# Integral Formulae of Bernoulli Polynomials

Authors: Dae San Kim; Dmitry V. Dolgy; Hyun-Mee Kim; Sang-Hun Lee; Taekyun Kim
Journal: Discrete Dynamics in Nature and Society
 (2012)

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## 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.


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## Body

## 1. Introduction

As is well known, the Bernoulli polynomials are defined by generating functions as follows:(1.1)tet-1ext=∑n=0∞‍Bn(x)tnn!,
(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
(1.2)2et+1ext=eE(x)t=∑n=0∞‍En(x)tnn!
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:
(1.3)tet-1ext=t2(2extet+1)+(tet-1)(2extet+1)=∑n=0∞‍(∑k≠1k=0n(nk)BkEn-k(x))tnn!.
By (1.1) and (1.3), we get
(1.4)Bn(x)=∑k≠1k=0n(nk)BkEn-k(x),(n∈ℤ+=ℕ∪{0}).From (1.1), we have
(1.5)Bn(x)=∑l=0n‍(nl)Blxn-l.
Thus, by (1.5), we get
(1.6)ddxBn(x)=n∑l=0n-1‍(n-1l)Blxn-1-l=nBn-1(x).
It is known that En(0)=En are called the nth Euler numbers (see [7]). The Euler polynomials are also given by
(1.7)En(x)=(E+x)n=∑l=0n‍(nl)Elxn-l,
(see [6]).  From (1.7), we can derive the following equation:
(1.8)ddxEn(x)=n∑l=0n-1‍(nl)Elxn-1-l=nEn-1(x).
By the definition of Bernoulli and Euler numbers, we get the following recurrence formulae:
(1.9)E0=1,En(1)+En=2δ0,n,B0=1,Bn(1)-Bn=δ1,n,
where δn,k is the kronecker symbol (see [5]). From (1.6), (1.8), and (1.9), we note that
(1.10)∫01‍Bn(x)dx=δ0,nn+1,∫01‍En(x)dx=-2En+1n+1,
where n∈ℤ+. The following identity is known in [5]:
(1.11)∑j=0m‍∑l=0k‍(a+b+1)m-j(a+1)k-l(mj)(kl)(-1)j+lj+l+1+∑j=0m‍∑l=0k‍((a+b+1)m-j(a+1)k-l-(a+b)m-jak-l)(mj)(kl)Bj+l+1(x)j+l+1=(x+a)k(x+a+b)m,wherea,b∈ℤ.
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
(2.1)2et+1ext=1t(2(et-1)et+1)(textet-1)=1t(2-22et+1)(textet-1)=1t(-∑l=1∞‍2Ell!tl)(∑m=0∞‍Bm(x)tmm!)=-2(∑l=0∞‍El+1l+1tll!)(∑m=0∞‍Bm(x)tmm!)=-2∑n=0∞‍(∑l=0n‍El+1l+1(nl)Bn-l(x))tnn!.Therefore, by (1.2) and (2.1), we obtain the following theorem.Theorem 2.1.
Forn∈ℤ+, one has
(2.2)En(x)=-2∑l=0n‍(nl)El+1l+1Bn-l(x).Let us take the definite integral from0 to 1 on both sides of (1.4): for n≥2,
(2.3)0=-2∑k≠1k=0n(nk)BkEn-k+1n-k+1=-2BnE1-2∑k≠1k=0n-1(nk)BkEn-k+1n-k+1.By (2.3), we get
(2.4)Bn=2∑k≠1k=0n-1(nk)BkEn-k+1n-k+1.Therefore, by (2.4), we obtain the following theorem.Theorem 2.2.
Forn∈ℕ, with n≥2, one has
(2.5)Bn=2∑k≠1k=0n-1(nk)BkEn-k+1n-k+1.Let us takek=m, a=0, and b=-2 in (1.11). Then we have
(2.6)∑j=0m‍∑l=0m‍(-1)m-j(mj)(ml)(-1)j+lj+l+1+∑j=0m‍∑l=0m‍(-1)m-j(mj)(ml)Bj+l+1(x)j+l+1-∑j=0m‍(-2)m-j(mj)Bj+m+1(x)j+m+1=xm(x-2)m.
