Some Properties and Identities of Bernoulli and Euler Polynomials Associated with p-adic Integral on ℤ_𝑝

Abstract

We investigate some properties and identities of Bernoulli and Euler polynomials. Further, we give some formulae on Bernoulli and Euler polynomials by using p-adic integral on ℤ_𝑝.

1. Introduction

Let 𝑝 be a fixed odd prime number. Throughout this paper, ℤ_𝑝, ℚ_𝑝, and ℂ_𝑝 will denote the ring of 𝑝-adic rational integers, the field of 𝑝-adic rational numbers, and the completion of the algebraic closure of ℚ_𝑝. Let 𝑣_𝑝 be the normalized exponential valuation of ℂ_𝑝 with |𝑝|_𝑝=𝑝^{−𝑣𝑝(𝑝)}=1/𝑝.

For 𝑓∈𝑈𝐷(ℤ_𝑝), the 𝑝-adic invariant integral on ℤ_𝑝 in the bosonic sense is defined by 𝐼𝑓=_ℤ𝑝𝑓(𝑥)𝑑𝜇(𝑥)=lim_𝑁→∞1𝑝^{𝑁𝑝𝑁−1}_𝑥=0𝑓(𝑥)𝑑𝜇(𝑥) (1.1) (see [1, 2]). The fermionic 𝑝-adic integral on ℤ_𝑝 is defined by Kim as follows: 𝐼₋₁𝑓=_ℤ𝑝𝑓(𝑥)𝑑𝜇₋₁(𝑥)=lim^𝑝_𝑁→∞^𝑁−1_𝑥=0𝑓(𝑥)(−1)^𝑥 (1.2) (see [3]). As is well known, Bernoulli polynomials are defined by 𝑡𝑒^𝑡𝑒−1^𝑥𝑡=𝑒^{𝐵(𝑥)𝑡}=^∞_𝑛=0𝐵_𝑛(𝑥)𝑡^𝑛,𝑛! (1.3) with the usual convention about replacing 𝐵^𝑛(𝑥) by 𝐵_𝑛(𝑥), symbolically (see [1–19]). In the special case 𝑥=0, 𝐵_𝑛(0)=𝐵_𝑛 is called the 𝑛th Bernoulli number.

The Euler polynomials are also defined by the generating function as follows: 2𝑒^𝑡𝑒+1^𝑥𝑡=𝑒^{𝐸(𝑥)𝑡}=^∞_𝑛=0𝐸_𝑛(𝑥)𝑡^𝑛,𝑛! (1.4) with the usual convention about replacing 𝐸^𝑛(𝑥) by 𝐸_𝑛(𝑥), symbolically (see [1–19]). In the special case 𝑥=0, 𝐸_𝑛(0)=𝐸_𝑛 is called the 𝑛-th Euler number.

By (1.3) and (1.4), we easily see that 𝐵_𝑛(𝑥)=(𝐵+𝑥)^𝑛=^𝑛_𝑙=0𝑛𝑙𝑥^𝑛−𝑙𝐵_𝑙,𝐸_𝑛(𝑥)=(𝐸+𝑥)^𝑛=^𝑛_𝑙=0𝑛𝑙𝑥^𝑛−𝑙𝐸_𝑙, (1.5) where (^𝑛_𝑙)=𝑛!/(𝑛−𝑙)!𝑙!=𝑛(𝑛−1)(𝑛−2)⋯(𝑛−𝑙+1)/𝑙! (see [14, 16, 19]).

The following properties of Bernoulli numbers and polynomials are well known (see [10, 11]).

For 𝑛∈ℤ₊=ℕ∪{0}, ^𝑛_𝑗=0𝑛𝑗𝑦^𝑛−𝑗𝐵_𝑗+1(𝑥)=𝐵𝑗+1^𝑛+1𝑥+𝑦−𝑦^𝑛+1,𝑛+1_{[𝑛/2]𝑗=0}𝑛𝑦2𝑗^𝑛−2𝑗𝐵_2𝑗+1(𝑥)=𝐵2𝑗+1^𝑛+1𝑥+𝑦+(−1)^𝑛𝐵_𝑛+1𝑥−𝑦,2𝑛+2_{[(𝑛+1)/2]𝑗=1}𝑛𝑦2𝑗−1^{𝑛+1−2𝑗}𝐵_2𝑗(𝑥)=𝐵2𝑗^𝑛+1𝑥+𝑦+(−1)^𝑛−1𝐵_𝑛+1𝑥−𝑦−2𝑦^𝑛+1,2𝑛+2 (1.6)(1.7)(1.8) where [⋅] is Gauss’ symbol.

