Integral Formulae of Bernoulli Polynomials

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: 𝑡𝑒^𝑡−1𝑒^𝑥𝑡=^∞_𝑛=0𝐵_𝑛(𝑥)𝑡^𝑛𝑛!,(1.1) (see [1–11]). In the special case, 𝑥=0,𝐵_𝑛(0)=𝐵_𝑛 are called the 𝑛th Bernoulli numbers. The Euler polynomials are also defined by 2𝑒^𝑡+1𝑒^𝑥𝑡=𝑒^{𝐸(𝑥)𝑡}=^∞_𝑛=0𝐸_𝑛(𝑥)𝑡^𝑛𝑛!(1.2) with the usual convention about replacing 𝐸^𝑛(𝑥) by 𝐸_𝑛(𝑥) (see [1–11]). From (1.1) and (1.2), we can easily derive the following equation: 𝑡𝑒^𝑡−1𝑒^𝑥𝑡=𝑡=2𝑒2^𝑡𝑒+^𝑥𝑡1+𝑒^𝑡−𝑡1𝑒2^𝑡𝑒+^𝑥𝑡1_∞_𝑛=0⎛⎜⎜⎝^𝑛_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠𝐵_𝑘𝐸_𝑛−𝑘(𝑥)⎞⎟⎟⎠𝑡^𝑛𝑛!.(1.3) By (1.1) and (1.3), we get 𝐵_𝑛(𝑥)=^𝑛_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠𝐵_𝑘𝐸_𝑛−𝑘(𝑥),𝑛∈ℤ₊=ℕ∪{0}.(1.4)

From (1.1), we have 𝐵_𝑛(𝑥)=^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐵_𝑙𝑥^𝑛−𝑙.(1.5) Thus, by (1.5), we get 𝑑𝑑𝑥𝐵^𝑛(𝑥)=𝑛^𝑛−1_𝑙=0⎛⎜⎜⎝𝑛−𝑙1⎞⎟⎟⎠𝐵_𝑙𝑥^{𝑛−1−𝑙}=𝑛𝐵_𝑛−1(𝑥).(1.6) It is known that 𝐸_𝑛(0)=𝐸_𝑛 are called the 𝑛th Euler numbers (see [7]). The Euler polynomials are also given by 𝐸_𝑛(𝑥)=(𝐸+𝑥)^𝑛=^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐸_𝑙𝑥^𝑛−𝑙,(1.7) (see [6]). From (1.7), we can derive the following equation: 𝑑𝑑𝑥𝐸^𝑛(𝑥)=𝑛^𝑛−1_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐸_𝑙𝑥^{𝑛−1−𝑙}=𝑛𝐸_𝑛−1(𝑥).(1.8) By the definition of Bernoulli and Euler numbers, we get the following recurrence formulae: 𝐸₀=1,𝐸_𝑛(1)+𝐸_𝑛=2𝛿_0,𝑛,𝐵₀=1,𝐵_𝑛(1)−𝐵_𝑛=𝛿_1,𝑛,(1.9) where 𝛿_𝑛,𝑘 is the kronecker symbol (see [5]). From (1.6), (1.8), and (1.9), we note that ¹⁰𝐵_𝑛(𝑥)𝑑𝑥=𝛿^0,𝑛𝑛+1,¹⁰𝐸_𝑛(𝑥)𝑑𝑥=−2𝐸^𝑛+1𝑛+1,(1.10) where 𝑛∈ℤ₊. The following identity is known in [5]: ^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏+1)^𝑚−𝑗(𝑎+1)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠(−1)^𝑗+𝑙+𝑗+𝑙+1_𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏+1)^𝑚−𝑗(𝑎+1)^𝑘−𝑙−(𝑎+𝑏)^𝑚−𝑗𝑎^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐵^𝑗+𝑙+1(𝑥)=(𝑥+𝑎)_𝑘𝑗+𝑙+1(𝑥+𝑎+𝑏)^𝑚,where𝑎,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 2𝑒^𝑡+1𝑒^𝑥𝑡=1𝑡2𝑒^𝑡−1=1𝑒^𝑡+1𝑒^𝑡𝑡𝑒−^𝑥𝑡1=1𝑡2−2𝑒^𝑡2+1𝑒^𝑡𝑡𝑒−^𝑥𝑡1𝑡−^∞_𝑙=12𝐸_𝑙𝑙!𝑡^𝑙^∞_𝑚=0𝐵_𝑚(𝑥)𝑡^𝑚=−2_∞𝑚!_𝑙=0𝐸_𝑙+1𝑙+1𝑙𝑡!^𝑙^∞_𝑚=0𝐵_𝑚(𝑥)𝑡^𝑚=−2_∞𝑚!_𝑛=0⎛⎜⎜⎝^𝑛_𝑙=0𝐸_𝑙+1𝑙+1⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐵^𝑛−𝑙(𝑥)⎞⎟⎟⎠𝑡^𝑛𝑛!.(2.1)

