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Bol. Soc. Mat. Mex. (2015) 21:149–162DOI 10.1007/s40590-015-0061-y
ORIGINAL ARTICLE
More properties on multi-poly-Euler polynomials
Hassan Jolany1 · Roberto B. Corcino2 ·Takao Komatsu3
Received: 23 January 2014 / Accepted: 4 May 2015 / Published online: 20 June 2015© Sociedad Matemática Mexicana 2015
Abstract In this paper, we establish more properties of generalized poly-Euler poly-nomialswith three parameters andwe investigate a kind of symmetrized generalizationof poly-Euler polynomials. Moreover, we introduce a more general form of multi-poly-Euler polynomials and obtain some identities parallel to those of the generalizedpoly-Euler polynomials.
Keywords Poly-Euler polynomials · Appell polynomials · Poly-logarithm ·Generating function
Mathematics Subject Classification 11B68 · 11B73 · 05A15
1 Introduction
The nature of introducing Bernoulli numbers is parallel to that of Euler numbers.Bernoulli numbers have been introduced by Jacob Bernoulli (1655–1705) in his effortto describe the coefficients of the polynomial representation of the sum
B Hassan [email protected]; [email protected]
Roberto B. [email protected]
Takao [email protected]
1 Laboratoire Paul Painlevé, UFR de Mathématiques, Université des Sciences etTechnologies de Lille, CNRS-UMR 8524, 59655 Villeneuve d’Ascq Cedex, France
2 Mathematics and ICT Department, Cebu Normal University, Osmena Blvd.,6000 Cebu, Philippines
3 School of Mathematics and Statistics, Wuhan University, Wuhan 430072, China
150 H. Jolany et al.
Sm(n) = 1m + 2m + · · · + nm,
while Euler numbers have been introduced by Leonard Euler (1707–1783) in his desireto evaluate the alternating sum
An(m) = mn − (m − 1)n + · · · + (−1)m−11n .
Moreover, Bernoulli and Euler numbers have been usually defined by means of thefollowing generating functions with parallel structures
x
ex − 1=
∞∑
n=0
Bnxn
n! (1)
2
ex + e−x=
∞∑
n=0
Enxn
n! , (2)
respectively. It is worth mentioning that Euler worked intensively on Bernoulli num-bers and gave great contributions in the development of the numbers. This made himknown as “godfather” of Bernoulli numbers (see [5]).
TheBernoulli andEuler polynomials are defined parallel to (1) and (2), respectively,as follows
text
et − 1=
∞∑
n=0
Bn(x)tn
n! (3)
2ext
et + 1=
∞∑
n=0
En(x)tn
n! . (4)
In particular, Bn = Bn(0) and En = 2n En(1/2).In 1997, Kaneko [15] introduced the poly-Bernoulli numbers B(k)
n by means of thefollowing exponential generating function
Lik(1 − e−x )
1 − e−x=
∞∑
n=0
B(k)n
xn
n!
where
Lik(z) =∞∑
n=0
zn
nk.
Ohno and Sasaki [17], on the other hand, defined poly-Euler numbers as
Lik(1 − e−4t )
4t cosh t=
∞∑
n=0
E (k)n
tn
n!
More properties on multi-poly-Euler polynomials 151
which have been recently extended by Jolany et al. [14] in polynomial form as
2Lik(1 − e−t )
1 + etext =
∞∑
n=0
E (k)n (x)
tn
n! . (5)
One can easily verify that B(1)n = Bn(1), E
(1)n = En = 2n En(1/2) and E (1)
n (x) =nEn−1(x). Further generalization and other properties of poly-Bernoulli and poly-Euler numbers and polynomials including their relations with other special numbersand functions are found in [1–4,6,7,9–13].
