Theory and Modern Applications

# Some results for the q-Bernoulli, q-Euler numbers and polynomials

## Abstract

The q-analogues of many well known formulas are derived by using several results of q-Bernoulli, q-Euler numbers and polynomials. The q-analogues of ζ-type functions are given by using generating functions of q-Bernoulli, q-Euler numbers and polynomials. Finally, their values at non-positive integers are also been computed.

2010 Mathematics Subject Classification: 11B68; 11S40; 11S80.

## 1. Introduction

Carlitz [1, 2] introduced q-analogues of the Bernoulli numbers and polynomials. From that time on these and other related subjects have been studied by various authors (see, e.g., [310]). Many recent studies on q-analogue of the Bernoulli, Euler numbers, and polynomials can be found in Choi et al. [11], Kamano [3], Kim [5, 6, 12], Luo [7], Satoh [9], Simsek [13, 14] and Tsumura [10].

For a fixed prime p, p , p , and p denote the ring of p-adic integers, the field of p-adic numbers, and the completion of the algebraic closure of p , respectively. Let | · | p be the p-adic norm on with |p| p = p -1. For convenience, | · | p will also be used to denote the extended valuation on p .

The Bernoulli polynomials, denoted by B n (x), are defined as

${B}_{n}\left(x\right)=\sum _{k=0}^{n}\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right){B}_{k}{x}^{n-k},\phantom{\rule{1em}{0ex}}n\ge 0,$
(1.1)

where B k are the Bernoulli numbers given by the coefficients in the power series

$\frac{t}{{e}^{t}-1}=\sum _{k=0}^{\infty }{B}_{k}\frac{{t}^{k}}{k!}.$
(1.2)

From the above definition, we see B k 's are all rational numbers. Since $\frac{t}{{e}^{t}-1}-1+\frac{t}{2}$ is an even function (i.e., invariant under x - x), we see that B k = 0 for any odd integer k not smaller than 3. It is well known that the Bernoulli numbers can also be expressed as follows

${B}_{k}=\underset{N\to \infty }{lim}\frac{1}{{p}^{N}}\sum _{a=0}^{{p}^{N}-1}{a}^{k}$
(1.3)

(see [15, 16]). Notice that, from the definition B k , and these integrals are independent of the prime p which used to compute them. The examples of (1.3) are:

$\begin{array}{c}\underset{N\to \infty }{lim}\frac{1}{{p}^{N}}\sum _{a=0}^{{p}^{N}-1}a=\underset{N\to \infty }{lim}\frac{1}{{p}^{N}}\frac{{p}^{N}\left({p}^{N}-1\right)}{2}=-\frac{1}{2}={B}_{1},\\ \underset{N\to \infty }{lim}\frac{1}{{p}^{N}}\sum _{a=0}^{{p}^{N}-1}{a}^{2}=\underset{N\to \infty }{lim}\frac{1}{{p}^{N}}\frac{{p}^{N}\left({p}^{N}-1\right)\left(2{p}^{N}-1\right)}{6}=\frac{1}{6}={B}_{2}.\end{array}$
(1.4)

Euler numbers E k , k ≥ 0 are integers given by (cf. [1719])

(1.5)

The Euler polynomial E k (x) is defined by (see [[20], p. 25]):

${E}_{k}\left(x\right)=\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\frac{{E}_{i}}{{2}^{i}}{\left(x-\frac{1}{2}\right)}^{k-i},$
(1.6)

which holds for all nonnegative integers k and all real x, and which was obtained by Raabe [21] in 1851. Setting x = 1/2 and normalizing by 2 k gives the Euler numbers

${E}_{k}={2}^{k}{E}_{k}\left(\frac{1}{2}\right),$
(1.7)

where E 0 = 1, E 2 = -1, E 4 = 5, E 6 = -61,.... Therefore, E k E k (0), in fact ([[19], p. 374 (2.1)])

${E}_{k}\left(0\right)=\frac{2}{k+1}\left(1-{2}^{k+1}\right){B}_{k+1},$
(1.8)

where B k are Bernoulli numbers. The Euler numbers and polynomials (so-named by Scherk in 1825) appear in Euler's famous book, Institutiones Calculi Differentialis (1755, pp. 487-491 and p. 522).

In this article, we derive q-analogues of many well known formulas by using several results of q-Bernoulli, q-Euler numbers, and polynomials. By using generating functions of q-Bernoulli, q-Euler numbers, and polynomials, we also present the q-analogues of ζ-type functions. Finally, we compute their values at non-positive integers.

In Section 2, we recall definitions and some properties for the q-Bernoulli, Euler numbers, and polynomials related to the bosonic and the fermionic p-adic integral on p .

In Section 3, we obtain the generating functions of the q-Bernoulli, q-Euler numbers, and polynomials. We shall provide some basic formulas for the q-Bernoulli and q-Euler polynomials which will be used to prove the main results of this article.

In Section 4, we construct the q-analogue of the Riemann's ζ-functions, the Hurwitz ζ-functions, and the Dirichlet's L-functions. We prove that the value of their functions at non-positive integers can be represented by the q-Bernoulli, q-Euler numbers, and polynomials.

## 2. q-Bernoulli, q-Euler numbers and polynomials related to the Bosonic and the Fermionic p-adic integral on ℤ p

In this section, we provide some basic formulas for p-adic q-Bernoulli, p-adic q-Euler numbers and polynomials which will be used to prove the main results of this article.

Let UD( p , p ) denote the space of all uniformly (or strictly) differentiable p -valued functions on p . The p-adic q-integral of a function f UD( p ) on p is defined by

${I}_{q}\left(f\right)=\underset{N\to \infty }{lim}\frac{1}{{\left[{p}^{N}\right]}_{q}}\sum _{a=0}^{{p}^{N}-1}f\left(a\right){q}^{a}={\int }_{{ℤ}_{p}}f\left(z\right)d{\mu }_{q}\left(z\right),$
(2.1)

where [x] q = (1 - q x )/(1 - q), and the limit taken in the p-adic sense. Note that

$\underset{q\to 1}{lim}{\left[x\right]}_{q}=x$
(2.2)

for x p , where q tends to 1 in the region 0 < |q - 1| p < 1 (cf. [22, 5, 12]). The bosonic p-adic integral on p is considered as the limit q → 1, i.e.,

${I}_{1}\left(f\right)=\underset{N\to \infty }{lim}\frac{1}{{p}^{N}}\sum _{a=0}^{{p}^{N}-1}f\left(a\right)={\int }_{{ℤ}_{p}}f\left(z\right)d{\mu }_{1}\left(z\right).$
(2.3)

From (2.1), we have the fermionic p-adic integral on p as follows:

