# 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., ). Many recent studies on q-analogue of the Bernoulli, Euler numbers, and polynomials can be found in Choi et al. , Kamano , Kim [5, 6, 12], Luo , Satoh , Simsek [13, 14] and Tsumura .

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

$B n ( x ) = ∑ k = 0 n n k B k x n - k , n ≥ 0 ,$
(1.1)

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

$t e t - 1 = ∑ k = 0 ∞ B k t k k ! .$
(1.2)

From the above definition, we see B k 's are all rational numbers. Since $t e t - 1 -1+ t 2$ 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

$B k = lim N → ∞ 1 p N ∑ a = 0 p N - 1 a k$
(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:

$lim N → ∞ 1 p N ∑ a = 0 p N - 1 a = lim N → ∞ 1 p N p N ( p N - 1 ) 2 = - 1 2 = B 1 , lim N → ∞ 1 p N ∑ a = 0 p N - 1 a 2 = lim N → ∞ 1 p N p N ( p N - 1 ) ( 2 p N - 1 ) 6 = 1 6 = B 2 .$
(1.4)

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

(1.5)

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

$E k ( x ) = ∑ i = 0 k k i E i 2 i x - 1 2 k - i ,$
(1.6)

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

$E k = 2 k E k 1 2 ,$
(1.7)

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

$E k ( 0 ) = 2 k + 1 ( 1 - 2 k + 1 ) B k + 1 ,$
(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

$I q ( f ) = lim N → ∞ 1 [ p N ] q ∑ a = 0 p N - 1 f ( a ) q a = ∫ ℤ p f ( z ) d μ q ( z ) ,$
(2.1)

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

$lim q → 1 [ x ] q = x$
(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.,

$I 1 ( f ) = lim N → ∞ 1 p N ∑ a = 0 p N - 1 f ( a ) = ∫ ℤ p f ( z ) d μ 1 ( z ) .$
(2.3)

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

$I - 1 ( f ) = lim q → - 1 I q ( f ) = lim N → ∞ ∑ a = 0 p N - 1 f ( a ) ( - 1 ) a = ∫ ℤ p f ( z ) d μ - 1 ( z ) .$
(2.4)

In particular, setting $f ( z ) = [ z ] q k$ in (2.3) and $f ( z ) = [ z + 1 2 ] q k$ 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

$B k ( q ) = ∫ ℤ p [ z ] q k d μ 1 ( z ) = lim N → ∞ 1 p N ∑ a = 0 p N - 1 [ a ] q k ,$
(2.5)
$E k ( q ) = 2 k ∫ ℤ p z + 1 2 q k d μ - 1 ( z ) = 2 k lim N → ∞ ∑ a = 0 p N - 1 a + 1 2 q k ( - 1 ) a .$
(2.6)

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

$B k ( q ) → B k = ∫ ℤ p z k d μ 1 ( z ) , E k ( q ) → E k = ∫ ℤ p ( 2 z + 1 ) k d μ - 1 ( z ) .$

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

$q i z = ∑ n = 0 ∞ ( q i - 1 ) n z n and | q i - 1 | p ≤ | q - 1 | p < 1 ,$
(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 ).

By (2.3) and (2.7), we obtain

$I 1 ( q i z ) = 1 q i − 1 lim N → ∞ ( q i ) p N − 1 p N = 1 q i − 1 lim N → ∞ 1 p N { ∑ m = 0 ∞ ( p N m ) ( q i − 1 ) m − 1 } = 1 q i − 1 lim N → ∞ 1 p N ∑ m = 1 ∞ ( p N m ) ( q i − 1 ) m = 1 q i − 1 lim N → ∞ ∑ m = 1 ∞ 1 m ( p N − 1 m − 1 ) ( q i − 1 ) m = 1 q i − 1 ∑ m = 1 ∞ 1 m ( − 1 m − 1 ) ( q i − 1 ) m = 1 q i − 1 ∑ m = 1 ∞ ( − 1 ) m − 1 ( q i − 1 ) m m = i log q q i − 1$
(2.8)

since the series log $log ( 1 + x ) = ∑ m = 1 ∞ ( - 1 ) m - 1 x m ∕m$ converges at |x| p < 1. Similarly, by (2.4), we obtain (see [, p. 4, (2.10)])