In [14], the generalized poly-Euler polynomials with parameters a, b and c aredefined by
2Lik(1 − (ab)−t )
a−t + btcxt =
∞∑
n=0
E (k)n (x; a, b, c)
tn
n! . (6)
Note that, the poly-Euler polynomials in [2,17] can be deduced from (6) by replacingt with 4t and taking x = 1/2. Moreover, when x = 0, (6) gives
E (k)n (0; a, b, c) = E (k)
n (a, b)
where
2Lik(1 − (ab)−t )
a−t + bt=
∞∑
n=0
E (k)n (a, b)
tn
n! ,
and when a = 1 and b = c = e with E (k)n (x; 1, e, e) = E (k)
n (x), we obtain Eq. (5).However, only one identity has been obtained in [14] for E (k)
n (x; a, b, c) which isgiven by
E (k)n (x; a, b, c) =
n∑
m=0
m∑
j=0
j∑
i=0
2(−1)m− j+i
j k
(j
i
)(x ln c − (m − j + i + 1)
ln a − (m − j + i + 1) ln b)n . (7)
In the same paper by Jolany et al. [14], they defined certain multi-poly-Euler poly-nomials as follows
2Li(k1,k2,...,kr )(1 − (ab)−t )
(a−t + bt )rer xt =
∞∑
n=0
E (k1,k2,...,kr )n (x; a, b)
tn
n! . (8)
where
Li(k1,k2,...,kr )(z) =∑
0≤m1≤m2≤···≤mr
zmr
mk11 mk2
2 · · ·mkrr
152 H. Jolany et al.
is the generalization of poly-logarithm, also known as multiple zeta values, whichhave been given much attention recently but until now, their precise structure remainsa mystery. Note that, when r = 1, (8) immediately yields (6). Several identities onE (k1,k2,...,kr )n (x; a, b) have been obtained in [14] including the recurrence relations
and certain explicit formula. However, this explicit formula is limited only to the casewhere a = 1 and b = e. In this present paper, more identities for E (k)
n (x; a, b, c)will be established and further generalization of E (k1,k2,...,kr )
n (x; a, b) will beinvestigated.
2 Some results on generalized poly-Euler polynomials
The main objective of this section is to establish more identities for E (k)n (x; a, b, c).
First, let us consider an expression for E (k)n (x; a, b, c) in terms of E (k)
i (a, b), i =0, 1, . . . , n.
Theorem 2.1 The generalized poly-Euler polynomials satisfy the following relation
E (k)n (x; a, b, c) =
n∑
i=0
(n
i
)(ln c)n−i E (k)
i (a, b)xn−i (9)
Proof Using (6), we have
∞∑
n=0
E (k)n (x; a, b, c)
tn
n! = 2Lik(1 − (ab)−t )
a−t + btcxt = ext ln c
∞∑
n=0
E (k)n (a, b)
tn
n!
=∞∑
n=0
n∑
i=0
(xt ln c)n−i
(n − i)! E (k)i (a, b)
t i
i !
=∞∑
n=0
(n∑
i=0
(n
i
)(ln c)n−i E (k)
i (a, b)xn−i
)tn
n! .
Comparing the coefficients of tnn! , we obtain the desired result. ��
The next identity gives a relation between E (k)n (x; a, b, c) and E (k)
n (x).
Theorem 2.2 The generalized poly-Euler polynomials satisfy the following relation
E (k)n (x; a, b, c) = (ln a + ln b)n E (k)
n
(x ln c + ln a
ln a + ln b
)(10)
Proof Using (6), we have
More properties on multi-poly-Euler polynomials 153
∞∑
n=0
E (k)n (x; a, b, c)
tn
n! = 2Lik(1 − (ab)−t )
a−t (1 + (ab)t )ext ln c
= 2ex ln c+ln a
ln ab t ln ab Lik(1 − e−t ln ab)
1 + et ln ab
=∞∑
n=0
(ln a + ln b)n E (k)n
(x ln c + ln a
ln a + ln b
)tn
n! .
Comparing the coefficients of tnn! , we obtain the desired result. ��
Theorem 2.3 The generalized poly-Euler polynomials satisfy the following relation
d
dxE (k)n+1(x; a, b, c) = (n + 1)(ln c)E (k)
n (x; a, b, c) (11)
Proof Using (6), we have
∞∑
n=0
d
dxE (k)n (x; a, b, c)
tn
n! = 2t (ln c)Lik(1 − (ab)−t )
(a−t + bt )ext ln c
∞∑
n=0
d
dxE (k)n (x; a, b, c)
tn−1
n! =∞∑
n=0
(ln c)E (k)n (x; a, b, c)
tn
n! .
