Hindawi Publishing Corporation
Discrete Dynamics in Nature and Society
Volume 2012, Article ID 269847, 15 pages
doi:10.1155/2012/269847
Research Article
Integral Formulae of Bernoulli Polynomials
Dae San Kim,1 Dmitry V. Dolgy,2 Hyun-Mee Kim,3
Sang-Hun Lee,3 and Taekyun Kim4
1
Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea
Hanrimwon, Kwangwoon University, Seoul 139-701, Republic of Korea
3
Division of General Education, Kwangwoon University, Seoul 139-701, Republic of Korea
4
Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea
2
Correspondence should be addressed to Taekyun Kim, tkkim@kw.ac.kr
Received 24 February 2012; Accepted 10 May 2012
Academic Editor: Lee Chae Jang
Copyright q 2012 Dae San Kim et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
Recently, some interesting and new identities are introduced in Hwang et al., Communicated.
From these identities, we derive some new and interesting integral formulae for the Bernoulli
polynomials.
1. Introduction
As is well known, the Bernoulli polynomials are defined by generating functions as follows:
∞
t
tn
xt
,
e
B
x
n
n!
et − 1
n0
1.1
see 1–11. In the special case, x 0, Bn 0 Bn are called the nth Bernoulli numbers. The
Euler polynomials are also defined by
∞
tn
2
xt
Ext
e
e
E
x
n
n!
et 1
n0
1.2
with the usual convention about replacing En x by En x see 1–11. From 1.1 and 1.2,
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we can easily derive the following equation:
xt t
t
2e
t 2ext
xt
e
t
t
t
2 e 1
e −1
e −1
et 1
⎛
⎞
∞
n n
n
⎜
⎟t
Bk En−k x⎠ .
⎝
k
n!
k0
n0
1.3
k/
1
By 1.1 and 1.3, we get
Bn x n n
B E x,
k k n−k
k0
n ∈ Z N ∪ {0}.
1.4
k/
1
From 1.1, we have
Bn x n n
l
l0
Bl xn−l .
1.5
Thus, by 1.5, we get
n−1 d
n−1
Bn x n
Bl xn−1−l nBn−1 x.
l
dx
l0
1.6
It is known that En 0 En are called the nth Euler numbers see 7. The Euler polynomials
are also given by
n
En x E x n n
l0
l
El xn−l ,
1.7
see 6. From 1.7, we can derive the following equation:
n−1 d
n
En x n
El xn−1−l nEn−1 x.
l
dx
l0
1.8
By the definition of Bernoulli and Euler numbers, we get the following recurrence formulae:
E0 1,
En 1 En 2δ0,n ,
B0 1,
Bn 1 − Bn δ1,n ,
1.9
where δn,k is the kronecker symbol see 5. From 1.6, 1.8, and 1.9, we note that
1
0
δ0,n
,
Bn xdx n1
1
0
En xdx −
2En1
,
n1
1.10
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where n ∈ Z . The following identity is known in 5:
jl
m k
k −1
m−j
k−l m
a b 1 a 1
j
l j l1
j0 l0
k m a b 1m−j a 1k−l − a bm−j ak−l
mk Bjl1 x
j0 l0
x ak x a bm ,
j
l
1.11
j l1
where a, b ∈ Z.
From the identities of Bernoulli polynomials, we derive some new and interesting integral
formulae of an arithmetical nature on the Bernoulli polynomials.
2. Integral Formulae of Bernoulli Polynomials
From 1.1 and 1.2, we note that
2
1
ext t
et 1
t
xt 2 e −1
te
et 1
et − 1
xt 1
2
te
2−2 t
t
e 1
et − 1
∞
∞
1
tm
2El l
t
−
Bm x
t
l!
m!
m0
l1
∞
∞
tm
El1 tl
Bm x
−2
l 1 l!
m!
m0
l0
∞
n
tn
El1 n
.
