# Some Inequalities for theq-Analogue of the Classical Riemann Zeta Functions and the q-Polygamma Functions

Authors: Banyat Sroysang
Journal: The Scientific World Journal
 (2014)

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## Abstract

We present the generalizations on some inequalities for theq-analogue of the classical Riemann zeta functions and the q-polygamma functions.


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## Body

## 1. Introduction and Preliminaries

Letc∈ℂ and q∈(0,1). The q-shifted factorial [1–5] is defined by
(1)(c;q)∞=∏k=0∞‍(1-cqk).
The q-Jackson integral [6] from 0 to c is defined by
(2)∫0c‍f(x)dqx=(1-q)c∑n=0∞‍f(cqn)qn.
The q-gamma function [6] is defined by
(3)Γq(x)=(q;q)∞(qx;q)∞(1-q)1-x,x≠0,-1,-2,….
The q-digamma [7–9] function is defined by
(4)ψq(x)=Γq′(x)Γq(x)
and may be represented [10] as
(5)ψq(x)=-log⁡(1-q)+log⁡q1-q∫0q‍tx-11-tdqt,
where x>0.Theq-polygamma function [7] is defined by
(6)ψq,n(x)=ψq(n)(x)
and may be represented as
(7)ψq,n(x)=log⁡q1-q∫0q‍(log⁡u)nux-11-udqu,
where x>0 and n∈ℕ.For anyx>0, we denote
(8)α(x)=log⁡xlog⁡q-E(log⁡xlog⁡q),[x]q=1-qx1-q,{x}q=[x]qqx+α([x]q),
where E(log⁡x/log⁡q) is the integer part of log⁡x/log⁡q.Theq-zeta function [11] is defined by
(9)ζq(s)=∑n=1∞‍1{n}qs,
where Re(s)>1. Moreover, in [11], the q-analogue of the classical Riemann Zeta function is
(10)ζq(s)=1Γ~q(s)∫0∞‍ts-1Zq(t)dqt,
where Re(s)>1 and for any t>0,
(11)Zq(t)=∑n=1∞‍1((q-1){n}qt;q)∞,Γ~q(s)=(1+(1-q)-1)((q-1)qt;q)∞((q-1)-1q1-t;q)∞((1-q)-s(q-1);q)∞((q-1)-1;q)∞Γq(s).In 2009, Brahim [12] proved the results as follows:
(12)ψq,m(x)ψq,n(x)≥ψq,(m+n)/22(x),
where x>0 and m,n,(m+n)/2∈ℕ. Consider
(13)[s+1]qζq(s)ζq(s+1)≥q[s]qζq(s+1)ζq(s+2),
where s>1.In 2010, Krasniqi et al. [13] gave an inequality as follows:
(14)Γ~q(x/a+y/b)Γ~q1/a(x)Γ~q1/b(y)≤ζq1/a(x)ζq1/b(y)ζq(x/a+y/b),
where x,y,a>1 and x/a+y/b>1=1/a+1/b.In 2012, Sulaiman [10] gave the inequalities as follows:
(15)ψq,m/a+n/b(xa+yb)≤ψq,m1/a(x)ψq,n1/b(y),
where x,y>0, a>1, 1/a+1/b=1, and m,n,x/a+y/b∈ℕ. 						(16)(ψq,m(x)+ψq,n(y))1/p≤ψq,m1/p(x)+ψq,n1/p(y),
where x,y>0, p≥1, and m,n∈ℕ.In this paper, we present the generalizations on the above inequalities.