Hence,
∞∑
n=0
1
n + 1
d
dxE (k)n+1(x; a, b, c)
tn
n! =∞∑
n=0
(ln c)E (k)n (x; a, b, c)
tn
n! .
Comparing the coefficients of tnn! , we obtain the desired result. ��
The following corollary immediately follows from Theorem 2.3 by taking c = e.For brevity, let us denote E (k)
n (x; a, b, e) by E (k)n (x; a, b).
Corollary 2.4 The generalized poly-Euler polynomials are Appell polynomials in thesense that
d
dxE (k)n+1(x; a, b) = (n + 1)E (k)
n (x; a, b) (12)
Consequently, using the characterization of Appell polynomials [16,18,19], thefollowing addition formula can easily be obtained.
Corollary 2.5 The generalized poly-Euler polynomials satisfy the following additionformula
E (k)n (x + y; a, b) =
n∑
i=0
(n
i
)E (k)i (x; a, b)yn−i (13)
154 H. Jolany et al.
Taking x = 0 in formula (13) and using the fact that E (k)n (0; a, b) = E (k)
n (a, b),Corollary 2.5 gives formula (9) in Theorem 2.1 with c = e.
Furthermore, using the fact that E (k)n (x; a, b) satisfies the relation
∞∑
n=0
E (k)n (x; a, b)
tn
n! = 2Lik(1 − (ab)−t )
a−t + btext , (14)
and the exponential generating function for Stirling numbers of the second kind [8]
∞∑
n=0
{nk
}tn
n! = (et − 1)k
k! , (15)
we have the following theorem:
Theorem 2.6 The generalized poly-Euler polynomials E (k)n (x; a, b) satisfy the fol-
lowing explicit formulas
E (k)n (x; a, b) =
∞∑
m=0
n∑
l=m
{lm
} (n
l
)E (k)n−l(−m; a, b)(x)(m) (16)
E (k)n (x; a, b) =
∞∑
m=0
n∑
l=m
{lm
} (n
l
)E (k)n−l(0; a, b)(x)m (17)
E (k)n (x; a, b) =
∞∑
m=0
(n
m
) n−m∑
l=0
(n−ml
)(l+s
l
){l + ss
}E (k)n−m−l(0; a, b)B(s)
m (x) (18)
E (k)n (x; a, b) =
∞∑
m=0
(nm
)
(1 − λ)s
s∑
j=0
(s
j
)(−λ)s− j E (k)
n−m( j; a, b)H (s)m (x; λ), (19)
where (x)(n) = x(x + 1) · · · (x + n − 1), (x)n = x(x − 1) · · · (x − n + 1),
(t
et − 1
)s
ext =∞∑
n=0
B(s)n (x)
tn
n! and
(1 − λ
et − λ
)s
ext =∞∑
n=0
H (s)n (x; λ)
tn
n! .
Proof First, we prove (16). From (14) and using Newton’s Binomial Theorem, wehave
∞∑
n=0
E (k)n (x; a, b)
tn
n! = 2Lik(1 − (ab)−t )
a−t + bt
∞∑
m=0
(x + m − 1
m
)(1 − e−t )m
=∞∑
m=0
(x)(m) (et − 1)m
m!2Lik(1 − (ab)−t )
a−t + bte−mt
More properties on multi-poly-Euler polynomials 155
=∞∑
m=0
(x)(m)
( ∞∑
n=0
{nm
}tn
n!
) ( ∞∑
n=0
E (k)n (−m; a, b)
tn
n!
)
=∞∑
m=0
(x)(m)∞∑
n=0
n∑
l=0
{lm
}t l
l! E(k)n−l(−m; a, b)
tn−l
(n − l)!