$I - 1 ( q i z ) = lim N → ∞ ∑ a = 0 p N - 1 ( q i ) a ( - 1 ) a = 2 q i + 1 .$
(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):

$B k ( q ) = log q ( 1 - q ) k ∑ i = 0 k k i ( - 1 ) i i q i - 1 ,$
(2.10)
$E k ( q ) = 2 k + 1 ( 1 - q ) k ∑ i = 0 k k i ( - 1 ) i q 1 2 i 1 q i + 1 ,$
(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

$B k ( x , q ) = ∑ i = 0 k k i [ x ] q k - i q i x B i ( q ) = ( q x B ( q ) + [ x ] q ) k$
(2.13)

and

$E k ( x , q ) = ∑ i = 0 k k i E i ( q ) 2 i x - 1 2 q k - i q i ( x - 1 2 ) = q x - 1 2 2 E ( q ) + x - 1 2 q k ,$
(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 $[ x + y ] q k = ( [ x ] q + q x [ y ] q ) k$ and

$[ x + z ] q k = 1 2 q k [ 2 x - 1 ] q 1 2 + q x - 1 2 1 2 q - 1 z + 1 2 q k = 1 2 q k ∑ i = 0 k k i [ 2 x - 1 ] q k - i q ( x - 1 2 ) i 1 2 q - i z + 1 2 q i$
(2.15)

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

(2.16)

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

$B k ( x + y , q ) = ∑ i = 0 k k i q i y B i ( x , q ) [ y ] q k - i ( k ≥ 0 ) , E k ( x + y , q ) = ∑ i = 0 k k i q i y E i ( x , q ) [ y ] q k - i ( k ≥ 0 ) .$

Proof. Applying the relationship $[ x + y - 1 2 ] q = [ y ] q + q y [ x - 1 2 ] q$ to (2.14) for x α x + y, we have

$E k ( x + y , q ) = q x + y - 1 2 2 E ( q ) + x + y - 1 2 q k = q y q x - 1 2 2 E ( q ) + x - 1 2 q + [ y ] q k = ∑ i = 0 k k i q i y q x - 1 2 2 E ( q ) + x - 1 2 q i [ y ] q k - i = ∑ i = 0 k k i q i y E i ( x , q ) [ y ] q k - i .$

Similarly, the first identity follows.□

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

$lim q → 1 B k ( x + y , q ) = ∑ i = 0 k k i B i ( x ) y k - i = ( B ( x ) + y ) k , lim q → 1 E k ( x + y , q ) = ∑ i = 0 k k i E i ( x ) y k - i = ( E ( x ) + y ) k ,$

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

$∑ i = 0 k k i q i [ n ] q i B i ( x , q n ) = [ n ] q k B k x + 1 n , q n , ∑ i = 0 k k i q i [ n ] q i E i ( x , q n ) = [ n ] q k E k x + 1 n , q n .$

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

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

$∑ i = 0 k k i n i B i ( x ) = n k B k x + 1 n , ∑ i = 0 k k i n i E i ( x ) = n k E k x + 1 n ,$

respectively.

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

From (2.3), we see that

$B k ( n x , q ) = ∫ ℤ p [ n x + z ] q k d μ 1 ( z ) = lim N → ∞ 1 n p N ∑ a = 0 n p N - 1 [ n x + a ] q k = 1 n lim N → ∞ 1 p N ∑ i = 0 n - 1 ∑ a = 0 p N - 1 [ n x + n a + i ] q k = [ n ] q k n ∑ i = 0 n - 1 ∫ ℤ p x + i n + z q n k d μ 1 ( z )$
(2.17)

is equivalent to

$B k ( x , q ) = [ n ] q k n ∑ i = 0 n - 1 B k x + i n , q n .$
(2.18)

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

$B k ( q ) = [ n ] q k n ∑ i = 0 n - 1 B k i n , q n = [ n ] q k n ∑ i = 0 n - 1 ∑ j = 0 k k j i n q n k - j q i j B j ( q n ) = 1 n ∑ j = 0 k [ n ] q j k j B j ( q n ) ∑ i = 0 n - 1 q i j [ i ] q k - j .$
(2.19)

