# On the families of $q$-Euler numbers and polynomials and their applications

Serkan Aracı University of Gaziantep, Faculty of Science and Arts, Department
of Mathematics, 27310 Gaziantep, TURKEY<EMAIL_ADDRESS>,
Mehmet Acikgoz
University of Gaziantep, Faculty of Science and Arts, Department of
Mathematics, 27310 Gaziantep, TURKEY<EMAIL_ADDRESS>and
Hassan Jolany School of Mathematics, Statistics and Computer Science,
University of Tehran, Iran<EMAIL_ADDRESS>
###### Abstract.

In the present paper, we investigate special generalized $q$-Euler numbers and
polynomials. Some earlier results of T. Kim in terms of $q$-Euler polynomials
with weight $\alpha$ can be deduced. For presentation of our formulas we apply
the method of generating function and $p$-adic $q$-integral representation on
$\mathbb{Z}_{p}$. We summarize our results as follows. In section 2, by using
combinatorial techniques we present two formulas for $q$-Euler numbers with
weight $\alpha$. In section 3, we derive distribution formula (Multiplication
Theorem) for Dirichlet type of $q$-Euler numbers and polynomials with weight
$\alpha$. Moreover we define partial Dirichlet type zeta function and
Dirichlet $q$-$L$-function, and obtain some interesting combinatorial
identities for interpolating our new definitions. In addition, we derive
behavior of the Dirichlet type of $q$-Euler $L$-function with weight $\alpha$,
$\mathcal{L}_{q}^{\chi}\left(s,x\mid\alpha\right)$ at $s=0$. Furthermore by
using second kind stirling numbers, we obtain an explicit formula for
Dirichlet type $q$-Euler numbers with weight $\alpha$, and $\beta$. Moreover a
novel formula for $q$-Euler-Zeta function with weight $\alpha$ in terms of
nested series of $\widetilde{\zeta}_{E,q}\left(n\mid\alpha\right)$ is derived
. In section 4, by introducing $p$-adic Dirichlet type of $q$-Euler measure
with weight $\alpha$, and $\beta$, we obtain some combinatorial relations,
which interpolate our previous results. In section 5, which is the main
section of our paper. As an application, we introduce a novel concept of
dynamics of the zeros of analytically continued $q$-Euler polynomials with
weight $\alpha$.

###### Key words and phrases:

Euler numbers and polynomials, $q$-Euler numbers and polynomials, weighted$\
q$-Euler numbers and polynomials, weighted $q$-Euler-Zeta function, $p$-adic
$q$-integral on $\mathbb{Z}_{p}$.

###### 2000 Mathematics Subject Classification:

Primary 05A10, 11B65; Secondary 11B68, 11B73.

##  1\. Introduction

In this paper, we use notations like $\mathbb{N}$, $\mathbb{R}$ and
$\mathbb{C}$, where $\mathbb{N}$ denotes the set of natural numbers,
$\mathbb{R}$ denotes the field of real numbers and $\mathbb{C}$ also denotes
the set of complex numbers. When one talks of $q$-extension, $q$ is variously
considered as an indeterminate, a complex number or a $p$-adic number.

Throughout this paper, we will assume that $q\in\mathbb{C}$ with
$\left|q\right|<1$. The $q$-symbol $\left[x:q\right]$ denotes as

$\left[x:q\right]=\frac{q^{x}-1}{q-1}\text{.}$

Originally, $q$-Euler numbers and polynomials were introduced by L. Carlitz in
1948 and gave properties of this polynomials (see [20], [21]). Recently,
Taekyun Kim, by using $p$-adic $q$-integral in the $p$-adic integers ring, has
added a weight to $q$-Bernoulli numbers and polynomials and gave surprising
and fascinating identities of them (for details, see [8]). The $q$-Bernoulli
numbers and polynomials with weight $\alpha$ are related to weighted
$q$-Bernstein polynomials which is shown by Kim (for details, see [7]). These
polynomials have surprising properties in Analytic Numbers Theory and in
$p$-adic analysis, especially, in Mathematical physics. Ryoo also constructed
$q$-Euler numbers and polynomials with weight $\alpha$ and introduced some
properties of $q$-Euler numbers and polynomials with weight $\alpha$ in ”A
note on the weighted $q$-Euler numbers and polynomials with weight $\alpha$,
Advanced Studies Contemporary Mathematics 21 (2011), No. 1, 47-54.”

Analytic continuation of $q$-Euler numbers and polynomials was investigated by
Kim in [1]. In previous paper, Araci $et$ $al$. also considered analytic
continuation of weighted $q$-Genocchi numbers and polynomials and introduced
some interesting ideas (for detail, see [26]). In this article, we also
specify analytic continuation of weighted $q$-Euler numbers and polynomials.
Also, we give some interesting identities by using generating function of
Ryoo’s weighted $q$-Euler polynomials.

Because in the literature of our present paper we use of $p$-adic Arithmetic
and $p$-adic numbers, so we need to give short review on $p$-adic numbers.
Historically the $p$-adic numbers were introduced by K. Hensel in 1908 in his
book Theorie der algebraíschen Zahlen, Leipzig, 1908 (for more informations on
this subject, see [19]).

Let $p$ be a prime number, fixed once and for all. If $x$ is any rational
number other than 0, we can write $x$ in the form $x=p^{n}\frac{a}{b}$ , where
$a,b\in\mathbb{Z}$ are relatively prime to $p$ and $n\in\mathbb{Z}$. We now
define

$|x|_{p}=p^{-n}\text{ and }|0|_{p}=0,\text{and }ord_{p}(x)=n\text{ and
}ord_{p}(0)=+\infty\text{.}$

They satisfy the following properties,

$\displaystyle|x|_{p}\geq 0,\text{ }|x|_{p}=0\text{ if and only if }x=0$
$\displaystyle|x+y|_{p}\leq\max\\{|x|_{p},|y|_{p}\\}\text{ }(\text{the strong
triangle inequality})$ with
$\displaystyle|x+y|_{p}=\max\\{|x|_{p},|y|_{p}\\}\text{ if
}|x|_{p}\neq|y|_{p}(\text{the isosceles triangle principle})$
$\displaystyle|x.y|_{p}=|x|_{p}.|y|_{p}$

$|x|_{p}$ is called the $p$-adic valuation. Ostrowski proved that each
nontrivial valuation on the field of rational numbers is equivalent either to
the absolute value function or to some $p$-adic valuation. The completion of
the field $\mathbb{Q}$ of rational numbers with respect to the p-adic
valuation $|.|_{p}$ is called the field of $p$-adic numbers and will be
denoted $\mathbb{Q}_{p}$. The set

$\mathbb{Z}_{p}=\\{x\in\mathbb{Q}_{p}\mid|x|_{p}\leq 1\\}$

is the ring of $p$-adic integers. It can be easily proved that each $p$-adic
number $x$ can be written in the form

$x=\sum_{n=-f}^{\infty}a_{n}p^{n}$

where each $a_{n}$ is one of the elements $0,1,\cdot\cdot\cdot,p-1$, and
$f\in\mathbb{Z}$. This is called the Hensel representation of $p$-adic
numbers. With this representation, one obtain for $x\in\mathbb{Q}_{p}$,
$ord_{p}(x)=+\infty$ if $a_{i}=0$ for all $i$ and
$ord_{p}(x)=min\\{s|a_{s}\neq 0\\}$, otherwise.
Moreover we can write

