Discrete Dynamics in Nature and Society

Discrete Dynamics in Nature and Society / 2012 / Article

Research Article | Open Access

Volume 2012 |Article ID 269847 | https://doi.org/10.1155/2012/269847

Dae San Kim, Dmitry V. Dolgy, Hyun-Mee Kim, Sang-Hun Lee, Taekyun Kim, "Integral Formulae of Bernoulli Polynomials", Discrete Dynamics in Nature and Society, vol. 2012, Article ID 269847, 15 pages, 2012. https://doi.org/10.1155/2012/269847

Integral Formulae of Bernoulli Polynomials

Academic Editor: Lee Chae Jang
Received24 Feb 2012
Accepted10 May 2012
Published27 Jun 2012

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: 𝑡𝑒𝑡𝑒−1𝑥𝑡=âˆžî“ğ‘›=0𝐵𝑛(𝑡𝑥)𝑛,𝑛!(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 2𝑒𝑡𝑒+1𝑥𝑡=𝑒𝐸(𝑥)𝑡=âˆžî“ğ‘›=0𝐸𝑛(𝑡𝑥)𝑛𝑛!(1.2) with the usual convention about replacing 𝐸𝑛(𝑥) by 𝐸𝑛(𝑥) (see [1–11]). From (1.1) and (1.2), we can easily derive the following equation: 𝑡𝑒𝑡𝑒−1𝑥𝑡=𝑡22𝑒𝑥𝑡𝑒𝑡+𝑡+1𝑒𝑡−12𝑒𝑥𝑡𝑒𝑡=+1âˆžî“ğ‘›=0âŽ›âŽœâŽœâŽğ‘›î“ğ‘˜=0𝑘≠1âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘˜ğ¸ğ‘›âˆ’ğ‘˜âŽžâŽŸâŽŸâŽ ğ‘¡(𝑥)𝑛.𝑛!(1.3) By (1.1) and (1.3), we get 𝐵𝑛(𝑥)=𝑛𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘˜ğ¸ğ‘›âˆ’ğ‘˜î€·(𝑥),𝑛∈ℤ+.=ℕ∪{0}(1.4)

From (1.1), we have 𝐵𝑛(𝑥)=𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘™ğ‘¥ğ‘›âˆ’ğ‘™.(1.5) Thus, by (1.5), we get 𝑑𝐵𝑑𝑥𝑛(𝑥)=𝑛𝑛−1𝑙=0âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’1𝑙𝑥𝑛−1−𝑙=𝑛𝐵𝑛−1(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=𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘™ğ‘¥ğ‘›âˆ’ğ‘™,(1.7) (see [6]). From (1.7), we can derive the following equation: 𝑑𝐸𝑑𝑥𝑛(𝑥)=𝑛𝑛−1𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘™ğ‘¥ğ‘›âˆ’1−𝑙=𝑛𝐸𝑛−1(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∈ℤ+. The following identity is known in [5]: 𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏+1)𝑚−𝑗(ğ‘Ž+1)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ (−1)𝑗+𝑙+𝑗+𝑙+1𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏+1)𝑚−𝑗(ğ‘Ž+1)𝑘−𝑙−(ğ‘Ž+𝑏)ğ‘šâˆ’ğ‘—ğ‘Žğ‘˜âˆ’ğ‘™î€¸âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘—+𝑙+1(𝑥)𝑗+𝑙+1=(𝑥+ğ‘Ž)𝑘(𝑥+ğ‘Ž+𝑏)𝑚,whereğ‘Ž,𝑏∈ℤ.(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=1𝑡2𝑒𝑡−1𝑒𝑡+1𝑡𝑒𝑥𝑡𝑒𝑡=1−1𝑡22−2𝑒𝑡+1𝑡𝑒𝑥𝑡𝑒𝑡=1−1ğ‘¡îƒ©âˆ’âˆžî“ğ‘™=12𝐸𝑙𝑡𝑙!ğ‘™îƒªîƒ©âˆžî“ğ‘š=0𝐵𝑚𝑡(𝑥)𝑚𝑚!=−2âˆžî“ğ‘™=0𝐸𝑙+1𝑡𝑙+1𝑙𝑙!îƒªîƒ©âˆžî“ğ‘š=0𝐵𝑚𝑡(𝑥)𝑚𝑚!=−2âˆžî“ğ‘›=0âŽ›âŽœâŽœâŽğ‘›î“ğ‘™=0𝐸𝑙+1âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘™+1ğ‘›âˆ’ğ‘™âŽžâŽŸâŽŸâŽ ğ‘¡(𝑥)𝑛.𝑛!(2.1)

