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Bernoulli polynomials

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Inmathematics, theBernoulli polynomials, named afterJacob Bernoulli, combine theBernoulli numbers andbinomial coefficients. They are used forseries expansion offunctions, and with theEuler–MacLaurin formula.

Bernoulli polynomials

Thesepolynomials occur in the study of manyspecial functions and, in particular, theRiemann zeta function and theHurwitz zeta function. They are anAppell sequence (i.e. aSheffer sequence for the ordinaryderivative operator). For the Bernoulli polynomials, the number of crossings of thex-axis in theunit interval does not go up with thedegree. In the limit of large degree, they approach, when appropriately scaled, thesine and cosine functions.

A similar set of polynomials, based on a generating function, is the family ofEuler polynomials.

Representations

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The Bernoulli polynomialsBn can be defined by agenerating function. They also admit a variety of derived representations.

Generating functions

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The generating function for the Bernoulli polynomials istextet1=n=0Bn(x)tnn!.{\displaystyle {\frac {te^{xt}}{e^{t}-1}}=\sum _{n=0}^{\infty }B_{n}(x){\frac {t^{n}}{n!}}.} The generating function for the Euler polynomials is2extet+1=n=0En(x)tnn!.{\displaystyle {\frac {2e^{xt}}{e^{t}+1}}=\sum _{n=0}^{\infty }E_{n}(x){\frac {t^{n}}{n!}}.} 

Explicit formula

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Bn(x)=k=0n(nk)Bnkxk,{\displaystyle B_{n}(x)=\sum _{k=0}^{n}{n \choose k}B_{n-k}x^{k},} Em(x)=k=0m(mk)Ek2k(x12)mk.{\displaystyle E_{m}(x)=\sum _{k=0}^{m}{m \choose k}{\frac {E_{k}}{2^{k}}}\left(x-{\tfrac {1}{2}}\right)^{m-k}.} forn0{\displaystyle n\geq 0} , whereBk{\displaystyle B_{k}}  are theBernoulli numbers, andEk{\displaystyle E_{k}}  are theEuler numbers. It follows thatBn(0)=Bn{\displaystyle B_{n}(0)=B_{n}}  andEm(12)=12mEm{\displaystyle E_{m}{\big (}{\tfrac {1}{2}}{\big )}={\tfrac {1}{2^{m}}}E_{m}} .

Representation by a differential operator

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The Bernoulli polynomials are also given by Bn(x)=D eD1  xn {\displaystyle \ B_{n}(x)={\frac {D}{\ e^{D}-1\ }}\ x^{n}\ } where Dd dx  {\displaystyle \ D\equiv {\frac {\mathrm {d} }{\ \mathrm {d} x\ }}\ }  is differentiation with respect tox and the fraction is expanded as aformal power series. It follows that ax Bn(u) d u= Bn+1(x)Bn+1(a) n+1 .{\displaystyle \ \int _{a}^{x}\ B_{n}(u)\ \mathrm {d} \ u={\frac {\ B_{n+1}(x)-B_{n+1}(a)\ }{n+1}}~.} cf.§ Integrals below. By the same token, the Euler polynomials are given by En(x)=2 eD+1  xn .{\displaystyle \ E_{n}(x)={\frac {2}{\ e^{D}+1\ }}\ x^{n}~.} 

Representation by an integral operator

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The Bernoulli polynomials are also the unique polynomials determined byxx+1Bn(u)du=xn.{\displaystyle \int _{x}^{x+1}B_{n}(u)\,du=x^{n}.} 

Theintegral transform(Tf)(x)=xx+1f(u)du{\displaystyle (Tf)(x)=\int _{x}^{x+1}f(u)\,du} on polynomialsf, simply amounts to(Tf)(x)=eD1Df(x)=n=0Dn(n+1)!f(x)=f(x)+f(x)2+f(x)6+f(x)24+.{\displaystyle {\begin{aligned}(Tf)(x)={e^{D}-1 \over D}f(x)&{}=\sum _{n=0}^{\infty }{D^{n} \over (n+1)!}f(x)\\&{}=f(x)+{f'(x) \over 2}+{f''(x) \over 6}+{f'''(x) \over 24}+\cdots .\end{aligned}}} This can be used to produce theinversion formulae below.

