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Discrete Dynamics in Nature and Society
Volume 2012 (2012), Article ID 269847, 15 pages
http://dx.doi.org/10.1155/2012/269847
Research Article

Integral Formulae of Bernoulli Polynomials

1Department of Mathematics, Sogang University, Seoul 121-742, Republic of Korea
2Hanrimwon, Kwangwoon University, Seoul 139-701, Republic of Korea
3Division of General Education, Kwangwoon University, Seoul 139-701, Republic of Korea
4Department of Mathematics, Kwangwoon University, Seoul 139-701, Republic of Korea

Received 24 February 2012; Accepted 10 May 2012

Academic Editor: Lee Chae Jang

Copyright © 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.

Abstract

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: 𝑡𝑒𝑡𝑒1𝑥𝑡=𝑛=0𝐵𝑛(𝑡𝑥)𝑛,𝑛!(1.1) (see [111]). In the special case, 𝑥=0,𝐵𝑛(0)=𝐵𝑛 are called the 𝑛th Bernoulli numbers. The Euler polynomials are also defined by 2𝑒𝑡𝑒+1𝑥𝑡=𝑒𝐸(𝑥)𝑡=𝑛=0𝐸𝑛(𝑡𝑥)𝑛𝑛!(1.2) with the usual convention about replacing 𝐸𝑛(𝑥) by 𝐸𝑛(𝑥) (see [111]). From (1.1) and (1.2), we can easily derive the following equation: 𝑡𝑒𝑡𝑒1𝑥𝑡=𝑡22𝑒𝑥𝑡𝑒𝑡+𝑡+1𝑒𝑡12𝑒𝑥𝑡𝑒𝑡=+1𝑛=0𝑛𝑘=0𝑘1𝑛𝑘𝐵𝑘𝐸𝑛𝑘𝑡(𝑥)𝑛.𝑛!(1.3) By (1.1) and (1.3), we get 𝐵𝑛(𝑥)=𝑛𝑘=0𝑘1𝑛𝑘𝐵𝑘𝐸𝑛𝑘(𝑥),𝑛+.={0}(1.4)

From (1.1), we have 𝐵𝑛(𝑥)=𝑛𝑙=0𝑛𝑙𝐵𝑙𝑥𝑛𝑙.(1.5) Thus, by (1.5), we get 𝑑𝐵𝑑𝑥𝑛(𝑥)=𝑛𝑛1𝑙=0𝑙𝐵𝑛1𝑙𝑥𝑛1𝑙=𝑛𝐵𝑛1(𝑥).(1.6) It is known that 𝐸𝑛(0)=𝐸𝑛 are called the 𝑛th Euler numbers (see [7]). The Euler polynomials are also given by 𝐸𝑛(𝑥)=(𝐸+𝑥)𝑛=𝑛𝑙=0𝑛𝑙𝐸𝑙𝑥𝑛𝑙,(1.7) (see [6]). From (1.7), we can derive the following equation: 𝑑𝐸𝑑𝑥𝑛(𝑥)=𝑛𝑛1𝑙=0𝑛𝑙𝐸𝑙𝑥𝑛1𝑙=𝑛𝐸𝑛1(𝑥).(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 10𝐵𝑛𝛿(𝑥)𝑑𝑥=0,𝑛,𝑛+110𝐸𝑛(𝑥)𝑑𝑥=2𝐸𝑛+1,𝑛+1(1.10) where 𝑛+. The following identity is known in [5]: 𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏+1)𝑚𝑗(𝑎+1)𝑘𝑙𝑚𝑗𝑘𝑙(1)𝑗+𝑙+𝑗+𝑙+1𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏+1)𝑚𝑗(𝑎+1)𝑘𝑙(𝑎+𝑏)𝑚𝑗𝑎𝑘𝑙𝑚𝑗𝑘𝑙𝐵𝑗+𝑙+1(𝑥)𝑗+𝑙+1=(𝑥+𝑎)𝑘(𝑥+𝑎+𝑏)𝑚,where𝑎,𝑏.(1.11) 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 2𝑒𝑡𝑒+1𝑥𝑡=1𝑡2𝑒𝑡1𝑒𝑡+1𝑡𝑒𝑥𝑡𝑒𝑡=11𝑡222𝑒𝑡+1𝑡𝑒𝑥𝑡𝑒𝑡=11𝑡𝑙=12𝐸𝑙𝑡𝑙!𝑙𝑚=0𝐵𝑚𝑡(𝑥)𝑚𝑚!=2𝑙=0𝐸𝑙+1𝑡𝑙+1𝑙𝑙!𝑚=0𝐵𝑚𝑡(𝑥)𝑚𝑚!=2𝑛=0𝑛𝑙=0𝐸𝑙+1𝑛𝑙𝐵𝑙+1𝑛𝑙𝑡(𝑥)𝑛.𝑛!(2.1)

Therefore, by (1.2) and (2.1), we obtain the following theorem.