$|x|_{p}=p^{-ord_{p}(x)}\text{.}$

##  2\. Properties of the $q$-Euler Numbers and polynomials with weight
$\alpha$

For $\alpha\in\mathbb{N}\mathop{\textstyle\bigcup}\left\\{0\right\\}$, the
weighted $q$-Euler polynomials are given as:

For $x\in\mathbb{C}$,

(2.1)
$\sum_{n=0}^{\infty}\widetilde{E}_{n,q}\left(x\mid\alpha\right)\frac{t^{n}}{n!}=\left[2:q\right]\sum_{n=0}^{\infty}\left(-1\right)^{n}q^{n}e^{t\left[n+x:q^{\alpha}\right]}\text{.}$

As a special case, substituting $x=0$ into (2.1),
$\widetilde{E}_{n,q}\left(0\mid\alpha\right):=\widetilde{E}_{n,q}\left(\alpha\right)$
are called weighted $q$-Euler numbers. By (2.1), we readily derive the
following

(2.2)
$\widetilde{E}_{n,q}\left(x\mid\alpha\right)=\frac{\left[2:q\right]}{\left[\alpha:q\right]^{n}\left(1-q\right)^{n}}\sum_{l=0}^{n}\binom{n}{l}\left(-1\right)^{l}\frac{q^{\alpha
lx}}{1+q^{\alpha l+1}}\text{,}$

where $\binom{n}{l}$ is the binomial coefficient. By expression (2.1), we see
that

(2.3) $\widetilde{E}_{n,q}\left(x\mid\alpha\right)=q^{-\alpha
x}\left(q^{\alpha
x}\widetilde{E}_{q}\left(\alpha\right)+\left[x:q^{\alpha}\right]\right)^{n}\text{,}$

with the usual convention of replacing
$\left(\widetilde{E}_{q}\left(\alpha\right)\right)^{n}$ by
$\widetilde{E}_{n,q}\left(\alpha\right)$ (for details, see [14]).

Let $\widetilde{H}_{q}^{\left(\alpha\right)}\left(x,t\right)$ be the
generating function of weighted $q$-Euler polynomials as follows:

(2.4)
$\widetilde{H}_{q}^{\left(\alpha\right)}\left(x,t\right)=\sum_{n=0}^{\infty}\widetilde{E}_{n,q}\left(x\mid\alpha\right)\frac{t^{n}}{n!}\text{.}$

Then, we easily notice that

(2.5)
$\widetilde{H}_{q}^{\left(\alpha\right)}\left(x,t\right)=\left[2:q\right]\sum_{n=0}^{\infty}\left(-1\right)^{n}q^{n}e^{t\left[n+x:q^{\alpha}\right]}\text{.}$

From expressions (2.4) and (2.5), we procure the followings:

For $k$ (=even) and
$n,\alpha\in\mathbb{N}\mathop{\textstyle\bigcup}\left\\{0\right\\}$, we have

(2.6)
$\widetilde{E}_{n,q}\left(\alpha\right)-q^{k}\widetilde{E}_{n,q}\left(k\mid\alpha\right)=\left[2:q\right]\sum_{l=0}^{k-1}\left(-1\right)^{l}q^{l}\left[l:q^{\alpha}\right]^{n}\text{.}$

For $k$ (=odd) and
$n,\alpha\in\mathbb{N}\mathop{\textstyle\bigcup}\left\\{0\right\\}$, we have

(2.7)
$q^{k}\widetilde{E}_{n,q}\left(k\mid\alpha\right)+\widetilde{E}_{n,q}\left(\alpha\right)=\left[2:q\right]\sum_{l=0}^{k-1}\left(-1\right)^{l}q^{l}\left[l:q^{\alpha}\right]^{n}\text{.}$

Via Eq. (2.5), we easily obtain the following:

(2.8) $\widetilde{E}_{n,q}\left(x\mid\alpha\right)=q^{-\alpha
x}\sum_{k=0}^{n}\binom{n}{k}q^{\alpha
kx}\widetilde{E}_{k,q}\left(\alpha\right)\left[x:q^{\alpha}\right]^{n-k}\text{.}$

From (2.6)-(2.8), we get the following theorem.

######  Theorem 1.

Let $k$ be even positive integer. Then we have

(2.9)
$\displaystyle\left[2:q\right]\sum_{l=0}^{k-1}\left(-1\right)^{l}q^{l}\left[l:q^{\alpha}\right]^{n}$
$\displaystyle=\left(1-q^{k\left(1-\alpha+\alpha
n\right)}\right)\widetilde{E}_{n,q}\left(\alpha\right)-q^{k\left(1-\alpha\right)}\sum_{l=0}^{n-1}\binom{n}{l}q^{\alpha
lk}\widetilde{E}_{l,q}\left(\alpha\right)\left[k:q^{\alpha}\right]^{n-l}\text{.}$

######  Theorem 2.

Let $k$ be an odd positive integer. Then, we procure the following

(2.10)
$\displaystyle\left[2:q\right]\sum_{l=0}^{k-1}\left(-1\right)^{l}q^{l}\left[l:q^{\alpha}\right]^{n}$
$\displaystyle=\left(q^{k\left(1-\alpha+\alpha
n\right)}+1\right)\widetilde{E}_{n,q}\left(\alpha\right)+q^{k\left(1-\alpha\right)}\sum_{l=0}^{n-1}\binom{n}{l}q^{\alpha
lk}\widetilde{E}_{l,q}\left(\alpha\right)\left[k:q^{\alpha}\right]^{n-l}\text{.}$

##  3\. $q$-Euler-Zeta function with weight $\alpha$

The familiar Euler polynomials are defined by

(3.1)
$\frac{2}{e^{t}+1}e^{xt}=\sum_{n=0}^{\infty}E_{n}\left(x\right)\frac{t^{n}}{n!},\text{
}\left|t\right|<\pi\text{ cf. \cite[cite]{[\@@bibref{}{kim 4}{}{}]}.}$

For $s\in\mathbb{C}$, $x\in\mathbb{R}$ with $0\leq x<1$, Euler-Zeta function
is given by

(3.2)
$\zeta_{E}\left(s,x\right)=2\sum_{m=0}^{\infty}\frac{\left(-1\right)^{m}}{\left(m+x\right)^{s}}\text{,
}$

and

(3.3)
$\zeta_{E}\left(s\right)=\sum_{m=1}^{\infty}\frac{\left(-1\right)^{m}}{m^{s}}\text{.}$

By expressions (3.1), (3.2) and (3.3), Euler-Zeta functions are related to the
Euler numbers as follows:

$\zeta_{E}\left(-n\right)=E_{n}\text{.}$

Moreover, it is simple to see

$\zeta_{E}\left(-n,x\right)=E_{n}\left(x\right)\text{.}$

The weighted $q$-Euler Hurwitz-Zeta type function is defined by

$\widetilde{\zeta}_{E,q}\left(s,x\mid\alpha\right)=\left[2:q\right]\sum_{m=0}^{\infty}\frac{\left(-1\right)^{m}q^{m}}{\left[m+x:q^{\alpha}\right]^{s}}\text{
.}$

Similarly, weighted $q$-Euler-Zeta function is given by

$\widetilde{\zeta}_{E,q}\left(s\mid\alpha\right)=\left[2:q\right]\sum_{m=1}^{\infty}\frac{\left(-1\right)^{m}q^{m}}{\left[m:q^{\alpha}\right]^{s}}\text{.}$

For $n,\alpha\in\mathbb{N}\mathop{\textstyle\bigcup}\left\\{0\right\\}$, we
have

$\widetilde{\zeta}_{E,q}\left(-n\mid\alpha\right)=\widetilde{E}_{n,q}\left(\alpha\right)\text{
(see \cite[cite]{[\@@bibref{}{Ryoo}{}{}]}).}$

We now consider the function $\widetilde{E}_{q}\left(n:\alpha\right)$ as the
analytic continuation of weighted $q$-Euler numbers. All the weighted
$q$-Euler numbers agree with $\widetilde{E}_{q}\left(n:\alpha\right)$, the
analytic continuation of weighted $q$-Euler numbers evaluated at $n$. For
$n\geq 0$,
$\widetilde{E}_{q}\left(n:\alpha\right)=\widetilde{E}_{n,q}\left(\alpha\right)$.

