Theory and Modern Applications

# q-Bernoulli numbers and q-Bernoulli polynomials revisited

## Abstract

This paper performs a further investigation on the q-Bernoulli numbers and q-Bernoulli polynomials given by Acikgöz et al. (Adv Differ Equ, Article ID 951764, 9, 2010), some incorrect properties are revised. It is point out that the generating function for the q-Bernoulli numbers and polynomials is unreasonable. By using the theorem of Kim (Kyushu J Math 48, 73-86, 1994) (see Equation 9), some new generating functions for the q-Bernoulli numbers and polynomials are shown.

Mathematics Subject Classification (2000) 11B68, 11S40, 11S80

## 1. Introduction

As well-known definition, the Bernoulli polynomials are given by

$\frac{t}{{e}^{t}-1}{e}^{xt}={e}^{B\left(x\right)t}=\sum _{n=0}^{\infty }{B}_{n}\left(x\right)\frac{{t}^{n}}{n!},$

(see ),

with usual convention about replacing B n(x) by B n (x). In the special case, x = 0, B n (0) = B n are called the n th Bernoulli numbers.

Let us assume that q with |q| < 1 as an indeterminate. The q-number is defined by

${\left[x\right]}_{q}=\frac{1-{q}^{x}}{1-q},$

(see ).

Note that lim q→1[x] q = x.

Since Carlitz brought out the concept of the q-extension of Bernoulli numbers and polynomials, many mathematicians have studied q-Bernoulli numbers and q-Bernoulli polynomials (see [1, 7, 5, 6, 812]). Recently, Acikgöz, Erdal, and Araci have studied to a new approach to q-Bernoulli numbers and q-Bernoulli polynomials related to q-Bernstein polynomials (see ). But, their generating function is unreasonable. The wrong properties are indicated by some counter-examples, and they are corrected.

It is point out that Acikgöz, Erdal and Araci's generating function for q-Bernoulli numbers and polynomials is unreasonable by counter examples, then the new generating function for the q-Bernoulli numbers and polynomials are given.

## 2. q-Bernoulli numbers and q-Bernoulli polynomials revisited

In this section, we perform a further investigation on the q-Bernoulli numbers and q-Bernoulli polynomials given by Acikgöz et al. , some incorrect properties are revised.

Definition 1 (Acikgöz et al. ). For q with |q| < 1, let us define q-Bernoulli polynomials as follows:

${D}_{q}\left(t,x\right)=-t\sum _{y=0}^{\infty }{q}^{y}{e}^{{\left[x+y\right]}_{q}t}=\sum _{n=0}^{\infty }{B}_{n,q}\left(x\right)\frac{{t}^{n}}{n!},\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}|t+logq|<2\pi .$
(1)

In the special case, x = 0, B n,q (0) = B n,q are called the n th q-Bernoulli numbers.

Let D q (t, 0) = D q (t). Then

${D}_{q}\left(t\right)=-t\sum _{y=0}^{\infty }{q}^{y}{e}^{{\left[y\right]}_{q}t}=\sum _{n=0}^{\infty }{B}_{n,q}\frac{{t}^{n}}{n!}.$
(2)

Remark 1. Definition 1 is unreasonable, since it is not the generating function of q-Bernoulli numbers and polynomials.

Indeed, by (2), we get

$\begin{array}{lll}\hfill {D}_{q}\left(t,x\right)& =-t\sum _{y=0}^{\infty }{q}^{y}{e}^{{\left[x+y\right]}_{q}t}=-t\sum _{y=0}^{\infty }{q}^{y}{e}^{{\left[x\right]}_{q}t}{e}^{{q}^{x}{\left[y\right]}_{q}t}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\left(-\frac{{q}^{x}t}{{q}^{x}}\sum _{y=0}^{\infty }{q}^{y}{e}^{{q}^{x}{\left[y\right]}_{q}t}\right){e}^{{\left[x\right]}_{q}t}\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =\frac{1}{{q}^{x}}{e}^{{\left[x\right]}_{q}t}{D}_{q}\left({q}^{x}t\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\left(\sum _{m=0}^{\infty }\frac{{\left[x\right]}_{q}^{m}}{m!}{t}^{m}\right)\left(\sum _{l=0}^{\infty }\frac{{q}^{\left(l-1\right)x}{B}_{l,q}}{l!}{t}^{l}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ =\sum _{n=0}^{\infty }\left(\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){\left[x\right]}_{q}^{n-l}{q}^{\left(l-1\right)x}{B}_{l,q}\right)\frac{{t}^{n}}{n!}.\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ \hfill \text{(6)}\end{array}$
(3)