Theorem 2.1. For 𝑛+, one has 𝐸𝑛(𝑥)=2𝑛𝑙=0𝑛𝑙𝐸𝑙+1𝐵𝑙+1𝑛𝑙(𝑥).(2.2)

Let us take the definite integral from 0 to 1 on both sides of (1.4): for 𝑛2, 0=2𝑛𝑘=0𝑘1𝑛𝑘𝐵𝑘𝐸𝑛𝑘+1𝑛𝑘+1=2𝐵𝑛𝐸12𝑛1𝑘=0𝑘1𝑛𝑘𝐵𝑘𝐸𝑛𝑘+1.𝑛𝑘+1(2.3)

By (2.3), we get 𝐵𝑛=2𝑛1𝑘=0𝑘1𝑛𝑘𝐵𝑘𝐸𝑛𝑘+1.𝑛𝑘+1(2.4)

Therefore, by (2.4), we obtain the following theorem.

Theorem 2.2. For 𝑛, with 𝑛2, one has 𝐵𝑛=2𝑛1𝑘=0𝑘1𝑛𝑘𝐵𝑘𝐸𝑛𝑘+1.𝑛𝑘+1(2.5)

Let us take 𝑘=𝑚, 𝑎=0, and 𝑏=2 in (1.11). Then we have 𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑗𝑚𝑗𝑚𝑙(1)𝑗+𝑙+𝑗+𝑙+1𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑗𝑚𝑗𝑚𝑙𝐵𝑗+𝑙+1(𝑥)𝑗+𝑙+1𝑚𝑗=0(2)𝑚𝑗𝑚𝑗𝐵𝑗+𝑚+1(𝑥)𝑗+𝑚+1=𝑥𝑚(𝑥2)𝑚.(2.6) It is easy to show that 10𝑥𝑚(𝑥2)𝑚𝑑𝑥=201/2(2𝑡2)𝑚(2𝑡)𝑚=𝑑𝑡(1)𝑚22𝑚201/2𝑡𝑚(1𝑡)𝑚=𝑑𝑡(1)𝑚22𝑚10𝑡𝑚(1𝑡)𝑚𝑑𝑡=(1)𝑚22𝑚𝑚!𝑚!=((2𝑚+1)!1)𝑚22𝑚12𝑚+1𝑚2𝑚.(2.7)

Let us consider the integral from 0 to 1 in (2.6): 𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑙𝑚𝑗𝑚𝑙1=(𝑗+𝑙+11)𝑚22𝑚(2𝑚+1)𝑚2𝑚(𝑚).(2.8) By (2.6) and (2.8), we get 𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑗𝑚𝑗𝑚𝑙𝐵𝑗+𝑙+1𝑗+𝑙+1=2𝑚𝑗=0(2)𝑚𝑗𝑚𝑗1𝑗+𝑚+1𝑗+𝑚𝑘=0𝑘1𝑘𝐵𝑗+𝑚+1𝑘𝐸𝑗+𝑚+2𝑘+𝑗+𝑚+2𝑘(1)𝑚+122𝑚12𝑚+1𝑚2𝑚,for𝑚.(2.9) Therefore, by (2.9), we obtain the following theorem.

Theorem 2.3. For 𝑚, one has 𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑗𝑚𝑗𝑚𝑙𝐵𝑗+𝑙+1𝑗+𝑙+1=2𝑚𝑗=0(2)𝑚𝑗𝑚𝑗1𝑗+𝑚+1𝑗+𝑚𝑘=0𝑘1𝑘𝐵𝑗+𝑚+1𝑘𝐸𝑗+𝑚+2𝑘+𝑗+𝑚+2𝑘(1)𝑚+122𝑚12𝑚+1𝑚2𝑚.(2.10)

Lemma 2.4. Let 𝑎,𝑏. For 𝑚,𝑘+, one has 𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏+1)𝑚𝑗(𝑎+1)𝑘𝑙𝑚𝑗𝑘𝑙𝐸𝑗+𝑙+(𝑥)𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏)𝑚𝑗𝑎𝑘𝑙𝑚𝑗𝑘𝑙𝐸𝑗+𝑙(𝑥)=2(𝑥+𝑎+𝑏)𝑚(𝑥+𝑎)𝑘,(2.11) (see [5]).