It is easy to show that
(2.7)∫01‍xm(x-2)mdx=2∫01/2‍(2t-2)m(2t)mdt=(-1)m22m(2∫01/2‍tm(1-t)mdt)=(-1)m22m∫01‍tm(1-t)mdt=(-1)m22mm!m!(2m+1)!=(-1)m22m2m+11(2mm).Let us consider the integral from0 to 1 in (2.6):
(2.8)∑j=0m‍∑l=0m‍(-1)m-l(mj)(ml)1j+l+1=(-1)m22m(2m+1)(2mm)(m∈ℕ).
By (2.6) and (2.8), we get
(2.9)∑j=0m‍∑l=0m‍(-1)m-j(mj)(ml)Bj+l+1j+l+1=2∑j=0m‍(-2)m-j(mj)1j+m+1∑k≠1k=0j+m(j+m+1k)BkEj+m+2-kj+m+2-k+(-1)m+122m2m+11(2mm),form∈ℕ.
Therefore, by (2.9), we obtain the following theorem.Theorem 2.3.
Form∈ℕ, one has
(2.10)∑j=0m‍∑l=0m‍(-1)m-j(mj)(ml)Bj+l+1j+l+1=2∑j=0m‍(-2)m-j(mj)1j+m+1∑k≠1k=0j+m(j+m+1k)BkEj+m+2-kj+m+2-k+(-1)m+122m2m+11(2mm).Lemma 2.4.
Leta,b∈ℤ. For m,k∈ℤ+, one has
(2.11)∑j=0m‍∑l=0k‍(a+b+1)m-j(a+1)k-l(mj)(kl)Ej+l(x)+∑j=0m‍∑l=0k‍(a+b)m-jak-l(mj)(kl)Ej+l(x)=2(x+a+b)m(x+a)k,
(see [5]).Let us takek=m, a=1, b=-2 in Lemma 2.4. Then we have
(2.12)∑l=0m‍2m-l(ml)Em+l(x)+∑j=0m‍∑l=0m‍(-1)m-j(mj)(ml)Ej+l(x)=2(x2-1)m.
Taking integral from 0 to 1 in (2.12), we get
(2.13)-2∑l=0m‍2m-l(ml)Em+l+1m+l+1-2∑j=0m‍∑l=0m‍(-1)m-l(mj)(ml)Ej+l+1j+l+1=2∫01‍(x2-1)mdx.It is easy to show that(2.14)∫01‍(x2-1)mdx=(-1)m∏k=1m‍(2k2k+1)=(-1)m22m(2m+1)(2mm).Thus, by (2.13) and (2.14), we get
(2.15)∑j=0m‍∑l=0m‍(-1)m-l(mj)(ml)Ej+l+1j+l+1=-∑l=0m‍2m-l(ml)Em+l+1m+l+1+(-1)m+122m(2m+1)(2mm).Therefore, by (2.2) and (2.15), we obtain the following theorem.Theorem 2.5.
Form∈ℤ+, one has
(2.16)∑j=0m‍∑l=0m‍(-1)m-l(mj)(ml)Ej+l+1j+l+1+(-1)m22m(2m+1)(2mm)=2∑l=0m‍2m-l(ml)1m+l+1∑k=0m+l+1‍(m+l+1k)Ek+1k+1Bm+l+1-k.


## 3.p-Adic Integral on ℤp Associated with Bernoulli and Euler 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 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
(3.1)I(f)=∫Zp‍f(x)dμ(x)=lim⁡n→∞1pn∑x=0pn-1‍f(x),
(see [8]).  Thus, by (3.1), we get
(3.2)∫Zp‍f1(x)dμ(x)=∫Zp‍f(x)dμ(x)+f′(0),
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
(3.3)∫Zp‍et(x+y)dμ(y)=tet-1etx=∑n=0∞‍Bn(x)tnn!.