We can now state $\widetilde{E}{\acute{}}_{q}\left(s:\alpha\right)$ in terms
of $\widetilde{\zeta}{\acute{}}_{E,q}\left(s\mid\alpha\right)$, the derivative
of $\widetilde{\zeta}_{E,q}\left(s:\alpha\right)$

$\widetilde{E}_{q}\left(s:\alpha\right)=\widetilde{\zeta}_{E,q}\left(-s\mid\alpha\right)\text{,
}\widetilde{E}{\acute{}}_{q}\left(s:\alpha\right)=\widetilde{\zeta}{\acute{}}_{E,q}\left(-s\mid\alpha\right)\text{.}$

For $n,\alpha\in\mathbb{N}\mathop{\textstyle\bigcup}\left\\{0\right\\}$

$\text{
}\widetilde{E}{\acute{}}_{q}\left(2n:\alpha\right)=\widetilde{\zeta}{\acute{}}_{E,q}\left(-2n\mid\alpha\right)\text{.}$

This is suitable for the differential of the functional equation and so
supports the coherence of $\widetilde{E}_{q}\left(s:\alpha\right)$ and
$\widetilde{E}{\acute{}}_{q}\left(s:\alpha\right)$ with
$\widetilde{E}_{n,q}\left(\alpha\right)$ and
$\widetilde{\zeta}_{E,q}\left(s\mid\alpha\right)$. From the analytic
continuation of weighted $q$-Euler numbers, we derive as follows:

$\widetilde{E}_{q}\left(s:\alpha\right)=\widetilde{\zeta}_{E,q}\left(-s\mid\alpha\right)\text{
and
}\widetilde{E}_{q}\left(-s:\alpha\right)=\widetilde{\zeta}_{E,q}\left(s\mid\alpha\right)\text{.}$

Moreover, we derive the following:

For $n\in\mathbb{N}$

$\widetilde{E}_{-n,q}\left(\alpha\right)=\widetilde{E}_{q}\left(-n:\alpha\right)=\widetilde{\zeta}_{E,q}\left(n\mid\alpha\right)\text{.}$

The curve $\widetilde{E}_{q}\left(s:a\right)$ runs through the points
$\widetilde{E}_{-s,q}\left(\alpha\right)$ and grows $\sim n$ asymptotically
$\left(-n\right)\rightarrow-\infty$. The curve
$\widetilde{E}_{q}\left(s:a\right)$ runs through the point
$\widetilde{E}_{q}\left(-n:a\right)$. Then, we procure the following:

$\displaystyle\lim_{n\rightarrow\infty}\widetilde{E}_{q}\left(-n:\alpha\right)$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\widetilde{\zeta}_{E,q}\left(n\mid\alpha\right)=\lim_{n\rightarrow\infty}\left(\left[2:q\right]\sum_{m=1}^{\infty}\frac{\left(-1\right)^{m}q^{m}}{\left[m:q^{\alpha}\right]^{n}}\right)$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\left(-q\left[2:q\right]+\left[2:q\right]\sum_{m=2}^{\infty}\frac{\left(-1\right)^{m}q^{m}}{\left[m:q^{\alpha}\right]^{n}}\right)=-q^{2}\left[2:q^{-1}\right]\text{.}$

From this, we note that

$\widetilde{E}_{q}\left(-n:\alpha\right)=\widetilde{\zeta}_{E,q}\left(n\mid\alpha\right)\mapsto\widetilde{E}_{q}\left(-s:\alpha\right)=\widetilde{\zeta}_{E,q}\left(s\mid\alpha\right)\text{.}$

Notations: Assume that $p$ be a fixed odd prime number. Throughout this paper
we use the following notations. By $\mathbb{Z}_{p}$ we denote the ring of
$p$-adic rational integers, $\mathbb{Q}$ denotes the field of rational
numbers, $\mathbb{Q}_{p}$ denotes the field of $p$-adic rational numbers, and
$\mathbb{C}_{p}$ denotes the completion of algebraic closure of
$\mathbb{Q}_{p}$. Let $\mathbb{N}$ be the set of natural numbers and
$\mathbb{N}^{\ast}=\mathbb{N}\cup\left\\{0\right\\}$. The $p$-adic absolute
value is defined by $\left|p\right|_{p}=\frac{1}{p}$. In this paper we assume
$\left|q-1\right|_{p}<1$ as an indeterminate. Let
$UD\left(\mathbb{Z}_{p}\right)$ be the space of uniformly differentiable
functions on $\mathbb{Z}_{p}$. For a positive integer $d$ with
$\left(d,p\right)=1$, set

$\displaystyle X$ $\displaystyle=$ $\displaystyle
X_{d}=\lim_{\overleftarrow{n}}\mathbb{Z}/dp^{n}\mathbb{Z}\text{,}$
$\displaystyle X^{\ast}$ $\displaystyle=$
$\displaystyle\underset{\underset{\left(a,p\right)=1}{0<a<dp}}{\cup}a+dp\mathbb{Z}_{p}$

and

$a+dp^{n}\mathbb{Z}_{p}=\left\\{x\in X\mid x\equiv
a\left(\mathop{\mathrm{m}od}dp^{n}\right)\right\\}\text{,}$

where $a\in\mathbb{Z}$ satisfies the condition $0\leq a<dp^{n}$.

Firstly, for introducing fermionic $p$-adic $q$-integral, we need some basic
information which we state here. A measure on $\mathbb{Z}_{p}$ with values in
a $p$-adic Banach space $B$ is a continuous linear map

$f\mapsto\int f(x)\mu=\int_{\mathbb{Z}_{p}}f(x)\mu(x)$

from $C^{0}(\mathbb{Z}_{p},\mathbb{C}_{p})$, (continuous function on
$\mathbb{Z}_{p}$ ) to $B$. We know that the set of locally constant functions
from $\mathbb{Z}_{p}$ to $\mathbb{Q}_{p}$ is dense in
$C^{0}(\mathbb{Z}_{p},\mathbb{C}_{p})$ so.