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

Theorem 2.1. For 𝑛∈ℤ₊, one has 𝐸_𝑛(𝑥)=−2^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐸_𝑙+1𝑙+1𝐵^𝑛−𝑙(𝑥).(2.2)

Let us take the definite integral from 0 to 1 on both sides of (1.4): for 𝑛≥2, 0=−2^𝑛_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠𝐵_𝑘𝐸_{𝑛−𝑘+1}𝑛−𝑘+1=−2𝐵^𝑛𝐸₁−2^𝑛−1_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠𝐵_𝑘𝐸_{𝑛−𝑘+1}𝑛−𝑘+1.(2.3)

By (2.3), we get 𝐵_𝑛=2^𝑛−1_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠𝐵_𝑘𝐸_{𝑛−𝑘+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 𝐵_𝑛=2^𝑛−1_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠𝐵_𝑘𝐸_{𝑛−𝑘+1}𝑛−𝑘+1.(2.5)

Let us take 𝑘=𝑚, 𝑎=0, and 𝑏=−2 in (1.11). Then we have ^𝑚_𝑚𝑗=0_𝑙=0(−1)^𝑚−𝑗⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠(−1)^𝑗+𝑙𝑗+𝑙+1+^𝑚_𝑚𝑗=0_𝑙=0(−1)^𝑚−𝑗⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐵_𝑗+𝑙+1(𝑥)−𝑗+𝑙+1_𝑚_𝑗=0(−2)^𝑚−𝑗⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠𝐵_𝑗+𝑚+1(𝑥)𝑗+𝑚+1=𝑥^𝑚(𝑥−2)^𝑚.(2.6) It is easy to show that ¹⁰𝑥^𝑚(𝑥−2)^𝑚𝑑𝑥==2(−1)^01/2_𝑚(2𝑡−2)^𝑚(2𝑡)^𝑚𝑑𝑡2^2𝑚2^01/2𝑡^𝑚(1−𝑡)^𝑚𝑑𝑡=(−1)^𝑚2^2𝑚¹⁰=(−1)_𝑚𝑡^𝑚(1−𝑡)^𝑚𝑑𝑡2^2𝑚𝑚!𝑚!(2𝑚+1)!=(−1)^𝑚2^2𝑚2𝑚+11^𝑚2𝑚.(2.7)

Let us consider the integral from 0 to 1 in (2.6): ^𝑚_𝑚𝑗=0_𝑙=0(−1)^𝑚−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠1𝑗+𝑙+1=(−1)^𝑚2^2𝑚(2𝑚+1)^𝑚2𝑚(𝑚∈ℕ).(2.8) By (2.6) and (2.8), we get ^𝑚_𝑚𝑗=0_𝑙=0(−1)^𝑚−𝑗⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐵_𝑗+𝑙+1=2_𝑚𝑗+𝑙+1_𝑗=0(−2)^𝑚−𝑗⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠1𝑗+𝑚+1^𝑗+𝑚_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑗+𝑚𝑘+1⎞⎟⎟⎠𝐵_𝑘𝐸_{𝑗+𝑚+2−𝑘}+(−1)_𝑚+1𝑗+𝑚+2−𝑘2^2𝑚2𝑚+1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)^𝑚−𝑗⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐵_𝑗+𝑙+1=2_𝑚𝑗+𝑙+1_𝑗=0(−2)^𝑚−𝑗⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠1𝑗+𝑚+1^𝑗+𝑚_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑗+𝑚𝑘+1⎞⎟⎟⎠𝐵_𝑘𝐸_{𝑗+𝑚+2−𝑘}+(−1)_𝑚+1𝑗+𝑚+2−𝑘2^2𝑚2𝑚+11^𝑚2𝑚.(2.10)

