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Certain Relationships Among Polygamma Functions, Riemann Zeta Function and Generalized Zeta Function Junesang Choi1* and Chao-Ping Chen2

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Certain Relationships Among Polygamma Functions, Riemann Zeta Function and Generalized Zeta Function Junesang Choi1* and Chao-Ping Chen2 Choi and Chen Journal of Inequalities and Applications 2013, 2013:75 http://www.journalofinequalitiesandapplications.com/content/2013/1/75 R E S E A R C H Open Access Certain relationships among polygamma functions, Riemann zeta function and generalized zeta function Junesang Choi1* and Chao-Ping Chen2 *Correspondence: [email protected] Abstract 1Department of Mathematics, Dongguk University, Gyeongju, Many useful and interesting properties, identities, and relations for the Riemann zeta 780-714, Republic of Korea function ζ (s) and the Hurwitz zeta function ζ (s, a) have been developed. Here, we Full list of author information is aim at giving certain (presumably) new and (potentially) useful relationships among available at the end of the article polygamma functions, Riemann zeta function, and generalized zeta function by modifying Chen’s method. We also present a double inequality approximating ζ (2r + 1) by a more rapidly convergent series. MSC: Primary 11M06; 33B15; secondary 40A05; 26D07 Keywords: Riemann zeta function; Hurwitz zeta function; gamma function; psi-function; polygamma functions; Bernoulli numbers; multiple Hurwitz zeta function; Stirling numbers of the first kind; asymptotic formulas 1 Introduction The Riemann zeta function ζ (s)isdefinedby ∞ ζ (s):= (s)> . (.) ns n= The Hurwitz (or generalized) zeta function ζ (s, a)isdefinedby ∞ –s ∈ C \ Z– ζ (s, a):= (k + a) (s)>;a ,(.) k= C Z– where and denote the sets of complex numbers and nonpositive integers, respec- tively. It is easy to see from the definitions (.)and(.)that – ζ (s)=ζ (s,)= s – ζ s, =+ζ (s, ). (.) It is noted that, in many different ways, the Riemann zeta function ζ (s) and the Hurwitz zeta function ζ (s, a) can be continued meromorphically to the whole complex s-plane hav- ing simple poles only at s = with their respective residues at this point. © 2013 Choi and Chen; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Choi and Chen Journal of Inequalities and Applications 2013, 2013:75 Page 2 of 11 http://www.journalofinequalitiesandapplications.com/content/2013/1/75 The polygamma functions ψ(n)(s)(n ∈ N)aredefinedby dn+ dn ψ(n)(s):= log (s)= ψ(s) n ∈ N ; s ∈ C \ Z– ,(.) dzn+ dsn where N denotes the set of positive integers and N := N ∪{},and(s)isthefamiliar gamma function and psi (or digamma) function ψ is defined by d ψ(s):= log (s)andψ()(s)=ψ(s). (.) ds Here is a well-known (useful) relationship between the polygamma functions ψ (n)(s)and the generalized zeta function ζ (s, a): ∞ ψ(n)(s)=(–)n+n! =(–)n+n!ζ (n +,s) (k + s)n+ k= ∈ N ∈ C \ Z– n ; s .(.) It is also easy to have the following expression (cf. [,Eq..()]and[, Eq. .()]): m ψ(n)(s + m)=ψ(n)(s)+(–)nn! (m, n ∈ N ), (.) (s + k –)n+ k= which, in view of (.), can be expressed in the following form: m ζ (n, s + m)=ζ (n, s)– n ∈ N \{}; m ∈ N ,(.) (s + k –)n k= where an empty sum is (elsewhere throughout this paper) understood to be nil. The Riemann zeta function ζ (s)in(.) plays a central role in the applications of complex analysis to number theory. The number-theoretic properties of ζ (s) are exhibited by the following result known as Euler’s formula, which gives a relationship between the set of primes and the set of positive integers: – ζ (s)= –p–s (s)> ,(.) p where the product is taken over all primes. It is readily seen that ζ (s) =( (s)=σ ), and the Riemann’s functional equation for ζ (s) ζ (s)=(π)s–( – s) sin πs ζ ( – s) (.) shows that ζ (s) =( σ ) except for the trivial zeros in ζ (–n)= (n ∈ N). (.) Choi and Chen Journal of Inequalities and Applications 2013, 2013:75 Page 3 of 11 http://www.journalofinequalitiesandapplications.com/content/2013/1/75 Furthermore, in view of the following known relation: ∞ (–)n– ζ (s)= (s)>;s = , (.) ––s ns n= we find that ζ (s)<(s ∈ R;<s < ). The assertion that all the non-trivial zeros of ζ (s) have real part is popularly known as the Riemann hypothesis which was conjectured (but not proven) in the memoir of Riemann []. This hypothesis is still one of the most chal- lenging mathematical problems today (see Edwards []), which was unanimously chosen to be one of the seven greatest unsolved mathematical puzzles of our time, the so-called millennium problems (see Devlin []). Leonhard Euler (-), in , computed the Basel problem: π + + + + + ···= ζ () = (.) to decimal places with only a few terms of his powerful summation formula discovered in the early s, now called the Euler-Maclaurin summation formula. This probably convinced him that the sum in (.)equalsπ /, which he proved in the same year (see []).Euleralsoproved (π)n ζ (n)=(–)n+ B (n ∈ N ), (.) (n)! n where Bn are the Bernoulli numbers (see [, Section .]; see also [, Section .]). Subse- quently, many authors have proved the Basel problem (.)andEq.