−2
Bn−l x
l1 l
n!
n0
l0
2.1
Therefore, by 1.2 and 2.1, we obtain the following theorem.
Theorem 2.1. For n ∈ Z , one has
n n El1
En x −2
B x.
l l 1 n−l
l0
2.2
Let us take the definite integral from 0 to 1 on both sides of 1.4: for n ≥ 2,
n n−1 En−k1
En−k1
n
n
−2Bn E1 − 2
.
0 −2
Bk
Bk
k
k
n
−
k
1
n
−k1
k0
k0
k
/1
k
/1
2.3
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By 2.3, we get
n−1 n Bk En−k1
Bn 2
.
k n−k1
k0
2.4
k/
1
Therefore, by 2.4, we obtain the following theorem.
Theorem 2.2. For n ∈ N, with n ≥ 2, one has
n−1 n Bk En−k1
Bn 2
.
k n−k1
k0
2.5
k
/1
Let us take k m, a 0, and b −2 in 1.11. Then we have
m m
−1m−j
j0 l0
−
m
jl
m m
m
m
m −1
m Bjl1 x
−1m−j
j
j
l j l 1 j0 l0
l j l1
−2
m−j
j0
m Bjm1 x
xm x − 2m .
j j m1
2.6
It is easy to show that
1
0
xm x − 2m dx 2
1/2
2t − 2m 2tm dt
0
m 2m
−1 2
1/2
1
m
m 2m
m
2
t 1 − t dt −1 2
tm 1 − tm dt
0
−1m 22m
2.7
0
m!m!
−1m 22m 1
.
2m 1 2m
2m 1!
m
Let us consider the integral from 0 to 1 in 2.6:
m m
1
−1m 22m
m
m−l m
−1
j
l j l 1 2m 1 2m
m
j0 l0
m ∈ N.
2.8
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By 2.6 and 2.8, we get
m
m m Bjl1
m−j m
−1
j
l j l1
j0 l0
jm
m
j m 1 Bk Ejm2−k
1
m−j m
2 −2
j j m 1 k0
k
j m2−k
j0
2.9
k
/1
−1m1 22m 1
2m ,
2m 1
m
for m ∈ N.
Therefore, by 2.9, we obtain the following theorem.
Theorem 2.3. For m ∈ N, one has
m m
m
m Bjl1
−1m−j
j
l j l1
j0 l0
jm
m
j m 1 Bk Ejm2−k
1
m
2 −2m−j
j j m 1 k0
k
j m2−k
j0
2.10
k/
1
−1m1 22m 1
2m .
2m 1
m
Lemma 2.4. Let a, b ∈ Z. For m, k ∈ Z , one has
k
m j0 l0
a b 1m−j a 1k−l
m
k
Ejl x
j
l
m k
m
k
Ejl x 2x a bm x ak ,
a bm−j ak−l
j
l
j0 l0
2.11
(see [5]).
Let us take k m, a 1, b −2 in Lemma 2.4. Then we have
m
m
m m
m
m−j m
m−l m
2
Eml x Ejl x 2 x2 − 1 .
−1
l
j
l
j0 l0
l0
2.12
Taking integral from 0 to 1 in 2.12, we get
1 m
m m
m
Eml1
m Ejl1
m−l m
m−l m
−2 2
x2 − 1 dx.
−2
2
−1
l ml1
j
l j l1
0
j0 l0
l0
2.13
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It is easy to show that
1 0
m m
2k
−1m 22m
m
x − 1 dx −1
.
2k 1
2m 1 2m
m
k1
2
2.14
Thus, by 2.13 and 2.14, we get
m m
m
Eml1
−1m1 22m
m Ejl1
m−l m
m−l m
−
2
.
−1
j
l j l1
l m l 1 2m 1 2m
m
j0 l0
l0
2.15
Therefore, by 2.2 and 2.15, we obtain the following theorem.