By comparing the coefficients on the both sides of (1) and (3), we obtain the following equation

${B}_{n,q}\left(x\right)=\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){\left[x\right]}_{q}^{n-l}{q}^{\left(l-1\right)x}{B}_{l,q}.$
(4)

From (1), we note that

$\begin{array}{lll}\hfill {D}_{q}\left(t,x\right)& =-t\sum _{y=0}^{\infty }{q}^{y}{e}^{{\left[x+y\right]}_{q}t}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\sum _{n=0}^{\infty }\left(-t\sum _{y=0}^{\infty }{q}^{y}{\left[x+y\right]}_{q}^{n}\right)\frac{{t}^{n}}{n!}\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =-\sum _{n=0}^{\infty }\left(\frac{n+1}{{\left(1-q\right)}^{n}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){\left(-1\right)}^{l}{q}^{lx}\sum _{y=0}^{\infty }{q}^{\left(l+1\right)y}\right)\frac{{t}^{n+1}}{\left(n+1\right)!}\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\sum _{n=1}^{\infty }\left(\frac{-n}{{\left(1-q\right)}^{n-1}}\sum _{l=0}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill l\hfill \end{array}\right){\left(-1\right)}^{l}{q}^{lx}\left(\frac{1}{1-{q}^{l+1}}\right)\right)\frac{{t}^{n}}{n!}.\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$
(5)

By comparing the coefficients on the both sides of (1) and (5), we obtain the following equation

$\begin{array}{c}{B}_{0,q}=0,\\ {B}_{n,q}=\frac{-n}{{\left(1-q\right)}^{n-1}}\sum _{l=0}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill l\hfill \end{array}\right){\left(-1\right)}^{l}{q}^{lx}\left(\frac{1}{1-{q}^{l+1}}\right)\phantom{\rule{1em}{0ex}}if\phantom{\rule{1em}{0ex}}n>0.\end{array}$
(6)

By (6), we see that Definition 1 is unreasonable because we cannot derive Bernoulli numbers from Definition 1 for any q.

In particular, by (1) and (2), we get

$q{D}_{q}\left(t,1\right)-{D}_{q}\left(t\right)=t.$
(7)

Thus, by (7), we have

$q{B}_{n,q}\left(1\right)-{B}_{n,q}=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill if\phantom{\rule{1em}{0ex}}n=1,\hfill \\ \hfill 0,\hfill & \hfill if\phantom{\rule{1em}{0ex}}n>1,\hfill \end{array}\right\$
(8)

and

${B}_{n,q}\left(1\right)=\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){q}^{l-1}{B}_{l,q}.$
(9)

Therefore, by (4) and (6)-(9), we see that the following three theorems are incorrect.

Theorem 1 (Acikgöz et al. ). For n *, one has

${B}_{0,q}=1,\phantom{\rule{1em}{0ex}}q{\left(qB+1\right)}^{n}-{B}_{n,q}=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill if\phantom{\rule{1em}{0ex}}n=0,\hfill \\ \hfill 0,\hfill & \hfill if\phantom{\rule{1em}{0ex}}n>0.\hfill \end{array}\right\$

Theorem 2 (Acikgöz et al. ). For n *, one has

${B}_{n,q}\left(x\right)=\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){q}^{lx}{B}_{l,q}{\left[x\right]}_{q}^{n-l}.$

Theorem 3 (Acikgöz et al. ). For n *, one has

${B}_{n,q}\left(x\right)=\frac{1}{{\left(1-q\right)}^{n}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){\left(-1\right)}^{l}{q}^{lx}\frac{l+1}{{\left[l+1\right]}_{q}}.$

In , Acikgöz, Erdal and Araci derived some results by using Theorems 1-3. Hence, the other results are incorrect.

Now, we redefine the generating function of q-Bernoulli numbers and polynomials and correct its wrong properties, and rebuild the theorems of q-Bernoulli numbers and polynomials.

Redefinition 1. For q with |q| < 1, let us define q-Bernoulli polynomials as follows:

$\begin{array}{lll}\hfill {F}_{q}\left(t,x\right)& =-t\sum _{m=0}^{\infty }{q}^{2m+x}{e}^{{\left[x+m\right]}_{q}t}+\left(1-q\right)\sum _{m=0}^{\infty }{q}^{m}{e}^{{\left[x+m\right]}_{q}t}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\sum _{n=0}^{\infty }{\beta }_{n,q}\left(x\right)\frac{{t}^{n}}{n!},\phantom{\rule{1em}{0ex}}where\phantom{\rule{1em}{0ex}}|t+logq|<2\pi .\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
(10)

In the special case, x = 0, β n,q (0) = β n,q are called the n th q-Bernoulli numbers.