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

Theorem 2.1. For 𝑛∈ℤ+, one has 𝐸𝑛(𝑥)=−2𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘™+1𝐵𝑙+1𝑛−𝑙(𝑥).(2.2)

Let us take the definite integral from 0 to 1 on both sides of (1.4): for 𝑛≥2, 0=−2𝑛𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘˜ğ¸ğ‘›âˆ’ğ‘˜+1𝑛−𝑘+1=−2𝐵𝑛𝐸1−2𝑛−1𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘˜ğ¸ğ‘›âˆ’ğ‘˜+1.𝑛−𝑘+1(2.3)

By (2.3), we get 𝐵𝑛=2𝑛−1𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘˜ğ¸ğ‘›âˆ’ğ‘˜+1.𝑛−𝑘+1(2.4)

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

Theorem 2.2. For 𝑛∈ℕ, with 𝑛≥2, one has 𝐵𝑛=2𝑛−1𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘˜ğ¸ğ‘›âˆ’ğ‘˜+1.𝑛−𝑘+1(2.5)

Let us take 𝑘=𝑚, ğ‘Ž=0, and 𝑏=−2 in (1.11). Then we have 𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘—âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ (−1)𝑗+𝑙+𝑗+𝑙+1𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘—âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘—+𝑙+1(𝑥)−𝑗+𝑙+1𝑚𝑗=0(−2)ğ‘šâˆ’ğ‘—âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ ğµğ‘—+𝑚+1(𝑥)𝑗+𝑚+1=𝑥𝑚(𝑥−2)𝑚.(2.6) It is easy to show that 10𝑥𝑚(𝑥−2)𝑚𝑑𝑥=201/2(2𝑡−2)𝑚(2𝑡)𝑚=𝑑𝑡(−1)𝑚22𝑚201/2𝑡𝑚(1−𝑡)𝑚=𝑑𝑡(−1)𝑚22𝑚10𝑡𝑚(1−𝑡)𝑚𝑑𝑡=(−1)𝑚22𝑚𝑚!𝑚!=((2𝑚+1)!−1)𝑚22𝑚12𝑚+1𝑚2𝑚.(2.7)

Let us consider the integral from 0 to 1 in (2.6): 𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ 1=(𝑗+𝑙+1−1)𝑚22𝑚(2𝑚+1)𝑚2𝑚(𝑚∈ℕ).(2.8) By (2.6) and (2.8), we get 𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘—âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘—+𝑙+1𝑗+𝑙+1=2𝑚𝑗=0(−2)ğ‘šâˆ’ğ‘—âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ 1𝑗+𝑚+1𝑗+𝑚𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘—+𝑚+1𝑘𝐸𝑗+𝑚+2−𝑘+𝑗+𝑚+2−𝑘(−1)𝑚+122𝑚12𝑚+1𝑚2𝑚,for𝑚∈ℕ.(2.9) Therefore, by (2.9), we obtain the following theorem.

Theorem 2.3. For 𝑚∈ℕ, one has 𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘—âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘—+𝑙+1𝑗+𝑙+1=2𝑚𝑗=0(−2)ğ‘šâˆ’ğ‘—âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ 1𝑗+𝑚+1𝑗+𝑚𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘—+𝑚+1𝑘𝐸𝑗+𝑚+2−𝑘+𝑗+𝑚+2−𝑘(−1)𝑚+122𝑚12𝑚+1𝑚2𝑚.(2.10)

Lemma 2.4. Let ğ‘Ž,𝑏∈ℤ. For 𝑚,𝑘∈ℤ+, one has 𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏+1)𝑚−𝑗(ğ‘Ž+1)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙+(𝑥)𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏)ğ‘šâˆ’ğ‘—ğ‘Žğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙(𝑥)=2(𝑥+ğ‘Ž+𝑏)𝑚(𝑥+ğ‘Ž)𝑘,(2.11) (see [5]).