First, we investigate some identities of Euler polynomials corresponding to (1.6), (1.7) and (1.8). From those identities, we derive some interesting identities and properties by using 𝑝-adic integral on ℤ_𝑝.

2. Some Identities of Bernoulli and Euler Polynomials

By (1.4), we get 𝐸_𝑘𝑥+𝑦=^𝑘_𝑗=0𝑘𝑗𝑦^𝑘−𝑗𝐸_𝑗(𝑥),for∈ℤ₊. (2.1) From (2.1), we note that 𝐸_𝑘𝑥+𝑦=^𝑘_𝑗=0𝑘𝑗𝑦^𝑘−𝑗𝐸_𝑗(𝑥)=𝑦^𝑘+^𝑘_𝑗=1𝑘𝑗𝑦𝑘−1𝑗−1^𝑘−𝑗𝐸_𝑗(𝑥). (2.2) Thus, we have ^{𝑘−1𝑗=0}𝑗𝑦𝑘−1_{𝑘−1−𝑗}𝐸_𝑗+1(𝑥)=𝐸𝑗+1^𝑘𝑥+𝑦−𝑦^𝑘𝑘. (2.3) Replacing 𝑘 by 𝑘+1 in (2.3), we obtain the following proposition.

Proposition 2.1. For 𝑘∈ℤ₊, one has ^𝑘_𝑗=0𝑘𝑗𝑦^𝑘−𝑗𝐸_𝑗+1(𝑥)=𝐸𝑗+1^𝑘+1𝑥+𝑦−𝑦^𝑘+1.𝑘+1 (2.4)

Let us replace 𝑦 by −𝑦 in Proposition 2.1. Then we have ^𝑘_𝑗=0𝑘𝑗(−1)^𝑘−𝑗𝑦^𝑘−𝑗𝐸_𝑗+1(𝑥)=𝐸𝑗+1^𝑘+1𝑥−𝑦−(−1)^𝑘+1𝑦^𝑘+1.𝑘+1 (2.5) Thus, we see that ^𝑘_𝑗=0𝑘𝑗(−1)^𝑗𝑦^𝑘−𝑗𝐸_𝑗+1(𝑥)=()𝑗+1−1^𝑘𝐸_𝑘+1𝑥−𝑦+𝑦^𝑘+1.𝑘+1 (2.6) Therefore, adding (2.4) and (2.6), we obtain the following proposition.

Proposition 2.2. For 𝑘∈ℤ₊, one has ^[𝑘/2]_𝑗=0𝑘𝑦2𝑗^𝑘−2𝑗𝐸_2𝑗+1(𝑥)=𝐸2𝑗+1^𝑘+1𝑥+𝑦+(−1)^𝑘𝐸_𝑘+1𝑥−𝑦.2𝑘+2 (2.7)

From (2.2), we note that ^𝑘_𝑗=11𝑗𝑦𝑘−1𝑗−1^𝑘−𝑗(−1)^𝑗𝐸_𝑗(𝑥)=(−1)^𝑘𝐸_𝑘𝑥−𝑦−𝑦^𝑘𝑘. (2.8) By (2.3) and (2.8), we get ^[𝑘/2]_𝑗=11𝑦2𝑗𝑘−12𝑗−1^𝑘−2𝑗𝐸_2𝑗(𝑥)=𝐸_𝑘𝑥+𝑦+(−1)^𝑘𝐸_𝑘𝑥−𝑦−2𝑦^𝑘.2𝑘 (2.9) Therefore, replacing 𝑘 by 𝑘+1, we obtain the following proposition.