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

$B k = 1 n ( 1 - n k ) ∑ j = 0 k - 1 n j k j B j ∑ i = 1 n - 1 i k - j ,$

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

From (2.4), we see that

$E k ( n x , q ) = ∫ ℤ p [ n x + z ] q k d μ − 1 ( z ) = lim N → ∞ ∑ i = 0 n − 1 ∑ a = 0 p N − 1 [ n x + n a + i ] q k ( − 1 ) n a + i = [ n ] q k ∑ i = 0 n − 1 ( − 1 ) i ∫ ℤ p [ x + i n + z ] q n k d μ ( − 1 ) n ( z ) .$
(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

$B k ( x , q ) = [ n ] q k n ∑ i = 0 n − 1 B k ( x + i n , q n ) , E k ( x , q ) = [ n ] q k ∑ i = 0 n − 1 ( − 1 ) i E k ( x + i n , q n ) if n o d d .$

## 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.

$F q ( t ) = e t 1 − q + t log q 1 − q ∑ m = 0 ∞ q m e [ m ] q t , G q ( t ) = 2 ∑ m = 0 ∞ ( − 1 ) m e 2 [ m + 1 2 ] q t .$

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

$F q ( t ) = ∑ k = 0 ∞ log q ( 1 - q ) k ∑ i = 0 k k i ( - 1 ) i i q i - 1 t k k ! = 1 + log q ∑ k = 1 ∞ 1 ( 1 - q ) k t k k ! 1 log q + ∑ i = 1 k k i ( - 1 ) i i q i - 1 .$

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

$k k - 1 i - 1 = i k i ( k ≥ i ≥ 1 ) .$

Thus, we find that

$F q ( t ) = 1 + log q ∑ k = 1 ∞ 1 ( 1 - q ) k t k k ! 1 log q + k ∑ i = 1 k k - 1 i - 1 ( - 1 ) i 1 q i - 1 = ∑ k = 0 ∞ 1 ( 1 - q ) k t k k ! + log q ∑ k = 1 ∞ k ( 1 - q ) k t k k ! ∑ m = 0 ∞ q m ∑ i = 0 k - 1 k - 1 i ( - 1 ) i q m i = e t 1 - q + log q 1 - q ∑ k = 1 ∞ k ( 1 - q ) k - 1 t k k ! ∑ m = 0 ∞ q m ( 1 - q m ) k - 1 = e t 1 - q + t log q 1 - q ∑ m = 0 ∞ q m ∑ k = 0 ∞ 1 - q m 1 - q k t k k ! .$

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

$G q ( t ) = ∑ k = 0 ∞ 2 k + 1 ( 1 − q ) k ∑ i = 0 k ( k i ) ( − 1 ) i q 1 2 i 1 q i + 1 t k k ! = 2 ∑ k = 0 ∞ 2 k ∑ m = 0 ∞ ( − 1 ) m ( 1 − q m + 1 2 1 − q ) k t k k ! = 2 ∑ m = 0 ∞ ( − 1 ) m ∑ k = 0 ∞ [ m + 1 2 ] q k ( 2 t ) k k ! = 2 ∑ m = 0 ∞ ( − 1 ) m e 2 [ m + 1 2 ] q t .$

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

$F q ( t , x ) = ∑ k = 0 ∞ B k ( x , q ) t k k ! = ∑ k = 0 ∞ ( q x B ( q ) + [ x ] q ) k t k k ! ,$
(3.2)
$G q ( t , x ) = ∑ k = 0 ∞ E k ( x , q ) t k k ! = ∑ k = 0 ∞ q x - 1 2 E ( q ) 2 + x - 1 2 q k t k k ! .$
(3.3)

Hence, we have

Lemma 3.3.

$F q ( t , x ) = e [ x ] q t F q ( q x t ) = e t 1 - q + t log q 1 - q ∑ m = 0 ∞ q m + x e [ m + x ] q t .$

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

$F q ( t , x ) = ∑ k = 0 ∞ ( q x B ( q ) + [ x ] q ) k t k k ! = e ( q x B ( q ) + [ x ] q ) t = e B ( q ) q x t e [ x ] q t = e [ x ] q t F q ( q x t ) .$

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

Lemma 3.4.

