related Functions, Kluwer Academic, Dordrecht, The Nether- ] (𝑠) =𝐶sin (𝜋 (𝑠0 −𝑠 )) sin (𝜋 (𝑠0 +𝑠 )) lands, 2001. ∗ ∗ (A.1) × sin (𝜋 ( 𝑠 −𝑠 0 )) sin (𝜋 (𝑠0 +𝑠 )) , [11] H. M. Srivastava and J. Choi, Zeta and q-Zeta Functions and Associated Series and Integrals, Elsevier, New York, NY, USA, ∗ where 𝑠0 isthecomplexconjugateof𝑠0, 𝑠0 =𝜎0 +𝑖𝜌0 is 2012. arbitrary complex, and 𝐶,givenby [12] A. Erdelyi,W.Magnus,F.Oberhettinger,andF.G.Tricomi,Eds.,´ 1 Higher Transcendental Functions,vol.1–3,McGraw-Hill,New 𝐶= , 2 York, NY, USA, 1953. 2 2 (A.2) (cosh (𝜋𝜌0)−cos (𝜋𝜎0)) [13] I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, Academic Press, New York, NY, USA, 1980. ](1) = 1 is a constant chosen such that .Then [14] A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev, Integrals andSeries:Volume3:MoreSpecialFunctions,Gordonand 𝜁] (𝑠) ≡ ] (𝑠) 𝜁 (𝑠) (A.3) Breach Science, New York, NY, USA, 1990. satisfies analyticity conditions (1–3) of [45](whichinclude [15] Encyclopedia of mathematics, http://www.encyclopediaofmath the functional equation (13)). Additionally, 𝜁](𝑠) has zeros at .org/index.php/Zeta-function. the zeros of 𝜁(𝑠) as well as complex zeros at 𝑠=±(𝑛+𝑠0) and [16] http://en.wikipedia.org/wiki/Riemann zeta function. 𝑠=±(𝑛+𝑠∗) 𝑠=1 0 and a pole at with residue equal to unity, [17] http://en.wikipedia.org/wiki/Dirichlet eta function. 𝑛 = 0, ±1, ±2, . where . [18] Mathworld, http://mathworld.wolfram.com/HankelFunction ThusthelastfewparagraphsofSection2of[41] beginning .html. (2.9) ... with the words “There are two cases where fails ” [19] Mathworld, http://mathworld.wolfram.com/RiemannZeta- cannot be valid. However, the remainder of that paper, Function.html. particularly those sections dealing with the properties of 𝜁(1/2 + 𝑖𝜌) [20] H. M. Srivastava, M. L. Glasser, and V. S. Adamchik, “Some as referenced here in Section 9,arevalidand definite integrals associated with the Riemann zeta function,” correct. I am grateful to Professor Gauthier for discussions Zeitschrift fur Analysis und ihre Anwendung,vol.19,no.3,pp. on this topic. 831–846, 2000. [21] P. Flajolet and B. Salvy, “Euler sums and contour integral Acknowledgments representations,” Experimental Mathematics,vol.7,no.1,pp.15– 35, 1998. The author wishes to thank Vinicius Anghel and Larry [22] R. G. Newton, The Complex J-Plane,TheMathematicalPhysics Glasser for commenting on a draft copy of this paper. Leslie Monograph Series, W. A. Benjamin, New York, NY, USA, 1964. M. Saunders obituary may be found in the journal Physics [ 2 3 ] J. L . W. V. Je n s e n , L’ Int e r m ediaire´ des mathematiciens´ ,vol.2, Today, issue of Dec. 1972, page 63, http://www.physicstoday 1895. .org/resource/1/phtoad/v25/i12/p63 s2?bypassSSO=1. [24] E. Lindelof, LeCalculDesResidusetsesApplicationsaLaTheorie Des Fonctions, Gauthier-Villars, Paris, France, 1905. References [25] S. N. M. Ruijsenaars, “Special functions defined by analytic difference equations,” in Special Functions 2000: Current Per- [1]F.W.J.Olver,D.W.Lozier,R.F.Boisvert,andC.W.Clark,NIST spective and Future Directions (Tempe, AZ), J. Bustoz, M. Ismail, Handbook of Mathematical Functions, Cambridge University andS.Suslov,Eds.,vol.30ofNATO Science Series, pp. 281–333, Press, Cambridge, UK, 2010, http://dlmf.nist.gov/25.5. Kluwer Academic, Dordrecht, The Netherlands, 2001. Journal of Mathematics 17