Proposition 2.3. For 𝑘∈ℕ, one has ^[(𝑘+1)/2]_𝑗=1𝑘𝑦2𝑗−1^{𝑘+1−2𝑗}𝐸_2𝑗(𝑥)=𝐸2𝑗^𝑘+1𝑥+𝑦+(−1)^𝑘+1𝐸_𝑘+1𝑥−𝑦−2𝑦^𝑘+1.2𝑘+2 (2.10)

Letting 𝑦=1 in Proposition 2.1, we have ^𝑘_𝑗=0𝑘𝑗𝐸_𝑗+1(𝑥)=𝐸𝑗+1^𝑘+1(𝑥+1)−1,𝑘+1𝐸_𝑘+1(𝑥+1)=^𝑘+1_𝑙=0𝑙()𝑘+1𝐸+1^𝑙𝑥^{𝑘+1−𝑙}=2−𝐸₀𝑥^𝑘+1−^𝑘+1_𝑙=1𝑙𝐸𝑘+1_𝑙𝑥^{𝑘+1−𝑙}=2𝑥^𝑘+1−^𝑘+1_𝑙=0𝑙𝐸𝑘+1_𝑙𝑥^{𝑘+1−𝑙}=2𝑥^𝑘+1−𝐸_𝑘+1(𝑥). (2.11)(2.12) Therefore, by (2.11) and (2.12), we obtain the following corollary.

Corollary 2.4. For 𝑘∈ℤ₊, one has ^𝑘_𝑗=0𝑘𝑗𝐸_𝑗+1(𝑥)𝐸𝑗+1=−^𝑘+1(𝑥)+𝑘+12𝑥^𝑘+1−1.𝑘+1 (2.13)

Replacing 𝑦 by 1 and 𝑘 by 2𝑘 in Proposition 2.2, we have 2𝑘𝐸^𝑘_𝑗=02𝑗^2𝑗+1(𝑥)=𝐸2𝑗+1^2𝑘+1(𝑥+1)+𝐸_2𝑘+1(𝑥−1)=𝐸4𝑘+2_2𝑘+1(𝑥+1)+𝐸_2𝑘+1(𝑥)+𝐸_2𝑘+1(𝑥)+𝐸_2𝑘+1(𝑥−1)−4𝑘+22𝐸_2𝑘+1(𝑥)=4𝑘+22𝑥^2𝑘+1+2(𝑥−1)^2𝑘+1−𝐸4𝑘+2^2𝑘+1(𝑥).2𝑘+1 (2.14) Therefore, by (2.14), we obtain the following corollary.

Corollary 2.5. For 𝑘∈ℤ₊, one has 2𝑘𝐸^𝑘_𝑗=02𝑗^2𝑗+1(𝑥)𝐸2𝑗+1=−^2𝑘+1(𝑥)+𝑥2𝑘+1^2𝑘+1+(𝑥−1)^2𝑘+1.2𝑘+1 (2.15)

Replacing 𝑦 by 1 and 𝑘 by 2𝑘 in Proposition 2.3, we have 2𝑘𝐸^𝑘_𝑗=12𝑗−1^2𝑗(𝑥)=𝐸2𝑗^2𝑘+1(𝑥+1)−𝐸_2𝑘+1(𝑥−1)−2=𝐸4𝑘+2_2𝑘+1(𝑥+1)+𝐸_2𝑘+1(𝑥)−𝐸_2𝑘+1(𝑥)+𝐸_2𝑘+1(𝑥−1)−14𝑘+2=2𝑘+12𝑥^2𝑘+1−2(𝑥−1)^2𝑘+1−14𝑘+2=𝑥2𝑘+1_2𝑘+1−(𝑥−1)^2𝑘+1−12𝑘+1.2𝑘+1 (2.16) Therefore, by (2.16), we obtain the following corollary.