${I}_{-1}\left(f\right)=\underset{q\to -1}{lim}{I}_{q}\left(f\right)=\underset{N\to \infty }{lim}\sum _{a=0}^{{p}^{N}-1}f\left(a\right){\left(-1\right)}^{a}={\int }_{{ℤ}_{p}}f\left(z\right)d{\mu }_{-1}\left(z\right).$
(2.4)

In particular, setting $f\left(z\right)={\left[z\right]}_{q}^{k}$ in (2.3) and $f\left(z\right)={\left[z+\frac{1}{2}\right]}_{q}^{k}$ in (2.4), respectively, we get the following formulas for the p-adic q-Bernoulli and p-adic q-Euler numbers, respectively, if q p with 0 < |q - 1| p < 1 as follows

${B}_{k}\left(q\right)={\int }_{{ℤ}_{p}}{\left[z\right]}_{q}^{k}d{\mu }_{1}\left(z\right)=\underset{N\to \infty }{lim}\frac{1}{{p}^{N}}\sum _{a=0}^{{p}^{N}-1}{\left[a\right]}_{q}^{k},$
(2.5)
${E}_{k}\left(q\right)={2}^{k}{\int }_{{ℤ}_{p}}{\left[z+\frac{1}{2}\right]}_{q}^{k}d{\mu }_{-1}\left(z\right)={2}^{k}\underset{N\to \infty }{lim}\sum _{a=0}^{{p}^{N}-1}{\left[a+\frac{1}{2}\right]}_{q}^{k}{\left(-1\right)}^{a}.$
(2.6)

Remark 2.1. The q-Bernoulli numbers (2.5) are first defined by Kamano [3]. In (2.5) and (2.6), take q → 1. Form (2.2), it is easy to that (see [[17], Theorem 2.5])

${B}_{k}\left(q\right)\to {B}_{k}={\int }_{{ℤ}_{p}}{z}^{k}d{\mu }_{1}\left(z\right),\phantom{\rule{1em}{0ex}}{E}_{k}\left(q\right)\to {E}_{k}={\int }_{{ℤ}_{p}}{\left(2z+1\right)}^{k}d{\mu }_{-1}\left(z\right).$

For |q - 1| p < 1 and z p , we have

${q}^{iz}=\sum _{n=0}^{\infty }{\left({q}^{i}-1\right)}^{n}\left(\begin{array}{c}\hfill z\hfill \\ \hfill n\hfill \end{array}\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}|{q}^{i}-1{|}_{p}\le \phantom{\rule{2.77695pt}{0ex}}|q-1{|}_{p}<1,$
(2.7)

where i . We easily see that if |q - 1| p < 1, then q x = 1 for x ≠ 0 if and only if q is a root of unity of order p N and x p N p (see [16]).

By (2.3) and (2.7), we obtain

$\begin{array}{c}{I}_{1}\left({q}^{iz}\right)=\frac{1}{{q}^{i}-1}\underset{N\to \infty }{\mathrm{lim}}\frac{{\left({q}^{i}\right)}^{{p}^{N}}-1}{{p}^{N}}\\ =\frac{1}{{q}^{i}-1}\underset{N\to \infty }{\mathrm{lim}}\frac{1}{{p}^{N}}\left\{\sum _{m=0}^{\infty }\left(\begin{array}{c}{p}^{N}\\ m\end{array}\right){\left({q}^{i}-1\right)}^{m}-1\right\}\\ =\frac{1}{{q}^{i}-1}\underset{N\to \infty }{\mathrm{lim}}\frac{1}{{p}^{N}}\sum _{m=1}^{\infty }\left(\begin{array}{c}{p}^{N}\\ m\end{array}\right){\left(}^{{q}^{i}-1\right)m}\\ =\frac{1}{{q}^{i}-1}\underset{N\to \infty }{\mathrm{lim}}\sum _{m=1}^{\infty }\frac{1}{m}\left(\begin{array}{c}{p}^{N}-1\\ m-1\end{array}\right){\left(}^{{q}^{i}-1\right)m}\\ =\frac{1}{{q}^{i}-1}\sum _{m=1}^{\infty }\frac{1}{m}\left(\begin{array}{c}-1\\ m-1\end{array}\right){\left(}^{{q}^{i}-1\right)m}\\ =\frac{1}{{q}^{i}-1}\sum _{m=1}^{\infty }{\left(-1\right)}^{m-1}\frac{{\left({q}^{i}-1\right)}^{m}}{m}\\ =\frac{i\mathrm{log}q}{{q}^{i}-1}\end{array}$
(2.8)

since the series log $log\left(1+x\right)={\sum }_{m=1}^{\infty }{\left(-1\right)}^{m-1}{x}^{m}∕m$ converges at |x| p < 1. Similarly, by (2.4), we obtain (see [[4], p. 4, (2.10)])

${I}_{-1}\left({q}^{iz}\right)=\underset{N\to \infty }{lim}\sum _{a=0}^{{p}^{N}-1}{\left({q}^{i}\right)}^{a}{\left(-1\right)}^{a}=\frac{2}{{q}^{i}+1}.$
(2.9)

From (2.5), (2.6), (2.8) and (2.9), we obtain the following explicit formulas of B k (q) and E k (q):

${B}_{k}\left(q\right)=\frac{logq}{{\left(1-q\right)}^{k}}\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right){\left(-1\right)}^{i}\frac{i}{{q}^{i}-1},$
(2.10)
${E}_{k}\left(q\right)=\frac{{2}^{k+1}}{{\left(1-q\right)}^{k}}\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right){\left(-1\right)}^{i}{q}^{\frac{1}{2}i}\frac{1}{{q}^{i}+1},$
(2.11)

where k ≥ 0 and log is the p-adic logarithm. Note that in (2.10), the term with i = 0 is understood to be 1/log q (the limiting value of the summand in the limit i → 0).

We now move on to the p-adic q-Bernoulli and p-adic q-Euler polynomials. The p-adic q-Bernoulli and p-adic q-Euler polynomials in q x are defined by means of the bosonic and the fermionic p-adic integral on p :

(2.12)

where q p with 0 < |q - 1| p < 1 and x p , respectively. We will rewrite the above equations in a slightly different way. By (2.5), (2.6), and (2.12), after some elementary calculations, we get

${B}_{k}\left(x,q\right)=\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right){\left[x\right]}_{q}^{k-i}{q}^{ix}{B}_{i}\left(q\right)={\left({q}^{x}B\left(q\right)+{\left[x\right]}_{q}\right)}^{k}$
(2.13)

and

${E}_{k}\left(x,q\right)=\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\frac{{E}_{i}\left(q\right)}{{2}^{i}}{\left[x-\frac{1}{2}\right]}_{q}^{k-i}{q}^{i\left(x-\frac{1}{2}\right)}={\left(\frac{{q}^{x-\frac{1}{2}}}{2}E\left(q\right)+{\left[x-\frac{1}{2}\right]}_{q}\right)}^{k},$
(2.14)

where the symbol B k (q) and E k (q) are interpreted to mean that (B(q)) k and (E(q)) k must be replaced by B k (q) and E k (q) when we expanded the one on the right, respectively, since ${\left[x+y\right]}_{q}^{k}={\left({\left[x\right]}_{q}+{q}^{x}{\left[y\right]}_{q}\right)}^{k}$ and