Explicitly, for all $f\in C^{0}(\mathbb{Z}_{p},\mathbb{C}_{p})$, the locally
constant functions

$f_{n}=\sum_{i=0}^{p^{n}-1}f(i)1_{i+p^{n}\mathbb{Z}_{p}}\rightarrow f\text{ in
}C^{0}$

Now, set
$\mu(i+p^{n}\mathbb{Z}_{p})=\int_{\mathbb{Z}_{p}}1_{i+p^{n}\mathbb{Z}_{p}}\mu$.
Then $\int_{\mathbb{Z}_{p}}f\mu$, is given by the following Riemannian sum

$\int_{\mathbb{Z}_{p}}f\mu=\lim_{n\rightarrow\infty}\sum_{i=0}^{p^{n}-1}f(i)\mu{(i+p^{n}\mathbb{Z}_{p})}$

T. Kim introduced $\mu$ as follows:

$\mu_{-q}(a+p^{n}\mathbb{Z}_{p})=\frac{(-q)^{a}}{[p^{n}]_{-q}}$

So, for $f\in UD\left(\mathbb{Z}_{p}\right)$, the fermionic $p$-adic
$q$-integral on $\mathbb{Z}_{p}$ is defined by Kim as follows:

$\displaystyle I_{-q}\left(f\right)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{Z}_{p}}f\left(\eta\right)d\mu_{-q}\left(\eta\right)$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\left[p^{n}:-q\right]}\sum_{\eta=0}^{p^{n}-1}q^{\eta}f\left(\eta\right)\left(-1\right)^{\eta}\text{.}$

Let $\chi$ be the Dirichlet’s character with conductor $d$ (=
odd)$\in\mathbb{N}$ and let us take
$f\left(\eta\right)=\chi\left(\eta\right)\left[x+\eta:q^{\alpha}\right]^{n}$,
then we define Dirichlet’s type of $q$-Euler numbers and polynomials with
weight $\alpha$ as follows:

(3.5)
$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha\right)=\int_{\mathbb{Z}_{p}}\chi\left(\eta\right)\left[x+\eta:q^{\alpha}\right]^{n}d\mu_{-q}\left(\eta\right)\text{.}$

From (3), we have the following well-known equality.

(3.6)
$q^{d}\int_{\mathbb{Z}_{p}}f\left(\eta+d\right)d\mu_{-q}\left(\eta\right)+\left(-1\right)^{d-1}\int_{\mathbb{Z}_{p}}f\left(\eta\right)d\mu_{-q}\left(\eta\right)=\left[2:q\right]\sum_{l=0}^{d-1}q^{l}\left(-1\right)^{d-1-l}f\left(l\right)\text{.}$

By expressions of (3.5) and (3.6), for $d$ $($=odd$)$ positive integer, we
have the following

(3.7)
$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha\right)=\frac{\left[2:q\right]}{\left[\alpha:q\right]^{n}\left(1-q\right)^{n}}\sum_{j=0}^{n}\binom{n}{j}\left(-1\right)^{j}q^{\alpha
jx}\sum_{l=0}^{d-1}\chi\left(l\right)q^{l}\left(-1\right)^{l}\frac{q^{\alpha
jl}}{q^{\left(\alpha j+1\right)d}+1}\text{.}$

Substituting $x=0$ in (3.7),
$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(0\mid\alpha\right):=$
$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(\alpha\right)$ are called Dirichlet
type of $q$-Euler numbers with weight $\alpha$. That is, we easily derive the
following

(3.8)
$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(\alpha\right)=\frac{\left[2:q\right]}{\left(1-q^{\alpha}\right)^{n}}\sum_{j=0}^{n}\binom{n}{j}\left(-1\right)^{j}\sum_{l=0}^{d-1}\chi\left(l\right)\left(-1\right)^{l}\frac{q^{\left(\alpha
j+1\right)l}}{q^{\left(\alpha j+1\right)d}+1}\text{.}$

######  Theorem 3.

Let $\chi$ be Dirichlet’s character and for any $n\in\mathbb{N}^{\ast}$. Then
we have

$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha\right)=\sum_{k=0}^{n}\binom{n}{k}q^{\alpha
kx}\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(\alpha\right)\left[x:q^{\alpha}\right]^{n-k}\text{.}$

###### Proof.

By using (3.5) and (3.8), becomes

$\displaystyle\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha\right)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{Z}_{p}}\chi\left(\eta\right)\left[x+\eta:q^{\alpha}\right]^{n}d\mu_{-q}\left(\eta\right)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{Z}_{p}}\chi\left(\eta\right)\left(\left[x:q^{\alpha}\right]+q^{\alpha
x}\left[\eta:q^{\alpha}\right]\right)^{n}d\mu_{-q}\left(\eta\right)\text{.}$

From this, by using binomial theorem, we can write the following

$\displaystyle\sum_{k=0}^{n}\binom{n}{k}q^{\alpha
kx}\left[x:q^{\alpha}\right]^{n-k}\int_{\mathbb{Z}_{p}}\chi\left(\eta\right)\left[\eta:q^{\alpha}\right]^{k}d\mu_{-q}\left(\eta\right)$
$\displaystyle=$ $\displaystyle\sum_{k=0}^{n}\binom{n}{k}q^{\alpha
kx}\left[x:q^{\alpha}\right]^{n-k}\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(\alpha\right)\text{.}$

Thus, we complete the proof of the theorem.

######  Theorem 4.

The following identity

$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(dx\mid\alpha\right)=\frac{\left[d:q^{\alpha}\right]}{\left[d:-q\right]}\sum_{a=0}^{d-1}\left(-1\right)^{a}\chi\left(a\right)q^{a}\widetilde{E}_{n,q^{d}}\left(x+\frac{a}{d}\mid\alpha\right)$

holds true.

###### Proof.

To prove this, we compute as follows:

$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\left[dp^{n}:-q\right]}\sum_{y=0}^{dp^{n}-1}\left(-q\right)^{y}\chi\left(y\right)\left[x+y:q^{\alpha}\right]^{n}$
$\displaystyle=$
$\displaystyle\frac{1}{\left[d:-q\right]}\lim_{n\rightarrow\infty}\frac{1}{\left[p^{n}:-q^{d}\right]}\sum_{y=0}^{p^{n}-1}\sum_{a=0}^{d-1}\left(-q\right)^{a+dy}\chi\left(a+dy\right)\left[x+a+dy:q^{\alpha}\right]^{n}$
$\displaystyle=$
$\displaystyle\frac{\left[d:q^{\alpha}\right]}{\left[d:-q\right]}\sum_{a=0}^{d-1}\left(-q\right)^{a}\chi\left(a\right)\lim_{n\rightarrow\infty}\frac{1}{\left[p^{n}\right]_{-q^{d}}}\sum_{y=0}^{p^{n}-1}\left(-q\right)^{dy}\left[\frac{x+a}{d}+y:q^{d\alpha}\right]^{n}$
$\displaystyle=$
$\displaystyle\frac{\left[d:q^{\alpha}\right]}{\left[d:-q\right]}\sum_{a=0}^{d-1}\left(-q\right)^{a}\chi\left(a\right)\widetilde{E}_{n,q^{d}}\left(\frac{x+a}{d}\mid\alpha\right)\text{.}$

So, we get the desired result and proof is complete.

By (3.7), we procure the following:

(3.9)
$\sum_{n=0}^{\infty}\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha\right)\frac{t^{n}}{n!}=\left[2:q\right]\sum_{m=0}^{\infty}q^{m}\chi\left(m\right)\left(-1\right)^{m}e^{t\left[x+m:q^{\alpha}\right]}\text{.}$

By applying derivative operator of order $k$ as
$\frac{d^{k}}{dt^{k}}\mid_{t=0}$, we have the following

$\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(x\mid\alpha\right)=\left[2:q\right]\sum_{m=0}^{\infty}q^{m}\chi\left(m\right)\left(-1\right)^{m}\left[x+m:q^{\alpha}\right]^{k}\text{.}$

That is, we can define Dirichlet $q$-$L$-function as follows:

(3.10)
$\mathcal{L}_{q}^{\chi}\left(s,x\mid\alpha\right)=\left[2:q\right]\sum_{m=0}^{\infty}\frac{q^{m}\chi\left(m\right)\left(-1\right)^{m}}{\left[x+m:q^{\alpha}\right]^{s}}.$

######  Lemma 1.

The following equality holds true:

$\mathcal{L}_{q}^{\chi}\left(-k,x\mid\alpha\right)=\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(x\mid\alpha\right)\text{.}$

###### Proof.

Substituting $s=-k$ into (3.10), we arrive at the desired result.