Lemma 2.4. Let 𝑎,𝑏∈ℤ. For 𝑚,𝑘∈ℤ₊, one has ^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏+1)^𝑚−𝑗(𝑎+1)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐸_𝑗+𝑙+(𝑥)_𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏)^𝑚−𝑗𝑎^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐸_𝑗+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 ^𝑚_𝑙=02^𝑚−𝑙⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐸_𝑚+𝑙(𝑥)+^𝑚_𝑚𝑗=0_𝑙=0(−1)^𝑚−𝑗⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐸_𝑗+𝑙(𝑥)=2𝑥²−1^𝑚.(2.12) Taking integral from 0 to 1 in (2.12), we get −2^𝑚_𝑙=02^𝑚−𝑙⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐸_𝑚+𝑙+1𝑚+𝑙+1−2^𝑚_𝑚𝑗=0_𝑙=0(−1)^𝑚−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐸_𝑗+𝑙+1𝑗+𝑙+1=2¹⁰𝑥²−1^𝑚𝑑𝑥.(2.13)

It is easy to show that ¹⁰𝑥²−1^𝑚𝑑𝑥=(−1)^𝑚𝑚_𝑘=12𝑘2+𝑘1=(−1)^𝑚2^2𝑚(2𝑚+1)^𝑚2𝑚.(2.14)

Thus, by (2.13) and (2.14), we get ^𝑚_𝑚𝑗=0_𝑙=0(−1)^𝑚−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐸_𝑗+𝑙+1𝑗+𝑙+1=−^𝑚_𝑙=02^𝑚−𝑙⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐸_𝑚+𝑙+1𝑚+𝑙+1+(−1)^𝑚+12^2𝑚(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)^𝑚−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠𝐸_𝑗+𝑙+1𝑗+𝑙+1+(−1)^𝑚2^2𝑚=2_𝑚(2𝑚+1)^𝑚2𝑚_𝑙=02^𝑚−𝑙⎛⎜⎜⎝𝑚𝑙⎞⎟⎟⎠1𝑚+𝑙+1^𝑚+𝑙+1_𝑘=0⎛⎜⎜⎝𝑚+𝑘𝑙+1⎞⎟⎟⎠𝐸_𝑘+1𝑘+1𝐵^{𝑚+𝑙+1−𝑘}.(2.16)