Corollary 2.6. For 𝑘∈ℕ, one has 2𝑘𝐸^𝑘_𝑗=12𝑗−1^2𝑗(𝑥)=𝑥2𝑗^2𝑘+1−(𝑥−1)^2𝑘+1−12𝑘+1.2𝑘+1 (2.17)

Replacing 𝑦 by 1/2 and 𝑘 by 2𝑘 in Proposition 2.3, we get 2𝑘1^𝑘_𝑗=12𝑗−12^{2𝑘+1−2𝑗}𝐸_2𝑗(𝑥)=𝐸2𝑗^2𝑘+1(𝑥+1/2)−𝐸_2𝑘+1(𝑥−1/2)−2^−2𝑘.4𝑘+2 (2.18) Thus, we have ^𝑘_𝑗=122𝑘2𝑗−1^2𝑗𝐸_2𝑗(𝑥)=22𝑗^2𝑘𝐸_2𝑘+1(𝑥+1/2)−𝐸_2𝑘+1(𝑥−1/2)−1,2𝑘+1 (2.19)𝐸_2𝑘+11𝑥+2=𝐸_2𝑘+11𝑥−2+1=^2𝑘+1_𝑙=0𝑙2𝑘+11𝑥−2^{2𝑘+1−𝑙}(𝐸+1)^𝑙1=2𝑥−2^2𝑘+1−^2𝑘+1_𝑙=0𝑙2𝑘+11𝑥−2^{2𝑘+1−𝑙}𝐸_𝑙1=2𝑥−2^2𝑘+1−𝐸_2𝑘+11𝑥−2. (2.20) Therefore, by (2.19) and (2.20), we obtain the following corollary.

Corollary 2.7. For 𝑘∈ℕ, we have ^𝑘_𝑗=122𝑘2𝑗−1^2𝑗𝐸_2𝑗(𝑥)22𝑗=−^2𝑘+1𝐸_2𝑘+1(𝑥−1/2)+22𝑘+1^2𝑘+1(𝑥−1/2)^2𝑘+1−12𝑘+1.2𝑘+1 (2.21)

Replacing 𝑦 by 1 and 𝑘 by 2𝑘+1 in Proposition 2.2, we get 2𝑘+1𝐸^𝑘_𝑗=02𝑗^2𝑗+1(𝑥)=𝐸2𝑗+1^2𝑘+2(𝑥+1)−𝐸_2𝑘+2(𝑥−1)=𝐸4𝑘+4_2𝑘+2(𝑥+1)+𝐸_2𝑘+2(𝑥)−𝐸_2𝑘+2(𝑥)+𝐸_2𝑘+2(𝑥−1)=4𝑘+42𝑥^2𝑘+2−2(𝑥−1)^2𝑘+2=𝑥4𝑘+4^2𝑘+2−(𝑥−1)^2𝑘+2.2𝑘+2 (2.22) Therefore, by (2.22), we obtain the following corollary.

Corollary 2.8. For 𝑘∈ℤ₊, one has 2𝑘+1𝐸^𝑘_𝑗=02𝑗^2𝑗+1(𝑥)=𝑥2𝑗+1^2𝑘+2−(𝑥−1)^2𝑘+2.2𝑘+2 (2.23)

Replacing 𝑘 by 2𝑘+1 and 𝑦 by 1 in Proposition 2.3, we get2𝑘+1𝐸^𝑘+1_𝑗=12𝑗−1^2𝑗(𝑥)=𝐸2𝑗^2𝑘+2(𝑥+1)+𝐸_2𝑘+2(𝑥−1)−2=𝐸4𝑘+4_2𝑘+2(𝑥+1)+𝐸_2𝑘+2(𝑥)+𝐸_2𝑘+2(𝑥)+𝐸_2𝑘+2(𝑥−1)−𝐸4𝑘+4^2𝑘+2(𝑥)+1=𝑥2𝑘+2_2𝑘+2+(𝑥−1)^2𝑘+2−𝐸2𝑘+2^2𝑘+2(𝑥)+1.2𝑘+2 (2.24) Therefore, by (2.24), we obtain the following corollary.