Let us take 𝑘=𝑚, 𝑎=1, 𝑏=2 in Lemma 2.4. Then we have 𝑚𝑙=02𝑚𝑙𝑚𝑙𝐸𝑚+𝑙(𝑥)+𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑗𝑚𝑗𝑚𝑙𝐸𝑗+𝑙𝑥(𝑥)=221𝑚.(2.12) Taking integral from 0 to 1 in (2.12), we get 2𝑚𝑙=02𝑚𝑙𝑚𝑙𝐸𝑚+𝑙+1𝑚+𝑙+12𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑙𝑚𝑗𝑚𝑙𝐸𝑗+𝑙+1𝑗+𝑙+1=210𝑥21𝑚𝑑𝑥.(2.13)

It is easy to show that 10𝑥21𝑚𝑑𝑥=(1)𝑚𝑚𝑘=12𝑘=2𝑘+1(1)𝑚22𝑚(2𝑚+1)𝑚2𝑚.(2.14)

Thus, by (2.13) and (2.14), we get 𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑙𝑚𝑗𝑚𝑙𝐸𝑗+𝑙+1𝑗+𝑙+1=𝑚𝑙=02𝑚𝑙𝑚𝑙𝐸𝑚+𝑙+1+(𝑚+𝑙+11)𝑚+122𝑚(2𝑚+1)𝑚2𝑚.(2.15)

Therefore, by (2.2) and (2.15), we obtain the following theorem.

Theorem 2.5. For 𝑚+, one has 𝑚𝑚𝑗=0𝑙=0(1)𝑚𝑙𝑚𝑗𝑚𝑙𝐸𝑗+𝑙+1+(𝑗+𝑙+11)𝑚22𝑚(2𝑚+1)𝑚2𝑚=2𝑚𝑙=02𝑚𝑙𝑚𝑙1𝑚+𝑙+1𝑚+𝑙+1𝑘=0𝑘𝐸𝑚+𝑙+1𝑘+1𝐵𝑘+1𝑚+𝑙+1𝑘.(2.16)