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

Let 𝑝 be a fixed odd prime number. Throughout this section, ℤ_𝑝, ℚ_𝑝, and ℂ_𝑝 will denote the ring of 𝑝-adic integers, the field of 𝑝-adic rational numbers, and the completion of algebraic closure of ℚ_𝑝, respectively. Let 𝜈_𝑝 be the normalized exponential valuation of ℂ_𝑝 with |𝑝|_𝑝=𝑝^{−𝜈𝑝(𝑝)}=1/𝑝. Let 𝑈𝐷(ℤ_𝑝) be the space of uniformly differentiable functions on ℤ_𝑝. For 𝑓∈𝑈𝐷(ℤ_𝑝), the bosonic 𝑝-adic integral on ℤ_𝑝 is defined by 𝐼(𝑓)=_ℤ𝑝𝑓(𝑥)𝑑𝜇(𝑥)=lim_𝑛→∞1𝑝^{𝑛𝑝𝑛−1}_𝑥=0𝑓(𝑥),(3.1) (see [8]). Thus, by (3.1), we get _ℤ𝑝𝑓₁(𝑥)𝑑𝜇(𝑥)=_ℤ𝑝𝑓(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)=𝑒^{𝑡(𝑥+𝑦)}. Then we have _ℤ𝑝𝑒^{𝑡(𝑥+𝑦)}𝑑𝜇(𝑦)=𝑡𝑒^𝑡−1𝑒^𝑡𝑥=^∞_𝑛=0𝐵_𝑛(𝑥)𝑡^𝑛𝑛!.(3.3) From (3.3), we have _ℤ𝑝(𝑥+𝑦)^𝑛𝑑𝜇(𝑦)=𝐵_𝑛(𝑥),_ℤ𝑝𝑦^𝑛𝑑𝜇(𝑦)=𝐵_𝑛.(3.4) From (1.2), we can derive the following integral equation: 𝐼𝑓_𝑛=𝐼(𝑓)+^𝑛−1_𝑖=0𝑓^(𝑖)(𝑛∈ℕ).(3.5) Thus, from (3.4) and (3.5), we get _ℤ𝑝(𝑥+𝑛)^𝑚𝑑𝜇(𝑥)=_ℤ𝑝𝑥^𝑚𝑑𝜇(𝑥)+𝑚^𝑛−1_𝑖=0𝑖^𝑚−1.(3.6) From (3.6), we have 𝐵_𝑚(𝑛)−𝐵_𝑚=𝑚^𝑛−1_𝑖=0𝑖^𝑚−1𝑚∈ℤ₊.(3.7) The fermionic 𝑝-adic integral on ℤ_𝑝 is defined by Kim as follows [6, 7]: 𝐼₋₁(𝑓)=_ℤ𝑝𝑓(𝑥)𝑑𝜇₋₁(𝑥)=lim_{𝑝𝑛→∞}^𝑛−1_𝑥=0𝑓(𝑥)(−1)^𝑥.(3.8) Let 𝑓₁(𝑥)=𝑓(𝑥+1). Then we have 𝐼𝐼₋₁𝑓₁=−𝐼₋₁(𝑓)+2𝑓(0),₋₁𝑓₂==(−−1𝐼)₋₁₂𝑓₁+2𝑓₁(0)=−𝐼₋₁𝑓₁+2𝑓(1)𝐼₋₁(𝑓)+2(−1)²⁻¹𝑓(0)+2𝑓(1).(3.9) Continuing this process, we obtain the following equation: 𝐼₋₁𝑓_𝑛=(−1)^𝑛𝐼₋₁(𝑓)+2^𝑛−1_𝑙=0(−1)^{𝑛−𝑙−1}𝑓(𝑙),where𝑓_𝑛(𝑥)=𝑓(𝑥+𝑛).(3.10) Thus, by (3.10), we have _ℤ𝑝(𝑥+𝑛)^𝑚𝑑𝜇₋₁(𝑥)=(−1)^𝑛_ℤ𝑝𝑥^𝑚𝑑𝜇₋₁(𝑥)+2^𝑛−1_𝑙=0(−1)^{𝑛−𝑙−1}𝑙^𝑚.(3.11) Let us take 𝑓(𝑦)=𝑒^{𝑡(𝑥+𝑦)}. By (3.9), we get _ℤ𝑝𝑒^{𝑡(𝑥+𝑦)}𝑑𝜇₋₁(𝑦)=2𝑒^𝑥𝑡𝑒^𝑡+1=^∞_𝑛=0𝐸_𝑛(𝑥)𝑡^𝑛𝑛!.(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)=𝐸_𝑛(𝑥),_ℤ𝑝𝑦^𝑛𝑑𝜇₋₁(𝑦)=𝐸_𝑛,where𝑛∈ℤ₊.(3.13) By (3.11) and (3.13), we get 𝐸_𝑚(𝑛)=(−1)^𝑛𝐸^𝑚+2^𝑛−1_𝑙=0(−1)^𝑙−1𝑙^𝑚,𝑚∈ℤ⁺,𝑛∈ℕ.(3.14)

Let us consider the following 𝑝-adic integral on ℤ_𝑝: 𝐾₁=_ℤ𝑝𝐵_𝑛(𝑥)𝑑𝜇(𝑥)=^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐵_𝑛−𝑙_ℤ𝑝=𝑥^𝑙𝑑𝜇(𝑥)_𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐵_𝑛−𝑙𝐵_𝑙.(3.15)

From (1.4) and (3.15), we have 𝐾₁=^𝑛_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠𝐵_{𝑘𝑛−𝑘}_𝑙=0𝐸_{𝑛−𝑘−𝑙}⎛⎜⎜⎝𝑛−𝑙𝑘⎞⎟⎟⎠_ℤ𝑝=𝑥^𝑙𝑑𝜇(𝑥)_𝑛_{𝑘=0𝑘≠1}^𝑛−𝑘_𝑙=0⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠⎛⎜⎜⎝𝑛−𝑙𝑘⎞⎟⎟⎠𝐵_𝑘𝐵_𝑙𝐸_{𝑛−𝑘−𝑙}.(3.16) Therefore, by (3.15) and (3.16), we obtain the following theorem.