Corollary 2.9. For 𝑘∈ℤ₊, we have 2𝑘+1𝐸^𝑘+1_𝑗=12𝑗−1^2𝑗(𝑥)=𝑥2𝑗^2𝑘+2+(𝑥−1)^2𝑘+2−𝐸2𝑘+2^2𝑘+2(𝑥)+1.2𝑘+2 (2.25)

Replacing 𝑘 by 2𝑘+1 and 𝑦 by 1/2 in Proposition 2.2, we have 2𝑘+11^𝑘_𝑗=02𝑗2^{2𝑘+1−2𝑗}𝐸_2𝑗+1(𝑥)=𝐸2𝑗+1^2𝑘+2(𝑥+1/2)−𝐸_2𝑘+2(𝑥−1/2).4𝑘+4 (2.26) Thus, by multipling 2^2𝑘+1 on both sides, we get ^𝑘_𝑗=022𝑘+12𝑗^2𝑗𝐸_2𝑗+1(𝑥)=22𝑗+1^2𝑘+1𝐸_2𝑘+2(𝑥+1/2)−𝐸_2𝑘+2(𝑥−1/2).4𝑘+4 (2.27) By (2.20) and (2.27), we see that ^𝑘_𝑗=022𝑘+12𝑗^2𝑗𝐸_2𝑗+1(𝑥)=22𝑗+1^2𝑘2(𝑥−1/2)^2𝑘+2−2𝐸_2𝑘+2(𝑥−1/2)=22𝑘+2_2𝑘(𝑥−1/2)^2𝑘+2−2^2𝑘𝐸_2𝑘+2(𝑥−1/2).𝑘+1 (2.28) Therefore, by (2.28), we obtain the following corollary.

Corollary 2.10. For 𝑘∈ℤ₊, we have ^𝑘_𝑗=022𝑘+12𝑗^2𝑗𝐸_2𝑗+1(𝑥)22𝑗+1=−^2𝑘𝐸_2𝑘+2(𝑥−1/2)+2𝑘+1^2𝑘(𝑥−1/2)^2𝑘+2.𝑘+1 (2.29)

From (1.6), we can derive the following equation: ^𝑘−1_𝑗=0𝑘𝑗𝐵_𝑗+1(𝑥)𝑗+1=𝑥^𝑘−1,𝑘+1for𝑘∈ℕ. (2.30) Let us take the 𝑝-adic integral on both sides in (2.30) as follows: for 𝑘∈ℕ, 𝐼₁=^𝑘−1_𝑗=0𝑘𝑗^ℤ𝑝𝐵_𝑗+1(𝑥)(𝑥)=𝑗+1𝑑𝜇^𝑘−1_𝑗=0𝑘𝑗1𝑗+1^𝑗+1_𝑙=0𝑙𝐵𝑗+1_{𝑗+1−𝑙ℤ𝑝}𝑥^𝑙(𝑥)=𝑑𝜇_{𝑘−1𝑗=0𝑗+1𝑙=0}1𝑘𝑗𝑙𝐵𝑗+1𝑗+1_{𝑗+1−𝑙}𝐵_𝑙. (2.31) On the other hand, 𝐼₁=_ℤ𝑝𝑥^𝑘(𝑥)−1𝑑𝜇𝑘+1_ℤ𝑝𝑑𝜇(𝑥)=𝐵_𝑘−1.𝑘+1 (2.32) Therefore, by (2.31) and (2.32), we obtain the following theorem.

Theorem 2.11. For 𝑘∈ℕ, one has ^𝑘−1_𝑗=0^𝑗+1_𝑙=01𝑘𝑗𝑙𝐵𝑗+1𝑗+1_{𝑗+1−𝑙}𝐵_𝑙=𝐵_𝑘−1.𝑘+1 (2.33)

In (2.30), let us take the fermionic 𝑝-adic integral on both sides as follows: 𝐼₂=^𝑘−1_𝑗=0𝑘𝑗1𝑗+1_ℤ𝑝𝐵_𝑗+1(𝑥)𝑑𝜇₋₁(𝑥)=^𝑘−1_𝑗=0^𝑘𝑗𝑗+1^𝑗+1_𝑙=0𝑙𝐵𝑗+1_{𝑗+1−𝑙ℤ𝑝}𝑥^𝑙𝑑𝜇₋₁(𝑥)=^𝑘−1_𝑗=0^𝑗+1_𝑙=01𝑘𝑗𝑙𝐵𝑗+1𝑗+1_{𝑗+1−𝑙}𝐸_𝑙. (2.34) On the other hand 𝐼₂=_ℤ𝑝𝑥^𝑘𝑑𝜇₋₁(𝑥)−1𝑘+1_ℤ𝑝𝑑𝜇₋₁(𝑥)=𝐸_𝑘−1.𝑘+1 (2.35) Therefore, by (2.34) and (2.35), we obtain the following theorem.