=∞∑
n=0
{ ∞∑
m=0
n∑
l=m
{lm
} (n
l
)E (k)n−l(−m; a, b)(x)(m)
}tn
n!
Comparing coefficients completes the proof of (16). To prove identity (17), we have
∞∑
n=0
E (k)n (x; a, b)
tn
n! = 2Lik(1 − (ab)−t )
a−t + bt
∞∑
m=0
(x
m
)(et − 1)m
=∞∑
m=0
(x)m(et − 1)m
m!2Lik(1 − (ab)−t )
a−t + bt
=∞∑
m=0
(x)m
( ∞∑
n=0
{nm
}tn
n!
) ( ∞∑
n=0
E (k)n (0; a, b)
tn
n!
)
=∞∑
m=0
(x)m
∞∑
n=0
n∑
l=0
{lm
}t l
l! E(k)n−l(0; a, b)
tn−l
(n − l)!
=∞∑
n=0
{ ∞∑
m=0
n∑
l=m
{lm
} (n
l
)E (k)n−l(0; a, b)(x)m
}tn
n!
Again, comparing coefficients completes the proof of (17).The proofs for identities (18) and (19) are left to the reader. ��Now, let us consider the following definition which contains certain symmetrized
generalization of poly-Euler polynomials with parameters a, b and c.
Definition 2.7 For m, n ≥ 0, we define
D(m)n (x, y; a, b, c)= 1
(ln a + ln b)n
m∑
k=0
(m
k
)E (−k)n (x; a, b, c)
(y ln c + ln a
ln a + ln b
)m−k
.
(20)
The following theorem contains the double generating function for D(m)n (x, y;
a, b, c).
Theorem 2.8 For n,m ≥ 0, we have
∞∑
n=0
∞∑
m=0
D(m)n (x, y; a, b, c)
tn
n!um
m! = 2e
(y ln c+ln aln a+ln b
)ue
(x ln c+ln aln a+ln b
)tet+u
(1 − e−t
)
(et + 1)(et + eu − et+u). (21)
156 H. Jolany et al.
Proof
∞∑
n=0
∞∑
m=0
D(m)n (x, y; a, b, c)
tn
n!um
m!
=∞∑
n=0
∞∑
m=0
1
(ln a + ln b)n
m∑
k=0
E (−k)n (x; a, b, c)
(y ln c + ln a
ln a + ln b
)m−k tn
n!um
k!(m − k)!
=∞∑
n=0
∞∑
k=0
∑
m≥k
1
(ln a + ln b)nE (−k)n (x; a, b, c)
(y ln c + ln a
ln a + ln b
)m−k tn
n!um
k!(m − k)! .
Replacing m − k with l and using Theorem 2.2, we obtain∞∑
n=0
∞∑
m=0
D(m)n (x, y; a, b, c)
tn
n!um
m!
=∞∑
n=0
∞∑
k=0
∞∑
l=0
1
(ln a + ln b)nE (−k)n (x; a, b, c)
(y ln c + ln a
ln a + ln b
)l tn
n!uk
k!ul
l!
= e
(y ln c+ln aln a+ln b
)u
∞∑
n=0
∞∑
k=0
E (−k)n
(x ln c + ln a
ln a + ln b
)tn
n!uk
k!
= e
(y ln c+ln aln a+ln b
)u
∞∑
k=0
2Lik(1 − e−t )
1 + e−te
(x ln c+ln aln a+ln b
)t uk
k!
= e
(y ln c+ln aln a+ln b
)ue
(x ln c+ln aln a+ln b
)t
∞∑
k=0
∞∑
n=0
E (−k)n (0)
tn
n!uk
k! .
Note that∞∑
k=0
∞∑
n=0
E (−k)n (0)
tn
n!uk
k! =∞∑
k=0
∑
m>0
2(1 − e−t
)mmk
1 + etuk
k!
=∑
m>0
2(1 − e−t
)m
1 + et
∞∑
k=0
(mu)k
k!
= 2
1 + et∑
m>0
(1 − e−t)m emu
= 2
1 + et
(1 − e−t
)eu
1 − (1 − e−t
)eu
= 2et+u(1 − e−t
)
(et + 1)(et + eu − et+u).