Let F q (t, 0) = F q (t). Then we have

$\begin{array}{lll}\hfill {F}_{q}\left(t\right)& =\sum _{n=0}^{\infty }{\beta }_{n,q}\frac{{t}^{n}}{n!}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =-t\sum _{m=0}^{\infty }{q}^{2m}{e}^{{\left[m\right]}_{q}t}+\left(1-q\right)\sum _{m=0}^{\infty }{q}^{m}{e}^{{\left[m\right]}_{q}t}.\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
(11)

By (10), we get

(12)

By (10) and (11), we get

$\begin{array}{lll}\hfill {F}_{q}\left(t,x\right)& ={e}^{{\left[x\right]}_{q}t}{F}_{q}\left({q}^{x}t\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\left(\sum _{m=0}^{\infty }{\left[x\right]}_{q}^{m}\frac{{t}^{m}}{m!}\right)\left(\sum _{l=0}^{\infty }\frac{{\beta }_{l,q}}{l!}{q}^{lx}{t}^{l}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ =\sum _{n=0}^{\infty }\left(\sum _{l=0}^{n}\frac{{q}^{lx}{\beta }_{l,q}{\left[x\right]}_{q}^{n-l}n!}{l!\left(n-l\right)!}\right)\frac{{t}^{n}}{n!}\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\sum _{n=0}^{\infty }\left(\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){q}^{lx}{\beta }_{l,q}{\left[x\right]}_{q}^{n-l}\right)\frac{{t}^{n}}{n!}.\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \hfill \text{(5)}\end{array}$
(13)

Thus, by (12) and (13), we have

$\begin{array}{lll}\hfill {\beta }_{n,q}\left(x\right)& =\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){q}^{lx}{\beta }_{l,q}{\left[x\right]}_{q}^{n-l}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =-n\sum _{m=0}^{\infty }{q}^{m}{\left[x+m\right]}_{q}^{n-1}+\left(1-q\right)\left(n+1\right)\sum _{m=0}^{\infty }{q}^{m}{\left[x+m\right]}_{q}^{n}.\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
(14)

From (10) and (11), we can derive the following equation:

$q{F}_{q}\left(t,1\right)-{F}_{q}\left(t\right)=t+\left(q-1\right).$
(15)

By (15), we get

$q{\beta }_{n,q}\left(1\right)-{\beta }_{n,q}=\left\{\begin{array}{cc}\hfill q-1,\hfill & \hfill if\phantom{\rule{1em}{0ex}}n=0,\hfill \\ \hfill 1,\hfill & \hfill if\phantom{\rule{1em}{0ex}}n=1,\hfill \\ \hfill 0\hfill & \hfill if\phantom{\rule{1em}{0ex}}n>1.\hfill \end{array}\right\$
(16)

Therefore, by (14) and (15), we obtain

${\beta }_{0,q}=1,q{\left(q{\beta }_{q}+1\right)}^{n}-{\beta }_{n,q}=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill if\phantom{\rule{1em}{0ex}}n=1,\hfill \\ \hfill 0\hfill & \hfill if\phantom{\rule{1em}{0ex}}n>1,\hfill \end{array}\right\$
(17)

with the usual convention about replacing ${\beta }_{q}^{n}$ by β n,q .

From (12), (14) and (16), Theorems 1-3 are revised by the following Theorems 1'-3'.

Theorem 1'. For n +, we have

${\beta }_{0,q}=1,\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}q{\left(q{\beta }_{q}+1\right)}^{n}-{\beta }_{n,q}=\left\{\begin{array}{cc}\hfill 1,\hfill & \hfill if\phantom{\rule{1em}{0ex}}n=1,\hfill \\ \hfill 0\hfill & \hfill if\phantom{\rule{1em}{0ex}}n>1.\hfill \end{array}\right\$

Theorem 2'. For n +, we have

${\beta }_{n,q}\left(x\right)=\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){q}^{lx}{\beta }_{l,q}{\left[x\right]}_{q}^{n-l}.$

Theorem 3'. For n +, we have

${\beta }_{n,q}\left(x\right)=\frac{1}{{\left(1-q\right)}^{n}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){\left(-1\right)}^{l}{q}^{lx}\frac{l+1}{{\left[l+1\right]}_{q}}.$

From (10), we note that

${F}_{q}\left(t,x\right)=\frac{1}{{\left[d\right]}_{q}}\sum _{a=0}^{d-1}{q}^{a}{F}_{{q}^{d}}\left({\left[d\right]}_{q}t,\frac{x+a}{d}\right),\phantom{\rule{1em}{0ex}}d\in ℕ.$
(18)