Theorem 3.1. For 𝑛∈ℤ₊, one has ^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐵_𝑛−𝑙𝐵_𝑙=^𝑛_{𝑘=0𝑘≠1}^𝑛−𝑘_𝑙=0⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠⎛⎜⎜⎝𝑛−𝑙𝑘⎞⎟⎟⎠𝐵_𝑘𝐵_𝑙𝐸_{𝑛−𝑘−𝑙}.(3.17)

Now, we set 𝐾₂=_ℤ𝑝𝐵_𝑛(𝑥)𝑑𝜇₋₁(𝑥)=^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐵_𝑛−𝑙𝐸_𝑙.(3.18)

By (1.4), we get 𝐾₂=^𝑛_{𝑘=0𝑘≠1}⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠𝐵_{𝑘𝑛−𝑘}_𝑙=0𝐸_{𝑛−𝑘−𝑙}⎛⎜⎜⎝𝑛−𝑙𝑘⎞⎟⎟⎠_ℤ𝑝=𝑥^𝑙𝑑𝜇₋₁(𝑥)_𝑛_{𝑘=0𝑘≠1}^𝑛−𝑘_𝑙=0⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠⎛⎜⎜⎝𝑛−𝑙𝑘⎞⎟⎟⎠𝐵_𝑘𝐸_{𝑛−𝑘−𝑙}𝐸_𝑙.(3.19) Therefore, by (3.18) and (3.19), we obtain the following theorem.

Theorem 3.2. For 𝑛∈ℤ₊, one has ^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐵_𝑛−𝑙𝐸_𝑙=^𝑛_{𝑘=0𝑘≠1}^𝑛−𝑘_𝑙=0⎛⎜⎜⎝𝑘𝑛⎞⎟⎟⎠⎛⎜⎜⎝𝑛−𝑙𝑘⎞⎟⎟⎠𝐵_𝑘𝐸_{𝑛−𝑘−𝑙}𝐸_𝑙.(3.20)

Let us consider the following integral on ℤ_𝑝: 𝐾₃=_ℤ𝑝𝐸_𝑛(𝑥)𝑑𝜇₋₁(𝑥)=^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐸_𝑛−𝑙_ℤ𝑝𝑥^𝑙𝑑𝜇₋₁(𝑥)=^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐸_𝑛−𝑙𝐸_𝑙.(3.21)

From (2.2), we have 𝐾₃=−2^𝑛_𝑙=0𝐸_𝑙+1𝑙+1⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠^𝑛−𝑙_𝑘=0⎛⎜⎜⎝𝑛𝑘−𝑙⎞⎟⎟⎠𝐵_{𝑛−𝑙−𝑘}_ℤ𝑝=−2_𝑛𝑥^𝑘𝑑𝜇₋₁(𝑥)_{𝑙=0𝑛−𝑙𝑘=0}⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠⎛⎜⎜⎝𝑛𝑘−𝑙⎞⎟⎟⎠𝐸_𝑙+1𝑙+1𝐵^{𝑛−𝑙−𝑘}𝐸_𝑘.(3.22) Therefore, by (3.21) and (3.22), we obtain the following theorem.

Theorem 3.3. For 𝑛∈ℤ₊, one has ^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐸_𝑛−𝑙𝐸_𝑙=−2^𝑛_{𝑙=0𝑛−𝑙𝑘=0}⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠⎛⎜⎜⎝𝑛𝑘−𝑙⎞⎟⎟⎠𝐸_𝑙+1𝑙+1𝐸^𝑘𝐵_{𝑛−𝑙−𝑘}.(3.23)

Now, we set 𝐾₄=_ℤ𝑝𝐸_𝑛(𝑥)𝑑𝜇(𝑥)=^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐸_𝑛−𝑙𝐵_𝑙.(3.24)

By (2.2), we get 𝐾₄=−2^𝑛_{𝑙=0𝑛−𝑙𝑘=0}⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠⎛⎜⎜⎝𝑛𝑘−𝑙⎞⎟⎟⎠𝐸_𝑙+1𝑙+1𝐵^{𝑛−𝑙−𝑘}𝐵_𝑘.(3.25) Therefore, by (3.24) and (3.25), we obtain the following corollary.