[26] B. Riemann, “Uber¨ die Anzahl der Primzahlen unter einer gegebenen Grosse,”¨ 1859. [27]T.Amdeberhan,M.L.Glasser,M.C.Jones,V.Moll,R. Posey, and D. Varela, “The Cauchy-Schlomilch¨ transformation,” http://arxiv.org/abs/1004.2445. [28] E. R. Hansen, Table of Series and Products, Prentice Hall, New York, NY, USA, 1975. [29] R. E. Crandall, Topics in Advanced Scientific Computation, Springer,NewYork,NY,USA,1996. [30]J.Choi,H.M.Srivastava,andV.S.Adamchick,“Multiple gamma and related functions,” Applied Mathematics and Com- putation,vol.134,no.2-3,pp.515–533,2003. [31] S. Tyagi and C. Holm, “A new integral representation for the Riemann zeta function,” http://arxiv.org/abs/math-ph/0703073. [32] N. G. de Bruijn, “Mathematica,” Zutphen,vol.B5,pp.170–180, 1937. [33] R. B. Paris and D. Kaminski, Asymptotics and Mellin-Barnes Integrals,vol.85ofEncyclopedia of Mathematics and Its Appli- cations, Cambridge University Press, Cambridge, UK, 2001. [34] Y. L. Luke, The Special Functions and Their Approximations,vol. 1, Academic Press, New York, NY, USA, 1969. [35] M. S. Milgram, “The generalized integro-exponential function,” Mathematics of Computation,vol.44,no.170,pp.443–458,1985. [36] N. N. Lebedev, Special Functions and Their Applications,Pren- ticeHall,NewYork,NY,USA,1965. [37] T.M. Dunster, “Uniform asymptotic approximations for incom- plete Riemann zeta functions,” Journal of Computational and Applied Mathematics,vol.190,no.1-2,pp.339–353,2006. [38] V. S. Adamchik and H. M. Srivastava, “Some series of the zeta and related functions,” Analysis, vol. 18, no. 2, pp. 131–144, 1998. [39] M. S. Milgram, “Notes on a paper of Tyagi and Holm: ‘a new integral representation for the Riemann Zeta function’,” http://arxiv.org/abs/0710.0037v1. [40] E. Y.Deeba and D. M. Rodriguez, “Stirling’s series and Bernoulli numbers,” American Mathematical Monthly,vol.98,no.5,pp. 423–426, 1991. [41] M. S. Milgram, “Notes on the zeros of Riemann’s zeta Function,” http://arxiv.org/abs/0911.1332. 22 [42] A. M. Odlyzko, “The 10 -nd zero of the Riemann zeta func- tion,” in Dynamical, Spectral, and Arithmetic Zeta Functions (San Antonio, TX, 1999),vol.290ofContemporary Mathematics Series, pp. 139–144, American Mathematical Society, Provi- dence, RI, USA, 2001. [43] V.S. Adamchik, “Contributions to the theory of the barnes func- tion, computer physics communication,” http://arxiv.org/abs/ math/0308086. [44] D. CvijovicandH.M.Srivastava,“Evaluationsofsomeclassesof´ the trigonometric moment integrals,” JournalofMathematical Analysis and Applications,vol.351,no.1,pp.244–256,2009. [45] P. M. Gauthier and E. S. Zeron, “Small perturbations of the Riemann zeta function and their zeros,” Computational Methods and Function Theory,vol.4,no.1,pp.143–150,2004, http://www.heldermann-verlag.de/cmf/cmf04/cmf0411.pdf. Advances in Advances in Journal of Journal of Operations Research Decision Sciences Applied Mathematics Algebra Probability and Statistics Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation Hindawi Publishing Corporation http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014 http://www.hindawi.com Volume 2014

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