Now also, we define partial Dirichlet type zeta function as follows:

(3.11)
$\mathcal{H}_{q}^{\chi}\left(s:x:a:F\mid\alpha\right)=\left[2:q\right]\sum_{m\equiv
a\left(\mathop{\mathrm{m}od}F\right)}^{\infty}\frac{q^{m}\chi\left(m\right)\left(-1\right)^{m}}{\left[x+m:q^{\alpha}\right]^{s}}\text{.}$

Now, for interpolating partial Dirichlet type zeta function, we rewrite it in
terms of weighted $q$-Euler Hurwitz-Zeta function as follows.

######  Theorem 5.

For $F\equiv 1(\mathop{\mathrm{m}od}2)$, then the following equality holds
true:

(3.12)
$\mathcal{H}_{q}^{\chi}\left(s:x:a:F\mid\alpha\right)=\frac{\left[2:q\right]q^{a}\left(-1\right)^{a}\chi\left(a\right)}{\left[F:q^{\alpha}\right]^{s}}\widetilde{\zeta}_{E,q^{F}}\left(s,\frac{x+a}{F}\mid\alpha\right)\text{.}$

###### Proof.

By expression of (3.11), we compute as follows:

$\displaystyle\mathcal{H}_{q}^{\chi}\left(s:x:a:F\mid\alpha\right)$
$\displaystyle=$ $\displaystyle\left[2:q\right]\sum_{m\equiv
a\left(\mathop{\mathrm{m}od}F\right)}^{\infty}\frac{q^{m}\chi\left(m\right)\left(-1\right)^{m}}{\left[x+m:q^{\alpha}\right]^{s}}$
$\displaystyle=$
$\displaystyle\left[2:q\right]\sum_{m=0}^{\infty}\frac{q^{mF+a}\chi\left(mF+a\right)\left(-1\right)^{mF+a}}{\left[x+mF+a:q^{\alpha}\right]^{s}}$
$\displaystyle=$
$\displaystyle\frac{\left[2:q\right]q^{a}\left(-1\right)^{a}\chi\left(a\right)}{\left[F:q^{\alpha}\right]^{s}}\sum_{m=0}^{\infty}\frac{\left(q^{F}\right)^{m}\left(-1\right)^{m}}{\left[\frac{x+a}{F}+m:q^{F\alpha}\right]^{s}}\text{.}$

Thus, we arrive at the desired result.

If we put $s=-n$ into (3.12), then, we can write partial Dirichlet type Zeta
function in terms of weighted $q$-Euler numbers

(3.13)
$\mathcal{H}_{q}^{\chi}\left(-n:x:a:F\mid\alpha\right)=\left[2:q\right]q^{a}\left(-1\right)^{a}\chi\left(a\right)\left[F:q^{\alpha}\right]^{n}\widetilde{E}_{n,q^{F}}\left(\frac{x+a}{F}\mid\alpha\right)\text{.}$

######  Theorem 6.

The following identity

(3.14)
$\mathcal{H}_{q}^{\chi}\left(s:x:a:F\mid\alpha\right)=\frac{\left[2:q\right]q^{a}\left(-1\right)^{a}\chi\left(a\right)}{\left[x+a:q^{\alpha}\right]^{s}}\sum_{k=0}^{\infty}q^{\alpha
k\left(x+a\right)}\binom{-s}{k}\left(\frac{\left[F:q^{\alpha}\right]}{\left[x+a:q^{\alpha}\right]}\right)^{k}\widetilde{E}_{k,q^{F}}$

holds true.

###### Proof.

Taking $n=-s$ into (3.13) and some manipulation by using combinatorial
techniques, we can reach to the proof of the theorem.

If we substitute $s=-n$ into (3.14), then, (3.14) reduces to (3.13). Now also,
we give the following theorem.

######  Theorem 7.

Let $d\equiv 1(\mathop{\mathrm{m}od}2)$, then, we have

$\mathcal{L}_{q}^{\chi}\left(s,x\mid\alpha\right)=\frac{\left[2:q\right]}{\left[2:q^{d}\right]\left[d:q^{\alpha}\right]^{s}}\sum_{l=0}^{d-1}\left(-1\right)^{l}\chi\left(l\right)q^{l}\widetilde{\zeta}_{E,q^{d}}\left(s,\frac{x+l}{d}\mid\alpha\right)\text{.}$

###### Proof.

By using (3.10), we compute as follows:

$\displaystyle\mathcal{L}_{q}^{\chi}\left(s,x\mid\alpha\right)$
$\displaystyle=$
$\displaystyle\left[2:q\right]\sum_{m=0}^{\infty}\sum_{l=0}^{d-1}\frac{q^{l+md}\chi\left(l+md\right)\left(-1\right)^{l+md}}{\left[x+l+md:q^{\alpha}\right]^{s}}$
$\displaystyle=$
$\displaystyle\frac{\left[2:q\right]}{\left[2:q^{d}\right]\left[d:q^{\alpha}\right]^{s}}\sum_{l=0}^{d-1}\left(-1\right)^{l}\chi\left(l\right)q^{l}\widetilde{\zeta}_{E,q^{d}}\left(s,\frac{x+l}{d}\mid\alpha\right)\text{.}$

Thus, we prove the above theorem.

By means of the above theorem and (3.12), we have the following corollary.

######  Corollary 1.

The following equality

(3.15)
$\mathcal{L}_{q}^{\chi}\left(s,x\mid\alpha\right)=\frac{1}{\left[2:q^{d}\right]}\sum_{l=0}^{d-1}\mathcal{H}_{q}^{\chi}\left(s:x:l:d\mid\alpha\right)$

holds true.

By (3.14) and (3.15), we have the following corollary.

######  Corollary 2.

The following nice identity

$\mathcal{L}_{q}^{\chi}\left(s,x\mid\alpha\right)=\frac{\left[2:q\right]}{\left[2:q^{d}\right]}\sum_{l=0}^{d-1}\frac{\left(-1\right)^{l}\chi\left(l\right)}{\left[x+l:q^{\alpha}\right]^{s}}\sum_{k=0}^{\infty}q^{\alpha
k\left(x+l\right)+l}\binom{-s}{k}\widetilde{E}_{k,q^{F}}\left(\frac{\left[F:q^{\alpha}\right]}{\left[x+l:q^{\alpha}\right]}\right)^{k}$

holds true.