$\begin{array}{ll}\hfill {\left[x+z\right]}_{q}^{k}& ={\left[\frac{1}{2}\right]}_{q}^{k}{\left({\left[2x-1\right]}_{{q}^{\frac{1}{2}}}+{q}^{x-\frac{1}{2}}{\left[\frac{1}{2}\right]}_{q}^{-1}{\left[z+\frac{1}{2}\right]}_{q}\right)}^{k}\phantom{\rule{2em}{0ex}}\\ ={\left[\frac{1}{2}\right]}_{q}^{k}\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right){\left[2x-1\right]}_{q}^{k-i}{q}^{\left(x-\frac{1}{2}\right)i}{\left[\frac{1}{2}\right]}_{q}^{-i}{\left[z+\frac{1}{2}\right]}_{q}^{i}\phantom{\rule{2em}{0ex}}\end{array}$
(2.15)

(cf. [4, 5]). The above formulas can be found in [7] which are the q-analogues of the corresponding classical formulas in [[17], (1.2)] and [23], etc. Obviously, put $x=\frac{1}{2}$ in (2.14). Then

(2.16)

where E k are Euler numbers (see (1.5) above).

$\begin{array}{c}{B}_{k}\left(x+y,q\right)=\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{q}^{iy}{B}_{i}\left(x,q\right){\left[y\right]}_{q}^{k-i}\phantom{\rule{1em}{0ex}}\left(k\ge 0\right),\\ {E}_{k}\left(x+y,q\right)=\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{q}^{iy}{E}_{i}\left(x,q\right){\left[y\right]}_{q}^{k-i}\phantom{\rule{1em}{0ex}}\left(k\ge 0\right).\end{array}$

Proof. Applying the relationship ${\left[x+y-\frac{1}{2}\right]}_{q}={\left[y\right]}_{q}+{q}^{y}{\left[x-\frac{1}{2}\right]}_{q}$ to (2.14) for x α x + y, we have

$\begin{array}{ll}\hfill {E}_{k}\left(x+y,q\right)& ={\left(\frac{{q}^{x+y-\frac{1}{2}}}{2}E\left(q\right)+{\left[x+y-\frac{1}{2}\right]}_{q}\right)}^{k}\phantom{\rule{2em}{0ex}}\\ ={\left({q}^{y}\left(\frac{{q}^{x-\frac{1}{2}}}{2}E\left(q\right)+{\left[x-\frac{1}{2}\right]}_{q}\right)+{\left[y\right]}_{q}\right)}^{k}\phantom{\rule{2em}{0ex}}\\ =\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{q}^{iy}{\left(\frac{{q}^{x-\frac{1}{2}}}{2}E\left(q\right)+{\left[x-\frac{1}{2}\right]}_{q}\right)}^{i}{\left[y\right]}_{q}^{k-i}\phantom{\rule{2em}{0ex}}\\ =\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{q}^{iy}{E}_{i}\left(x,q\right){\left[y\right]}_{q}^{k-i}.\phantom{\rule{2em}{0ex}}\end{array}$

Similarly, the first identity follows.□

Remark 2.3. From (2.12), we obtain the not completely trivial identities

$\begin{array}{c}\underset{q\to 1}{lim}{B}_{k}\left(x+y,q\right)=\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right){B}_{i}\left(x\right){y}^{k-i}={\left(B\left(x\right)+y\right)}^{k},\\ \underset{q\to 1}{lim}{E}_{k}\left(x+y,q\right)=\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right){E}_{i}\left(x\right){y}^{k-i}={\left(E\left(x\right)+y\right)}^{k},\end{array}$

where q p tends to 1 in |q - 1| p < 1. Here B i (x) and E i (x) denote the classical Bernoulli and Euler polynomials, see [17, 15] and see also the references cited in each of these earlier works.

Lemma 2.4. Let n be any positive integer. Then

$\begin{array}{c}\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{q}^{i}{\left[n\right]}_{q}^{i}{B}_{i}\left(x,{q}^{n}\right)={\left[n\right]}_{q}^{k}{B}_{k}\left(x+\frac{1}{n},{q}^{n}\right),\\ \sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{q}^{i}{\left[n\right]}_{q}^{i}{E}_{i}\left(x,{q}^{n}\right)={\left[n\right]}_{q}^{k}{E}_{k}\left(x+\frac{1}{n},{q}^{n}\right).\end{array}$

Proof. Use Lemma 2.2, the proof can be obtained by the similar way to [[7], Lemma 2.3]. □

We note here that similar expressions to those of Lemma 2.4 are given by Luo [[7], Lemma 2.3]. Obviously, Lemma 2.4 are the q-analogues of

$\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{n}^{i}{B}_{i}\left(x\right)={n}^{k}{B}_{k}\left(x+\frac{1}{n}\right),\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{n}^{i}{E}_{i}\left(x\right)={n}^{k}{E}_{k}\left(x+\frac{1}{n}\right),$

respectively.

We can now obtain the multiplication formulas by using p-adic integrals.

From (2.3), we see that

$\begin{array}{ll}\hfill {B}_{k}\left(nx,q\right)& ={\int }_{{ℤ}_{p}}{\left[nx+z\right]}_{q}^{k}d{\mu }_{1}\left(z\right)\phantom{\rule{2em}{0ex}}\\ =\underset{N\to \infty }{lim}\frac{1}{n{p}^{N}}\sum _{a=0}^{n{p}^{N}-1}{\left[nx+a\right]}_{q}^{k}\phantom{\rule{2em}{0ex}}\\ =\frac{1}{n}\underset{N\to \infty }{lim}\frac{1}{{p}^{N}}\sum _{i=0}^{n-1}\sum _{a=0}^{{p}^{N}-1}{\left[nx+na+i\right]}_{q}^{k}\phantom{\rule{2em}{0ex}}\\ =\frac{{\left[n\right]}_{q}^{k}}{n}\sum _{i=0}^{n-1}{\int }_{{ℤ}_{p}}{\left[x+\frac{i}{n}+z\right]}_{{q}^{n}}^{k}d{\mu }_{1}\left(z\right)\phantom{\rule{2em}{0ex}}\end{array}$
(2.17)

is equivalent to

${B}_{k}\left(x,q\right)=\frac{{\left[n\right]}_{q}^{k}}{n}\sum _{i=0}^{n-1}{B}_{k}\left(\frac{x+i}{n},{q}^{n}\right).$
(2.18)