Integral Recurrence

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In,[1][2] it is deduced and proved that the Bernoulli polynomials can be obtained by the following integral recurrenceBm(x)=m0xBm1(t)dtm010tBm1(s)dsdt.{\displaystyle B_{m}(x)=m\int _{0}^{x}B_{m-1}(t)\,dt-m\int _{0}^{1}\int _{0}^{t}B_{m-1}(s)\,dsdt.} 

Another explicit formula

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An explicit formula for the Bernoulli polynomials is given byBn(x)=k=0n[1k+1=0k(1)(k)(x+)n].{\displaystyle B_{n}(x)=\sum _{k=0}^{n}{\biggl [}{\frac {1}{k+1}}\sum _{\ell =0}^{k}(-1)^{\ell }{k \choose \ell }(x+\ell )^{n}{\biggr ]}.} 

That is similar to the series expression for theHurwitz zeta function in the complex plane. Indeed, there is the relationshipBn(x)=nζ(1n,x){\displaystyle B_{n}(x)=-n\zeta (1-n,\,x)} whereζ(s,q){\displaystyle \zeta (s,\,q)}  is theHurwitz zeta function. The latter generalizes the Bernoulli polynomials, allowing for non-integer valuesofn.

The inner sum may be understood to be thenthforward difference ofxm,{\displaystyle x^{m},}  that is,Δnxm=k=0n(1)nk(nk)(x+k)m{\displaystyle \Delta ^{n}x^{m}=\sum _{k=0}^{n}(-1)^{n-k}{n \choose k}(x+k)^{m}} whereΔ{\displaystyle \Delta }  is theforward difference operator. Thus, one may writeBn(x)=k=0n(1)kk+1Δkxn.{\displaystyle B_{n}(x)=\sum _{k=0}^{n}{\frac {(-1)^{k}}{k+1}}\Delta ^{k}x^{n}.} 

This formula may be derived from an identity appearing above as follows. Since the forward difference operatorΔ equalsΔ=eD1{\displaystyle \Delta =e^{D}-1} whereD is differentiation with respect tox, we have, from theMercator series,DeD1=log(Δ+1)Δ=n=0(Δ)nn+1.{\displaystyle {\frac {D}{e^{D}-1}}={\frac {\log(\Delta +1)}{\Delta }}=\sum _{n=0}^{\infty }{\frac {(-\Delta )^{n}}{n+1}}.} 

As long as this operates on anmth-degree polynomial such asxm,{\displaystyle x^{m},}  one may letn go from0 only uptom.

An integral representation for the Bernoulli polynomials is given by theNörlund–Rice integral, which follows from the expression as a finite difference.

An explicit formula for the Euler polynomials is given byEn(x)=k=0n[12k=0n(1)(k)(x+)n].{\displaystyle E_{n}(x)=\sum _{k=0}^{n}\left[{\frac {1}{2^{k}}}\sum _{\ell =0}^{n}(-1)^{\ell }{k \choose \ell }(x+\ell )^{n}\right].} 

The above follows analogously, using the fact that2eD+1=11+12Δ=n=0(12Δ)n.{\displaystyle {\frac {2}{e^{D}+1}}={\frac {1}{1+{\tfrac {1}{2}}\Delta }}=\sum _{n=0}^{\infty }{\bigl (}{-{\tfrac {1}{2}}}\Delta {\bigr )}^{n}.} 

Sums ofpth powers

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Main article:Faulhaber's formula

Using either the aboveintegral representation ofxn{\displaystyle x^{n}}  or theidentityBn(x+1)Bn(x)=nxn1{\displaystyle B_{n}(x+1)-B_{n}(x)=nx^{n-1}} , we havek=0xkp=0x+1Bp(t)dt=Bp+1(x+1)Bp+1p+1{\displaystyle \sum _{k=0}^{x}k^{p}=\int _{0}^{x+1}B_{p}(t)\,dt={\frac {B_{p+1}(x+1)-B_{p+1}}{p+1}}} (assuming 00 = 1).