Thus,
∞∑
n=0
∞∑
m=0
D(m)n (x, y; a, b, c)
tn
n!um
m! = 2e
(y ln c+ln aln a+ln b
)ue
(x ln c+ln aln a+ln b
)tet+u
(1 − e−t
)
(et + 1)(et + eu − et+u).
��
More properties on multi-poly-Euler polynomials 157
The following is an explicit formula for D(m)n (x, y; a, b, c).
Theorem 2.9 For n,m ≥ 0, we have
D(m)n (x, y; a, b, c)
= 2∞∑
j=0
( j !)2(
n∑
l=0
∞∑
i=0
(−1)i(ln cxai+2bi+1
)n−l − (ln cxai+1bi
)n−l
(ln a + ln b)n−l
(n
l
){lj
})
×(
m∑
r=0
(y ln c + 2 ln a + ln b
ln a + ln b
)m−r (m
r
){rj
})
Proof Using Theorem 2.8,
∞∑
n=0
∞∑
m=0
D(m)n (x, y; a, b, c)
tn
n!um
m!
= 2e
(y ln c+ln aln a+ln b
)ue
(x ln c+ln aln a+ln b
)tet+u
(1 − e−t
)
(et + 1)(1 − (et − 1)(eu − 1))
= 2e
(y ln c+2 ln a+ln b
ln a+ln b
)ue
(x ln c+2 ln a+ln b
ln a+ln b
)t (1 − e−t)
∞∑
i=0
(−1)iei t∞∑
j=0
(et − 1) j (eu − 1) j
= 2e
(y ln c+2 ln a+ln b
ln a+ln b
)u
∞∑
i=0
(−1)ie
(x ln c+(i+2) ln a+(i+1) ln b
ln a+ln b
)t
∞∑
j=0
(et − 1) j (eu − 1) j
−2e
(y ln c+2 ln a+ln b
ln a+ln b
)u
∞∑
i=0
(−1)ie
(x ln c+(i+1) ln a+i ln b
ln a+ln b
)t
∞∑
j=0
(et − 1) j (eu − 1) j .
Applying the exponential generating function for
{nk
}in (15), we obtain
∞∑
n=0
∞∑
m=0
D(m)n (x, y; a, b, c)
tn
n!um
m!
= 2∞∑
j=0
⎛
⎜⎝ j !∞∑
i=0
(−1)i∞∑
n=0
(x ln c+(i+2) ln a+(i+1) ln b
ln a+ln b
)ntn
n!∞∑
m=0
{mj
}tm
m!
⎞
⎟⎠
×⎛
⎜⎝ j !∞∑
n=0
(y ln c+2 ln a+ln b
ln a+ln b
)nun
n!∞∑
m=0
{mj
}um
m!
⎞
⎟⎠
−2∞∑
j=0
⎛
⎜⎝ j !∞∑
i=0
(−1)i∞∑
n=0
(x ln c+(i+1) ln a+i ln b
ln a+ln b
)ntn
n!∞∑
m=0
{mj
}tm
m!
⎞
⎟⎠
158 H. Jolany et al.
×⎛
⎜⎝ j !∞∑
n=0
(y ln c+2 ln a+ln b
ln a+ln b
)nun
n!∞∑
m=0
{mj
}um
m!
⎞
⎟⎠
= 2∞∑
j=0
(j !
∞∑
i=0
∞∑
l=0
l∑
m=0
(−1)i(x ln c+(i+2) ln a+(i+1) ln b
ln a+ln b
)l−m(l
m
) {mj
}t l
l!
)
×⎛
⎝ j !∞∑
p=0
p∑
r=0
(y ln c + 2 ln a + ln b
ln a + ln b
)p−r (p
r
) {rj
}u p
p!
⎞
⎠
−2∞∑
j=0
(j !
∞∑
i=0
∞∑
l=0
l∑
m=0
(−1)i(x ln c + (i + 1) ln a + i ln b
ln a + ln b
)l−m (l
m
) {mj
}t l
l!