From (3.3), we have
(3.4)∫Zp‍(x+y)ndμ(y)=Bn(x),∫Zp‍yndμ(y)=Bn.
From (1.2), we can derive the following integral equation:
(3.5)I(fn)=I(f)+∑i=0n-1‍f′(i)(n∈N).
Thus, from (3.4) and (3.5), we get
(3.6)∫Zp‍(x+n)mdμ(x)=∫Zp‍xmdμ(x)+m∑i=0n-1‍im-1.
From (3.6), we have
(3.7)Bm(n)-Bm=m∑i=0n-1‍im-1(m∈Z+).
The fermionic p-adic integral on ℤp is defined by Kim as follows [6, 7]:
(3.8)I-1(f)=∫Zp‍f(x)dμ-1(x)=lim⁡n→∞∑x=0pn-1‍f(x)(-1)x.
Let f1(x)=f(x+1). Then we have
(3.9)I-1(f1)=-I-1(f)+2f(0),I-1(f2)=-I-1(f1)+2f1(0)=-I-1(f1)+2f(1)=(-1)2I-1(f)+2(-1)2-1f(0)+2f(1).
Continuing this process, we obtain the following equation:
(3.10)I-1(fn)=(-1)nI-1(f)+2∑l=0n-1‍(-1)n-l-1f(l),wherefn(x)=f(x+n).
Thus, by (3.10), we have
(3.11)∫Zp‍(x+n)mdμ-1(x)=(-1)n∫Zp‍xmdμ-1(x)+2∑l=0n-1‍(-1)n-l-1lm.
Let us take f(y)=et(x+y). By (3.9), we get
(3.12)∫Zp‍et(x+y)dμ-1(y)=2extet+1=∑n=0∞‍En(x)tnn!.
From (3.2), we have the Witt's formula for the nth Euler polynomials and numbers as follows:
(3.13)∫Zp‍(x+y)ndμ-1(y)=En(x),∫Zp‍yndμ-1(y)=En,wheren∈Z+.
By (3.11) and (3.13), we get
(3.14)Em(n)=(-1)n(Em+2∑l=0n-1‍(-1)l-1lm),(m∈Z+,n∈N).Let us consider the followingp-adic integral on ℤp:
(3.15)K1=∫Zp‍Bn(x)dμ(x)=∑l=0n‍(nl)Bn-l∫ℤp‍xldμ(x)=∑l=0n‍(nl)Bn-lBl.From (1.4) and (3.15), we have
(3.16)K1=∑k≠1k=0n(nk)Bk∑l=0n-k‍En-k-l(n-kl)∫ℤp‍xldμ(x)=∑k≠1k=0n∑l=0n-k‍(nk)(n-kl)BkBlEn-k-l.
Therefore, by (3.15) and (3.16), we obtain the following theorem.Theorem 3.1.
Forn∈ℤ+, one has
(3.17)∑l=0n‍(nl)Bn-lBl=∑k≠1k=0n∑l=0n-k‍(nk)(n-kl)BkBlEn-k-l.Now, we set(3.18)K2=∫Zp‍Bn(x)dμ-1(x)=∑l=0n‍(nl)Bn-lEl.By (1.4), we get
(3.19)K2=∑k≠1k=0n(nk)Bk∑l=0n-k‍En-k-l(n-kl)∫ℤpxldμ-1(x)=∑k≠1k=0n∑l=0n-k‍(nk)(n-kl)BkEn-k-lEl.
Therefore, by (3.18) and (3.19), we obtain the following theorem.Theorem 3.2.
Forn∈ℤ+, one has
(3.20)∑l=0n‍(nl)Bn-lEl=∑k≠1k=0n∑l=0n-k‍(nk)(n-kl)BkEn-k-lEl.Let us consider the following integral onℤp:
(3.21)K3=∫Zp‍En(x)dμ-1(x)=∑l=0n‍(nl)En-l∫ℤp‍xldμ-1(x)=∑l=0n‍(nl)En-lEl.From (2.2), we have
(3.22)K3=-2∑l=0n‍El+1l+1(nl)∑k=0n-l‍(n-lk)Bn-l-k∫ℤp‍xkdμ-1(x)=-2∑l=0n‍∑k=0n-l‍(nl)(n-lk)El+1l+1Bn-l-kEk.
