# Integral Formulae of Bernoulli and Genocchi Polynomials

Authors: Seog-Hoon Rim; Joung-Hee Jin; Joohee Jeong
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

Recently, some interesting and new identities are introduced in the work of Kim et al. (2012). From these identities, we derive some new and interesting integral formulae for Bernoulli and Genocchi polynomials.


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## Body

## 1. Introduction

As it is well known, the Bernoulli polynomials are defined by generating functions as follows:(1.1)tet-1ext=eB(x)t=∑n=0∞Bn(x)tnn!
(see [1–5]) with the usual convention about replacing Bn(x) by Bn(x). In the special case, x=0, Bn(0)=Bn are called the nth Bernoulli numbers.The Genocchi polynomials are also defined by(1.2)2tet+1ext=eG(x)t=∑n=0∞Gn(x)tnn!
(see [1, 6–10]) with the usual convention about replacing Gn(x) by Gn(x). In the special case, x=0, Gn(0)=Gn are called the nth Genocchi numbers.From (1.1), we note that
(1.3)Bn(x)=∑l=0n‍(nl)Blxn-l
(see [1–5]). Thus, by (1.3), we get
(1.4)ddxBn(x)=n∑l=0n-1(n-1l)Blxn-1-l=nBn-1(x)
(see [2]). From (1.2), we note that
(1.5)Gn(x)=∑l=0n(nl)Glxn-l.
From (1.5), we can derive the following equation:
(1.6)ddxGn(x)=n∑l=0n-1(n-1l)Glxn-1-l=nGn-1(x).
By the definition of Bernoulli and Genocchi numbers, we get the following recurrence formulae:
(1.7)B0=1,Bn(1)-Bn=δ1,n,G0=0,Gn(1)+Gn=2δ1,n,
where δn,k is the Kronecker symbol (see [2]). From (1.4), (1.6), and (1.7), we note that
(1.8)∫01Bn(x)dx=δ0,nn+1(n≥0),∫01Gn(x)dx=-2Gn+1n+1(n≥1).
From the identities of Bernoulli and Genocchi polynomials, we derive some new and interesting integral formulae of an arithmetical nature on the Bernoulli and Genocchi polynomials.


## 2. Integral Formula of Bernoulli and Genocchi Polynomials

From (1.1) and (1.2), we note that
(2.1)tet-1ext=12(2textet+1)+1t(tet-1)(2textet+1)=12(∑n=0∞Gn(x)tnn!)+1t(∑l=0∞Bltll!)(∑m=0∞Gm(x)tmm!)=12∑n=0∞Gn(x)tnn!+1t∑n=0∞∑l=0n(nl)Gl(x)Bn-ltnn!=12∑n=0∞Gn(x)tnn!+∑n=0∞(-12Gn(x)+∑l=0l≠nn+1‍(n+1l)Gl(x)Bn+1-ln+1)tnn!=∑n=0∞(∑l=0l≠nn+1‍(n+1l)Gl(x)Bn+1-ln+1)tnn!.
By comparing the coefficients on the both sides of (2.1), we obtain the following theorem.Theorem 2.1.
Forn∈ℤ+, one has
(2.2)Bn(x)=∑l=0l≠nn+1(n+1l)Gl(x)Bn+1-ln+1.From (1.1) and (1.2), also notes that
(2.3)2tet+1ext=1t(2t(et-1)et+1)(textet-1)=1t(2t-22tet+1)(textet-1)=1t(2t-2∑l=0∞Gltll!)(∑m=0∞Bm(x)tmm!)=1t(-2∑l=1∞Gl+1l+1tl+1l!)(∑m=0∞Bm(x)tmm!)=∑n=1∞(-2∑l=1n(nl)Gl+1l+1Bn-l(x))tnn!.
By comparing the coefficients on the both sides of (2.3), we obtain the following theorem.Theorem 2.2.
Forn∈ℕ, one has
(2.4)Gn(x)=-2∑l=1n(nl)Gl+1l+1Bn-l(x).Let one take the definite integral from 0 to 1 on both sides of Theorem2.1. For n≥2,
(2.5)0=-2∑l=1l≠nn+1(n+1l)Gl+1l+1Bn+1-ln+1=-BnG2-2∑l=1l≠n-1n(nl)Bn-lGl+2(l+1)(l+2).
Therefore, by (2.3), we obtain the following theorem.Theorem 2.3.
Forn∈ℕ with n≥2, one has
(2.6)Bn=2∑l=1l≠n-1n(nl)Bn-lGl+2(l+1)(l+2).