Theorem 2.12. For 𝑘∈ℕ, one has ^𝑘−1_𝑗=0^𝑗+1_𝑙=01𝑘𝑗𝑙𝐵𝑗+1𝑗+1_{𝑗+1−𝑙}𝐸_𝑙=𝐸_𝑘−1.𝑘+1 (2.36)

From (1.7), we can easily derive the following equation: 2𝑘𝐵^𝑘−1_𝑗=02𝑗^2𝑗+1(𝑥)=𝑥2𝑗+1^2𝑘−(𝑥−1)^2𝑘2. (2.37) Let us take ∫_ℤ𝑝𝑑𝜇(𝑥) on both sides in (2.37). Then we have 𝐼₃=^𝑘−1_𝑗=02𝑘12𝑗2𝑗+1_ℤ𝑝𝐵_2𝑗+1(𝑥)(𝑥)=𝑑𝜇_{𝑘−1𝑗=0}2𝑘12𝑗2𝑗+1^2𝑗+1_𝑙=0𝑙𝐵2𝑗+1_{2𝑗+1−𝑙ℤ𝑝}𝑥^𝑙(𝑥)=𝑑𝜇_{𝑘−1𝑗=02𝑗+1𝑙=0}1𝑙𝐵2𝑗+12𝑘2𝑗2𝑗+1_{2𝑗+1−𝑙}𝐵_𝑙. (2.38) On the other hand, 𝐼₃=12_ℤ𝑝𝑥^2𝑘𝑑𝜇(𝑥)−_ℤ𝑝(𝑥−1)^2𝑘(𝑥)=1𝑑𝜇2𝐵_2𝑘−𝐵_2𝑘()=1−12𝐵_2𝑘−𝐵_2𝑘(2)=12𝐵_2𝑘−2𝑘+𝛿_1,2𝑘+𝐵_2𝑘, (2.39) where 𝛿_𝑛,𝑘 is a Kronecker symbol.

Therefore, by (2.38) and (2.39), we obtain the following theorem.

Theorem 2.13. For 𝑘∈ℕ, one has ^𝑘−1_𝑗=0^2𝑗+1_𝑙=01𝑙𝐵2𝑗+12𝑘2𝑗2𝑗+1_{2𝑗+1−𝑙}𝐵_𝑙=−𝑘. (2.40)

Taking ∫_ℤ𝑝𝑑𝜇₋₁(𝑥) on both sides in (2.37), we get 𝐼₄=^𝑘−1_𝑗=02𝑘12𝑗2𝑗+1^2𝑗+1_𝑙=0𝑙𝐵2𝑗+1_{2𝑗+1−𝑙ℤ𝑝}𝑥^𝑙𝑑𝜇₋₁(𝑥)=^𝑘−1_𝑗=0^2𝑗+1_𝑙=01𝑙𝐵2𝑗+12𝑘2𝑗2𝑗+1_{2𝑗+1−𝑙}𝐸_𝑙. (2.41) On the other hand𝐼₄=12_ℤ𝑝𝑥^𝑘𝑑𝜇₋₁(𝑥)−_ℤ𝑝(𝑥−1)^2𝑘𝑑𝜇₋₁(𝑥)=12_ℤ𝑝𝑥^2𝑘𝑑𝜇₋₁(𝑥)−_ℤ𝑝(𝑥+2)^2𝑘𝑑𝜇₋₁(𝑥)=12𝐸_2𝑘−𝐸_2𝑘(2)=12𝐸_2𝑘−2+𝐸_2𝑘−2𝛿_0,2𝑘=−1+𝛿_0,𝑘. (2.42) Therefore, by (2.41) and (2.42), we obtain the following theorem.