By using (3.13) and (3.15), we derive behavior of the Dirichlet type of
$q$-Euler $L$-function with weight $\alpha$ at $s=0$ as follows:

######  Theorem 8.

The following identity holds true:

$\mathcal{L}_{q}^{\chi}\left(0,x\mid\alpha\right)=\frac{\left[2:q\right]}{\left[2:q^{d}\right]}\sum_{l=0}^{d-1}\left(-1\right)^{l}\chi\left(l\right)q^{l}\text{.}$

Now also, we define Dirichlet type $q$-Euler polynomials with weight $\alpha$
and $\beta$ with the following expression

(3.16)
$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha:\beta\right)=\int_{\mathbb{Z}_{p}}\left[x+\eta:q^{\alpha}\right]^{n}\chi\left(\eta\right)d\mu_{-q^{\beta}}\left(\eta\right)\text{.}$

Taking $x=0$ into (3.16), we have
$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(0\mid\alpha:\beta\right):=\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(\alpha:\beta\right)$
which is called Dirichlet type $q$-Euler numbers. Then, by (3.16), we easily
derive the following

(3.17)
$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha:\beta\right)=\sum_{l=0}^{n}\binom{n}{l}q^{\alpha
lx}\widetilde{\mathcal{E}}_{l,q}^{\chi}\left(\alpha:\beta\right)\left[x:q^{\alpha}\right]^{n-l}\text{.}$

######  Theorem 9.

The following equality holds true:

$\displaystyle\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha:\beta\right)$
$\displaystyle=$
$\displaystyle\sum_{l=0}^{n}\binom{n}{l}\widetilde{\mathcal{E}}_{l,q}^{\chi}\left(\alpha:\beta\right)\sum_{j=0}^{l}\binom{l}{j}\left(q^{\alpha}-1\right)^{j}\left(n-l+j\right)!\left(-1\right)^{n-l+j}$
$\displaystyle\times\sum_{m,n=0}^{\infty}\binom{n-l+j+m-1}{m}\alpha^{n}q^{\alpha
m}\left(\log
q\right)^{n}\mathcal{S}\left(n,n-l+j\right)\frac{x^{n}}{n!}\text{.}$

###### Proof.

To prove this, by applying (3.17), we easily discover the following assertion

$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha:\beta\right)=\sum_{l=0}^{n}\binom{n}{l}\widetilde{\mathcal{E}}_{l,q}^{\chi}\left(\alpha:\beta\right)\sum_{j=0}^{l}\binom{l}{j}\left(q^{\alpha}-1\right)^{j}\left[x:q^{\alpha}\right]^{n-l+j}\text{.}$

The Second kind Stirling numbers are defined by means of the following
generating function.

(3.18)
$\sum_{n=0}^{\infty}\mathcal{S}\left(n,k\right)\frac{t^{n}}{n!}=\frac{\left(e^{t}-1\right)^{k}}{k!}$

(for details on this subject, see [12]). $t$ replace by $\alpha x\log q$ in
(3.18), then, we easily derive the following

(3.19)
$\left[x:q^{\alpha}\right]^{k}=k!\left(-1\right)^{k}\sum_{m,n=0}^{\infty}\binom{k+m-1}{m}\alpha^{n}q^{\alpha
m}\left(\log q\right)^{n}\mathcal{S}\left(n,k\right)\frac{x^{n}}{n!}\text{,}$

where $\sum_{m,n=0}^{\infty}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}$. Thus, by
(3.18) and (3.19), we get the desired result and proof is complete.

######  Theorem 10.

The following equality

$\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha:\beta\right)=\frac{\left[d:q^{\alpha}\right]^{n}}{\left[d:-q^{\beta}\right]}\sum_{a=0}^{d-1}\left(-q\right)^{a}\chi\left(a\right)\widetilde{E}_{n,q^{d}}\left(\frac{x+a}{d}\mid\alpha:\beta\right)$

holds true.

###### Proof.

By applying the $p$-adic integral representation on the Dirichlet type of
$q$-Euler polynomials with weight $\alpha$ and $\beta$, we compute as follows:

$\displaystyle\widetilde{\mathcal{E}}_{n,q}^{\chi}\left(x\mid\alpha:\beta\right)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{Z}_{p}}\chi\left(\eta\right)\left[x+\eta:q^{\alpha}\right]^{n}d\mu_{-q^{\beta}}\left(\eta\right)$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\left[dp^{n}:-q^{\beta}\right]}\sum_{y=0}^{dp^{n}-1}\left(-q\right)^{y}\chi\left(y\right)\left[x+y:q^{\alpha}\right]$
$\displaystyle=$
$\displaystyle\frac{1}{\left[d:-q^{\beta}\right]}\lim_{n\rightarrow\infty}\frac{1}{\left[p^{n}:-q^{d\beta}\right]}\sum_{y=0}^{p^{n}-1}\sum_{a=0}^{d-1}\left(-q\right)^{a+dy}\chi\left(a+dy\right)\left[x+a+dy:q^{\alpha}\right]^{n}$
$\displaystyle=$
$\displaystyle\frac{\left[d:q^{\alpha}\right]^{n}}{\left[d:-q^{\beta}\right]}\sum_{a=0}^{d-1}\left(-q\right)^{a}\chi\left(a\right)\lim_{n\rightarrow\infty}\frac{1}{\left[p^{n}:-q^{d}\right]}\sum_{y=0}^{p^{n}-1}\left(-q^{d\beta}\right)^{y}\left[\frac{x+a}{d}+y:q^{d\alpha}\right]^{n}$
$\displaystyle=$
$\displaystyle\frac{\left[d:q^{\alpha}\right]^{n}}{\left[d:-q^{\beta}\right]}\sum_{a=0}^{d-1}\left(-q\right)^{a}\chi\left(a\right)\widetilde{E}_{n,q^{d}}\left(\frac{x+a}{d}\mid\alpha:\beta\right)\text{.}$

Here, $\widetilde{E}_{n,q^{d}}\left(\frac{x+a}{d}\mid\alpha:\beta\right)$ is
defined by Ryoo in [16], which is called $q$-Euler polynomials with weight
$\left(\alpha,\beta\right)$. As a result, we have the proof of the theorem.

##  4\. On $p$-adic Dirichlet type of $q$-Euler measure with weight $\alpha$
and $\beta$

Now, we introduce a map
$\mu_{k,q}^{\left(\alpha,\beta\right)}\left(a+p^{n}\mathbb{Z}_{p}\right)$ on
the balls in $\mathbb{Z}_{p}$ as follows:

(4.1)
$\mathcal{\mu}_{k,q}^{\left(\alpha,\beta\right)}\left(a+p^{n}\mathbb{Z}_{p}\mid\chi\right)=\frac{\left[p^{n}:q^{\alpha}\right]^{k}}{\left[p^{n}:-q^{\beta}\right]}\chi\left(a\right)\left(-1\right)^{a}q^{a}f_{k,p^{n}}\left(\frac{\left\\{a\right\\}_{n}}{p^{n}}\mid\alpha:\beta\right)$

where $\left\\{a\right\\}_{n}\equiv a\left(\mathop{\mathrm{m}od}p^{n}\right)$.

######  Theorem 11.

Let $\alpha$, $k\in\mathbb{N}$. Then we specify that
$\mathcal{\mu}_{k,q}^{\left(\alpha,\beta\right)}$ is $p$-adic measure on
$\mathbb{Z}_{p}$ if and only if

$f_{k,q^{p^{n}}}\left(\frac{a}{p^{n}}\mid\alpha:\beta\right)=\frac{\left[p^{n}:q^{p\alpha}\right]^{k}}{\left[p^{n}:-q^{p\beta}\right]}\sum_{b=0}^{p-1}\left(-1\right)^{b}q^{bp^{n}}f_{k,\left(q^{p^{n}}\right)^{p}}\left(\frac{\frac{a}{p^{n}}+b}{p}\mid\alpha:\beta\right)\text{.}$

###### Proof.

By similar method in [23], we can state the proof of this theorem. Therefore,
we omit it.