3. 𝑝-Adic Integral on 𝑝 Associated with Bernoulli and Euler Numbers

Let 𝑝 be a fixed odd prime number. Throughout this section, 𝑝, 𝑝, and 𝑝 will denote the ring of 𝑝-adic integers, the field of 𝑝-adic rational numbers, and the completion of algebraic closure of 𝑝, respectively. Let 𝜈𝑝 be the normalized exponential valuation of 𝑝 with |𝑝|𝑝=𝑝𝜈𝑝(𝑝)=1/𝑝. Let 𝑈𝐷(𝑝) be the space of uniformly differentiable functions on 𝑝. For 𝑓𝑈𝐷(𝑝), the bosonic 𝑝-adic integral on 𝑝 is defined by 𝐼(𝑓)=𝑝𝑓(𝑥)𝑑𝜇(𝑥)=lim𝑛1𝑝𝑛𝑝𝑛1𝑥=0𝑓(𝑥),(3.1) (see [8]). Thus, by (3.1), we get 𝑝𝑓1(𝑥)𝑑𝜇(𝑥)=𝑝𝑓(𝑥)𝑑𝜇(𝑥)+𝑓(0),(3.2) where 𝑓1(𝑥)=𝑓(𝑥+1), and 𝑓(0)=𝑑𝑓(𝑥)/𝑑𝑥|𝑥=0. Let us take 𝑓(𝑦)=𝑒𝑡(𝑥+𝑦). Then we have 𝑝𝑒𝑡(𝑥+𝑦)𝑡𝑑𝜇(𝑦)=𝑒𝑡𝑒1𝑡𝑥=𝑛=0𝐵𝑛(𝑡𝑥)𝑛.𝑛!(3.3) From (3.3), we have 𝑝(𝑥+𝑦)𝑛𝑑𝜇(𝑦)=𝐵𝑛(𝑥),𝑝𝑦𝑛𝑑𝜇(𝑦)=𝐵𝑛.(3.4) From (1.2), we can derive the following integral equation: 𝐼𝑓𝑛=𝐼(𝑓)+𝑛1𝑖=0𝑓(𝑖)(𝑛).(3.5) Thus, from (3.4) and (3.5), we get 𝑝(𝑥+𝑛)𝑚𝑑𝜇(𝑥)=𝑝𝑥𝑚𝑑𝜇(𝑥)+𝑚𝑛1𝑖=0𝑖𝑚1.(3.6) From (3.6), we have 𝐵𝑚(𝑛)𝐵𝑚=𝑚𝑛1𝑖=0𝑖𝑚1𝑚+.(3.7) The fermionic 𝑝-adic integral on 𝑝 is defined by Kim as follows [6, 7]: 𝐼1(𝑓)=𝑝𝑓(𝑥)𝑑𝜇1(𝑥)=lim𝑝𝑛𝑛1𝑥=0𝑓(𝑥)(1)𝑥.(3.8) Let 𝑓1(𝑥)=𝑓(𝑥+1). Then we have 𝐼1𝑓1=𝐼1𝐼(𝑓)+2𝑓(0),1𝑓2=𝐼1𝑓1+2𝑓1(0)=𝐼1𝑓1+2𝑓(1)=(1)2𝐼1(𝑓)+2(1)21𝑓(0)+2𝑓(1).(3.9) Continuing this process, we obtain the following equation: 𝐼1𝑓𝑛=(1)𝑛𝐼1(𝑓)+2𝑛1𝑙=0(1)𝑛𝑙1𝑓(𝑙),where𝑓𝑛(𝑥)=𝑓(𝑥+𝑛).(3.10) Thus, by (3.10), we have 𝑝(𝑥+𝑛)𝑚𝑑𝜇1(𝑥)=(1)𝑛𝑝𝑥𝑚𝑑𝜇1(𝑥)+2𝑛1𝑙=0(1)𝑛𝑙1𝑙𝑚.(3.11) Let us take 𝑓(𝑦)=𝑒𝑡(𝑥+𝑦). By (3.9), we get 𝑝𝑒𝑡(𝑥+𝑦)𝑑𝜇1(𝑦)=2𝑒𝑥𝑡𝑒𝑡=+1𝑛=0𝐸𝑛(𝑡𝑥)𝑛.𝑛!(3.12) From (3.2), we have the Witt's formula for the 𝑛th Euler polynomials and numbers as follows: 𝑝(𝑥+𝑦)𝑛𝑑𝜇1(𝑦)=𝐸𝑛(𝑥),𝑝𝑦𝑛𝑑𝜇1(𝑦)=𝐸𝑛,where𝑛+.(3.13) By (3.11) and (3.13), we get 𝐸𝑚(𝑛)=(1)𝑛𝐸𝑚+2𝑛1𝑙=0(1)𝑙1𝑙𝑚,𝑚+.,𝑛(3.14)

Let us consider the following 𝑝-adic integral on 𝑝: 𝐾1=𝑝𝐵𝑛(𝑥)𝑑𝜇(𝑥)=𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝑝𝑥𝑙=𝑑𝜇(𝑥)𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝐵𝑙.(3.15)

From (1.4) and (3.15), we have 𝐾1=𝑛𝑘=0𝑘1𝑛𝑘𝐵𝑘𝑛𝑘𝑙=0𝐸𝑛𝑘𝑙𝑙𝑛𝑘𝑝𝑥𝑙=𝑑𝜇(𝑥)𝑛𝑘=0𝑘1𝑛𝑘𝑙=0𝑛𝑘𝑙𝐵𝑛𝑘𝑘𝐵𝑙𝐸𝑛𝑘𝑙.(3.16) Therefore, by (3.15) and (3.16), we obtain the following theorem.