Thus, by (10) and (18), we have

${\beta }_{n,q}\left(x\right)={\left[d\right]}_{q}^{n-1}\sum _{a=0}^{d-1}{q}^{a}{\beta }_{n,{q}^{d}}\left(\frac{x+a}{d}\right),\phantom{\rule{1em}{0ex}}n\in {ℤ}_{+}.$

For d , let χ be Dirichlet's character with conductor d. Then, we consider the generalized q-Bernoulli polynomials attached to χ as follows:

$\begin{array}{lll}\hfill {F}_{q,\chi }\left(t,x\right)& =-t\sum _{m=0}^{\infty }\chi \left(m\right){q}^{2m+x}{e}^{{\left[x+m\right]}_{q}t}+\left(1-q\right)\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m}{e}^{{\left[x+m\right]}_{q}t}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\sum _{n=0}^{\infty }{\beta }_{n,\chi ,q}\left(x\right)\frac{{t}^{n}}{n!}.\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$

In the special case, x = 0, β n,χ,q (0) = β n,χ,q are called the n th generalized Carlitz q-Bernoulli numbers attached to χ (see ).

Let F q,χ (t, 0) = F q,χ (t). Then we have

$\begin{array}{lll}\hfill {F}_{q,\chi }\left(t\right)& =-t\sum _{m=0}^{\infty }\chi \left(m\right){q}^{2m}{e}^{{\left[m\right]}_{q}t}+\left(1-q\right)\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m}{e}^{{\left[m\right]}_{q}t}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =\sum _{n=0}^{\infty }{\beta }_{n,\chi ,q}\frac{{t}^{n}}{n!}.\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$
(20)

From (20), we note that

$\begin{array}{lll}\hfill {\beta }_{n,\chi ,q}& =-n\sum _{m=0}^{\infty }{q}^{2m}\chi \left(m\right){\left[m\right]}_{q}^{n-1}+\left(1-q\right)\sum _{m=0}^{\infty }{q}^{m}\chi \left(m\right){\left[m\right]}_{q}^{n}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =-n\sum _{a=0}^{d-1}\sum _{m=0}^{\infty }{q}^{2a+2dm}\chi \left(a+dm\right){\left[a+dm\right]}_{q}^{n-1}\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \phantom{\rule{1em}{0ex}}+\sum _{a=0}^{d-1}\sum _{m=0}^{\infty }{q}^{a+dm}\chi \left(a+dm\right){\left[a+dm\right]}_{q}^{n}\phantom{\rule{2em}{0ex}}& \hfill \text{(3)}\\ =\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}\left(\frac{-n}{{\left(1-q\right)}^{n-1}}\sum _{l=0}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill l\hfill \end{array}\right)\frac{{\left(-1\right)}^{l}{q}^{\left(l+1\right)a}}{\left(1-{q}^{d\left(l+2\right)}\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(4)}\\ \phantom{\rule{1em}{0ex}}+\left(1-q\right)\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}\left(\frac{1}{{\left(1-q\right)}^{n}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right)\frac{{\left(-1\right)}^{l}{q}^{la}}{\left(1-{q}^{d\left(l+1\right)}\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(5)}\\ =\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}\left(\frac{-n}{{\left(1-q\right)}^{n-1}}\sum _{l=0}^{n-1}\left(\begin{array}{c}\hfill n-1\hfill \\ \hfill l\hfill \end{array}\right)\frac{{\left(-1\right)}^{l}{q}^{\left(l+1\right)a}}{\left(1-{q}^{d\left(l+2\right)}\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(6)}\\ \phantom{\rule{1em}{0ex}}+\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}\left(\frac{1}{{\left(1-q\right)}^{n-1}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right)\frac{{\left(-1\right)}^{l}{q}^{la}}{\left(1-{q}^{d\left(l+1\right)}\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(7)}\\ =\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}\left(\frac{1}{{\left(1-q\right)}^{n-1}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right)\frac{{\left(-1\right)}^{l}{q}^{la}l}{\left(1-{q}^{d\left(l+1\right)}\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(8)}\\ \phantom{\rule{1em}{0ex}}+\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}\left(\frac{1}{{\left(1-q\right)}^{n-1}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right)\frac{{\left(-1\right)}^{l}{q}^{la}}{\left(1-{q}^{d\left(l+1\right)}\right)}\right)\phantom{\rule{2em}{0ex}}& \hfill \text{(9)}\\ =\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}\frac{1-q}{{\left(1-q\right)}^{n}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){\left(-1\right)}^{l}{q}^{la}\left(\frac{l+1}{1-{q}^{d\left(l+1\right)}}\right).\phantom{\rule{2em}{0ex}}& \hfill \text{(10)}\\ \hfill \text{(11)}\end{array}$

Therefore, by (20) and (21), we obtain the following theorem.