Theorem 2.5. For m ∈ Z , one has
m m
−1m−l
j0 l0
−1m 22m
m
m Ejl1
j
l j l 1 2m 1 2m
m
m
ml1
m l 1 Ek1
1
m
Bml1−k .
2 2m−l
l m l 1 k0
k
k1
l0
2.16
3. p-Adic Integral on Zp Associated with Bernoulli and Euler Numbers
Let p be a fixed odd prime number. Throughout this section, Zp , Qp , and Cp will denote the
ring of p-adic integers, the field of p-adic rational numbers, and the completion of algebraic
closure of Qp , respectively. Let νp be the normalized exponential valuation of Cp with |p|p p−νp p 1/p. Let UDZp be the space of uniformly differentiable functions on Zp . For f ∈
UDZp , the bosonic p-adic integral on Zp is defined by
I f pn −1
1 fxdμx lim n
fx,
n→∞p
Zp
x0
3.1
see 8. Thus, by 3.1, we get
Zp
f1 xdμx Zp
fxdμx f 0,
3.2
where f1 x fx 1, and f 0 dfx/dx|x0 . Let us take fy etxy . Then we have
Zp
etxy dμ y ∞
tn
t
tx
.
e
B
x
n
n!
et − 1
n0
3.3
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From 3.3, we have
n x y dμ y Bn x,
Zp
Zp
yn dμ y Bn .
3.4
From 1.2, we can derive the following integral equation:
n−1
I fn I f f i n ∈ N.
3.5
i0
Thus, from 3.4 and 3.5, we get
m
Zp
x n dμx Zp
xm dμx m
n−1
im−1 .
3.6
i0
From 3.6, we have
Bm n − Bm m
n−1
im−1
m ∈ Z .
3.7
i0
The fermionic p-adic integral on Zp is defined by Kim as follows 6, 7:
I−1 f n
−1
p
fx−1x .
3.8
I−1 f1 − I−1 f 2f0,
I−1 f2 − I−1 f1 2f1 0 −I−1 f1 2f1
−12 I−1 f 2−12−1 f0 2f1.
3.9
Zp
fxdμ−1 x lim
n→∞
x0
Let f1 x fx 1. Then we have
Continuing this process, we obtain the following equation:
n−1
I−1 fn −1n I−1 f 2 −1n−l−1 fl,
where fn x fx n.
3.10
l0
Thus, by 3.10, we have
m
Zp
x n dμ−1 x −1
n
Zp
n−1
xm dμ−1 x 2 −1n−l−1 lm .
l0
3.11
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Let us take fy etxy . By 3.9, we get
∞
tn
2ext
etxy dμ−1 y t
En x .
n!
e 1 n0
Zp
3.12
From 3.2, we have the Witt’s formula for the nth Euler polynomials and numbers as follows:
Zp
n
x y dμ−1 y En x,
Zp
yn dμ−1 y En ,
where n ∈ Z .
3.13
By 3.11 and 3.13, we get
Em n −1
n
n−1
l−1 m
Em 2 −1 l
,
m ∈ Z , n ∈ N.
3.14
l0
Let us consider the following p-adic integral on Zp :
K1 Zp
Bn xdμx n n
l0
l
Bn−l
Zp
xl dμx
n n
Bn−l Bl .
l
l0
3.15
From 1.4 and 3.15, we have
K1 n−k
E
xl dμx
B
k k l0 n−k−l
l
Zp
n n
k0
k/
1
n−k
n n−k n
n−k
Bk Bl En−k−l .
k
l
k0 l0
3.16
k/
1
Therefore, by 3.15 and 3.16, we obtain the following theorem.
Theorem 3.1. For n ∈ Z , one has
n n n−k n
n
n−k
Bn−l Bl Bk Bl En−k−l .
l
k
l
k0 l0
l0
3.17
k/
1
Now, we set
K2 Zp
Bn xdμ−1 x n n
Bn−l El .
l
l0
3.18
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By 1.4, we get
K2 n n
k
k0
k/
1
Bk
n−k
En−k−l
l0
n n−k n
n−k
k0
k/
1
k
l0
l
n−k
l
Zp
xl dμ−1 x
3.19
Bk En−k−l El .
