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., [3–10]). 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, ℤ _p , ℚ _p , and ℂ _p denote the ring of p-adic integers, the field of p-adic numbers, and the completion of the algebraic closure of ℚ _p , respectively. Let | · | _p be the p-adic norm on ℚ with |p| _p = p ^-1. For convenience, | · | _p will also be used to denote the extended valuation on ℂ _p .

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

$$B_n\left(x\right)=\sum _{k=0}^n\left(\begin{array}{c}\hfill n\hfill \\ \hfill k\hfill \end{array}\right)B_k{x}^{n-k},n\ge 0,$$

(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 ∈ ℚ, and these integrals are independent of the prime p which used to compute them. The examples of (1.3) are:

$$\begin{array}{c}\underset{N\to \infty }{lim}\frac{1}{p^N}\sum _{a=0}^{p^N-1}a=\underset{N\to \infty }{lim}\frac{1}{p^N}\frac{p^N\left(p^N-1\right)}{2}=-\frac{1}{2}=B_1,\\ \underset{N\to \infty }{lim}\frac{1}{p^N}\sum _{a=0}^{p^N-1}{a}^2=\underset{N\to \infty }{lim}\frac{1}{p^N}\frac{p^N\left(p^N-1\right)\left(2p^N-1\right)}{6}=\frac{1}{6}=B_2.\end{array}$$

(1.4)

Euler numbers E _k , k ≥ 0 are integers given by (cf. [17–19])

$$E_0=1,E_k=-\sum _{\begin{array}{c}i=0\\ 2|k-i\end{array}}^{k-1}\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)E_i\mathsf{\text{ for }}k=1,2,\dots .$$

(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.

This article is organized as follows.

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

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

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

$$I_1\left(f\right)=\underset{N\to \infty }{lim}\frac{1}{p^N}\sum _{a=0}^{p^N-1}f\left(a\right)={\int }_{{\mathbb{Z}}_p}f\left(z\right)d{\mu }_1\left(z\right).$$

(2.3)

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

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

(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 ∈ ℂ _p with 0 < |q - 1| _p < 1 as follows

$$B_k\left(q\right)={\int }_{{\mathbb{Z}}_p}{\left[z\right]}_q^{k}d{\mu }_1\left(z\right)=\underset{N\to \infty }{lim}\frac{1}{p^N}\sum _{a=0}^{p^N-1}{\left[a\right]}_q^k,$$

(2.5)

$$E_k\left(q\right)=2^k{\int }_{{\mathbb{Z}}_p}{\left[z+\frac{1}{2}\right]}_q^{k}d{\mu }_{-1}\left(z\right)=2^k\underset{N\to \infty }{lim}\sum _{a=0}^{p^N-1}{\left[a+\frac{1}{2}\right]}_q^k{\left(-1\right)}^a.$$

(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])

$$B_k\left(q\right)\to B_k={\int }_{{\mathbb{Z}}_p}{z}^{k}d{\mu }_1\left(z\right),E_k\left(q\right)\to E_k={\int }_{{\mathbb{Z}}_p}{\left(2z+1\right)}^{k}d{\mu }_{-1}\left(z\right).$$

For |q - 1| _p < 1 and z ∈ ℤ _p , we have

$$q^{iz}=\sum _{n=0}^{\infty }{\left(q^i-1\right)}^n\left(\begin{array}{c}\hfill z\hfill \\ \hfill n\hfill \end{array}\right)\mathsf{\text{and}}|q^i-1{|}_p\le |q-1{|}_p<1,$$

(2.7)

where i ∈ ℤ. 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 ℤ _p (see [16]).