If we put x = 0 in (2.18) and use (2.13), we find easily that

$\begin{array}{ll}\hfill {B}_{k}\left(q\right)& =\frac{{\left[n\right]}_{q}^{k}}{n}\sum _{i=0}^{n-1}{B}_{k}\left(\frac{i}{n},{q}^{n}\right)\phantom{\rule{2em}{0ex}}\\ =\frac{{\left[n\right]}_{q}^{k}}{n}\sum _{i=0}^{n-1}\sum _{j=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill j\hfill \end{array}\right){\left[\frac{i}{n}\right]}_{{q}^{n}}^{k-j}{q}^{ij}{B}_{j}\left({q}^{n}\right)\phantom{\rule{2em}{0ex}}\\ =\frac{1}{n}\sum _{j=0}^{k}{\left[n\right]}_{q}^{j}\left(\begin{array}{c}\hfill k\hfill \\ \hfill j\hfill \end{array}\right){B}_{j}\left({q}^{n}\right)\sum _{i=0}^{n-1}{q}^{ij}{\left[i\right]}_{q}^{k-j}.\phantom{\rule{2em}{0ex}}\end{array}$
(2.19)

Obviously, Equation (2.19) is the q-analogue of

${B}_{k}=\frac{1}{n\left(1-{n}^{k}\right)}\sum _{j=0}^{k-1}{n}^{j}\left(\begin{array}{c}\hfill k\hfill \\ \hfill j\hfill \end{array}\right){B}_{j}\sum _{i=1}^{n-1}{i}^{k-j},$

which is true for any positive integer k and any positive integer n > 1 (see [[24], (2)]).

From (2.4), we see that

$\begin{array}{c}{E}_{k}\left(nx,q\right)={\int }_{{ℤ}_{p}}{\left[nx+z\right]}_{q}^{k}d{\mu }_{-1}\left(z\right)\\ =\underset{N\to \infty }{\mathrm{lim}}\sum _{i=0}^{n-1}\sum _{a=0}^{{p}^{N}-1}{\left[nx+na+i\right]}_{q}^{k}{\left(-1\right)}^{na+i}\\ ={\left[n\right]}_{q}^{k}\sum _{i=0}^{n-1}{\left(-1\right)}^{i}{\int }_{{ℤ}_{p}}{\left[x+\frac{i}{n}+z\right]}_{{q}^{n}}^{k}d{\mu }_{{\left(-1\right)}^{n}}\left(z\right).\end{array}$
(2.20)

By (2.12) and (2.20), we find easily that

(2.21)

From (2.18) and (2.21), we can obtain Proposition 2.5 below.

Proposition 2.5 (Multiplication formulas). Let n be any positive integer. Then

$\begin{array}{l}{B}_{k}\left(x,q\right)=\frac{{\left[n\right]}_{q}^{k}}{n}\sum _{i=0}^{n-1}{B}_{k}\left(\frac{x+i}{n},{q}^{n}\right),\\ {E}_{k}\left(x,q\right)={\left[n\right]}_{q}^{k}\sum _{i=0}^{n-1}{\left(-1\right)}^{i}{E}_{k}\left(\frac{x+i}{n},{q}^{n}\right)\phantom{\rule{0.5em}{0ex}}\text{if}\phantom{\rule{0.5em}{0ex}}n\phantom{\rule{0.5em}{0ex}}odd.\end{array}$

## 3. Construction generating functions of q-Bernoulli, q-Euler numbers, and polynomials

In the complex case, we shall explicitly determine the generating function F q (t) of q-Bernoulli numbers and the generating function G q (t) of q-Euler numbers:

(3.1)

where the symbol B k (q) and E k (q) are interpreted to mean that (B(q)) k and (E(q)) k must be replaced by B k (q) and E k (q) when we expanded the one on the right, respectively.

Lemma 3.1.

$\begin{array}{l}{F}_{q}\left(t\right)={e}^{\frac{t}{1-q}}+\frac{t\mathrm{log}q}{1-q}\sum _{m=0}^{\infty }{q}^{m}{e}^{{\left[m\right]}_{q}t},\\ {G}_{q}\left(t\right)=2\sum _{m=0}^{\infty }{\left(-1\right)}^{m}{e}^{2{\left[m+\frac{1}{2}\right]}_{q}t}.\end{array}$

Proof. Combining (2.10) and (3.1), F q (t) may be written as

$\begin{array}{ll}\hfill {F}_{q}\left(t\right)& =\sum _{k=0}^{\infty }\frac{logq}{{\left(1-q\right)}^{k}}\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right){\left(-1\right)}^{i}\frac{i}{{q}^{i}-1}\frac{{t}^{k}}{k!}\phantom{\rule{2em}{0ex}}\\ =1+logq\sum _{k=1}^{\infty }\frac{1}{{\left(1-q\right)}^{k}}\frac{{t}^{k}}{k!}\left(\frac{1}{logq}+\sum _{i=1}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right){\left(-1\right)}^{i}\frac{i}{{q}^{i}-1}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Here, the term with i = 0 is understood to be 1/log q (the limiting value of the summand in the limit i → 0). Specifically, by making use of the following well-known binomial identity

$k\left(\begin{array}{c}\hfill k-1\hfill \\ \hfill i-1\hfill \end{array}\right)=i\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{1em}{0ex}}\left(k\ge i\ge 1\right).$

Thus, we find that

$\begin{array}{ll}\hfill {F}_{q}\left(t\right)& =1+logq\sum _{k=1}^{\infty }\frac{1}{{\left(1-q\right)}^{k}}\frac{{t}^{k}}{k!}\left(\frac{1}{logq}+k\sum _{i=1}^{k}\left(\begin{array}{c}\hfill k-1\hfill \\ \hfill i-1\hfill \end{array}\right){\left(-1\right)}^{i}\frac{1}{{q}^{i}-1}\right)\phantom{\rule{2em}{0ex}}\\ =\sum _{k=0}^{\infty }\frac{1}{{\left(1-q\right)}^{k}}\frac{{t}^{k}}{k!}+logq\sum _{k=1}^{\infty }\frac{k}{{\left(1-q\right)}^{k}}\frac{{t}^{k}}{k!}\sum _{m=0}^{\infty }{q}^{m}\sum _{i=0}^{k-1}\left(\begin{array}{c}\hfill k-1\hfill \\ \hfill i\hfill \end{array}\right){\left(-1\right)}^{i}{q}^{mi}\phantom{\rule{2em}{0ex}}\\ ={e}^{\frac{t}{1-q}}+\frac{logq}{1-q}\sum _{k=1}^{\infty }\frac{k}{{\left(1-q\right)}^{k-1}}\frac{{t}^{k}}{k!}\sum _{m=0}^{\infty }{q}^{m}{\left(1-{q}^{m}\right)}^{k-1}\phantom{\rule{2em}{0ex}}\\ ={e}^{\frac{t}{1-q}}+\frac{tlogq}{1-q}\sum _{m=0}^{\infty }{q}^{m}\sum _{k=0}^{\infty }{\left(\frac{1-{q}^{m}}{1-q}\right)}^{k}\frac{{t}^{k}}{k!}.\phantom{\rule{2em}{0ex}}\end{array}$