Explicit expressions for low degrees

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The first few Bernoulli polynomials are:B0(x)=1,B4(x)=x42x3+x2130,B1(x)=x12,B5(x)=x552x4+53x316x,B2(x)=x2x+16,B6(x)=x63x5+52x412x2+142,B3(x)=x332x2+12x|, {\displaystyle {\begin{aligned}B_{0}(x)&=1,&B_{4}(x)&=x^{4}-2x^{3}+x^{2}-{\tfrac {1}{30}},\\[4mu]B_{1}(x)&=x-{\tfrac {1}{2}},&B_{5}(x)&=x^{5}-{\tfrac {5}{2}}x^{4}+{\tfrac {5}{3}}x^{3}-{\tfrac {1}{6}}x,\\[4mu]B_{2}(x)&=x^{2}-x+{\tfrac {1}{6}},&B_{6}(x)&=x^{6}-3x^{5}+{\tfrac {5}{2}}x^{4}-{\tfrac {1}{2}}x^{2}+{\tfrac {1}{42}},\\[-2mu]B_{3}(x)&=x^{3}-{\tfrac {3}{2}}x^{2}+{\tfrac {1}{2}}x{\vphantom {\Big |}},\qquad &&\ \,\,\vdots \end{aligned}}} 

The first few Euler polynomials are:E0(x)=1,E4(x)=x42x3+x,E1(x)=x12,E5(x)=x552x4+52x212,E2(x)=x2x,E6(x)=x63x5+5x33x,E3(x)=x332x2+14,   {\displaystyle {\begin{aligned}E_{0}(x)&=1,&E_{4}(x)&=x^{4}-2x^{3}+x,\\[4mu]E_{1}(x)&=x-{\tfrac {1}{2}},&E_{5}(x)&=x^{5}-{\tfrac {5}{2}}x^{4}+{\tfrac {5}{2}}x^{2}-{\tfrac {1}{2}},\\[4mu]E_{2}(x)&=x^{2}-x,&E_{6}(x)&=x^{6}-3x^{5}+5x^{3}-3x,\\[-1mu]E_{3}(x)&=x^{3}-{\tfrac {3}{2}}x^{2}+{\tfrac {1}{4}},\qquad \ \ &&\ \,\,\vdots \end{aligned}}} 

Maximum and minimum

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At highern the amount of variation inBn(x){\displaystyle B_{n}(x)}  betweenx=0{\displaystyle x=0}  andx=1{\displaystyle x=1}  gets large. For instance,B16(0)=B16(1)={\displaystyle B_{16}(0)=B_{16}(1)={}} 36175107.09,{\displaystyle -{\tfrac {3617}{510}}\approx -7.09,}  butB16(12)={\displaystyle B_{16}{\bigl (}{\tfrac {1}{2}}{\bigr )}={}} 11851823933423367.09.{\displaystyle {\tfrac {118518239}{3342336}}\approx 7.09.} Lehmer (1940)[3] showed that the maximum value (Mn) ofBn(x){\displaystyle B_{n}(x)}  between0 and1 obeysMn<2n!(2π)n{\displaystyle M_{n}<{\frac {2n!}{(2\pi )^{n}}}} unlessn is2 modulo 4, in which caseMn=2ζ(n)n!(2π)n{\displaystyle M_{n}={\frac {2\zeta (n)\,n!}{(2\pi )^{n}}}} (whereζ(x){\displaystyle \zeta (x)}  is theRiemann zeta function), while the minimum (mn) obeysmn>2n!(2π)n{\displaystyle m_{n}>{\frac {-2n!}{(2\pi )^{n}}}} unlessn = 0 modulo 4 , in which casemn=2ζ(n)n!(2π)n.{\displaystyle m_{n}={\frac {-2\zeta (n)\,n!}{(2\pi )^{n}}}.} 

These limits are quite close to the actual maximum and minimum, and Lehmer gives more accurate limits as well.