Let us take 𝑘=𝑚, ğ‘Ž=1, 𝑏=−2 in Lemma 2.4. Then we have 𝑚𝑙=02ğ‘šâˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘š+𝑙(𝑥)+𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘—âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙𝑥(𝑥)=22−1𝑚.(2.12) Taking integral from 0 to 1 in (2.12), we get −2𝑚𝑙=02ğ‘šâˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘š+𝑙+1𝑚+𝑙+1−2𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙+1𝑗+𝑙+1=210𝑥2−1𝑚𝑑𝑥.(2.13)

It is easy to show that 10𝑥2−1𝑚𝑑𝑥=(−1)𝑚𝑚𝑘=12𝑘=2𝑘+1(−1)𝑚22𝑚(2𝑚+1)𝑚2𝑚.(2.14)

Thus, by (2.13) and (2.14), we get 𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙+1𝑗+𝑙+1=−𝑚𝑙=02ğ‘šâˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘š+𝑙+1+(𝑚+𝑙+1−1)𝑚+122𝑚(2𝑚+1)𝑚2𝑚.(2.15)

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

Theorem 2.5. For 𝑚∈ℤ+, one has 𝑚𝑚𝑗=0𝑙=0(−1)ğ‘šâˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙+1+(𝑗+𝑙+1−1)𝑚22𝑚(2𝑚+1)𝑚2𝑚=2𝑚𝑙=02ğ‘šâˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘™âŽžâŽŸâŽŸâŽ 1𝑚+𝑙+1𝑚+𝑙+1𝑘=0âŽ›âŽœâŽœâŽğ‘˜âŽžâŽŸâŽŸâŽ ğ¸ğ‘š+𝑙+1𝑘+1𝐵𝑘+1𝑚+𝑙+1−𝑘.(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 𝑈𝐷(ℤ𝑝) be the space of uniformly differentiable functions on ℤ𝑝. For 𝑓∈𝑈𝐷(ℤ𝑝), the bosonic 𝑝-adic integral on ℤ𝑝 is defined by 𝐼(𝑓)=ℤ𝑝𝑓(𝑥)𝑑𝜇(𝑥)=limğ‘›â†’âˆž1𝑝𝑛𝑝𝑛−1𝑥=0𝑓(𝑥),(3.1) (see [8]). Thus, by (3.1), we get ℤ𝑝𝑓1(𝑥)𝑑𝜇(𝑥)=ℤ𝑝𝑓(𝑥)𝑑𝜇(𝑥)+ğ‘“î…ž(0),(3.2) where 𝑓1(𝑥)=𝑓(𝑥+1), and ğ‘“î…ž(0)=𝑑𝑓(𝑥)/𝑑𝑥|𝑥=0. Let us take 𝑓(𝑦)=𝑒𝑡(𝑥+𝑦). Then we have ℤ𝑝𝑒𝑡(𝑥+𝑦)𝑡𝑑𝜇(𝑦)=𝑒𝑡𝑒−1𝑡𝑥=âˆžî“ğ‘›=0𝐵𝑛(𝑡𝑥)𝑛.𝑛!(3.3) From (3.3), we have ℤ𝑝(𝑥+𝑦)𝑛𝑑𝜇(𝑦)=𝐵𝑛(𝑥),ℤ𝑝𝑦𝑛𝑑𝜇(𝑦)=𝐵𝑛.