Theorem 3.1. For 𝑛+, one has 𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝐵𝑙=𝑛𝑘=0𝑘1𝑛𝑘𝑙=0𝑛𝑘𝑙𝐵𝑛𝑘𝑘𝐵𝑙𝐸𝑛𝑘𝑙.(3.17)

Now, we set 𝐾2=𝑝𝐵𝑛(𝑥)𝑑𝜇1(𝑥)=𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝐸𝑙.(3.18)

By (1.4), we get 𝐾2=𝑛𝑘=0𝑘1𝑛𝑘𝐵𝑘𝑛𝑘𝑙=0𝐸𝑛𝑘𝑙𝑙𝑛𝑘𝑝𝑥𝑙𝑑𝜇1=(𝑥)𝑛𝑘=0𝑘1𝑛𝑘𝑙=0𝑛𝑘𝑙𝐵𝑛𝑘𝑘𝐸𝑛𝑘𝑙𝐸𝑙.(3.19) Therefore, by (3.18) and (3.19), we obtain the following theorem.

Theorem 3.2. For 𝑛+, one has 𝑛𝑙=0𝑛𝑙𝐵𝑛𝑙𝐸𝑙=𝑛𝑘=0𝑘1𝑛𝑘𝑙=0𝑛𝑘𝑙𝐵𝑛𝑘𝑘𝐸𝑛𝑘𝑙𝐸𝑙.(3.20)

Let us consider the following integral on 𝑝: 𝐾3=𝑝𝐸𝑛(𝑥)𝑑𝜇1(𝑥)=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝑝𝑥𝑙𝑑𝜇1(𝑥)=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝐸𝑙.(3.21)

From (2.2), we have 𝐾3=2𝑛𝑙=0𝐸𝑙+1𝑛𝑙𝑙+1𝑛𝑙𝑘=0𝑘𝐵𝑛𝑙𝑛𝑙𝑘𝑝𝑥𝑘𝑑𝜇1(𝑥)=2𝑛𝑙=0𝑛𝑙𝑘=0𝑛𝑙𝑘𝐸𝑛𝑙𝑙+1𝐵𝑙+1𝑛𝑙𝑘𝐸𝑘.(3.22) Therefore, by (3.21) and (3.22), we obtain the following theorem.

Theorem 3.3. For 𝑛+, one has 𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝐸𝑙=2𝑛𝑙=0𝑛𝑙𝑘=0𝑛𝑙𝑘𝐸𝑛𝑙𝑙+1𝐸𝑙+1𝑘𝐵𝑛𝑙𝑘.(3.23)

Now, we set 𝐾4=𝑝𝐸𝑛(𝑥)𝑑𝜇(𝑥)=𝑛𝑙=0𝑛𝑙𝐸𝑛𝑙𝐵𝑙.(3.24)

By (2.2), we get 𝐾4=2𝑛𝑙=0𝑛𝑙𝑘=0𝑛𝑙𝑘𝐸𝑛𝑙𝑙+1𝐵𝑙+1𝑛𝑙𝑘𝐵𝑘.(3.25) Therefore, by (3.24) and (3.25), we obtain the following corollary.

Corollary 3.4. For 𝑛+, we have 𝑛𝑙=0𝑛𝑙𝐸𝑙𝐵𝑙=2𝑛𝑙=0𝑛𝑙𝑘=0𝑛𝑙𝑘𝐸𝑛𝑙𝑙+1𝐵𝑙+1𝑛𝑙𝑘𝐵𝑘.(3.26)

Let us assume that 𝑎,𝑏,𝑐,𝑑. From Lemma 2.4 and (3.13), we note that 𝑝((𝑎+𝑏+1)+(𝑥+𝑦))𝑚((𝑎+1)+(𝑥+𝑦))𝑘𝑑𝜇1(+𝑦)𝑝((𝑎+𝑏)+(𝑥+𝑦))𝑚((𝑎+(𝑥+𝑦))𝑘𝑑𝜇1(𝑦)=2(𝑥+𝑎+𝑏)𝑚(𝑥+𝑎)𝑘.(3.27)

By (3.27), we get 2(𝑥+𝑎+𝑏)𝑚(𝑥+𝑎)𝑘=𝑝((𝑎+𝑏𝑐+1)+(𝑥+𝑐+𝑦))𝑚((𝑎𝑐+1)+(𝑥+𝑐+𝑦))𝑘𝑑𝜇1(+𝑦)𝑝((𝑎+𝑏𝑑)+(𝑥+𝑦+𝑑))𝑚((𝑎𝑑)+(𝑥+𝑦+𝑑))𝑘𝑑𝜇1(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. For 𝑚,𝑘+, one has 𝑚𝑘𝑗=0𝑙=0𝑚𝑗𝑘𝑙(𝑎+𝑏𝑐+1)𝑚𝑗(𝑎𝑐+1)𝑘𝑙𝐸𝑗+𝑙+(𝑥+𝑐)𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑑)𝑚𝑗(𝑎𝑑)𝑘𝑙𝑚𝑗𝑘𝑙𝐸𝑗+𝑙(𝑥+𝑑)=2(𝑥+𝑎+𝑏)𝑚(𝑥+𝑎)𝑘.(3.29)