By (2.3) and (2.7), we obtain

$$\begin{array}{c}{I}_1(q^{iz})=\frac{1}{q^i-1}\underset{N\to \infty }{\lim }\frac{{(q^i)}^{p^N}-1}{p^N}\\ =\frac{1}{q^i-1}\underset{N\to \infty }{\lim }\frac{1}{p^N}\left\{{\displaystyle \sum _{m=0}^{\infty }\left(\begin{array}{c}{p}^N\\ m\end{array}\right)}{(q^i-1)}^m-1\right\}\\ =\frac{1}{q^i-1}\underset{N\to \infty }{\lim }\frac{1}{p^N}{\displaystyle \sum _{m=1}^{\infty }\left(\begin{array}{c}{p}^N\\ m\end{array}\right)}{(}^{q^i-1)m}\\ =\frac{1}{q^i-1}\underset{N\to \infty }{\lim }{\displaystyle \sum _{m=1}^{\infty }\frac{1}{m}}\left(\begin{array}{c}{p}^N-1\\ m-1\end{array}\right){(}^{q^i-1)m}\\ =\frac{1}{q^i-1}{\displaystyle \sum _{m=1}^{\infty }\frac{1}{m}}\left(\begin{array}{c}-1\\ m-1\end{array}\right){(}^{q^i-1)m}\\ =\frac{1}{q^i-1}{\displaystyle \sum _{m=1}^{\infty }{(-1)}^{m-1}}\frac{{(q^i-1)}^m}{m}\\ =\frac{i\log q}{q^i-1}\end{array}$$

(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 ℤ _p :

$$B_k\left(x,q\right)={\int }_{{\mathbb{Z}}_p}{\left[x+z\right]}_q^{k}d{\mu }_1\left(z\right)\mathsf{\text{ and }}{E}_k\left(x,q\right)={\int }_{{\mathbb{Z}}_p}{\left[x+z\right]}_q^{k}d{\mu }_{-1}\left(z\right),$$

(2.12)

where q ∈ ℂ _p with 0 < |q - 1| _p < 1 and x ∈ ℤ _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

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

(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

$$E_k\left(q\right)=2^k{E}_k\left(\frac{1}{2},q\right)\ne E_k\left(0,q\right)\mathsf{\text{and }}\underset{q\to 1}{lim}{E}_k\left(q\right)=E_k,$$

(2.16)

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

Lemma 2.2 (Addition theorem).

$$\begin{array}{c}{B}_k\left(x+y,q\right)=\sum _{i=0}^k\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)q^{iy}{B}_i\left(x,q\right){\left[y\right]}_q^{k-i}\left(k\ge 0\right),\\ E_k\left(x+y,q\right)=\sum _{i=0}^k\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)q^{iy}{E}_i\left(x,q\right){\left[y\right]}_q^{k-i}\left(k\ge 0\right).\end{array}$$

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

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

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 ∈ ℂ _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

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

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

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

respectively.

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

From (2.3), we see that

$$\begin{array}{ll}\hfill B_k\left(nx,q\right)& ={\int }_{{\mathbb{Z}}_p}{\left[nx+z\right]}_q^{k}d{\mu }_1\left(z\right)\\ =\underset{N\to \infty }{lim}\frac{1}{n{p}^N}\sum _{a=0}^{n{p}^N-1}{\left[nx+a\right]}_q^k\\ =\frac{1}{n}\underset{N\to \infty }{lim}\frac{1}{p^N}\sum _{i=0}^{n-1}\sum _{a=0}^{p^N-1}{\left[nx+na+i\right]}_q^k\\ =\frac{{\left[n\right]}_q^k}{n}\sum _{i=0}^{n-1}{\int }_{{\mathbb{Z}}_p}{\left[x+\frac{i}{n}+z\right]}_{q^n}^{k}d{\mu }_1\left(z\right)\end{array}$$

(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

$$\begin{array}{ll}\hfill B_k\left(q\right)& =\frac{{\left[n\right]}_q^k}{n}\sum _{i=0}^{n-1}{B}_k\left(\frac{i}{n},q^n\right)\\ =\frac{{\left[n\right]}_q^k}{n}\sum _{i=0}^{n-1}\sum _{j=0}^k\left(\begin{array}{c}\hfill k\hfill \\ \hfill j\hfill \end{array}\right){\left[\frac{i}{n}\right]}_{q^n}^{k-j}{q}^{ij}{B}_j\left(q^n\right)\\ =\frac{1}{n}\sum _{j=0}^k{\left[n\right]}_q^j\left(\begin{array}{c}\hfill k\hfill \\ \hfill j\hfill \end{array}\right)B_j\left(q^n\right)\sum _{i=0}^{n-1}{q}^{ij}{\left[i\right]}_q^{k-j}.\end{array}$$