(3.4) From (1.2), we can derive the following integral equation: 𝐼𝑓𝑛=𝐼(𝑓)+𝑛−1𝑖=0ğ‘“î…ž(𝑖)(𝑛∈ℕ).(3.5) Thus, from (3.4) and (3.5), we get ℤ𝑝(𝑥+𝑛)𝑚𝑑𝜇(𝑥)=ℤ𝑝𝑥𝑚𝑑𝜇(𝑥)+𝑚𝑛−1𝑖=0𝑖𝑚−1.(3.6) From (3.6), we have 𝐵𝑚(𝑛)−𝐵𝑚=𝑚𝑛−1𝑖=0𝑖𝑚−1𝑚∈ℤ+.(3.7) The fermionic 𝑝-adic integral on ℤ𝑝 is defined by Kim as follows [6, 7]: 𝐼−1(𝑓)=ℤ𝑝𝑓(𝑥)𝑑𝜇−1(𝑥)=limğ‘ğ‘›â†’âˆžğ‘›âˆ’1𝑥=0𝑓(𝑥)(−1)𝑥.(3.8) Let 𝑓1(𝑥)=𝑓(𝑥+1). Then we have 𝐼−1𝑓1=−𝐼−1𝐼(𝑓)+2𝑓(0),−1𝑓2=−𝐼−1𝑓1+2𝑓1(0)=−𝐼−1𝑓1+2𝑓(1)=(−1)2𝐼−1(𝑓)+2(−1)2−1𝑓(0)+2𝑓(1).(3.9) Continuing this process, we obtain the following equation: 𝐼−1𝑓𝑛=(−1)𝑛𝐼−1(𝑓)+2𝑛−1𝑙=0(−1)𝑛−𝑙−1𝑓(𝑙),where𝑓𝑛(𝑥)=𝑓(𝑥+𝑛).(3.10) Thus, by (3.10), we have ℤ𝑝(𝑥+𝑛)𝑚𝑑𝜇−1(𝑥)=(−1)𝑛ℤ𝑝𝑥𝑚𝑑𝜇−1(𝑥)+2𝑛−1𝑙=0(−1)𝑛−𝑙−1𝑙𝑚.(3.11) Let us take 𝑓(𝑦)=𝑒𝑡(𝑥+𝑦). By (3.9), we get ℤ𝑝𝑒𝑡(𝑥+𝑦)𝑑𝜇−1(𝑦)=2𝑒𝑥𝑡𝑒𝑡=+1âˆžî“ğ‘›=0𝐸𝑛(𝑡𝑥)𝑛.𝑛!(3.12) From (3.2), we have the Witt's formula for the 𝑛th Euler polynomials and numbers as follows: ℤ𝑝(𝑥+𝑦)𝑛𝑑𝜇−1(𝑦)=𝐸𝑛(𝑥),ℤ𝑝𝑦𝑛𝑑𝜇−1(𝑦)=𝐸𝑛,where𝑛∈ℤ+.(3.13) By (3.11) and (3.13), we get 𝐸𝑚(𝑛)=(−1)𝑛𝐸𝑚+2𝑛−1𝑙=0(−1)𝑙−1𝑙𝑚,𝑚∈ℤ+.,𝑛∈ℕ(3.14)

Let us consider the following 𝑝-adic integral on ℤ𝑝: 𝐾1=ℤ𝑝𝐵𝑛(𝑥)𝑑𝜇(𝑥)=𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘™î€œâ„¤ğ‘ğ‘¥ğ‘™=𝑑𝜇(𝑥)𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘™ğµğ‘™.(3.15)

From (1.4) and (3.15), we have 𝐾1=𝑛𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘˜ğ‘›âˆ’ğ‘˜î“ğ‘™=0ğ¸ğ‘›âˆ’ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ î€œğ‘›âˆ’ğ‘˜â„¤ğ‘ğ‘¥ğ‘™=𝑑𝜇(𝑥)𝑛𝑘=0𝑘≠1𝑛−𝑘𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘˜ğ‘˜ğµğ‘™ğ¸ğ‘›âˆ’ğ‘˜âˆ’ğ‘™.(3.16) Therefore, by (3.15) and (3.16), we obtain the following theorem.