(.)invariousways (see, e.g.,[]). We get no information about ζ (n+)(n ∈ N) from Riemann’s functional equation, since both members of (.) vanish upon setting s =n +(n ∈ N). In fact, until now no simple formula analogous to (.)isknownforζ (n +) or even for any special case such as ζ (). It is not even known whether ζ (n + ) is rational or irrational, except that the irrationality of ζ () was proved recently by Apéry []. But it is known that there are infinitely many ζ (n +)whichareirrational(see[]and[]). The following formulae involving ζ (k + ) were given by Ramanujan (see []): (i) If k ∈ N \{}, ∞ ∞ nk– nk– αk ζ ( – k)+ =(–β)k ζ ( – k)+ ; (.) enα – enβ – n= n= (ii) If k ∈ N, ∞ = ζ (k +)+ (α)k nk+(enα –) n= ∞ – ζ (k +)+ (–β)k nk+(enβ –) n= [ k+ ] j j (–) π BjBk–j+ + αk–j+ +(–β)k–j+ , (.) (j)!(k –j +)! j= Choi and Chen Journal of Inequalities and Applications 2013, 2013:75 Page 4 of 11 http://www.journalofinequalitiesandapplications.com/content/2013/1/75 where Bj is the jth Bernoulli number, α >andβ >satisfyαβ = π ,and means that when k is an odd number m – , the last term of the right-hand side in (.)istakenas m m (–) π Bm (m!) . In , Hardy []proved(.). In , Grosswald []proved(.). In , Gross- wald [] gave another expression of ζ (k + ). In , Zhang [] not only proved Ra- manujan formulae (.)and(.), but also gave an explicit expression of ζ (k +).For various series representations for ζ (n +),see[] and also see [, Chapter ] and [, Chapter ]. Many useful and interesting properties, identities, and relations for ζ (s)andζ (s, a)have been developed, for example, the formulas recalled above. Here, we aim at presenting cer- tain (presumably) new and (potentially) useful relationships among polygamma functions, Riemann zeta function, and generalized zeta function by mainly modifying Chen’s method []. We also give a double inequality approximating ζ (r +)(r ∈ N) by a more rapidly convergent series, that is, the Dirichlet lambda function λ(s)definedby ∞ λ(s)= (s)> (.) (k –)s k= (see Theorem ). 2 Main results We first prove a relationship between polygamma functions and Riemann zeta functions asserted by Theorem . Theorem For each r ∈ N, the following formula holds true: n ψ(r)(j +)=ζ (r +)+ (n +)ψ(r)(n +) (r)! (r)! j= + ψ(r–)(n +)–ζ (r)(n ∈ N). (.) (r –)! Proof We prove Eq. (.) by using the principle of mathematical induction on n ∈ N.For n =,inviewofEq.(.), it is found that both sides of (.)equal–ζ (r +).Assume that (.) holds true for some n ∈ N. Then we will show that (.)istrueforn +,i.e., n+ ψ(r)(j +)=ζ (r +)+ (n +)ψ(r)(n +) (r)! (r)! j= + ψ(r–)(n +)–ζ (r). (.) (r –)! Let denote the left-hand side of (.)byL. Then, by the induction hypothesis, we have n L = ψ(r)(j +)+ ψ(r)(n +) (r)! (r)! j= = ζ (r +)+ (n +)ψ(r)(n +)+ ψ(r–)(n +)–ζ (r) (r)! (r –)! Choi and Chen Journal of Inequalities and Applications 2013, 2013:75 Page 5 of 11 http://www.journalofinequalitiesandapplications.com/content/2013/1/75 + ψ(r)(n +) (r)! = ζ (r +)+ (n +)ψ(r)(n +)+ ψ(r–)(n +)–ζ (r). (r)! (r –)! By using (.), we find (r)! L = ζ (r +)+ (n +) ψ(r)(n +)– (r)! (n +)r+ (r –)! + ψ(r–)(n +)+ – ζ (r) (r –)! (n +)r = ζ (r +)+ (n +)ψ(r)(n +)+ ψ(r–)(n +)–ζ (r), (r)! (r –)! which is equal to the right-hand side of (.). Hence, by the principle of mathematical induction, (.) is true for all n ∈ N. We recall the following asymptotic formula for the polygamma function ψ (n)(s)(see[, p., Entry ..]): ∞ (n –)! n! (k + n –)! ψ(n)(s) ∼ (–)n+ + + B sn sn+ k (k)!sk+n k= |s|→∞; arg(s) π – ( < < π); n ∈ N .(.) Next we give an interesting series representation for ζ (r) in terms of the generalized zeta functions ζ (r +,j +)(j ∈ N). Theorem The following formula holds true: ∞ ζ (r)= ζ (r +,j +) (r ∈ N). (.) j= Proof Applying the series representation (.)forψ(n)(s) to the left-hand side of (.), we have ∞ n – = ζ (r +)–ζ (r) ( + j +)r+ j= = + (n +)ψ(r)(n +)+ ψ(r–)(n +). (r)! (r –)! Taking the limit on each side of the last resulting identity as n →∞,inviewof(.), we obtain the following identity: ∞ ∞ ζ (r)=ζ (r +)+ (r ∈ N). (.) ( + j)r+ j= = Now it is easy to find from (.)and(.)that(.)equals(.). Choi and Chen Journal of Inequalities and Applications 2013, 2013:75 Page 6 of 11 http://www.journalofinequalitiesandapplications.com/content/2013/1/75 (r) We present a relationship between ζ (r +)andψ (n + ) asserted by the following theorem.
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