Therefore, by 3.18 and 3.19, we obtain the following theorem.
Theorem 3.2. For n ∈ Z , one has
n n
l0
l
Bn−l El n n−k n
n−k
k0
k/
1
l0
k
l
Bk En−k−l El .
3.20
Let us consider the following integral on Zp :
n n n
n
En xdμ−1 x xl dμ−1 x En−l
En−l El .
K3 l
l
Zp
Zp
l0
l0
3.21
From 2.2, we have
n−l n
El1 n n−l
K3 −2
xk dμ−1 x
Bn−l−k
l
k
l
1
Z
p
l0
k0
n−l n n
n − l El1
Bn−l−k Ek .
−2
l
k
l1
l0 k0
3.22
Therefore, by 3.21 and 3.22, we obtain the following theorem.
Theorem 3.3. For n ∈ Z , one has
n n n−l n
n
n − l El1
Ek Bn−l−k .
En−l El −2
l
l
k
l1
l0
l0 k0
3.23
Now, we set
K4 Zp
En xdμx n n
l0
l
En−l Bl .
3.24
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By 2.2, we get
n−l n n
n − l El1
K4 −2
Bn−l−k Bk .
l
k
l1
l0 k0
3.25
Therefore, by 3.24 and 3.25, we obtain the following corollary.
Corollary 3.4. For n ∈ Z , we have
n n
n n−l n
n − l El1
Bn−l−k Bk .
El Bl −2
l
l
k
l1
l0 k0
l0
3.26
Let us assume that a, b, c, d ∈ Z. From Lemma 2.4 and 3.13, we note that
Zp
m k
a b 1 x y
a 1 x y dμ−1 y
Zp
m k
a x y dμ−1 y
a b x y
3.27
2x a bm x ak .
By 3.27, we get
2x a bm x ak Zp
m k
a b − c 1 x c y
a − c 1 x c y dμ−1 y
Zp
m k
a b − d x y d
a − d x y d dμ−1 y .
3.28
Thus, by 3.28 and 3.13, we obtain the following lemma see 5.
Lemma 3.5. Let a, b, c, d ∈ Z. For m, k ∈ Z , one has
m k m
k
a b − c 1m−j a − c 1k−l Ejl x c
j
l
j0 l0
m k
k
m−j
k−l m
Ejl x d 2x a bm x ak .
a b − d a − d
j
l
j0 l0
3.29
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Let us consider the formula in Lemma 3.5 with d c − 1. Then we have
k
m k m−j
k−l m
b
−
c
1
−
c
1
Ejl x c Ejl x c − 1
a
a
j
l
j0 l0
3.30
2x a bm x ak .
Taking
Zp
dμx on both sides of 3.30,
jl m k
j l
k
m−j
k−l m
LHS a b − c 1 a − c 1
j
l s0 s
j0 l0
× Ejl−s
Zp
x cs x c − 1s dμx
jl m k
m
j l
k
2
a b − c 1m−j a − c 1k−l
j
l s0 s
j0 l0
× Ejl−s Bs c − 1 3.31
k
m m
k
a b − c 1m−j a − c 1k−l
j
l
j0 l0
× j l Ejl−1 c − 1.
By the same method, we get
m m m−s
RHS 2
b
x ask dμx
s
Zp
s0
m m m−s
2
b Bsk a.
s
s0
3.32
Therefore, by 3.31 and 3.32, we obtain the following proposition.