Theorem 3.1. For 𝑛∈ℤ+, one has 𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘™ğµğ‘™=𝑛𝑘=0𝑘≠1𝑛−𝑘𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘˜ğ‘˜ğµğ‘™ğ¸ğ‘›âˆ’ğ‘˜âˆ’ğ‘™.(3.17)

Now, we set 𝐾2=ℤ𝑝𝐵𝑛(𝑥)𝑑𝜇−1(𝑥)=𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘™ğ¸ğ‘™.(3.18)

By (1.4), we get 𝐾2=𝑛𝑘=0𝑘≠1âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘˜ğ‘›âˆ’ğ‘˜î“ğ‘™=0ğ¸ğ‘›âˆ’ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ î€œğ‘›âˆ’ğ‘˜â„¤ğ‘ğ‘¥ğ‘™ğ‘‘ğœ‡âˆ’1=(𝑥)𝑛𝑘=0𝑘≠1𝑛−𝑘𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘˜ğ‘˜ğ¸ğ‘›âˆ’ğ‘˜âˆ’ğ‘™ğ¸ğ‘™.(3.19) Therefore, by (3.18) and (3.19), we obtain the following theorem.

Theorem 3.2. For 𝑛∈ℤ+, one has 𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘™ğ¸ğ‘™=𝑛𝑘=0𝑘≠1𝑛−𝑘𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘˜âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘˜ğ‘˜ğ¸ğ‘›âˆ’ğ‘˜âˆ’ğ‘™ğ¸ğ‘™.(3.20)

Let us consider the following integral on ℤ𝑝: 𝐾3=ℤ𝑝𝐸𝑛(𝑥)𝑑𝜇−1(𝑥)=𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘›âˆ’ğ‘™î€œâ„¤ğ‘ğ‘¥ğ‘™ğ‘‘ğœ‡âˆ’1(𝑥)=𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘›âˆ’ğ‘™ğ¸ğ‘™.(3.21)

From (2.2), we have 𝐾3=−2𝑛𝑙=0𝐸𝑙+1âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ‘™+1𝑛−𝑙𝑘=0âŽ›âŽœâŽœâŽğ‘˜âŽžâŽŸâŽŸâŽ ğµğ‘›âˆ’ğ‘™ğ‘›âˆ’ğ‘™âˆ’ğ‘˜î€œâ„¤ğ‘ğ‘¥ğ‘˜ğ‘‘ğœ‡âˆ’1(𝑥)=−2𝑛𝑙=0𝑛−𝑙𝑘=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜âŽžâŽŸâŽŸâŽ ğ¸ğ‘›âˆ’ğ‘™ğ‘™+1𝐵𝑙+1𝑛−𝑙−𝑘𝐸𝑘.(3.22) Therefore, by (3.21) and (3.22), we obtain the following theorem.

Theorem 3.3. For 𝑛∈ℤ+, one has 𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘›âˆ’ğ‘™ğ¸ğ‘™=−2𝑛𝑙=0𝑛−𝑙𝑘=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜âŽžâŽŸâŽŸâŽ ğ¸ğ‘›âˆ’ğ‘™ğ‘™+1𝐸𝑙+1𝑘𝐵𝑛−𝑙−𝑘.(3.23)

Now, we set 𝐾4=ℤ𝑝𝐸𝑛(𝑥)𝑑𝜇(𝑥)=𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘›âˆ’ğ‘™ğµğ‘™.(3.24)

By (2.2), we get 𝐾4=−2𝑛𝑙=0𝑛−𝑙𝑘=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜âŽžâŽŸâŽŸâŽ ğ¸ğ‘›âˆ’ğ‘™ğ‘™+1𝐵𝑙+1𝑛−𝑙−𝑘𝐵𝑘.(3.25) Therefore, by (3.24) and (3.25), we obtain the following corollary.