Next, by (2.11) and (3.1), we obtain the result

$\begin{array}{c}{G}_{q}\left(t\right)=\sum _{k=0}^{\infty }\frac{{2}^{k+1}}{{\left(1-q\right)}^{k}}\sum _{i=0}^{k}\left(\begin{array}{c}k\\ i\end{array}\right){\left(-1\right)}^{i}{q}^{\frac{1}{2}i}\frac{1}{{q}^{i}+1}\frac{{t}^{k}}{k!}\\ =2\sum _{k=0}^{\infty }{2}^{k}\sum _{m=0}^{\infty }{\left(-1\right)}^{m}{\left(\frac{1-{q}^{m+\frac{1}{2}}}{1-q}\right)}^{k}\frac{{t}^{k}}{k!}\\ =2\sum _{m=0}^{\infty }{\left(-1\right)}^{m}{\sum _{k=0}^{\infty }\left[m+\frac{1}{2}\right]}_{q}^{k}\frac{{\left(2t\right)}^{k}}{k!}\\ =2\sum _{m=0}^{\infty }{\left(-1\right)}^{m}{e}^{2{\left[m+\frac{1}{2}\right]}_{q}t}.\end{array}$

This completes the proof. □

Remark 3.2. The remarkable point is that the series on the right-hand side of Lemma 3.1 is uniformly convergent in the wider sense.

From (2.13)and (2.14), we define the q-Bernoulli and q-Euler polynomials by

${F}_{q}\left(t,x\right)=\sum _{k=0}^{\infty }{B}_{k}\left(x,q\right)\frac{{t}^{k}}{k!}=\sum _{k=0}^{\infty }{\left({q}^{x}B\left(q\right)+{\left[x\right]}_{q}\right)}^{k}\frac{{t}^{k}}{k!},$
(3.2)
${G}_{q}\left(t,x\right)=\sum _{k=0}^{\infty }{E}_{k}\left(x,q\right)\frac{{t}^{k}}{k!}=\sum _{k=0}^{\infty }{\left({q}^{x-\frac{1}{2}}\frac{E\left(q\right)}{2}+{\left[x-\frac{1}{2}\right]}_{q}\right)}^{k}\frac{{t}^{k}}{k!}.$
(3.3)

Hence, we have

Lemma 3.3.

${F}_{q}\left(t,x\right)={{e}^{{\left[x\right]}_{q}}}^{t}{F}_{q}\left({q}^{x}t\right)={e}^{\frac{t}{1-q}}+\frac{tlogq}{1-q}\sum _{m=0}^{\infty }{q}^{m+x}{e}^{{\left[m+x\right]}_{q}t}.$

Proof. From (3.1) and (3.2), we note that

$\begin{array}{c}{F}_{q}\left(t,x\right)=\sum _{k=0}^{\infty }{\left({q}^{x}B\left(q\right)+{\left[x\right]}_{q}\right)}^{k}\frac{{t}^{k}}{k!}\\ ={e}^{\left({q}^{x}B\left(q\right)+{\left[x\right]}_{q}\right)t}\\ ={e}^{B\left(q\right){q}^{x}t}{e}^{{\left[x\right]}_{q}t}\\ ={{e}^{{\left[x\right]}_{q}}}^{t}{F}_{q}\left({q}^{x}t\right).\end{array}$

The second identity leads at once to Lemma 3.1. Hence, the lemma follows. □

Lemma 3.4.

${G}_{q}\left(t,x\right)={e}^{{\left[x-\frac{1}{2}\right]}_{q}t}{G}_{q}\left(\frac{{q}^{x-\frac{1}{2}}}{2}t\right)=2\sum _{m=0}^{\infty }{\left(-1\right)}^{m}{e}^{{\left[m+x\right]}_{q}t}.$

Proof. By similar method of Lemma 3.3, we prove this lemma by (3.1), (3.3), and Lemma 3.1. □

Corollary 3.5 (Difference equations).

$\begin{array}{c}{B}_{k+1}\left(x+1,q\right)-{B}_{k+1}\left(x,q\right)=\frac{{q}^{x}logq}{q-1}\left(k+1\right){\left[x\right]}_{q}^{k}\left(k\ge 0\right),\\ {E}_{k}\left(x+1,q\right)+{E}_{k}\left(x,q\right)=2{\left[x\right]}_{q}^{k}\phantom{\rule{1em}{0ex}}\left(k\ge 0\right).\end{array}$

Proof. By applying (3.2) and Lemma 3.3, we obtain (3.4)

$\begin{array}{ll}\hfill {F}_{q}\left(t,x\right)& =\sum _{k=0}^{\infty }{B}_{k}\left(x,q\right)\frac{{t}^{k}}{k!}\phantom{\rule{2em}{0ex}}\\ =1+\sum _{k=0}^{\infty }\left(\frac{1}{{\left(1-q\right)}^{k+1}}+\left(k+1\right)\frac{logq}{1-q}\sum _{m=0}^{\infty }{q}^{m+x}{\left[m+x\right]}_{q}^{k}\right)\frac{{t}^{k+1}}{\left(k+1\right)!}.\phantom{\rule{2em}{0ex}}\end{array}$
(3.4)

By comparing the coefficients of both sides of (3.4), we have B 0(x, q) = 1 and

${B}_{k}\left(x,q\right)=\frac{1}{{\left(1-q\right)}^{k}}+k\frac{logq}{1-q}\sum _{m=0}^{\infty }{q}^{m+x}{\left[m+x\right]}_{q}^{k-1}\phantom{\rule{1em}{0ex}}\left(k\ge 1\right).$
(3.5)

Hence,

${B}_{k}\left(x+1,q\right)-{B}_{k}\left(x,q\right)=k\frac{{q}^{x}logq}{q-1}{\left[x\right]}_{q}^{k-1}\phantom{\rule{1em}{0ex}}\left(k\ge 1\right).$

Similarly we prove the second part by (3.3) and Lemma 3.4. This proof is complete.