Theorem 4. For n +, we have

$\begin{array}{lll}\hfill {\beta }_{n,\chi ,q}& =\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}\frac{1}{{\left(1-q\right)}^{n}}\sum _{l=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill l\hfill \end{array}\right){\left(-1\right)}^{l}{q}^{la}\frac{l+1}{{\left[d\left(l+1\right)\right]}_{q}}\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\\ =-n\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m}{\left[m\right]}_{q}^{n-1}+\left(1-q\right)\left(1+n\right)\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m}{\left[m\right]}_{q}^{n},\phantom{\rule{2em}{0ex}}& \hfill \text{(2)}\\ \hfill \text{(3)}\end{array}$

and

${\beta }_{n,\chi ,q}\left(x\right)=-n\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m}{\left[m+x\right]}_{q}^{n-1}+\left(1-q\right)\left(1+n\right)\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m}{\left[m+x\right]}_{q}^{n}.$

From (19), we note that

${F}_{q,\chi }\left(t,x\right)=\frac{1}{{\left[d\right]}_{q}}\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}{F}_{{q}^{d}}\left({\left[d\right]}_{q}t,\frac{x+a}{d}\right).$
(22)

Thus, by (22), we obtain the following theorem.

Theorem 5. For n +, we have

${\beta }_{n,\chi ,q}\left(x\right)={\left[d\right]}_{q}^{n-1}\sum _{a=0}^{d-1}\chi \left(a\right){q}^{a}{\beta }_{n,{q}^{d}}\left(\frac{x+a}{d}\right).$

For s , we now consider the Mellin transform for F q (t, x) as follows:

$\frac{1}{\Gamma \left(s\right)}\underset{0}{\overset{\infty }{\int }}{F}_{q}\left(-t,x\right){t}^{s-2}dt=\sum _{m=0}^{\infty }\frac{{q}^{2m+x}}{{\left[m+x\right]}_{q}^{s}}+\frac{1-q}{s-1}\sum _{m=0}^{\infty }\frac{{q}^{m}}{{\left[m+x\right]}_{q}^{s-1}},$
(23)

where x ≠ 0, -1, -2,....

From (23), we note that

$\begin{array}{c}\frac{1}{\Gamma \left(s\right)}\underset{0}{\overset{\infty }{\int }}{F}_{q}\left(-t,x\right){t}^{s-2}dt\\ \phantom{\rule{1em}{0ex}}=\sum _{m=0}^{\infty }\frac{{q}^{m}}{{\left[m+x\right]}_{q}^{s}}+\left(1-q\right)\left(\frac{2-s}{s-1}\right)\sum _{m=0}^{\infty }\frac{{q}^{m}}{{\left[m+x\right]}_{q}^{s-1}},\end{array}$
(24)

where s , and x ≠ 0, -1, -2,....

Thus, we define q-zeta function as follows:

Definition 2. For s , q-zeta function is defined by

${\zeta }_{q}\left(s,x\right)=\sum _{m=0}^{\infty }\frac{{q}^{m}}{{\left[m+x\right]}_{q}^{s}}+\left(1-q\right)\left(\frac{2-s}{s-1}\right)\sum _{m=0}^{\infty }\frac{{q}^{m}}{{\left[m+x\right]}_{q}^{s-1}},\phantom{\rule{1em}{0ex}}Re\left(s\right)>1,$

where x ≠ 0, -1, -2,....

By (24) and Definition 2, we note that

${\zeta }_{q}\left(1-n,x\right)={\left(-1\right)}^{n-1}\frac{{\beta }_{n,q}\left(x\right)}{n},\phantom{\rule{1em}{0ex}}n\in ℕ.$

Note that

$\underset{q\to 1}{lim}{\zeta }_{q}\left(1-n,x\right)=-\frac{{B}_{n}\left(x\right)}{n},$

where B n (x) are the n th ordinary Bernoulli polynomials.

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## Acknowledgements

The authors express their gratitude to the referee for his/her valuable comments.

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Correspondence to Cheon Seoung Ryoo.

### Competing interests

The authors declare that they have no competing interests.

### Authors' contributions

All authors contributed equally to the manuscript and read and approved the finial manuscript.

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Ryoo, C.S., Kim, T. & Lee, B. q-Bernoulli numbers and q-Bernoulli polynomials revisited. Adv Differ Equ 2011, 33 (2011). https://doi.org/10.1186/1687-1847-2011-33 