Let us consider the formula in Lemma 3.5 with 𝑑=𝑐1. Then we have 𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐+1)𝑚𝑗(𝑎𝑐+1)𝑘𝑙𝑚𝑗𝑘𝑙𝐸𝑗+𝑙(𝑥+𝑐)+𝐸𝑗+𝑙(𝑥+𝑐1)=2(𝑥+𝑎+𝑏)𝑚(𝑥+𝑎)𝑘.(3.30) Taking 𝑝𝑑𝜇(𝑥) on both sides of (3.30), LHS=𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐+1)𝑚𝑗(𝑎𝑐+1)𝑘𝑙𝑚𝑗𝑘𝑙𝑗+𝑙𝑠=0𝑠𝑗+𝑙×𝐸𝑗+𝑙𝑠𝑝((𝑥+𝑐)𝑠+(𝑥+𝑐1)𝑠)𝑑𝜇(𝑥)=2𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐+1)𝑚𝑗(𝑎𝑐+1)𝑘𝑙𝑚𝑗𝑘𝑙𝑗+𝑙𝑠=0𝑠𝑗+𝑙×𝐸𝑗+𝑙𝑠𝐵𝑠(𝑐1)+𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐+1)𝑚𝑗(𝑎𝑐+1)𝑘𝑙𝑚𝑗𝑘𝑙×(𝑗+𝑙)𝐸𝑗+𝑙1(𝑐1).(3.31) By the same method, we get RHS=2𝑚𝑠=0𝑚𝑠𝑏𝑚𝑠𝑝(𝑥+𝑎)𝑠+𝑘𝑑𝜇(𝑥)=2𝑚𝑠=0𝑚𝑠𝑏𝑚𝑠𝐵𝑠+𝑘(𝑎).(3.32) Therefore, by (3.31) and (3.32), we obtain the following proposition.

Proposition 3.6. Let 𝑎,𝑏,𝑐. Then one has 2𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐+1)𝑚𝑗(𝑎𝑐+1)𝑘𝑙𝑚𝑗𝑘𝑙𝑗+𝑙𝑠=0𝑠𝐸𝑗+𝑙𝑗+𝑙𝑠×𝐵𝑠(𝑐1)+𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐+1)𝑚𝑗(𝑎𝑐+1)𝑘𝑙𝑚𝑗𝑘𝑙(𝑗+𝑙)𝐸𝑗+𝑙1(𝑐1)=2𝑚𝑠=0𝑚𝑠𝑏𝑚𝑠𝐵𝑠+𝑘(𝑎).(3.33)

Replacing 𝑐 by 𝑐+1, we have 2𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙𝑗+𝑙𝑠=0𝑠𝐸𝑗+𝑙𝑗+𝑙𝑠𝐵𝑠+(𝑐)𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙𝐸𝑗+𝑙𝑠(𝑐)=2𝑚𝑠=0𝑚𝑠𝑏𝑚𝑠𝐵𝑠+𝑘(𝑎).(3.34)