From Lemma 2.2 and Corollary 3.5, we obtain for any integer k ≥ 0,

$\begin{array}{c}{\left[x\right]}_{q}^{k}=\frac{1}{k+1}\frac{q-1}{{q}^{x}logq}\left(\sum _{i=0}^{k+1}\left(\begin{array}{c}\hfill k+1\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{q}^{i}{B}_{i}\left(x,q\right)-{B}_{k+1}\left(x,q\right)\right),\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{\left[x\right]}_{q}^{k}=\frac{1}{2}\left(\sum _{i=0}^{k}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)\phantom{\rule{2.77695pt}{0ex}}{q}^{i}{E}_{i}\left(x,q\right)+{E}_{k}\left(x,q\right)\right)\end{array}$

which are the q-analogues of the following familiar expansions (see, e.g., [[7], p. 9]):

respectively.

Corollary 3.6 (Difference equations). Let k ≥ 0 and n ≥ 1. Then

$\begin{array}{c}{B}_{k+1}\left(x+\frac{1}{n},{q}^{n}\right)-{B}_{k+1}\left(x+\frac{1-n}{n},{q}^{n}\right)\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}=\frac{n{q}^{n\left(x-1\right)+1}logq}{q-1}\frac{k+1}{{\left[n\right]}_{q}^{k+1}}{\left(1+q{\left[nx-n\right]}_{q}\right)}^{k},\\ {E}_{k}\left(x+\frac{1}{n},{q}^{n}\right)+{E}_{k}\left(x+\frac{1-n}{n},{q}^{n}\right)=\frac{2}{{\left[n\right]}_{q}^{k}}{\left(1+q{\left[nx-n\right]}_{q}\right)}^{k}.\end{array}$

Proof. Use Lemma 2.4 and Corollary 3.5, the proof can be obtained by the similar way to [[7], Lemma 2.4]. □

Letting n = 1, Corollary 3.6 reduces to Corollary 3.5. Clearly, the above difference formulas in Corollary 3.6 become the following difference formulas when q → 1:

${B}_{k}\left(x+\frac{1}{n}\right)-{B}_{k}\left(x+\frac{1-n}{n}\right)=k{\left(x+\frac{1-n}{n}\right)}^{k-1}\left(k\ge 1,n\ge 1\right),$
(3.6)
${E}_{k}\left(x+\frac{1}{n}\right)+{E}_{k}\left(x+\frac{1-n}{n}\right)=2{\left(x+\frac{1-n}{n}\right)}^{k}\left(k\ge 0,n\ge 1\right),$
(3.7)

respectively (see [[7], (2.22), (2.23)]). If we now let n = 1 in (3.6) and (3.7), we get the ordinary difference formulas

for k ≥ 0.

In Corollary 3.5, let x = 0. We arrive at the following proposition.

Proposition 3.7.

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{B}_{0}\left(q\right)=1,\phantom{\rule{2.77695pt}{0ex}}{\left(qB\left(q\right)+1\right)}^{k}-{B}_{k}\left(q\right)=\left\{\begin{array}{cc}\hfill \frac{{log}_{p}q}{q-1}\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}k=1\hfill \\ \hfill 0\hfill & \hfill if\phantom{\rule{2.77695pt}{0ex}}k>1,\hfill \end{array}\right\\\ {E}_{0}\left(q\right)=1,\phantom{\rule{1em}{0ex}}{\left({q}^{-\frac{1}{2}}\frac{E\left(q\right)}{2}+{\left[-\frac{1}{2}\right]}_{q}\right)}^{k}+{\left({q}^{\frac{1}{2}}\frac{E\left(q\right)}{2}+{\left[\frac{1}{2}\right]}_{q}\right)}^{k}=0\phantom{\rule{1em}{0ex}}if\phantom{\rule{2.77695pt}{0ex}}k\ge 1.\end{array}$

Proof. The first identity follows from (2.13). To see the second identity, setting x = 0 and x = 1 in (2.14) we have

This proof is complete. □

Remark 3.8. (1). We note here that quite similar expressions to the first identity of Proposition 3.7 are given by Kamano [[3], Proposition 2.4], Rim et al. [[8], Theorem 2.7] and Tsumura [[10], (1)].

(2). Letting q → 1 in Proposition 3.7, the first identity is the corresponding classical formulas in [[8], (1.2)]:

${B}_{0}=1,\phantom{\rule{1em}{0ex}}{\left(B+1\right)}^{k}-{B}_{k}=\left\{\begin{array}{c}\hfill 1\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}k=1\hfill \\ \hfill 0\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}k>1\hfill \end{array}\right\$

and the second identity is the corresponding classical formulas in [[25], (1.1)]:

${E}_{0}=1,\phantom{\rule{1em}{0ex}}{\left(E+1\right)}^{k}+{\left(E-1\right)}^{k}=0\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}k\ge 1.$

## 4. q-analogues of Riemann's ζ-functions, the Hurwitz ζ-functions and the Didichlet's L-functions

Now, by evaluating the k th derivative of both sides of Lemma 3.1 at t = 0, we obtain the following

(4.1)
(4.2)

for k ≥ 0.

Definition 4.1 (q-analogues of the Riemann's ζ-functions). For s ≤, define

$\begin{array}{c}{\zeta }_{q}\left(s\right)=\frac{1}{s-1}\frac{1}{{\left(\frac{1}{1-q}\right)}^{s-1}}+\frac{logq}{q-1}\sum _{m=1}^{\infty }\frac{{q}^{m}}{{\left[m\right]}_{q}^{s}},\\ {\zeta }_{q,E}\left(s\right)=\frac{2}{{2}^{s}}\sum _{m=0}^{\infty }\frac{{\left(-1\right)}^{m}}{{\left[m+\frac{1}{2}\right]}_{q}^{s}}.\end{array}$

Note that ζ q (s) is a meromorphic function on ≤ with only one simple pole at s = 1 and ζ q ,E(s) is a analytic function on ≤.

Also, we have

$\underset{q\to 1}{lim}{\zeta }_{q}\left(s\right)=\sum _{m=1}^{\infty }\frac{1}{{m}^{s}}=\zeta \left(s\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{and}}\phantom{\rule{1em}{0ex}}\underset{q\to 1}{lim}{\zeta }_{q,E}\left(s\right)=2\sum _{m=0}^{\infty }\frac{{\left(-1\right)}^{m}}{{\left(m+1\right)}^{s}}={\zeta }_{E}\left(s\right).$
(4.3)

(In [[26], p. 1070], our ζ E (s) is denote ϕ(s).)

The values of ζ q (s) and ζ q ,E(s) at non-positive integers are obtained by the following proposition.

Proposition 4.2. For k ≥ 1, we have

${\zeta }_{q}\left(1-k\right)=-\frac{{B}_{k}\left(q\right)}{k}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}{\zeta }_{q,E}\left(1-k\right)={E}_{k-1}\left(q\right).$

Proof. It is clear by (4.1) and (4.2). □

We can investigate the generating functions F q (t, x) and G q (t, x) by using a method similar to the method used to treat the q-analogues of Riemann's ζ-functions in Definition 4.1.