Therefore, by (3.21) and (3.22), we obtain the following theorem.Theorem 3.3.
Forn∈ℤ+, one has
(3.23)∑l=0n‍(nl)En-lEl=-2∑l=0n‍∑k=0n-l‍(nl)(n-lk)El+1l+1EkBn-l-k.Now, we set(3.24)K4=∫Zp‍En(x)dμ(x)=∑l=0n‍(nl)En-lBl.By (2.2), we get
(3.25)K4=-2∑l=0n‍∑k=0n-l‍(nl)(n-lk)El+1l+1Bn-l-kBk.
Therefore, by (3.24) and (3.25), we obtain the following corollary.Corollary 3.4.
Forn∈ℤ+, we have
(3.26)∑l=0n‍(nl)ElBl=-2∑l=0n‍∑k=0n-l‍(nl)(n-lk)El+1l+1Bn-l-kBk.Let us assume thata,b,c,d∈ℤ. From Lemma 2.4 and (3.13), we note that
(3.27)∫Zp‍((a+b+1)+(x+y))m((a+1)+(x+y))kdμ-1(y)+∫Zp‍((a+b)+(x+y))m((a+(x+y))kdμ-1(y)=2(x+a+b)m(x+a)k.By (3.27), we get
(3.28)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).Thus, by (3.28) and (3.13), we obtain the following lemma (see [5]).Lemma 3.5.
Leta,b,c,d∈ℤ. For m,k∈ℤ+, one has
(3.29)∑j=0m‍∑l=0k‍(mj)(kl)(a+b-c+1)m-j(a-c+1)k-lEj+l(x+c)+∑j=0m‍∑l=0k‍(a+b-d)m-j(a-d)k-l(mj)(kl)Ej+l(x+d)=2(x+a+b)m(x+a)k.Let us consider the formula in Lemma3.5 with d=c-1. Then we have
(3.30)∑j=0m‍∑l=0k‍(a+b-c+1)m-j(a-c+1)k-l(mj)(kl)(Ej+l(x+c)+Ej+l(x+c-1))=2(x+a+b)m(x+a)k.
Taking ∫ℤp‍dμ(x) on both sides of (3.30),
(3.31)LHS=∑j=0m‍∑l=0k‍(a+b-c+1)m-j(a-c+1)k-l(mj)(kl)∑s=0j+l‍(j+ls)×Ej+l-s∫ℤp‍((x+c)s+(x+c-1)s)dμ(x)=2∑j=0m‍∑l=0k‍(a+b-c+1)m-j(a-c+1)k-l(mj)(kl)∑s=0j+l‍(j+ls)×Ej+l-sBs(c-1)+∑j=0m‍∑l=0k‍(a+b-c+1)m-j(a-c+1)k-l(mj)(kl)×(j+l)Ej+l-1(c-1).
By the same method, we get
(3.32)RHS=2∑s=0m‍(ms)bm-s∫ℤp‍(x+a)s+kdμ(x)=2∑s=0m‍(ms)bm-sBs+k(a).
Therefore, by (3.31) and (3.32), we obtain the following proposition.Proposition 3.6.