## 2. Results

Theorem 1.
Letn1,n2,…,nk∈ℕ, x1,x2,…,xk>0, and p1,p2,…,pk>1 be such that ∑i=1k‍1/pi=1 and M=∑i=1k‍ni/pi∈ℕ. Then
(17)ψq,M(∑i=1k‍xipi)≤∏i=1k‍ψq,ni1/pi(xi).Proof.
By the generalized Hölder inequality,(18)ψq,M(∑i=1k‍xipi)=log⁡q1-q∫0q‍(log⁡u)Mu∑i=1kxi/pi-11-udqu=log⁡q1-q∫0q‍(log⁡u)∑i=1kni/piu∑i=1k(xi-1)/pi(1-u)∑i=1k(1/pi)dqu=log⁡q1-q∫0q‍∏i=1k‍((log⁡u)ni/piu(xi-1)/pi(1-u)1/pi)dqu≤log⁡q1-q∏i=1k‍(∫0q‍(log⁡u)niuxi-11-udqu)1/pi=∏i=1k‍ψq,ni1/pi(xi).One can easily check that if we putk=2 in Theorem 1 then we get the following.Corollary 2 (see [10]).
Letn1,n2∈ℕ, x1,x2>0, and p1,p2>1 be such that 1/p1+1/p2=1 and n1/p1+n2/p2∈ℕ. Then
(19)ψq,n1/p1+n2/p2s(x1p1+x2p2)≤ψq,n11/p1(x1)ψq,n21/p2(x2).It is easy to notice that if we putk=p1=p2=2 and x1=x2=x in Theorem 1 then we get the following.Corollary 3 (see [12]).
Letn1,n2∈ℕ and x>0 be such that (n1+n2)/2∈ℕ. Then
(20)ψq,(n1+n2)/22(x)≤ψq,n1(x)ψq,n2(x).Theorem 4.
Letn1,n2,…,nk∈ℕ, x1,x2,…,xk>0, and p≥1. Then
(21)(∑i=1k‍ψq,ni(xi))1/p≤∑i=1k‍ψq,ni1/p(xi).Proof.
First, we note that(22)∑i=1k‍aip≤(∑i=1k‍ai)p
for all a1,a2,…,ak>0. Then
(23)(∑i=1k‍ψq,ni(xi))1/p=(∑i=1k‍(log⁡q1-q∫0q‍(log⁡u)niuxi-11-udqu))1/p=(log⁡q1-q)1/p(∫0q‍∑i=1k‍((log⁡u)ni/pu(xi-1)/p(1-u)1/p)pdqu)1/p≤(log⁡q1-q)1/p(∫0q‍(∑i=1k‍(log⁡u)ni/pu(xi-1)/p(1-u)1/p)pdqu)1/p.
By Minkowski's inequality,
(24)(∑i=1k‍ψq,ni(xi))1/p≤(log⁡q1-q)1/p∑i=1k‍(∫0q‍((log⁡u)ni/pu(xi-1)/p(1-u)1/p)pdqu)1/p=∑i=1k‍(log⁡q1-q∫0q‍((log⁡u)niuxi-11-u)pdqu)1/p=∑i=1k‍ψq,ni1/p(xi).One can easily check that if we putk=2 in Theorem 4 then we get the following.Corollary 5 (see [10]).
Letn1,n2∈ℕ, x1,x2>0, and p≥1. Then
(25)(ψq,n1(x1)+ψq,n2(x2))1/p≤ψq,n11/p(x1)+ψq,n21/p(x2).Theorem 6.
Letx1,x2,…,xk,p1,p2,…,pk>1 be such that ∑i=1k‍xi/pi>1=∑i=1k‍1/pi. Then
(26)ζq(∑i=1k‍xipi)≤∏i=1k‍Γ~q1/pi(xi)Γ~q(∑i=1k‍xi/pi)∏i=1k‍ζq1/pi(xi).Proof.
By the generalized Hölder inequality,(27)ζq(∑i=1k‍xipi)=1Γ~q(∑i=1kxi/pi)∫0∞‍t∑i=1kxi/pi-1Zq(t)dqt=1Γ~q(∑i=1kxi/pi)∫0∞‍t∑i=1k(xi-1)/piZq∑i=1k1/pi(t)dqt=1Γ~q(∑i=1kxi/pi)∫0∞‍∏i=1k‍(txi-1Zq(t))1/pidqt≤1Γ~q(∑i=1kxi/pi)∏i=1k‍(∫0∞‍txi-1Zq(t)dqt)1/pi=1Γ~q(∑i=1kxi/pi)×∏i=1k‍(Γ~q1/pi(xi)(1Γ~q(xi)∫0∞‍txi-1Zq(t)dqt)1/pi)=∏i=1kΓ~q1/pi(xi)Γ~q(∑i=1kxi/pi)∏i=1k‍ζq1/pi(xi).It is easy to notice that if we putk=2 in Theorem 6 then we get the following.Corollary 7 (see [13]).
Letx1,x2,p1,p2>1 be such that x1/p1+x2/p2>1=1/p1+1/p2. Then
(28)Γ~q(x1/p1+x2/p2)Γ~q1/p1(x1)Γ~q1/p2(x2)≤ζq1/p1(x1)ζq1/p2(x2)ζq(x1/p1+x2/p2).



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*Source: 543593-2014-01-22.xml*