## 3.p-Adic Integral on ℤp Associated with Bernoulli and Genocchi Numbers

Letp 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 vp be the normalized exponential valuation of ℂp with |p|p=p-vp(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
(3.1)I(f)=∫Zpf(x)dμ(x)=lim⁡N→∞⁡1pN∑x=0pN-1‍f(x)
(see [2, 5, 11]). From (3.1), we can derive the following integral equation:
(3.2)I(fn)=I(f)+∑i=0n-1f′(i)(n∈N),
where fn(x)=f(x+n) and f′(i)=((df(x))/dx)|x=i (see [2]). Let us take f(y)=et(x+y). Then we have
(3.3)∫Zpet(x+y)dμ(y)=tet-1ext=∑n=0∞Bn(x)tnn!
(see [2, 5]). From (3.3), we have
(3.4)∫Zp(x+n)ndμ(y)=Bn(x),∫Zpyndμ(y)=Bn
(see [2, 5]). Thus, by (3.2) and (3.4), we get
(3.5)∫Zp(x+n)mdμ(x)=∫Zpxmdμ(x)+m∑i=0n-1im-1,
(see [2]). From (3.5), we have
(3.6)Bm(n)-Bm=m∑i=0n-1im-1(n∈Z+)
(see [2]). The fermionic p-adic integral on ℤp is defined by Kim as follows [2, 8, 9]:
(3.7)I-1(f)=∫Zpf(x)dμ-1(x)=lim⁡N→∞⁡1pN∑x=0pN-1f(x)(-1)x.
From (3.7), we obtain the following integral equation:
(3.8)I-1(fn)=(-1)nI-1(f)+2∑l=0n-1(-1)n-l-1f(l)
(see [2]), where fn(x)=f(x+n). Thus, by (3.8), we have
(3.9)∫Zp(x+n)mdμ-1(x)=(-1)n∫Zpxmdμ-1(x)+2∑l=0n-1(-1)n-l-1lm
(see [2]). Let us take f(y)=et(x+y). Then we have
(3.10)t∫Zpet(x+y)dμ-1(y)=2textet+1=∑n=0∞Gn(x)tnn!.
From (3.10), we have
(3.11)∫Zp(x+y)ndμ-1(y)=Gn+1(x)n+1,∫Zpyndμ-1(y)=Gn+1n+1.
Thus, by (3.9) and (3.11), we get
(3.12)Gm+1(n)m+1=(-1)n(Gn+1n+1+2∑l=0n-1(-1)l-1lm).
Let us consider the following p-adic integral on ℤp:
(3.13)K1=∫ZpBn(x)dμ(x)=∑l=0n(nl)Bn-l∫Zp‍xldμ(x)=∑l=0n(nl)Bn-lBl.
From Theorem 2.1 and (3.13), one has
(3.14)K1=∑k=0k≠nn+1(n+1k)Bn+1-kn+1∑l=0k(kl)Gk-l∫Zpxldμ(x)=∑k=0k≠nn+1∑l=0k(n+1k)(kl)Bn+1-kBlGk-ln+1.
Therefore, by (3.13) and (3.14), we obtain the following theorem.Theorem 3.1.
Forn∈ℤ+, one has
(3.15)∑l=0n(nl)Bn-lBl=∑k=0k≠nn+1∑l=0k(n+1k)(kl)Bn+1-kBlGk-ln+1.Now, one sets(3.16)K2=∫ZpBn(x)dμ-1(x)=∑l=0n(nl)Bn-lGl+1l+1.
By Theorem 2.1, one gets
(3.17)K2=∑k=0k≠nn+1(n+1k)Bn+1-kn+1∑l=0k(kl)Gk-l∫Zpxldμ-1(x)=∑k=0k≠nn+1∑l=0k(n+1k)(kl)Bn+1-kGk-lGl+1(n+1)(l+1).
Therefore, by (3.16) and (3.17), we obtain the following theorem.Theorem 3.2.
Forn∈ℤ+, one has
(3.18)∑l=0n(nl)Bn-lGl+1l+1=∑k=0k≠nn+1∑l=0k(n+1k)(kl)Bn+1-kGk-lGl+1(n+1)(l+1).Let us consider the followingp-adic integral on ℤp:
(3.19)K3=∫ZpGn(x)dμ-1(x)=∑l=0n(nl)Gn-l∫Zpxldμ-1(x)=∑l=0n(nl)Gn-lGl+1l+1.
From Theorem 2.2, one has
(3.20)K3=-2∑l=1n(nl)Gl+1l+1∑k=0n-l(n-lk)Bn-l-k∫Zpxkdμ-1(x)=-2∑l=1n∑k=0n-l(nl)(n-lk)Bn-l-kGl+1Gk+1(l+1)(k+1).
Therefore, by (3.19) and (3.20), we obtain the following theorem.Theorem 3.3.
Forn∈ℤ+, one has
(3.21)∑l=0n(nl)Gn-lGl+1l+1=-2∑l=1n∑k=0n-l(nl)(n-lk)Bn-l-kGl+1Gk+1(l+1)(k+1).Now, one sets(3.22)K4=∫ZpGn(x)dμ(x)=∑l=0n(nl)Gn-lBl.
By Theorem 2.2, one gets
(3.23)K4=-2∑l=1n∑k=0n-l(nl)(n-lk)Gl+1l+1Bn-l-kBk.
Therefore, by (3.22) and (3.23), we obtain the following corollary.Corollary 3.4.
Forn∈ℤ+, one has
(3.24)∑l=0n(nl)Gn-lBl=-2∑l=1n∑k=0n-l(nl)(n-lk)Gl+1Bn-l-kBkl+1.



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