Corollary 3.4. For 𝑛∈ℤ+, we have 𝑛𝑙=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘™ğµğ‘™=−2𝑛𝑙=0𝑛−𝑙𝑘=0âŽ›âŽœâŽœâŽğ‘›ğ‘™âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜âŽžâŽŸâŽŸâŽ ğ¸ğ‘›âˆ’ğ‘™ğ‘™+1𝐵𝑙+1𝑛−𝑙−𝑘𝐵𝑘.(3.26)

Let us assume that ğ‘Ž,𝑏,𝑐,𝑑∈ℤ. From Lemma 2.4 and (3.13), we note that ℤ𝑝((ğ‘Ž+𝑏+1)+(𝑥+𝑦))𝑚((ğ‘Ž+1)+(𝑥+𝑦))𝑘𝑑𝜇−1(+𝑦)ℤ𝑝((ğ‘Ž+𝑏)+(𝑥+𝑦))𝑚((ğ‘Ž+(𝑥+𝑦))𝑘𝑑𝜇−1(𝑦)=2(𝑥+ğ‘Ž+𝑏)𝑚(𝑥+ğ‘Ž)𝑘.(3.27)

By (3.27), we get 2(𝑥+ğ‘Ž+𝑏)𝑚(𝑥+ğ‘Ž)𝑘=ℤ𝑝((ğ‘Ž+𝑏−𝑐+1)+(𝑥+𝑐+𝑦))𝑚((ğ‘Žâˆ’ğ‘+1)+(𝑥+𝑐+𝑦))𝑘𝑑𝜇−1(+𝑦)ℤ𝑝((ğ‘Ž+𝑏−𝑑)+(𝑥+𝑦+𝑑))𝑚((ğ‘Žâˆ’ğ‘‘)+(𝑥+𝑦+𝑑))𝑘𝑑𝜇−1(𝑦).(3.28)

Thus, by (3.28) and (3.13), we obtain the following lemma (see [5]).

Lemma 3.5. Let ğ‘Ž,𝑏,𝑐,𝑑∈ℤ. For 𝑚,𝑘∈ℤ+, one has 𝑚𝑘𝑗=0𝑙=0âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ (ğ‘Ž+𝑏−𝑐+1)𝑚−𝑗(ğ‘Žâˆ’ğ‘+1)𝑘−𝑙𝐸𝑗+𝑙+(𝑥+𝑐)𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑑)𝑚−𝑗(ğ‘Žâˆ’ğ‘‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙(𝑥+𝑑)=2(𝑥+ğ‘Ž+𝑏)𝑚(𝑥+ğ‘Ž)𝑘.(3.29)

Let us consider the formula in Lemma 3.5 with 𝑑=𝑐−1. Then we have 𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐+1)𝑚−𝑗(ğ‘Žâˆ’ğ‘+1)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ î€·ğ¸ğ‘—+𝑙(𝑥+𝑐)+𝐸𝑗+𝑙(𝑥+𝑐−1)=2(𝑥+ğ‘Ž+𝑏)𝑚(𝑥+ğ‘Ž)𝑘.(3.30) Taking ∫ℤ𝑝𝑑𝜇(𝑥) on both sides of (3.30), LHS=𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐+1)𝑚−𝑗(ğ‘Žâˆ’ğ‘+1)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ‘—+𝑙𝑠=0âŽ›âŽœâŽœâŽğ‘ âŽžâŽŸâŽŸâŽ ğ‘—+𝑙×𝐸𝑗+𝑙−𝑠ℤ𝑝((𝑥+𝑐)𝑠+(𝑥+𝑐−1)𝑠)𝑑𝜇(𝑥)=2𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐+1)𝑚−𝑗(ğ‘Žâˆ’ğ‘+1)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ‘—+𝑙𝑠=0âŽ›âŽœâŽœâŽğ‘ âŽžâŽŸâŽŸâŽ ğ‘—+𝑙×𝐸𝑗+𝑙−𝑠𝐵𝑠(𝑐−1)+𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐+1)𝑚−𝑗(ğ‘Žâˆ’ğ‘+1)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ Ã—(𝑗+𝑙)𝐸𝑗+𝑙−1(𝑐−1).(3.31) By the same method, we get RHS=2𝑚𝑠=0âŽ›âŽœâŽœâŽğ‘šğ‘ âŽžâŽŸâŽŸâŽ ğ‘ğ‘šâˆ’ğ‘ î€œâ„¤ğ‘(𝑥+ğ‘Ž)𝑠+𝑘𝑑𝜇(𝑥)=2𝑚𝑠=0âŽ›âŽœâŽœâŽğ‘šğ‘ âŽžâŽŸâŽŸâŽ ğ‘ğ‘šâˆ’ğ‘ ğµğ‘ +𝑘(ğ‘Ž).(3.32) Therefore, by (3.31) and (3.32), we obtain the following proposition.