(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

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

(2.20)

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

$$E_k(x,q)={[n]}_q^k{\displaystyle \sum _{i=0}^{n-1}{(-1)}^i}{E}_k\left(\frac{x+i}{n},q^n\right)\text{ if }n\text{odd}.$$

(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

$$\begin{array}{l}{B}_k(x,q)=\frac{{[n]}_q^k}{n}{\displaystyle \sum _{i=0}^{n-1}{B}_k}\left(\frac{x+i}{n},q^n\right),\\ E_k(x,q)={[n]}_q^k{\displaystyle \sum _{i=0}^{n-1}{(-1)}^i}{E}_k\left(\frac{x+i}{n},q^n\right)\text{if}nodd.\end{array}$$

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:

$$F_q\left(t\right)=\sum _{k=0}^{\infty }{B}_k\left(q\right)\frac{t^k}{k!}=e^{B\left(q\right)t}\mathsf{\text{ and }}{G}_q\left(t\right)=\sum _{k=0}^{\infty }{E}_k\left(q\right)\frac{t^k}{k!}=e^{E\left(q\right)t},$$

(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.

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

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

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

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

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

Thus, we find that

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

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

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

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).

$$\begin{array}{c}{B}_{k+1}\left(x+1,q\right)-B_{k+1}\left(x,q\right)=\frac{q^{x}logq}{q-1}\left(k+1\right){\left[x\right]}_q^k\left(k\ge 0\right),\\ E_k\left(x+1,q\right)+E_k\left(x,q\right)=2{\left[x\right]}_q^k\left(k\ge 0\right).\end{array}$$

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

$$\begin{array}{ll}\hfill F_q\left(t,x\right)& =\sum _{k=0}^{\infty }{B}_k\left(x,q\right)\frac{t^k}{k!}\\ =1+\sum _{k=0}^{\infty }\left(\frac{1}{{\left(1-q\right)}^{k+1}}+\left(k+1\right)\frac{logq}{1-q}\sum _{m=0}^{\infty }{q}^{m+x}{\left[m+x\right]}_q^k\right)\frac{t^{k+1}}{\left(k+1\right)!}.\end{array}$$

(3.4)

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

$$B_k\left(x,q\right)=\frac{1}{{\left(1-q\right)}^k}+k\frac{logq}{1-q}\sum _{m=0}^{\infty }{q}^{m+x}{\left[m+x\right]}_q^{k-1}\left(k\ge 1\right).$$

(3.5)

Hence,

$$B_k\left(x+1,q\right)-B_k\left(x,q\right)=k\frac{q^{x}logq}{q-1}{\left[x\right]}_q^{k-1}\left(k\ge 1\right).$$

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,

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

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

$$x^k=\frac{1}{k+1}\sum _{i=0}^k\left(\begin{array}{c}\hfill k+1\hfill \\ \hfill i\hfill \end{array}\right)B_i\left(x\right)\mathsf{\text{and }}{x}^k=\frac{1}{2}\left(\sum _{i=0}^k\left(\begin{array}{c}\hfill k\hfill \\ \hfill i\hfill \end{array}\right)E_i\left(x\right)+E_k\left(x\right)\right),$$

respectively.