$G q ( t , x ) = e [ x - 1 2 ] q t G q q x - 1 2 2 t =2 ∑ m = 0 ∞ ( - 1 ) m e [ m + x ] q t .$

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).

$B k + 1 ( x + 1 , q ) - B k + 1 ( x , q ) = q x log q q - 1 ( k + 1 ) [ x ] q k ( k ≥ 0 ) , E k ( x + 1 , q ) + E k ( x , q ) = 2 [ x ] q k ( k ≥ 0 ) .$

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

$F q ( t , x ) = ∑ k = 0 ∞ B k ( x , q ) t k k ! = 1 + ∑ k = 0 ∞ 1 ( 1 - q ) k + 1 + ( k + 1 ) log q 1 - q ∑ m = 0 ∞ q m + x [ m + x ] q k t k + 1 ( k + 1 ) ! .$
(3.4)

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

$B k ( x , q ) = 1 ( 1 - q ) k + k log q 1 - q ∑ m = 0 ∞ q m + x [ m + x ] q k - 1 ( k ≥ 1 ) .$
(3.5)

Hence,

$B k ( x + 1 , q ) - B k ( x , q ) =k q x log q q - 1 [ x ] q k - 1 ( k ≥ 1 ) .$

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,

$[ x ] q k = 1 k + 1 q - 1 q x log q ∑ i = 0 k + 1 k + 1 i q i B i ( x , q ) - B k + 1 ( x , q ) , [ x ] q k = 1 2 ∑ i = 0 k k i q i E i ( x , q ) + E k ( x , q )$

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

respectively.

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

$B k + 1 x + 1 n , q n - B k + 1 x + 1 - n n , q n = n q n ( x - 1 ) + 1 log q q - 1 k + 1 [ n ] q k + 1 ( 1 + q [ n x - n ] q ) k , E k x + 1 n , q n + E k x + 1 - n n , q n = 2 [ n ] q k ( 1 + q [ n x - n ] q ) k .$

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

$B k x + 1 n - B k x + 1 - n n = k x + 1 - n n k - 1 ( k ≥ 1 , n ≥ 1 ) ,$
(3.6)
$E k x + 1 n + E k x + 1 - n n = 2 x + 1 - n n k ( k ≥ 0 , n ≥ 1 ) ,$
(3.7)

respectively (see [, (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.

$B 0 ( q ) = 1 , ( q B ( q ) + 1 ) k - B k ( q ) = log p q q - 1 i f k = 1 0 i f k > 1 , E 0 ( q ) = 1 , q - 1 2 E ( q ) 2 + - 1 2 q k + q 1 2 E ( q ) 2 + 1 2 q k = 0 i f k ≥ 1 .$

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 [, Proposition 2.4], Rim et al. [, Theorem 2.7] and Tsumura [, (1)].

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

$B 0 = 1 , ( B + 1 ) k - B k = 1 if k = 1 0 if k > 1$

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

$E 0 = 1 , ( E + 1 ) k + ( E - 1 ) k = 0 if k ≥ 1 .$

## 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

$ζ q ( s ) = 1 s - 1 1 1 1 - q s - 1 + log q q - 1 ∑ m = 1 ∞ q m [ m ] q s , ζ q , E ( s ) = 2 2 s ∑ m = 0 ∞ ( - 1 ) m [ m + 1 2 ] q s .$

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

$lim q → 1 ζ q ( s ) = ∑ m = 1 ∞ 1 m s = ζ ( s ) and lim q → 1 ζ q , E ( s ) = 2 ∑ m = 0 ∞ ( - 1 ) m ( m + 1 ) s = ζ E ( s ) .$
(4.3)

(In [, 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

$ζ q ( 1 - k ) = - B k ( q ) k a n d ζ q , E ( 1 - k ) = E k - 1 ( q ) .$

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

$ζ q ( s , x ) = 1 s - 1 1 ( 1 1 - q ) s - 1 + log q q - 1 ∑ m = 0 ∞ q m + x [ m + x ] q s , ζ q , E ( s , x ) = 2 ∑ m = 0 ∞ ( - 1 ) m [ m + x ] q s .$

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

$ζ q ( 1 - k , x ) = - B k ( x , q ) k a n d ζ q , E ( 1 - k , x ) = E k - 1 ( x , q ) .$