Proposition 3.6. Let ğ‘Ž,𝑏,𝑐∈ℤ. Then one has 2𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐+1)𝑚−𝑗(ğ‘Žâˆ’ğ‘+1)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ‘—+𝑙𝑠=0âŽ›âŽœâŽœâŽğ‘ âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙𝑗+𝑙−𝑠×𝐵𝑠(𝑐−1)+𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐+1)𝑚−𝑗(ğ‘Žâˆ’ğ‘+1)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ (𝑗+𝑙)𝐸𝑗+𝑙−1(𝑐−1)=2𝑚𝑠=0âŽ›âŽœâŽœâŽğ‘šğ‘ âŽžâŽŸâŽŸâŽ ğ‘ğ‘šâˆ’ğ‘ ğµğ‘ +𝑘(ğ‘Ž).(3.33)

Replacing 𝑐 by 𝑐+1, we have 2𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ‘—+𝑙𝑠=0âŽ›âŽœâŽœâŽğ‘ âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙𝑗+𝑙−𝑠𝐵𝑠+(𝑐)𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙−𝑠(𝑐)=2𝑚𝑠=0âŽ›âŽœâŽœâŽğ‘šğ‘ âŽžâŽŸâŽŸâŽ ğ‘ğ‘šâˆ’ğ‘ ğµğ‘ +𝑘(ğ‘Ž).(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)=2𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ‘—+𝑙𝑠=0âŽ›âŽœâŽœâŽğ‘ âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙𝑗+𝑙−𝑠×𝐵𝑠+𝑠𝑐−1𝑖=0𝑖𝑠−1=2𝑚𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ‘—+𝑙𝑠=0âŽ›âŽœâŽœâŽğ‘ âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙𝑗+𝑙−𝑠𝐵𝑠+2𝑐−1𝑚𝑖=0𝑘𝑗=0𝑙=0(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ (𝑗+𝑙)(−1)0×𝐸𝑗+𝑙−1+2𝑖−1𝑒=0(−1)𝑒−1𝑒𝑗+𝑙−1=2𝑚𝑘𝑗=0𝑙=0𝑗+𝑙𝑠=0âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘ âŽžâŽŸâŽŸâŽ ğ‘—+𝑙(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)𝑘−𝑙×𝐸𝑗+𝑙−𝑠𝐵𝑠+2𝑚𝑚−1𝑘𝑗=0𝑙=0âŽ›âŽœâŽœâŽğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ‘šâˆ’1(ğ‘Ž+𝑏−𝑐)𝑚−1−𝑗(ğ‘Žâˆ’ğ‘)𝑘−𝑙𝐸𝑗+𝑙𝛿𝑐≡1+2𝑘𝑚𝑗=0𝑘−1𝑙=0âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ ğ‘˜âˆ’1(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)𝑘−1−𝑙𝐸𝑗+𝑙𝛿𝑐≡1+4𝑚𝑐−2𝑒=0(ğ‘Ž+𝑏−𝑐+𝑒)𝑚−1(ğ‘Žâˆ’ğ‘+𝑒)𝑘𝛿𝑐≡𝑒+4𝑘𝑐−2𝑒=0(ğ‘Ž+𝑏−𝑐+𝑒)𝑚(ğ‘Žâˆ’ğ‘+𝑒)𝑘−1𝛿𝑐≡𝑒,(3.35) where 𝛿𝑐≡𝑘=1if𝑐≡𝑘(mod2),0if𝑐≠𝑘(mod2).(3.36) The second term of the LHS of (3.34) =𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ (−1)𝑐𝐸𝑗+𝑙−1+2𝑐−1𝑖=0(−1)𝑖−1𝑖𝑗+𝑙−1=(−1)𝑐𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙−1+2(−1)𝑐𝑐−1𝑖=0(−1)𝑖−1𝑚(ğ‘Ž+𝑏−𝑐+𝑖)𝑚−1(ğ‘Žâˆ’ğ‘+𝑖)𝑘+2(−1)𝑐𝑐−1𝑖=0(−1)𝑖−1(ğ‘Ž+𝑏−𝑐+𝑖)𝑚𝑘(ğ‘Žâˆ’ğ‘+𝑖)𝑘−1=(−1)𝑐𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙−1+2𝑚(−1)𝑐𝑐−1𝑖=0(−1)𝑖−1(ğ‘Ž+𝑏−𝑐+𝑖)𝑚−1(ğ‘Žâˆ’ğ‘+𝑖)𝑘+2𝑘(−1)𝑐𝑐−1𝑖=0(−1)𝑖−1(ğ‘Ž+𝑏−𝑐+𝑖)𝑚(a−𝑐+𝑖)𝑘−1.(3.37)