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

$$\begin{array}{c}{B}_{k+1}\left(x+\frac{1}{n},q^n\right)-B_{k+1}\left(x+\frac{1-n}{n},q^n\right)\\ =\frac{n{q}^{n\left(x-1\right)+1}logq}{q-1}\frac{k+1}{{\left[n\right]}_q^{k+1}}{\left(1+q{\left[nx-n\right]}_q\right)}^k,\\ E_k\left(x+\frac{1}{n},q^n\right)+E_k\left(x+\frac{1-n}{n},q^n\right)=\frac{2}{{\left[n\right]}_q^k}{\left(1+q{\left[nx-n\right]}_q\right)}^k.\end{array}$$

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

$$B_{k+1}\left(x+1\right)-B_{k+1}\left(x\right)=\left(k+1\right)x^{k-1}\mathsf{\text{ and }}{E}_k\left(x+1\right)+E_k\left(x\right)=2x^k$$

for k ≥ 0.

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

Proposition 3.7.

$$\begin{array}{c}{B}_0\left(q\right)=1,{\left(qB\left(q\right)+1\right)}^k-B_k\left(q\right)=\left\{\begin{array}{cc}\hfill \frac{{log}_{p}q}{q-1}\hfill & \hfill ifk=1\hfill \\ \hfill 0\hfill & \hfill ifk>1,\hfill \end{array}\right.\\ E_0\left(q\right)=1,{\left(q^{-\frac{1}{2}}\frac{E\left(q\right)}{2}+{\left[-\frac{1}{2}\right]}_q\right)}^k+{\left(q^{\frac{1}{2}}\frac{E\left(q\right)}{2}+{\left[\frac{1}{2}\right]}_q\right)}^k=0ifk\ge 1.\end{array}$$

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

$$E_k\left(0,q\right)={\left(\frac{q^{-\frac{1}{2}}}{2}E\left(q\right)+{\left[-\frac{1}{2}\right]}_q\right)}^k\mathsf{\text{ and }}{E}_k\left(1,q\right)={\left(\frac{q^{\frac{1}{2}}}{2}E\left(q\right)+{\left[\frac{1}{2}\right]}_q\right)}^k.$$

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)]:

$$B_0=1,{\left(B+1\right)}^k-B_k=\left\{\begin{array}{c}\hfill 1\mathsf{\text{if}}k=1\hfill \\ \hfill 0\mathsf{\text{if}}k>1\hfill \end{array}\right.$$

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

$$E_0=1,{\left(E+1\right)}^k+{\left(E-1\right)}^k=0\mathsf{\text{if}}k\ge 1.$$

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

$$B_k\left(q\right)={\left.{\left(\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\right)}^k{F}_q\left(t\right)\right|}_{t=0}={\left(\frac{1}{1-q}\right)}^k-\frac{klogq}{q-1}\sum _{m=0}^{\infty }{q}^m{\left[m\right]}_q^{k-1},$$

(4.1)

$$E_k\left(q\right)={\left.{\left(\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\right)}^k{G}_q\left(t\right)\right|}_{t=0}=2^{k+1}\sum _{m=0}^{\infty }{\left(-1\right)}^m{\left[m+\frac{1}{2}\right]}_q^k$$

(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

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

(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

$${\zeta }_q\left(1-k\right)=-\frac{B_k\left(q\right)}{k}and{\zeta }_{q,E}\left(1-k\right)=E_{k-1}\left(q\right).$$

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

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

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

$${\zeta }_q\left(1-k,x\right)=-\frac{B_k\left(x,q\right)}{k}and{\zeta }_{q,E}\left(1-k,x\right)=E_{k-1}\left(x,q\right).$$

Proof. From Lemma 3.3 and Definition 4.3, we have

$${\left.{\left(\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\right)}^k{F}_q\left(t,x\right)\right|}_{t=0}=-k{\zeta }_q\left(1-k,x\right)$$

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

$$\begin{array}{ll}\hfill F_q\left(t,x\right)& =\frac{1}{d}\sum _{i=0}^{d-1}{F}_{q^d}\left({\left[d\right]}_{q}t,\frac{x+i}{d}\right),\\ \hfill G_q\left(t,x\right)& =\sum _{i=0}^{d-1}{\left(-1\right)}^i{G}_{q^d}\left({\left[d\right]}_{q}t,\frac{x+i}{d}\right)ifdodd.\end{array}$$