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)=2𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙𝑗+𝑙𝑠=0𝑠𝐸𝑗+𝑙𝑗+𝑙𝑠×𝐵𝑠+𝑠𝑐1𝑖=0𝑖𝑠1=2𝑚𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙𝑗+𝑙𝑠=0𝑠𝐸𝑗+𝑙𝑗+𝑙𝑠𝐵𝑠+2𝑐1𝑚𝑖=0𝑘𝑗=0𝑙=0(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙(𝑗+𝑙)(1)0×𝐸𝑗+𝑙1+2𝑖1𝑒=0(1)𝑒1𝑒𝑗+𝑙1=2𝑚𝑘𝑗=0𝑙=0𝑗+𝑙𝑠=0𝑚𝑗𝑘𝑙𝑠𝑗+𝑙(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙×𝐸𝑗+𝑙𝑠𝐵𝑠+2𝑚𝑚1𝑘𝑗=0𝑙=0𝑗𝑘𝑙𝑚1(𝑎+𝑏𝑐)𝑚1𝑗(𝑎𝑐)𝑘𝑙𝐸𝑗+𝑙𝛿𝑐1+2𝑘𝑚𝑗=0𝑘1𝑙=0𝑚𝑗𝑙𝑘1(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘1𝑙𝐸𝑗+𝑙𝛿𝑐1+4𝑚𝑐2𝑒=0(𝑎+𝑏𝑐+𝑒)𝑚1(𝑎𝑐+𝑒)𝑘𝛿𝑐𝑒+4𝑘𝑐2𝑒=0(𝑎+𝑏𝑐+𝑒)𝑚(𝑎𝑐+𝑒)𝑘1𝛿𝑐𝑒,(3.35) where 𝛿𝑐𝑘=1if𝑐𝑘(mod2),0if𝑐𝑘(mod2).(3.36) The second term of the LHS of (3.34) =𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙(1)𝑐𝐸𝑗+𝑙1+2𝑐1𝑖=0(1)𝑖1𝑖𝑗+𝑙1=(1)𝑐𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙𝐸𝑗+𝑙1+2(1)𝑐𝑐1𝑖=0(1)𝑖1𝑚(𝑎+𝑏𝑐+𝑖)𝑚1(𝑎𝑐+𝑖)𝑘+2(1)𝑐𝑐1𝑖=0(1)𝑖1(𝑎+𝑏𝑐+𝑖)𝑚𝑘(𝑎𝑐+𝑖)𝑘1=(1)𝑐𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙𝐸𝑗+𝑙1+2𝑚(1)𝑐𝑐1𝑖=0(1)𝑖1(𝑎+𝑏𝑐+𝑖)𝑚1(𝑎𝑐+𝑖)𝑘+2𝑘(1)𝑐𝑐1𝑖=0(1)𝑖1(𝑎+𝑏𝑐+𝑖)𝑚(a𝑐+𝑖)𝑘1.(3.37)

Therefore, by (3.34), (3.35), and (3.37), we obtain the following theorem.

Theorem 3.7. Let 𝑎,𝑏,𝑐 with 𝑐1. Then one has 2𝑚𝑘𝑗=0𝑙=0𝑗+𝑙𝑠=0𝑚𝑗𝑘𝑙𝑠𝑗+𝑙(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝐸𝑗+𝑙𝑠𝐵𝑠+2𝑚𝛿𝑚𝑐1𝑘𝑗=0𝑙=0𝑗𝑘𝑙(𝑚1𝑎+𝑏𝑐)𝑚1𝑗(𝑎𝑐)𝑘𝑙𝐸𝑗+𝑙+2𝑘𝛿𝑚𝑐1𝑘𝑗=0𝑙=0𝑚𝑗𝑙𝑘1(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘1𝑙𝐸𝑗+𝑙+(1)𝑐𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(𝑎+𝑏𝑐)𝑚𝑗(𝑎𝑐)𝑘𝑙𝑚𝑗𝑘𝑙𝐸𝑗+𝑙1+2𝑚𝑐1𝑒=0(𝑎+𝑏𝑐+𝑒)𝑚1(𝑎𝑐+𝑒)𝑘+2𝑘𝑐1𝑒=0(𝑎+𝑏𝑐+𝑒)𝑚(𝑎𝑐+𝑒)𝑘1=2𝑚𝑠=0𝑚𝑠𝑏𝑚𝑠𝐵𝑠+𝑘(𝑎),(3.38) where 𝛿𝑐𝑘=1if𝑐𝑘(mod2),0if𝑐𝑘(mod2).(3.39)

Remark 3.8. Here, we note that 4𝑚𝑐2𝑒=0(𝑎+𝑏𝑐+𝑒)𝑚1(𝑎𝑐+𝑒)𝑘𝛿𝑐𝑒+2𝑚(1)𝑐𝑐1𝑖=0(1)𝑖1(𝑎+𝑏𝑐+𝑖)𝑚1(𝑎𝑐+𝑖)𝑘=2𝑚𝑐1𝑒=0(𝑎+𝑏𝑐+𝑒)𝑚1(𝑎𝑐+𝑒)𝑘.(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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