Proposition 3.6. Let a, b, c ∈ Z. Then one has
jl k
m m
k
j l
Ejl−s
2
a b − c 1m−j a − c 1k−l
j
l s0 s
j0 l0
× Bs c − 1 m k
m
k j l Ejl−1 c − 1
a b − c 1m−j a − c 1k−l
j
l
j0 l0
m m m−s
2
b Bsk a.
s
s0
3.33
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Replacing c by c 1, we have
jl m k
m
j l
k
2
Ejl−s Bs c
a b − cm−j a − ck−l
j
l s0 s
j0 l0
m
k
j l a b − cm−j a − ck−l
Ejl−s c
j
l
m k
j0 l0
3.34
m m m−s
2
b Bsk a.
s
s0
From 3.4 and 3.7, we derive some identity for the first term of the LHS of 3.34.
The first term of the LHS of 3.34
jl m k
j l
k
m−j
k−l m
2
Ejl−s
a b − c a − c
j
l s0 s
j0 l0
c−1
Bs s is−1
×
i0
jl m k
m
j l
k
2
Ejl−s Bs
a b − cm−j a − ck−l
j
l s0 s
j0 l0
m k
c−1 m
k 2
j l −10
a b − cm−j a − ck−l
j
l
i0 j0 l0
×
i−1
Ejl−1 2 −1e−1 ejl−1
e0
jl m k m
k
j l
2
a b − cm−j a − ck−l
j
l
s
j0 l0 s0
× Ejl−s Bs 2m
m−1
k j0 l0
2k
m−1
j
k−1 m m
k−1
j0 l0
4m
j
l
k
a b − cm−1−j a − ck−l Ejl δc≡1
l
a b − cm−j a − ck−1−l Ejl δc≡1
c−2
a b − c em−1 a − c ek δc≡e
e0
4k
c−2
e0
a b − c em a − c ek−1 δc≡e ,
3.35
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where
δc≡k
1
0
if c ≡ kmod 2,
if c / kmod 2.
3.36
The second term of the LHS of 3.34
k
m m−j
j l a b − c
k−l
a − c
j0 l0
−1c
c−1
m
k
c
i−1 jl−1
−1 Ejl−1 2 −1 i
j
l
i0
m
k
Ejl−1
j l a b − cm−j a − ck−l
j
l
m k
j0 l0
2−1c
c−1
−1i−1 ma b − c im−1 a − c ik
i0
2−1c
c−1
−1i−1 a b − c im ka − c ik−1
3.37
i0
−1c
m
k
j l a b − cm−j a − ck−l
Ejl−1
j
l
m k
j0 l0
2m−1c
c−1
−1i−1 a b − c im−1 a − c ik
i0
2k−1c
c−1
−1i−1 a b − c im a − c ik−1 .
i0
Therefore, by 3.34, 3.35, and 3.37, we obtain the following theorem.
Theorem 3.7. Let a, b, c ∈ Z with c ≥ 1. Then one has
jl k m m
k
j l
2
a b − cm−j a − ck−l Ejl−s Bs
j
l
s
j0 l0 s0
m k m−1
k
2mδc≡1
a b − cm−1−j a − ck−l Ejl
j
l
j0 l0
2kδc≡1
k m m
k−1
j0 l0
j
−1
c
m k
l
a b − cm−j a − ck−1−l Ejl
m−j
j l a b − c
j0 l0
k−l
a − c
m
k
Ejl−1
j
l
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2m
c−1
a b − c em−1 a − c ek
e0
2k
c−1
a b − c em a − c ek−1
e0
m 2
s0
m m−s
b Bsk a,
s
3.38
where
δc≡k
1 if c ≡ kmod 2,
0 if c /
≡ kmod 2.
3.39
Remark 3.8. Here, we note that
4m
c−2
a b − c em−1 a − c ek δc≡e 2m−1c
e0
2m
c−1
−1i−1 a b − c im−1 a − c ik
i0
c−1
a b − c em−1 a − c ek .
e0
3.40
Acknowledgment
The first author was supported by National Research Foundation of Korea Grant funded by
the Korean Government 2011-0002486.
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