)
×⎛
⎝ j !∞∑
p=0
p∑
r=0
(y ln c + 2 ln a + ln b
ln a + ln b
)p−r (p
r
) {rj
}u p
p!
⎞
⎠
=∞∑
n=0
∞∑
m=0
tn
n!um
m! 2∞∑
j=0
( j !)2(
n∑
l=0
∞∑
i=0
(−1)i(x ln c+(i+2) ln a+(i+1) ln b
ln a+ln b
)n−l
×(n
l
) {lj
}) (m∑
r=0
(y ln c + 2 ln a + ln b
ln a + ln b
)m−r (m
r
) {rj
})
−∞∑
n=0
∞∑
m=0
tn
n!um
m! 2∞∑
j=0
( j !)2(
n∑
l=0
∞∑
i=0
(−1)i(x ln c + (i + 1) ln a + i ln b
ln a + ln b
)n−l
×(n
l
) {lj
}) (m∑
r=0
(y ln c + 2 ln a + ln b
ln a + ln b
)m−r (m
r
) {rj
})
Comparing the coefficients yields the desired result. ��
3 Generalized multi-poly-Euler polynomials
Let us define a more general form of multi-poly-Euler polynomials.
Definition 3.1 The generalized multi-poly-Euler polynomials with parameters a, band c are defined by
2Li(k1,k2,...,kr )(1 − (ab)−t )
(a−t + bt )rcrxt =
∞∑
n=0
E (k1,k2,...,kr )n (x; a, b, c)
tn
n! . (22)
In particular,
E (k1,k2,...,kr )n (x) = E (k1,k2,...,kr )
n (x; 1, e, e)E (k1,k2,...,kr )n (a, b) = E (k1,k2,...,kr )
n (0; a, b)
More properties on multi-poly-Euler polynomials 159
The following theorem contains some identities for E (k1,k2,...,kr )n (x; a, b, c) which
can be derived using the process in deriving the identities in Theorems 2.1–2.3.
Theorem 3.2 The generalized multi-poly-Euler polynomials satisfy the followingrelations
E (k1,k2,...,kr )n (x; a, b, c) =
n∑
i=0
(n
i
)(r ln c)n−i E (k1,k2,...,kr )
i (a, b)xn−i
E (k1,k2,...,kr )n (x; a, b, c) = (ln a + ln b)n E (k1,k2,...,kr )
n
(r x ln c + ln a
ln a + ln b
)
d
dxE (k1,k2,...,kr )n+1 (x; a, b, c) = (n + 1)(r ln c)E (k1,k2,...,kr )
n (x; a, b, c) (23)
The characterization of Appell polynomials is supposed to be used to estab-lish the addition formula for the generalized multi-poly-Euler polynomials using(23). But the constant r ln c that appeared in the expression of (23) disqualifiesE (k1,k2,...,kr )n (x; a, b, c) to be an Appell polynomial and to, consequently, satisfy
any of the conditions of the said characterization. However, we can derive theaddition formula using the same method in deriving the addition formula forE (k1,k2,...,kr )n (x; a, b) in [14]. The following theorem contains the addition formula
for E (k1,k2,...,kr )n (x; a, b, c).
Theorem 3.3 The generalized poly-Euler polynomials satisfy the following additionformula
E (k1,k2,...,kr )n (x + y; a, b, c) =
n∑
i=0
(n
i
)(r ln c)n−i E (k1,k2,...,kr )
i (x; a, b, c)yn−i .
Proof
∞∑
n=0
E (k1,k2,...,kr )n (x + y; a, b, c)
tn
n! = 2Li(k1,k2,...,kr )(1 − (ab)−t )
(a−t + bt )rc(x+y)r t
= 2Li(k1,k2,...,kr )(1 − (ab)−t )
(a−t + bt )rcxrt cyrt
=( ∞∑
n=0
E (k1,k2,...,kr )n (x; a, b, c)
tn
n!
)( ∞∑
n=0
(yr ln c)ntn
n!