Corollary 3.4. For 𝑛∈ℤ₊, we have ^𝑛_𝑙=0⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠𝐸_𝑙𝐵_𝑙=−2^𝑛_{𝑙=0𝑛−𝑙𝑘=0}⎛⎜⎜⎝𝑛𝑙⎞⎟⎟⎠⎛⎜⎜⎝𝑛𝑘−𝑙⎞⎟⎟⎠𝐸_𝑙+1𝑙+1𝐵^{𝑛−𝑙−𝑘}𝐵_𝑘.(3.26)

Let us assume that 𝑎,𝑏,𝑐,𝑑∈ℤ. From Lemma 2.4 and (3.13), we note that _ℤ𝑝((𝑎++𝑏+1)+(𝑥+𝑦))^𝑚((𝑎+1)+(𝑥+𝑦))^𝑘𝑑𝜇₋₁(𝑦)_ℤ𝑝=2(𝑥+((𝑎𝑎++𝑏𝑏))_𝑚+(𝑥+𝑦))^𝑚((𝑎+(𝑥+𝑦))^𝑘𝑑𝜇₋₁(𝑦)(𝑥+𝑎)^𝑘.(3.27)

By (3.27), we get 2(𝑥+𝑎+𝑏)^𝑚(𝑥+𝑎)^𝑘=+_ℤ𝑝((𝑎+𝑏−𝑐+1)+(𝑥+𝑐+𝑦))^𝑚((𝑎−𝑐+1)+(𝑥+𝑐+𝑦))^𝑘𝑑𝜇₋₁(𝑦)_ℤ𝑝((𝑎+𝑏−𝑑)+(𝑥+𝑦+𝑑))^𝑚((𝑎−𝑑)+(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 𝑎,𝑏,𝑐,𝑑∈ℤ. For 𝑚,𝑘∈ℤ₊, one has ^𝑚_𝑘𝑗=0_𝑙=0⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠(𝑎+𝑏−𝑐+1)_𝑚−𝑗+(𝑎−𝑐+1)^𝑘−𝑙𝐸_𝑗+𝑙(𝑥+𝑐)_𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑑)^𝑚−𝑗(𝑎−𝑑)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐸_𝑗+𝑙(𝑥+𝑑)=2(𝑥+𝑎+𝑏)^𝑚(𝑥+𝑎)^𝑘.(3.29)

Let us consider the formula in Lemma 3.5 with 𝑑=𝑐−1. Then we have ^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐+1)^𝑚−𝑗(𝑎−𝑐+1)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐸_𝑗+𝑙=2(𝑥+𝑎+𝑏)_𝑚(𝑥+𝑐)+𝐸_𝑗+𝑙(𝑥+𝑐−1)(𝑥+𝑎)^𝑘.(3.30) Taking ∫_ℤ𝑝𝑑𝜇(𝑥) on both sides of (3.30), LHS=^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐+1)^𝑚−𝑗(𝑎−𝑐+1)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠^𝑗+𝑙_𝑠=0×𝐸_{𝑗+𝑙−𝑠}⎛⎜⎜⎝𝑗+𝑠𝑙⎞⎟⎟⎠_ℤ𝑝=2_𝑚((𝑥+𝑐)^𝑠+(𝑥+𝑐−1)^𝑠)𝑑𝜇(𝑥)_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐+1)^𝑚−𝑗(𝑎−𝑐+1)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠^𝑗+𝑙_𝑠=0×𝐸_{𝑗+𝑙−𝑠}⎛⎜⎜⎝𝑗+𝑠𝑙⎞⎟⎟⎠𝐵_𝑠(𝑐−1)+^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐+1)^𝑚−𝑗(𝑎−𝑐+1)^𝑘−𝑙×(𝑗+𝑙)𝐸_{𝑗+𝑙−1}⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠(𝑐−1).(3.31) By the same method, we get RHS=2^𝑚_𝑠=0⎛⎜⎜⎝𝑚𝑠⎞⎟⎟⎠𝑏_𝑚−𝑠_ℤ𝑝=2_𝑚(𝑥+𝑎)^𝑠+𝑘𝑑𝜇(𝑥)_𝑠=0⎛⎜⎜⎝𝑚𝑠⎞⎟⎟⎠𝑏_𝑚−𝑠𝐵_𝑠+𝑘(𝑎).(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 2^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐+1)^𝑚−𝑗(𝑎−𝑐+1)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠^𝑗+𝑙_𝑠=0⎛⎜⎜⎝𝑗+𝑠𝑙⎞⎟⎟⎠𝐸_{𝑗+𝑙−𝑠}×𝐵_𝑠(𝑐−1)+^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐+1)^𝑚−𝑗(𝑎−𝑐+1)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠(𝑗+𝑙)𝐸_{𝑗+𝑙−1}=2_𝑚(𝑐−1)_𝑠=0⎛⎜⎜⎝𝑚𝑠⎞⎟⎟⎠𝑏_𝑚−𝑠𝐵_𝑠+𝑘(𝑎).(3.33)