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

Theorem 3.7. Let ğ‘Ž,𝑏,𝑐∈ℤ with 𝑐≥1. Then one has 2𝑚𝑘𝑗=0𝑙=0𝑗+𝑙𝑠=0âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘ âŽžâŽŸâŽŸâŽ ğ‘—+𝑙(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)𝑘−𝑙𝐸𝑗+𝑙−𝑠𝐵𝑠+2𝑚𝛿𝑚𝑐≡1𝑘𝑗=0𝑙=0âŽ›âŽœâŽœâŽğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ (𝑚−1ğ‘Ž+𝑏−𝑐)𝑚−1−𝑗(ğ‘Žâˆ’ğ‘)𝑘−𝑙𝐸𝑗+𝑙+2𝑘𝛿𝑚𝑐≡1𝑘𝑗=0𝑙=0âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘™âŽžâŽŸâŽŸâŽ ğ‘˜âˆ’1(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)𝑘−1−𝑙𝐸𝑗+𝑙+(−1)𝑐𝑚𝑘𝑗=0𝑙=0(𝑗+𝑙)(ğ‘Ž+𝑏−𝑐)𝑚−𝑗(ğ‘Žâˆ’ğ‘)ğ‘˜âˆ’ğ‘™âŽ›âŽœâŽœâŽğ‘šğ‘—âŽžâŽŸâŽŸâŽ âŽ›âŽœâŽœâŽğ‘˜ğ‘™âŽžâŽŸâŽŸâŽ ğ¸ğ‘—+𝑙−1+2𝑚𝑐−1𝑒=0(ğ‘Ž+𝑏−𝑐+𝑒)𝑚−1(ğ‘Žâˆ’ğ‘+𝑒)𝑘+2𝑘𝑐−1𝑒=0(ğ‘Ž+𝑏−𝑐+𝑒)𝑚(ğ‘Žâˆ’ğ‘+𝑒)𝑘−1=2𝑚𝑠=0âŽ›âŽœâŽœâŽğ‘šğ‘ âŽžâŽŸâŽŸâŽ ğ‘ğ‘šâˆ’ğ‘ ğµğ‘ +𝑘(ğ‘Ž),(3.38) where 𝛿𝑐≡𝑘=1if𝑐≡𝑘(mod2),0if𝑐≢𝑘(mod2).(3.39)

Remark 3.8. Here, we note that 4𝑚𝑐−2𝑒=0(ğ‘Ž+𝑏−𝑐+𝑒)𝑚−1(ğ‘Žâˆ’ğ‘+𝑒)𝑘𝛿𝑐≡𝑒+2𝑚(−1)𝑐𝑐−1𝑖=0(−1)𝑖−1(ğ‘Ž+𝑏−𝑐+𝑖)𝑚−1(ğ‘Žâˆ’ğ‘+𝑖)𝑘=2𝑚𝑐−1𝑒=0(ğ‘Ž+𝑏−𝑐+𝑒)𝑚−1(ğ‘Žâˆ’ğ‘+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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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.


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