)
=∞∑
n=0
(n∑
i=0
(n
i
)(yr ln c)n−i E (k1,k2,...,kr )
i (x; a, b, c)
)tn
n! .
Comparing the coefficients of tnn! yields the desired result. ��
The next theorem contains an explicit formula for E (k1,k2,...,kr )n (x; a, b, c).
160 H. Jolany et al.
Theorem 3.4 The generalized multi-poly-Euler polynomials have the followingexplicit formula
E (k1,k2,...,kr )n (x; a, b, c)
=n∑
i=0
∑
0≤m1≤m2≤···≤mrc1+c2+···=r
mr∑
j=0
2(r x ln c − j ln ab)n−i r !(−1) j+s(s ln ab+r ln a)i(mr
j
)(ni
)
(c1!c2! · · · )(mk11 mk2
2 · · ·mkrr )
,
(24)
where s = c1 + 2c2 + · · ·Proof From Definition 3.1, we have
Li(k1,k2,...,kr )(1 − (ab)−t )crxt =∑
0≤m1≤m2≤···≤mr
(1 − (ab)−t )mr
mk11 mk2
2 · · ·mkrr
er xt ln c
=∑
0≤m1≤m2≤···≤mr
1
mk11 mk2
2 · · ·mkrr
mr∑
j=0
(−1) j(mr
j
) ∞∑
n=0
(r x ln c − j ln ab)ntn
n!
=∞∑
n=0
⎛
⎝∑
0≤m1≤m2≤···≤mr
mr∑
j=0
(−1) j (r x ln c − j ln ab)n(mr
j
)
mk11 mk2
2 · · ·mkrr
⎞
⎠ tn
n! .
On the other hand,
(1
a−t + bt
)r
= art(
1
1 + (ab)t
)r
= art
⎛
⎝∑
n≥0
(−1)n(ab)nt
⎞
⎠r
=∑
c1+c2+···=r
r !(−1)c1+2c2+···
c1!c2! · · · et[r ln a+(c1+2c2+··· ) ln ab
=∑
c1+c2+···=r
r !(−1)c1+2c2+···
c1!c2! · · ·∞∑
n=0
(r ln a + (c1 + 2c2 + · · · ) ln ab)n tn
n!
=∞∑
n=0
(∑
c1+c2+···=r
r !(−1)c1+2c2+···(r ln a+(c1+2c2+· · · ) ln ab)nc1!c2! · · ·
)tn
n! .
Hence,
2Li(k1,k2,...,kr )(1 − (ab)−t )
(a−t + bt )rcrxt =2Li(k1,k2,...,kr )(1 − (ab)−t )er xt ln cart
(1
1+(ab)t
)r
= 2∞∑
n=0
n∑
i=0
⎛
⎝∑
0≤m1≤m2≤···≤mr
mr∑
j=0
(−1) j (r x ln c − j ln ab)n−i(mr
j
)
mk11 mk2
2 · · ·mkrr
⎞
⎠ tn−i
(n − i)!
More properties on multi-poly-Euler polynomials 161
×(
∑
c1+c2+···=r
r !(−1)c1+2c2+···(r ln a + (c1 + 2c2 + · · · ) ln ab)ic1!c2! · · ·
)t i
i !
= 2∞∑
n=0
n∑
i=0
∑
0≤m1≤m2≤···≤mrc1+c2+···=r
mr∑
j=0
H(r, i, j, n, a, b)
(c1!c2! · · · )(mk11 mk2
2 · · ·mkrr )
tn
n!
where
H(r, i, j, n, a, b) = (r x ln c − j ln ab)n−i r !(−1) j+c1+2c2+···
×(r ln a + (c1 + 2c2 + · · · ) ln ab)i(mr
j
)(n
i
).
By comparing the coefficient of tn/n!, we obtain the desired explicit formula. ��
Remark 3.5 Identities parallel to those in Theorems 2.8 and 2.9 can possibly be estab-lished. However, the derivation is quite complicated that it may not be of interest tomany. Interested reader may try to obtain the said identities.
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