Replacing 𝑐 by 𝑐+1, we have 2^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠^𝑗+𝑙_𝑠=0⎛⎜⎜⎝𝑗+𝑠𝑙⎞⎟⎟⎠𝐸_{𝑗+𝑙−𝑠}+𝐵_𝑠(𝑐)_𝑚_𝑘𝑗=0_𝑙=0(𝑗+𝑙)(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐸_{𝑗+𝑙−𝑠}=2_𝑚(𝑐)_𝑠=0⎛⎜⎜⎝𝑚𝑠⎞⎟⎟⎠𝑏_𝑚−𝑠𝐵_𝑠+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)=2^𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠^𝑗+𝑙_𝑠=0⎛⎜⎜⎝𝑗+𝑠𝑙⎞⎟⎟⎠𝐸_{𝑗+𝑙−𝑠}×𝐵_𝑠+𝑠^𝑐−1_𝑖=0𝑖^𝑠−1=2_𝑚_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠^𝑗+𝑙_𝑠=0⎛⎜⎜⎝𝑗+𝑠𝑙⎞⎟⎟⎠𝐸_{𝑗+𝑙−𝑠}𝐵_𝑠+2^𝑐−1_𝑚𝑖=0_𝑘𝑗=0_𝑙=0(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠(𝑗+𝑙)(−1)₀×𝐸_{𝑗+𝑙−1}+2^𝑖−1_𝑒=0(−1)^𝑒−1𝑒^{𝑗+𝑙−1}=2_𝑚_𝑘𝑗=0_{𝑙=0𝑗+𝑙𝑠=0}⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠⎛⎜⎜⎝𝑗+𝑠𝑙⎞⎟⎟⎠(𝑎+𝑏−𝑐)_𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙×𝐸_{𝑗+𝑙−𝑠}𝐵_𝑠+2𝑚^𝑚−1_𝑘𝑗=0_𝑙=0⎛⎜⎜⎝𝑚−𝑗1⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠(𝑎+𝑏−𝑐)_{𝑚−1−𝑗}(𝑎−𝑐)^𝑘−𝑙𝐸_𝑗+𝑙𝛿_𝑐≡1+2𝑘^𝑚_{𝑗=0𝑘−1𝑙=0}⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘−𝑙1⎞⎟⎟⎠(𝑎+𝑏−𝑐)_𝑚−𝑗(𝑎−𝑐)^{𝑘−1−𝑙}𝐸_𝑗+𝑙𝛿_𝑐≡1+4𝑚^𝑐−2_𝑒=0(𝑎+𝑏−𝑐+𝑒)^𝑚−1(𝑎−𝑐+𝑒)^𝑘𝛿_𝑐≡𝑒+4𝑘^𝑐−2_𝑒=0(𝑎+𝑏−𝑐+𝑒)^𝑚(𝑎−𝑐+𝑒)^𝑘−1𝛿_𝑐≡𝑒,(3.35) where 𝛿_𝑐≡𝑘=10iiff𝑐𝑐≠≡𝑘𝑘((mmoodd22)).,(3.36) The second term of the LHS of (3.34) =^𝑚_𝑘𝑗=0_𝑙=0(𝑗+𝑙)(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠(−1)_𝑐𝐸^{𝑗+𝑙−1}+2^𝑐−1_𝑖=0(−1)^𝑖−1𝑖^{𝑗+𝑙−1}=(−1)_𝑐𝑚_𝑘𝑗=0_𝑙=0(𝑗+𝑙)(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐸_{𝑗+𝑙−1}+2(−1)^𝑐𝑐−1_𝑖=0(−1)^𝑖−1𝑚(𝑎+𝑏−𝑐+𝑖)^𝑚−1(𝑎−𝑐+𝑖)^𝑘+2(−1)^𝑐𝑐−1_𝑖=0(−1)^𝑖−1(𝑎+𝑏−𝑐+𝑖)^𝑚𝑘(𝑎−𝑐+𝑖)^𝑘−1=(−1)^𝑐𝑚_𝑘𝑗=0_𝑙=0(𝑗+𝑙)(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐸_{𝑗+𝑙−1}+2𝑚(−1)^𝑐𝑐−1_𝑖=0(−1)^𝑖−1(𝑎+𝑏−𝑐+𝑖)^𝑚−1(𝑎−𝑐+𝑖)^𝑘+2𝑘(−1)^𝑐𝑐−1_𝑖=0(−1)^𝑖−1(𝑎+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 𝑎,𝑏,𝑐∈ℤ with 𝑐≥1. Then one has 2^𝑚_𝑘𝑗=0_{𝑙=0𝑗+𝑙𝑠=0}⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠⎛⎜⎜⎝𝑗+𝑠𝑙⎞⎟⎟⎠(𝑎+𝑏−𝑐)_𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙𝐸_{𝑗+𝑙−𝑠}𝐵_𝑠+2𝑚𝛿_𝑚𝑐≡1_𝑘𝑗=0_𝑙=0⎛⎜⎜⎝𝑚−𝑗1⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠(𝑎+𝑏−𝑐)_{𝑚−1−𝑗}(𝑎−𝑐)^𝑘−𝑙𝐸_𝑗+𝑙+2𝑘𝛿_𝑚𝑐≡1_𝑘𝑗=0_𝑙=0⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘−𝑙1⎞⎟⎟⎠(𝑎+𝑏−𝑐)_𝑚−𝑗(𝑎−𝑐)^{𝑘−1−𝑙}𝐸_𝑗+𝑙+(−1)^𝑐𝑚_𝑘𝑗=0_𝑙=0(𝑗+𝑙)(𝑎+𝑏−𝑐)^𝑚−𝑗(𝑎−𝑐)^𝑘−𝑙⎛⎜⎜⎝𝑚𝑗⎞⎟⎟⎠⎛⎜⎜⎝𝑘𝑙⎞⎟⎟⎠𝐸_{𝑗+𝑙−1}+2𝑚^𝑐−1_𝑒=0(𝑎+𝑏−𝑐+𝑒)^𝑚−1(𝑎−𝑐+𝑒)^𝑘+2𝑘^𝑐−1_𝑒=0(𝑎+𝑏−𝑐+𝑒)^𝑚(𝑎−𝑐+𝑒)^𝑘−1=2^𝑚_𝑠=0⎛⎜⎜⎝𝑚𝑠⎞⎟⎟⎠𝑏_𝑚−𝑠𝐵_𝑠+𝑘(𝑎),(3.38) where 𝛿_𝑐≡𝑘=10iiff𝑐𝑐≡≢𝑘𝑘((mmoodd22)),.(3.39)

Remark 3.8. Here, we note that 4𝑚^𝑐−2_𝑒=0(𝑎+𝑏−𝑐+𝑒)^𝑚−1(𝑎−𝑐+𝑒)^𝑘𝛿_𝑐≡𝑒+2𝑚(−1)^𝑐𝑐−1_𝑖=0(−1)^𝑖−1(𝑎+𝑏−𝑐+𝑖)^𝑚−1(𝑎−𝑐+𝑖)^𝑘=2𝑚^𝑐−1_𝑒=0(𝑎+𝑏−𝑐+𝑒)^𝑚−1(𝑎−𝑐+𝑒)^𝑘.(3.40)

Acknowledgment

The first author was supported by National Research Foundation of Korea Grant funded by the Korean Government 2011-0002486.