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

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

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

$$\begin{array}{c}{\zeta }_q\left(1-k,x\right)=\frac{{\left[d\right]}_q^k}{d}\sum _{i=0}^{d-1}{\zeta }_{q^d}\left(1-k,\frac{x+i}{d}\right),\\ {\zeta }_{q,E}\left(-k,x\right)={\left[d\right]}_q^k\sum _{i=0}^{d-1}{\left(-1\right)}^i{\zeta }_{q^d,E}\left(-k,\frac{x+i}{d}\right)ifdodd.\end{array}$$

Let χ be a primitive Dirichlet character of conductor f ∈ ℕ. 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:

$$\begin{array}{ll}\hfill F_{q,\chi }\left(t,x\right)& =\sum _{k=0}^{\infty }{B}_{k,\chi }\left(x,q\right)\frac{t^k}{k!}\\ =\frac{1}{f}\sum _{a=1}^f\chi \left(a\right)F_{q^f}\left({\left[f\right]}_{q}t,\frac{a+x}{f}\right)\end{array}$$

(4.4)

and

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

(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

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

(4.6)

$$G_{q,\chi }\left(t,x\right)=2\sum _{k=0}^{\infty }{\left(-1\right)}^m\chi \left(m\right)e^{{\left[m+x\right]}_{q}t}\mathsf{\text{if}}f\mathsf{\text{odd}},$$

(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]:

$$F_{\chi }\left(t,x\right)=\sum _{a=1}^f\frac{\chi \left(a\right)t{e}^{\left(a+x\right)t}}{e^{ft}-1}=\sum _{k=0}^{\infty }{B}_{k,\chi }\left(x\right)\frac{t^k}{k!},$$

(4.8)

$$G_{\chi }\left(t,x\right)=2\sum _{a=1}^f\frac{{\left(-1\right)}^a\chi \left(a\right)e^{\left(a+x\right)t}}{e^{ft}+1}=\sum _{k=0}^{\infty }{G}_{k,\chi }\left(x\right)\frac{t^k}{k!}\mathsf{\text{if}}f\mathsf{\text{odd}}.$$

(4.9)

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

$$B_{k,\chi }\left(x,q\right)=\frac{{\left[f\right]}_q^k}{f}\sum _{a=1}^f\chi \left(a\right)B_k\left(\frac{a+x}{f},q^f\right),$$

(4.10)

$$E_{k,\chi }\left(x,q\right)={\left[f\right]}_q^k\sum _{a=1}^f{\left(-1\right)}^a\chi \left(a\right)E_k\left(\frac{a+x}{f},q^f\right)\mathsf{\text{if}}f\mathsf{\text{odd}}.$$

(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 ∈ ℂ and 0 < x ≤ 1,

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

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

Theorem 4.8. For k ≥ 1, we have

$$L_q\left(1-k,x,\chi \right)=-\frac{B_{k,\chi }\left(x,q\right)}{k}and{\ell }_q\left(1-k,x,\chi \right)=E_{k-1,\chi }\left(x,q\right).$$

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

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

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

$$\begin{array}{ll}\hfill B_{k,\chi }\left(x,q\right)& ={\left.{\left(\frac{\mathsf{\text{d}}}{\mathsf{\text{d}}t}\right)}^k\left(\sum _{k=0}^{\infty }{B}_{k,\chi }\left(x,q\right)\frac{t^k}{k!}\right)\right|}_{t=0}\\ =\frac{klogq}{1-q}\sum _{m=0}^{\infty }\chi \left(m\right)q^{m+x}{\left[m+x\right]}_q^{k-1}.\end{array}$$

Hence for k ≥ 1

$$\begin{array}{ll}\hfill -\frac{B_{k,\chi }\left(x,q\right)}{k}& =\frac{logq}{q-1}\sum _{m=0}^{\infty }\chi \left(m\right)q^{m+x}{\left[m+x\right]}_q^{k-1}\\ =L_q\left(1-k,x,\chi \right).\end{array}$$

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