We now set as follows:

(4.2)
$f_{k,q^{p^{n}}}\left(\frac{a}{p^{n}}\mid\alpha:\beta\right)=\widetilde{E}_{n,q^{p^{n}}}\left(\frac{a}{p^{n}}\mid\alpha:\beta\right)\text{.}$

From (4.1) and (4.2), we easily see

(4.3)
$\mathcal{\mu}_{k,q}^{\left(\alpha,\beta\right)}\left(a+p^{n}\mathbb{Z}_{p}\mid\chi\right)=\frac{\left[p^{n}:q^{\alpha}\right]^{k}}{\left[p^{n}:-q^{\beta}\right]}\chi\left(a\right)\left(-1\right)^{a}q^{a}\widetilde{E}_{k,q^{p^{n}}}\left(\frac{a}{p^{n}}\mid\alpha:\beta\right)\text{.}$

By (2.1) and (4.3), then, we have the following theorem.

######  Theorem 12.

For $\alpha$,$k\in\mathbb{N}$, we have

$\int_{X}d\mu_{k,q}^{\left(\alpha,\beta\right)}\left(x\mid\chi\right)=\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(\alpha:\beta\right)\text{.}$

###### Proof.

By using combinatorial techniques, we compute as follows:

$\displaystyle\int_{X}d\mu_{k,q}^{\left(\alpha,\beta\right)}\left(x\mid\chi\right)$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\frac{\left[dp^{n}:q^{\alpha}\right]^{k}}{\left[dp^{n}:-q^{\beta}\right]}\sum_{x=0}^{dp^{n}-1}\chi\left(x\right)\left(-1\right)^{x}q^{x}\widetilde{E}_{k,q^{dp^{n}}}\left(\frac{x}{dp^{n}}\mid\alpha:\beta\right)$
$\displaystyle=$
$\displaystyle\frac{\left[d:q^{\alpha}\right]^{k}}{\left[d:-q^{\beta}\right]}\sum_{a=0}^{d-1}\left(-1\right)^{a}q^{a}\chi\left(a\right)\lim_{n\rightarrow\infty}\frac{\left[p^{n}:q^{d\alpha}\right]^{k}}{\left[p^{n}:-q^{d\beta}\right]}\sum_{x=0}^{p^{n}-1}\left(-1\right)^{x}q^{dx}\widetilde{E}_{k,\left(q^{d}\right)^{p^{n}}}\left(\frac{\frac{a}{d}+x}{p^{n}}\mid\alpha:\beta\right)$
$\displaystyle=$
$\displaystyle\frac{\left[d:q^{\alpha}\right]^{k}}{\left[d:-q^{\beta}\right]}\sum_{a=0}^{d-1}\left(-1\right)^{a}q^{a}\chi\left(a\right)\widetilde{E}_{n,q^{d}}\left(\frac{a}{d}\mid\alpha:\beta\right)$

so,we obtain the desired result.

######  Theorem 13.

For any $k\in\mathbb{N}$, we get

$\int_{pX}d\mu_{k,q}^{\left(\alpha,\beta\right)}\left(x\mid\chi\right)=\chi\left(p\right)\frac{\left[p:q^{\alpha}\right]}{\left[p:-q^{\beta}\right]}\widetilde{\mathcal{E}}_{k,q^{p}}^{\chi}\left(\alpha:\beta\right)\text{.}$

###### Proof.

From (2.1) and (4.3), we derive the followings assertions

$\displaystyle\int_{pX}d\mu_{k,q}^{\left(\alpha,\beta\right)}\left(x\mid\chi\right)$
$\displaystyle=$
$\displaystyle\lim_{n\rightarrow\infty}\frac{\left[dp^{n+1}:q^{\alpha}\right]^{k}}{\left[dp^{n+1}:-q^{\beta}\right]}\sum_{x=0}^{dp^{n}-1}\chi\left(px\right)\left(-1\right)^{px}q^{px}\widetilde{E}_{k,q^{dp^{n}}}\left(\frac{px}{dp^{n+1}}\mid\alpha:\beta\right)$
$\displaystyle=$
$\displaystyle\chi\left(p\right)\frac{\left[p:q^{\alpha}\right]}{\left[p:-q^{\beta}\right]}\frac{\left[d:q^{p\alpha}\right]^{k}}{\left[d:-q^{p\beta}\right]}\sum_{a=0}^{d-1}\left\\{\begin{array}[]{c}\left(-1\right)^{a}q^{pa}\chi\left(a\right)\lim_{n\rightarrow\infty}\frac{\left[p^{n}:q^{dp\alpha}\right]^{k}}{\left[p^{n}:-q^{pd\beta}\right]}\\\
\times\sum_{x=0}^{p^{n}-1}\left(-1\right)^{x}q^{pdx}\widetilde{E}_{k,\left(q^{d}\right)^{p^{n}}}\left(\frac{dp\left(\frac{a}{d}+x\right)}{pdp^{n}}\mid\alpha:\beta\right)\end{array}\right\\}$
$\displaystyle=$
$\displaystyle\chi\left(p\right)\frac{\left[p:q^{\alpha}\right]}{\left[p:-q^{\beta}\right]}\frac{\left[d:q^{p\alpha}\right]^{k}}{\left[d:-q^{p\beta}\right]}\sum_{a=0}^{d-1}\left(-1\right)^{a}q^{pa}\chi\left(a\right)\widetilde{E}_{k,q^{pd}}\left(\frac{a}{d}\mid\alpha:\beta\right)$
$\displaystyle=$
$\displaystyle\chi\left(p\right)\frac{\left[p:q^{\alpha}\right]}{\left[p:-q^{\beta}\right]}\widetilde{\mathcal{E}}_{k,q^{p}}^{\chi}\left(\alpha:\beta\right)\text{.}$

Thus, we get the desired result and proof is complete.

By the same method which we used in above theorem, by a little bit
manipulations we can state the following theorem.

######  Theorem 14.

For $c\left(\neq 1\right)\in X^{\ast}$

, we have

$\int_{pX}d\mu_{k,q^{\frac{1}{c}}}^{\left(\alpha,\beta\right)}\left(cx\mid\chi\right)=\chi\left(\frac{p}{c}\right)\frac{\left[p:q^{\frac{\alpha}{c}}\right]}{\left[p:\left(-q^{\beta}\right)^{\frac{1}{c}}\right]}\widetilde{\mathcal{E}}_{k,q^{\frac{p}{c}}}^{\chi}\left(\alpha:\beta\right)\text{.}$

######  Theorem 15.

For $c\left(\neq 1\right)\in X^{\ast}$

, we have

$\int_{X}d\mu_{k,q^{\frac{1}{c}}}^{\left(\alpha,\beta\right)}\left(cx\mid\chi\right)=\chi\left(\frac{1}{c}\right)\widetilde{\mathcal{E}}_{k,q^{\frac{1}{c}}}^{\chi}\left(\alpha:\beta\right)\text{.}$

We can define the following identity:

$\mathcal{\mu}_{k,c,q}^{\left(\alpha,\beta\right)}\left(U\mid\chi\right)=\mathcal{\mu}_{k,q}^{\left(\alpha,\beta\right)}\left(U\mid\chi\right)-c^{-1}\frac{\left[c^{-1}:q^{\alpha}\right]^{k}}{\left[c^{-1}:-q^{\beta}\right]}\mathcal{\mu}_{k,q^{\frac{1}{c}}}^{\left(\alpha,\beta\right)}\left(cU\mid\chi\right)$

here $U$ is any compact open subset of $\mathbb{Z}_{p}$, it can be written as
a finite disjoint union of sets

$U=\overset{k}{\underset{j=1}{\cup}}\left(a_{j}+p^{n}\mathbb{Z}_{p}\right),$

where $n\in\mathbb{N}$ and $a_{1},a_{2},...,a_{k}\in\mathbb{Z}$ with $0\leq
a_{i}<p^{n}$ for $i=1,2,...,k$.