Definition 4.3 (q-analogues of the Hurwitz ζ-functions). For s ≤ and 0 < x ≤ 1, define

$\begin{array}{c}{\zeta }_{q}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)=\frac{1}{s-1}\frac{1}{{\left(\frac{1}{1-q}\right)}^{s-1}}+\frac{logq}{q-1}\sum _{m=0}^{\infty }\frac{{q}^{m+x}}{{\left[m+x\right]}_{q}^{s}},\\ {\zeta }_{q,E}\left(s,\phantom{\rule{2.77695pt}{0ex}}x\right)=2\sum _{m=0}^{\infty }\frac{{\left(-1\right)}^{m}}{{\left[m+x\right]}_{q}^{s}}.\end{array}$

Note that ζ q (s, x) is a meromorphic function on ≤ with only one simple pole at s = 1 and ζ q ,E(s, x) is a analytic function on ≤.

The values of ζ q (s, x) and ζ q ,E(s, x) at non-positive integers are obtained by the following proposition.

Proposition 4.4. For k ≥ 1, we have

${\zeta }_{q}\left(1-k,\phantom{\rule{2.77695pt}{0ex}}x\right)=-\frac{{B}_{k}\left(x,q\right)}{k}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}{\zeta }_{q,E}\left(1-k,\phantom{\rule{2.77695pt}{0ex}}x\right)={E}_{k-1}\left(x,\phantom{\rule{2.77695pt}{0ex}}q\right).$

Proof. From Lemma 3.3 and Definition 4.3, we have

for k ≥ 1. We obtain the desired result by (3.2). Similarly the second form follows by Lemma 3.4 and (3.3). □

Proposition 4.5. Let d be any positive integer. Then

$\begin{array}{ll}\hfill {F}_{q}\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)& =\frac{1}{d}\sum _{i=0}^{d-1}{F}_{{q}^{d}}\left({\left[d\right]}_{q}t,\phantom{\rule{2.77695pt}{0ex}}\frac{x+i}{d}\right),\phantom{\rule{2em}{0ex}}\\ \hfill {G}_{q}\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)& =\sum _{i=0}^{d-1}{\left(-1\right)}^{i}{G}_{{q}^{d}}\left({\left[d\right]}_{q}t,\phantom{\rule{2.77695pt}{0ex}}\frac{x+i}{d}\right)\phantom{\rule{1em}{0ex}}if\phantom{\rule{2.77695pt}{0ex}}d\phantom{\rule{2.77695pt}{0ex}}odd.\phantom{\rule{2em}{0ex}}\end{array}$

Proof. Substituting m = nd + i with n = 0, 1,... and i = 0,..., d - 1 into Lemma 3.3, we have

$\begin{array}{ll}\hfill {F}_{q}\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)& ={e}^{\frac{t}{1-q}}+\frac{tlogq}{1-q}\sum _{m=0}^{\infty }{q}^{m+x}{e}^{{\left[m+x\right]}_{q}t}\phantom{\rule{2em}{0ex}}\\ ={e}^{\frac{{\left[d\right]}_{q}t}{1-{q}^{d}}}+\frac{1}{d}\sum _{i=0}^{d-1}\frac{{\left[d\right]}_{q}tlog{q}^{d}}{1-{q}^{d}}\sum _{n=0}^{\infty }{q}^{nd+x+i}{e}^{{\left[nd+x+i\right]}_{q}t}\phantom{\rule{2em}{0ex}}\\ =\frac{1}{d}\sum _{i=0}^{d-1}\left({e}^{\frac{{\left[d\right]}_{q}t}{1-{q}^{d}}}+\frac{{\left[d\right]}_{q}tlog{q}^{d}}{1-{q}^{d}}\sum _{n=0}^{\infty }{\left({q}^{d}\right)}^{n+\frac{x+i}{d}}{e}^{{\left[n+\frac{x+i}{d}\right]}_{{q}^{d}}{\left[d\right]}_{q}t}\right),\phantom{\rule{2em}{0ex}}\end{array}$

where we use ${\left[n+\left(x+i\right)/d\right]}_{{q}^{d}}{\left[d\right]}_{q}={\left[nd+x+i\right]}_{q}$. So we have the first form. Similarly the second form follows by Lemma 3.4. □

From (3.2), (3.3), Propositions 4.4 and 4.5, we obtain the following:

Corollary 4.6. Let d and k be any positive integer. Then

$\begin{array}{c}{\zeta }_{q}\left(1-k,\phantom{\rule{2.77695pt}{0ex}}x\right)=\frac{{\left[d\right]}_{q}^{k}}{d}\sum _{i=0}^{d-1}{\zeta }_{{q}^{d}}\left(1-k,\phantom{\rule{2.77695pt}{0ex}}\frac{x+i}{d}\right),\\ {\zeta }_{q,E}\left(-k,\phantom{\rule{2.77695pt}{0ex}}x\right)={\left[d\right]}_{q}^{k}\sum _{i=0}^{d-1}{\left(-1\right)}^{i}{\zeta }_{{q}^{d},E}\left(-k,\phantom{\rule{2.77695pt}{0ex}}\frac{x+i}{d}\right)\phantom{\rule{1em}{0ex}}if\phantom{\rule{2.77695pt}{0ex}}d\phantom{\rule{2.77695pt}{0ex}}odd.\end{array}$

Let χ be a primitive Dirichlet character of conductor f . We define the generating function F q,χ (x, t) and G q,χ (x, t) of the generalized q-Bernoulli and q-Euler polynomials as follows:

$\begin{array}{ll}\hfill {F}_{q,\chi }\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)& =\sum _{k=0}^{\infty }{B}_{k,\chi }\left(x,\phantom{\rule{2.77695pt}{0ex}}q\right)\frac{{t}^{k}}{k!}\phantom{\rule{2em}{0ex}}\\ =\frac{1}{f}\sum _{a=1}^{f}\chi \left(a\right){F}_{{q}^{f}}\left({\left[f\right]}_{q}t,\phantom{\rule{2.77695pt}{0ex}}\frac{a+x}{f}\right)\phantom{\rule{2em}{0ex}}\end{array}$
(4.4)

and

$\begin{array}{ll}\hfill {G}_{q,\chi }\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)& =\sum _{k=0}^{\infty }{E}_{k,\chi }\left(x,\phantom{\rule{2.77695pt}{0ex}}q\right)\frac{{t}^{k}}{k!}\phantom{\rule{2em}{0ex}}\\ =\sum _{a=1}^{f}{\left(-1\right)}^{a}\chi \left(a\right){G}_{{q}^{f}}\left({\left[f\right]}_{q}t,\phantom{\rule{2.77695pt}{0ex}}\frac{a+x}{f}\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}f\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{odd}},\phantom{\rule{2em}{0ex}}\end{array}$
(4.5)

where B k,χ (x, q) and E k,χ (x, q) are the generalized q-Bernoulli and q-Euler polynomials, respectively. Clearly (4.4) and (4.5) are equal to