Leta,b,c∈ℤ. Then one has
(3.33)2∑j=0m‍∑l=0k‍(a+b-c+1)m-j(a-c+1)k-l(mj)(kl)∑s=0j+l‍(j+ls)Ej+l-s×Bs(c-1)+∑j=0m‍∑l=0k‍(a+b-c+1)m-j(a-c+1)k-l(mj)(kl)(j+l)Ej+l-1(c-1)=2∑s=0m‍(ms)bm-sBs+k(a).Replacingc by c+1, we have
(3.34)2∑j=0m‍∑l=0k‍(a+b-c)m-j(a-c)k-l(mj)(kl)∑s=0j+l‍(j+ls)Ej+l-sBs(c)+∑j=0m‍∑l=0k‍(j+l)(a+b-c)m-j(a-c)k-l(mj)(kl)Ej+l-s(c)=2∑s=0m‍(ms)bm-sBs+k(a).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)(3.35)=2∑j=0m‍∑l=0k‍(a+b-c)m-j(a-c)k-l(mj)(kl)∑s=0j+l‍(j+ls)Ej+l-s×(Bs+s∑i=0c-1‍is-1)=2∑j=0m‍∑l=0k‍(a+b-c)m-j(a-c)k-l(mj)(kl)∑s=0j+l‍(j+ls)Ej+l-sBs+2∑i=0c-1‍∑j=0m‍∑l=0k‍(a+b-c)m-j(a-c)k-l(mj)(kl)(j+l)(-1)0×(Ej+l-1+2∑e=0i-1‍(-1)e-1ej+l-1)=2∑j=0m‍∑l=0k‍∑s=0j+l‍(mj)(kl)(j+ls)(a+b-c)m-j(a-c)k-l×Ej+l-sBs+2m∑j=0m-1‍∑l=0k‍(m-1j)(kl)(a+b-c)m-1-j(a-c)k-lEj+lδc≡1+2k∑j=0m‍∑l=0k-1‍(mj)(k-1l)(a+b-c)m-j(a-c)k-1-lEj+lδc≡1+4m∑e=0c-2‍(a+b-c+e)m-1(a-c+e)kδc≡e+4k∑e=0c-2‍(a+b-c+e)m(a-c+e)k-1δc≡e,
where
(3.36)δc≡k={1ifc≡k(mod2),0ifc≠k(mod2).
The second term of the LHS of  (3.34)
(3.37)=∑j=0m‍∑l=0k‍(j+l)(a+b-c)m-j(a-c)k-l(mj)(kl)(-1)c(Ej+l-1+2∑i=0c-1‍(-1)i-1ij+l-1)=(-1)c∑j=0m‍∑l=0k‍(j+l)(a+b-c)m-j(a-c)k-l(mj)(kl)Ej+l-1+2(-1)c∑i=0c-1‍(-1)i-1m(a+b-c+i)m-1(a-c+i)k+2(-1)c∑i=0c-1‍(-1)i-1(a+b-c+i)mk(a-c+i)k-1=(-1)c∑j=0m‍∑l=0k‍(j+l)(a+b-c)m-j(a-c)k-l(mj)(kl)Ej+l-1+2m(-1)c∑i=0c-1‍(-1)i-1(a+b-c+i)m-1(a-c+i)k+2k(-1)c∑i=0c-1‍(-1)i-1(a+b-c+i)m(a-c+i)k-1.Therefore, by (3.34), (3.35), and (3.37), we obtain the following theorem.Theorem 3.7.
Leta,b,c∈ℤ with c≥1. Then one has
(3.38)2∑j=0m‍∑l=0k‍∑s=0j+l‍(mj)(kl)(j+ls)(a+b-c)m-j(a-c)k-lEj+l-sBs+2mδc≡1∑j=0m‍∑l=0k‍(m-1j)(kl)(a+b-c)m-1-j(a-c)k-lEj+l+2kδc≡1∑j=0m‍∑l=0k‍(mj)(k-1l)(a+b-c)m-j(a-c)k-1-lEj+l+(-1)c∑j=0m‍∑l=0k‍(j+l)(a+b-c)m-j(a-c)k-l(mj)(kl)Ej+l-1+2m∑e=0c-1‍(a+b-c+e)m-1(a-c+e)k+2k∑e=0c-1‍(a+b-c+e)m(a-c+e)k-1=2∑s=0m‍(ms)bm-sBs+k(a),
where
(3.39)δc≡k={1ifc≡k(mod2),0ifc≢k(mod2).Remark 3.8.
Here, we note that(3.40)4m∑e=0c-2‍(a+b-c+e)m-1(a-c+e)kδc≡e+2m(-1)c∑i=0c-1‍(-1)i-1(a+b-c+i)m-1(a-c+i)k=2m∑e=0c-1‍(a+b-c+e)m-1(a-c+e)k.



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