######  Theorem 16.

For $c\left(\neq 1\right)\in X^{\ast}$

, we procure the following

$\int_{X^{\ast}}d\mu_{k,c,q}^{\left(\alpha,\beta\right)}\left(cx\mid\chi\right)=(1-\chi^{p})\left(1-c^{-1}\chi^{c^{-1}}\right)\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(\alpha:\beta\right)$

where the operator $\chi^{y}:=\chi^{y,k,\alpha;q}$ on $f\left(q\right)$ is
defined by

$\chi^{y}f\left(q\right)=\chi^{y,k,\alpha;q}f\left(q\right)=\frac{\left[y:q^{\alpha}\right]}{\left[y:-q^{\beta}\right]}\chi\left(y\right)f\left(q^{y}\right)$

That is, we can write

$\chi^{x,k,\alpha;q}\circ\chi^{y,k,\alpha;q}f\left(q\right)=\chi^{xy,k,\alpha;q}f\left(q\right)=\chi^{xy}f\left(q\right)\text{.}$

###### Proof.

To prove this, we assume that
$f\left(q\right)=\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(\alpha:\beta\right)$.
Then, we get

$\left\\{\begin{array}[]{c}\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(\alpha:\beta\right)-\chi\left(p\right)\frac{\left[p:q^{\alpha}\right]}{\left[p:-q^{\beta}\right]}\widetilde{\mathcal{E}}_{k,q^{p}}^{\chi}\left(\alpha:\beta\right)-c^{-1}\frac{\left[c^{-1}:q^{\alpha}\right]^{k}}{\left[c^{-1}:-q^{\beta}\right]}\\\
\times\chi\left(\frac{1}{c}\right)\widetilde{\mathcal{E}}_{k,q^{\frac{1}{c}}}^{\chi}\left(\alpha:\beta\right)+\chi\left(\frac{p}{c}\right)\frac{\left[p:q^{\frac{\alpha}{c}}\right]}{\left[p:\left(-q^{\beta}\right)^{\frac{1}{c}}\right]}\widetilde{\mathcal{E}}_{k,q^{\frac{p}{c}}}^{\chi}\left(\alpha:\beta\right)\end{array}\right\\}$

From this, we derive the following

$(1-\chi^{p})\left(1-c^{-1}\chi^{c^{-1}}\right)\widetilde{\mathcal{E}}_{k,q}^{\chi}\left(\alpha:\beta\right)$

Then, we complete the proof of theorem.

##  5\.
Analytic continuation of $q$-Euler Polynomials with weight $\alpha$

The concept of analytic continuation just means enlarging the domain without
giving up the property of being differentiable, i.e. holomorphic or
meromorphic. More precisely Let $f_{1}$ and $f_{2}$ be analytic functions on
domains $\Omega_{1}$ and $\Omega_{2},$ respectively, and suppose that the
intersection $\Omega_{1}\cap\Omega_{2}$ is not empty and that$f_{1}=f_{2}$ on
$\Omega_{1}\cap\Omega_{2}$. Then $f_{2}$ is called an analytic continuation of
$f_{1}$ to $\Omega_{2}$, and vice versa . Moreover, if it exists, the analytic
continuation of $f_{1}$ to $\Omega_{2}$ is unique. By means of analytic
continuation, starting from a representation of a function by any one power
series, any number of other power series can be found which together define
the value of the function at all points of the domain. Furthermore, any point
can be reached from a point without passing through a singularity of the
function, and the aggregate of all the power series thus obtained constitutes
the analytic expression of the function. So we are ready to state analytic
continuation of $q$-Euler polynomials with weight $\alpha$ as follows.

For coherence with the redefinition of
$\widetilde{E}_{n,q}\left(\alpha\right)=\widetilde{E}_{q}\left(n:\alpha\right)$,
we have

$\widetilde{E}_{n,q}\left(x\mid\alpha\right)=q^{-\alpha
x}\sum_{k=0}^{n}\binom{n}{k}q^{\alpha
kx}\widetilde{E}_{k,q}\left(\alpha\right)\left[x:q^{\alpha}\right]^{n-k}\text{.}$

Let $\Gamma\left(s\right)$ be Euler-gamma function. Then the analytic
continuation can be get as

$\displaystyle n$ $\displaystyle\mapsto$ $\displaystyle s\in\mathbb{R}\text{,
}x\mapsto w\in\mathbb{C}\text{,}$
$\displaystyle\widetilde{E}_{n,q}\left(\alpha\right)$ $\displaystyle\mapsto$
$\displaystyle\widetilde{E}_{q}\left(k+s-\left[s\right]:\alpha\right)=\widetilde{\zeta}_{E,q}\left(-\left(k+s-\left[s\right]\right)\mid\alpha\right)\text{,}$
$\displaystyle\binom{n}{k}$ $\displaystyle=$
$\displaystyle\frac{\Gamma\left(n+1\right)}{\Gamma\left(n-k+1\right)\Gamma\left(k+1\right)}\mapsto\frac{\Gamma\left(s+1\right)}{\Gamma\left(1+k+\left(s-\left[s\right]\right)\right)\Gamma\left(1+\left[s\right]-k\right)}$
$\displaystyle\widetilde{E}_{s,q}\left(w\mid\alpha\right)$
$\displaystyle\mapsto$
$\displaystyle\widetilde{E}_{q}\left(s,w:\alpha\right)=q^{-\alpha
w}\sum_{k=-1}^{\left[s\right]}\frac{\Gamma\left(s+1\right)\widetilde{E}_{q}\left(k+\left(s-\left[s\right]\right):\alpha\right)q^{\alpha
w\left(k+\left(s-\left[s\right]\right)\right)}}{\Gamma\left(1+k+\left(s-\left[s\right]\right)\right)\Gamma\left(1+\left[s\right]-k\right)}\left[w:q^{\alpha}\right]^{\left[s\right]-k}$
$\displaystyle=$ $\displaystyle q^{-\alpha
w}\sum_{k=0}^{\left[s\right]+1}\frac{\Gamma\left(s+1\right)\widetilde{E}_{q}\left(-1+k+\left(s-\left[s\right]\right):\alpha\right)q^{\alpha
w\left(k-1+\left(s-\left[s\right]\right)\right)}}{\Gamma\left(k+\left(s-\left[s\right]\right)\right)\Gamma\left(2+\left[s\right]-k\right)}\left[w:q^{\alpha}\right]^{\left[s\right]+1-k}\text{.}$

Here $\left[s\right]$ gives the integer part of s, and so $s-\left[s\right]$
gives the fractional part.

Deformation of the curve $\widetilde{E}_{q}\left(1,w:\alpha\right)$ into the
curve of $\widetilde{E}_{q}\left(2,w:\alpha\right)$ is by means of the real
analytic cotinuation $\widetilde{E}_{q}\left(s,w:\alpha\right)$, $1\leq s\leq
2$, $-0.5\leq w\leq 0.5$.

######  Acknowledgement 1.

The third author would like to thank the Association SARA-GHU à Marseille for
their hospitality during his stay there, when the work for this paper was done
and dedicated this paper to Neda Agha-Soltan.

## References

  * [1] T. Kim, Analytic continuation of $q$-Euler numbers and polynomials, Applied Mathematics Letters 21 (2008) 1320-1323.
  * [2] T. Kim, On explicit formulas of $p$-adic $q$-$L$-functions, Kyushu J. Math. 43 (1994) 73–86.
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