${F}_{q,\chi }\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)=\frac{tlogq}{1-q}\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m+x}{e}^{{\left[m+x\right]}_{q}t},$
(4.6)
${G}_{q,\chi }\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)=2\sum _{k=0}^{\infty }{\left(-1\right)}^{m}\chi \left(m\right){e}^{{\left[m+x\right]}_{q}t}\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}f\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{odd}},$
(4.7)

respectively. As q → 1 in (4.6) and (4.7), we have F q,χ (t, x) → F χ (t, x) and G q,χ (t, x) → G χ (t, x), where F χ (t, x) and G χ (t, x) are the usual generating function of generalized Bernoulli and Euler numbers, respectively, which are defined as follows [13]:

${F}_{\chi }\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)=\sum _{a=1}^{f}\frac{\chi \left(a\right)t{e}^{\left(a+x\right)t}}{{e}^{ft}-1}=\sum _{k=0}^{\infty }{B}_{k,\chi }\left(x\right)\frac{{t}^{k}}{k!},$
(4.8)
${G}_{\chi }\left(t,\phantom{\rule{2.77695pt}{0ex}}x\right)=2\sum _{a=1}^{f}\frac{{\left(-1\right)}^{a}\chi \left(a\right){e}^{\left(a+x\right)t}}{{e}^{ft}+1}=\sum _{k=0}^{\infty }{G}_{k,\chi }\left(x\right)\frac{{t}^{k}}{k!}\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}f\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{odd}}.$
(4.9)

From (3.2), (3.3), (4.4) and (4.5), we can easily see that

${B}_{k,\chi }\left(x,\phantom{\rule{2.77695pt}{0ex}}q\right)=\frac{{\left[f\right]}_{q}^{k}}{f}\sum _{a=1}^{f}\chi \left(a\right){B}_{k}\left(\frac{a+x}{f},\phantom{\rule{2.77695pt}{0ex}}{q}^{f}\right),$
(4.10)
${E}_{k,\chi }\left(x,\phantom{\rule{2.77695pt}{0ex}}q\right)={\left[f\right]}_{q}^{k}\sum _{a=1}^{f}{\left(-1\right)}^{a}\chi \left(a\right){E}_{k}\left(\frac{a+x}{f},\phantom{\rule{2.77695pt}{0ex}}{q}^{f}\right)\phantom{\rule{1em}{0ex}}\mathsf{\text{if}}\phantom{\rule{2.77695pt}{0ex}}f\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{odd}}.$
(4.11)

By using the definitions of ζ q (s, x) and ζq,E (s, x), we can define the q-analogues of Dirichlet's L-function.

Definition 4.7 (q-analogues of the Dirichlet's L-functions). For s and 0 < x ≤ 1,

$\begin{array}{c}{L}_{q}\left(s,\phantom{\rule{2.77695pt}{0ex}}x,\phantom{\rule{2.77695pt}{0ex}}\chi \right)=\frac{logq}{q-1}\sum _{m=0}^{\infty }\frac{\chi \left(m\right){q}^{m+x}}{{\left[m+x\right]}_{q}^{s}},\\ {\ell }_{q}\left(s,\phantom{\rule{2.77695pt}{0ex}}x,\phantom{\rule{2.77695pt}{0ex}}\chi \right)=2\sum _{m=0}^{\infty }\frac{{\left(-1\right)}^{m}\chi \left(m\right)}{{\left[m+x\right]}_{q}^{s}}.\end{array}$

Similarly, we can compute the values of L q (s, x, χ) at non-positive integers.

Theorem 4.8. For k ≥ 1, we have

${L}_{q}\left(1-k,\phantom{\rule{2.77695pt}{0ex}}x,\phantom{\rule{2.77695pt}{0ex}}\chi \right)=-\frac{{B}_{k,\chi }\left(x,q\right)}{k}\phantom{\rule{1em}{0ex}}and\phantom{\rule{1em}{0ex}}{\ell }_{q}\left(1-k,\phantom{\rule{2.77695pt}{0ex}}x,\phantom{\rule{2.77695pt}{0ex}}\chi \right)={E}_{k-1,\chi }\left(x,\phantom{\rule{2.77695pt}{0ex}}q\right).$

Proof. Using Lemma 3.3 and (4.4), we obtain

$\begin{array}{ll}\hfill \sum _{k=0}^{\infty }{B}_{k,\chi }\left(x,\phantom{\rule{2.77695pt}{0ex}}q\right)\frac{{t}^{k}}{k!}& =\frac{1}{f}\sum _{a=1}^{f}\chi \left(a\right)\left({e}^{\frac{{\left[f\right]}_{q}t}{1-{q}^{f}}}+\frac{{\left[f\right]}_{q}tlog{q}^{f}}{1-{q}^{f}}\sum _{n=0}^{\infty }{\left({q}^{f}\right)}^{n+\frac{x+a}{f}}{e}^{{\left[n+\frac{x+a}{f}\right]}_{{q}^{f}}{\left[f\right]}_{q}t}\right)\phantom{\rule{2em}{0ex}}\\ =\frac{tlogq}{1-q}\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m+x}{e}^{{\left[m+x\right]}_{q}t},\phantom{\rule{2em}{0ex}}\end{array}$

where we use ${\left[n\phantom{\rule{2.77695pt}{0ex}}+\left(a+x\right)/f\right]}_{{q}^{f}}{\left[f\right]}_{q}={\left[nf+a+x\right]}_{q}$ and ${\sum }_{a=1}^{f}\chi \left(a\right)=0$. Therefore, we obtain

Hence for k ≥ 1

$\begin{array}{ll}\hfill -\frac{{B}_{k,\chi }\left(x,q\right)}{k}& =\frac{logq}{q-1}\sum _{m=0}^{\infty }\chi \left(m\right){q}^{m+x}{\left[m+x\right]}_{q}^{k-1}\phantom{\rule{2em}{0ex}}\\ ={L}_{q}\left(1-k,\phantom{\rule{2.77695pt}{0ex}}x,\phantom{\rule{2.77695pt}{0ex}}\chi \right).\phantom{\rule{2em}{0ex}}\end{array}$

Similarly the second identity follows. This completes the proof. □

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## Acknowledgements

This study was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MEST) (2011-0001184).

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Kim, D., Kim, MS. Some results for the q-Bernoulli, q-Euler numbers and polynomials. Adv Differ Equ 2011, 68 (2011). https://doi.org/10.1186/1687-1847-2011-68

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