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

$F q ( t , x ) = 1 d ∑ i = 0 d - 1 F q d [ d ] q t , x + i d , G q ( t , x ) = ∑ i = 0 d - 1 ( - 1 ) i G q d [ d ] q t , x + i d i f d o d d .$

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

$F q ( t , x ) = e t 1 - q + t log q 1 - q ∑ m = 0 ∞ q m + x e [ m + x ] q t = e [ d ] q t 1 - q d + 1 d ∑ i = 0 d - 1 [ d ] q t log q d 1 - q d ∑ n = 0 ∞ q n d + x + i e [ n d + x + i ] q t = 1 d ∑ i = 0 d - 1 e [ d ] q t 1 - q d + [ d ] q t log q d 1 - q d ∑ n = 0 ∞ ( q d ) n + x + i d e [ n + x + i d ] q d [ d ] q t ,$

where we use $[ n + ( x + i ) / d ] q d [ d ] q = [ n d + x + i ] q$. 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

$ζ q ( 1 - k , x ) = [ d ] q k d ∑ i = 0 d - 1 ζ q d 1 - k , x + i d , ζ q , E ( - k , x ) = [ d ] q k ∑ i = 0 d - 1 ( - 1 ) i ζ q d , E - k , x + i d i f d o d d .$

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:

$F q , χ ( t , x ) = ∑ k = 0 ∞ B k , χ ( x , q ) t k k ! = 1 f ∑ a = 1 f χ ( a ) F q f [ f ] q t , a + x f$
(4.4)

and

$G q , χ ( t , x ) = ∑ k = 0 ∞ E k , χ ( x , q ) t k k ! = ∑ a = 1 f ( - 1 ) a χ ( a ) G q f [ f ] q t , a + x f if f odd ,$
(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

$F q , χ ( t , x ) = t log q 1 - q ∑ m = 0 ∞ χ ( m ) q m + x e [ m + x ] q t ,$
(4.6)
$G q , χ ( t , x ) = 2 ∑ k = 0 ∞ ( - 1 ) m χ ( m ) e [ m + x ] q t if f odd ,$
(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 :

$F χ ( t , x ) = ∑ a = 1 f χ ( a ) t e ( a + x ) t e f t - 1 = ∑ k = 0 ∞ B k , χ ( x ) t k k ! ,$
(4.8)
$G χ ( t , x ) = 2 ∑ a = 1 f ( - 1 ) a χ ( a ) e ( a + x ) t e f t + 1 = ∑ k = 0 ∞ G k , χ ( x ) t k k ! if f odd .$
(4.9)

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

$B k , χ ( x , q ) = [ f ] q k f ∑ a = 1 f χ ( a ) B k a + x f , q f ,$
(4.10)
$E k , χ ( x , q ) = [ f ] q k ∑ a = 1 f ( - 1 ) a χ ( a ) E k a + x f , q f if f odd .$
(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,

$L q ( s , x , χ ) = log q q - 1 ∑ m = 0 ∞ χ ( m ) q m + x [ m + x ] q s , ℓ q ( s , x , χ ) = 2 ∑ m = 0 ∞ ( - 1 ) m χ ( m ) [ m + x ] q s .$

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

Theorem 4.8. For k ≥ 1, we have

$L q ( 1 - k , x , χ ) = - B k , χ ( x , q ) k a n d ℓ q ( 1 - k , x , χ ) = E k - 1 , χ ( x , q ) .$

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

$∑ k = 0 ∞ B k , χ ( x , q ) t k k ! = 1 f ∑ a = 1 f χ ( a ) e [ f ] q t 1 - q f + [ f ] q t log q f 1 - q f ∑ n = 0 ∞ ( q f ) n + x + a f e [ n + x + a f ] q f [ f ] q t = t log q 1 - q ∑ m = 0 ∞ χ ( m ) q m + x e [ m + x ] q t ,$

where we use $[ n + ( a + x ) / f ] q f [ f ] q = [ n f + a + x ] q$ and $∑ a = 1 f χ ( a ) =0$. Therefore, we obtain

Hence for k ≥ 1

$- B k , χ ( x , q ) k = log q q - 1 ∑ m = 0 ∞ χ ( m ) q m + x [ m + x ] q k - 1 = L q ( 1 - k , x , χ ) .$

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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