Theorem 2.14. For 𝑘∈ℕ, one has ^𝑘−1_𝑗=0^2𝑗+1_𝑙=01𝑙𝐵2𝑗+12𝑘2𝑗2𝑗+1_{2𝑗+1−𝑙}𝐸_𝑙=−1. (2.43)

From (1.8), we can also derive the following equation: 2𝑘𝐵^𝑘_𝑗=12𝑗−1^2𝑗(𝑥)=𝑥2𝑗^2𝑘+(𝑥−1)^2𝑘2−1.2𝑘+1 (2.44) Let us take the bosonic 𝑝-adic integral on both sides in (2.44). Then we get𝐼₅=^𝑘_𝑗=12𝑘12𝑗−12𝑗_ℤ𝑝𝐵_2𝑗(𝑥)(𝑥)=𝑑𝜇_𝑘𝑗=12𝑘12𝑗−12𝑗^2𝑗_𝑙=0𝑙𝐵2𝑗_{2𝑗−𝑙ℤ𝑝}𝑥^𝑙(𝑥)=𝑑𝜇_{𝑘𝑗=12𝑗𝑙=0}1𝑙𝐵2𝑗2𝑘2𝑗−12𝑗_2𝑗−𝑙𝐵_𝑙. (2.45) On the other hand, 𝐼₅=12_ℤ𝑝𝑥^2𝑘+(−1+𝑥)^2𝑘(𝑥)−1𝑑𝜇2𝑘+1_ℤ𝑝(𝑥)=1𝑑𝜇2_ℤ𝑝𝑥^2𝑘+(𝑥+2)^2𝑘(𝑥)−1𝑑𝜇=12𝑘+12𝐵_2𝑘+𝐵_2𝑘(2)−1=12𝑘+12𝐵_2𝑘+2𝑘+𝐵_2𝑘+𝛿_1,2𝑘−1.2𝑘+1 (2.46) Therefore, by (2.45) and (2.46), we obtain the following theorem.

Theorem 2.15. For 𝑘∈ℕ, one has ^𝑘_𝑗=1^2𝑗_𝑙=01𝑙𝐵2𝑗2𝑘2𝑗−12𝑗_2𝑗−𝑙𝐵_𝑙=𝐵_2𝑘1+𝑘−.2𝑘+1 (2.47)

Now, let us consider the fermionic 𝑝-adic integral on both sides in (2.44): 𝐼₆=^𝑘_𝑗=12𝑘12𝑗−12𝑗^2𝑗_𝑙=0𝑙𝐵2𝑗_{2𝑗−𝑙ℤ𝑝}𝑥^𝑙𝑑𝜇₋₁(𝑥)=^𝑘_𝑗=1^2𝑗_𝑙=01𝑙𝐵2𝑗2𝑘2𝑗−12𝑗_2𝑗−𝑙𝐸_𝑙. (2.48) On the other hand, 𝐼₆=12_ℤ𝑝𝑥^2𝑘+(𝑥−1)^2𝑘𝑑𝜇₋₁(𝑥)−12𝑘+1_ℤ𝑝𝑑𝜇₋₁(𝑥)=12_ℤ𝑝𝑥^2𝑘+(𝑥+2)^2𝑘𝑑𝜇₋₁(𝑥)−1=12𝑘+12𝐸_2𝑘+𝐸_2𝑘(2)−1=12𝑘+12𝐸_2𝑘+2+𝐸_2𝑘−2𝛿_0,2𝑘−12𝑘+1=𝐸_2𝑘+1−𝛿_0,2𝑘−1=2𝑘+12𝑘.2𝑘+1 (2.49) Therefore, by (2.48) and (2.49), we obtain the following theorem.

Theorem 2.16. For 𝑘∈ℕ, one has ^𝑘_𝑗=1^2𝑗_𝑙=01𝑙𝐵2𝑗2𝑘2𝑗−12𝑗_2𝑗−𝑙𝐸_𝑙=2𝑘.2𝑘+1 (2.50)

Acknowledgment

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