Differences and derivatives

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The Bernoulli and Euler polynomials obey many relations fromumbral calculus:ΔBn(x)=Bn(x+1)Bn(x)=nxn1,ΔEn(x)=En(x+1)En(x)=2(xnEn(x)).{\displaystyle {\begin{aligned}\Delta B_{n}(x)&=B_{n}(x+1)-B_{n}(x)=nx^{n-1},\\[3mu]\Delta E_{n}(x)&=E_{n}(x+1)-E_{n}(x)=2(x^{n}-E_{n}(x)).\end{aligned}}} (Δ is theforward difference operator). Also,En(x+1)+En(x)=2xn.{\displaystyle E_{n}(x+1)+E_{n}(x)=2x^{n}.} Thesepolynomial sequences areAppell sequences:Bn(x)=nBn1(x),En(x)=nEn1(x).{\displaystyle {\begin{aligned}B_{n}'(x)&=nB_{n-1}(x),\\[3mu]E_{n}'(x)&=nE_{n-1}(x).\end{aligned}}} 

Translations

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Bn(x+y)=k=0n(nk)Bk(x)ynkEn(x+y)=k=0n(nk)Ek(x)ynk{\displaystyle {\begin{aligned}B_{n}(x+y)&=\sum _{k=0}^{n}{n \choose k}B_{k}(x)y^{n-k}\\[3mu]E_{n}(x+y)&=\sum _{k=0}^{n}{n \choose k}E_{k}(x)y^{n-k}\end{aligned}}} These identities are also equivalent to saying that these polynomial sequences areAppell sequences. (Hermite polynomials are another example.)

Symmetries

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Bn(1x)=(1)nBn(x),n0, and in particular for n1, Bn(0)=Bn(1)En(1x)=(1)nEn(x)(1)nBn(x)=Bn(x)+nxn1(1)nEn(x)=En(x)+2xnBn(12)=(12n11)Bn,n0 from the multiplication theorems below.{\displaystyle {\begin{aligned}B_{n}(1-x)&=\left(-1\right)^{n}B_{n}(x),&&n\geq 0,{\text{ and in particular for }}n\neq 1,~B_{n}(0)=B_{n}(1)\\[3mu]E_{n}(1-x)&=\left(-1\right)^{n}E_{n}(x)\\[1ex]\left(-1\right)^{n}B_{n}(-x)&=B_{n}(x)+nx^{n-1}\\[3mu]\left(-1\right)^{n}E_{n}(-x)&=-E_{n}(x)+2x^{n}\\[1ex]B_{n}{\bigl (}{\tfrac {1}{2}}{\bigr )}&=\left({\frac {1}{2^{n-1}}}-1\right)B_{n},&&n\geq 0{\text{ from the multiplication theorems below.}}\end{aligned}}} Zhi-Wei Sun and Hao Pan[4] established the following surprising symmetry relation: Ifr +s +t =n andx +y +z = 1, thenr[s,t;x,y]n+s[t,r;y,z]n+t[r,s;z,x]n=0,{\displaystyle r[s,t;x,y]_{n}+s[t,r;y,z]_{n}+t[r,s;z,x]_{n}=0,} where[s,t;x,y]n=k=0n(1)k(sk)(tnk)Bnk(x)Bk(y).{\displaystyle [s,t;x,y]_{n}=\sum _{k=0}^{n}(-1)^{k}{s \choose k}{t \choose {n-k}}B_{n-k}(x)B_{k}(y).} 

Fourier series

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TheFourier series of the Bernoulli polynomials is also aDirichlet series, given by the expansionBn(x)=n!(2πi)nk0e2πikxkn=2n!k=1cos(2kπxnπ2)(2kπ)n.{\displaystyle B_{n}(x)=-{\frac {n!}{(2\pi i)^{n}}}\sum _{k\not =0}{\frac {e^{2\pi ikx}}{k^{n}}}=-2n!\sum _{k=1}^{\infty }{\frac {\cos \left(2k\pi x-{\frac {n\pi }{2}}\right)}{(2k\pi )^{n}}}.} Note the simple largen limit to suitably scaled trigonometric functions.

This is a special case of the analogous form for theHurwitz zeta functionBn(x)=Γ(n+1)k=1exp(2πikx)+eiπnexp(2πik(1x))(2πik)n.{\displaystyle B_{n}(x)=-\Gamma (n+1)\sum _{k=1}^{\infty }{\frac {\exp(2\pi ikx)+e^{i\pi n}\exp(2\pi ik(1-x))}{(2\pi ik)^{n}}}.} 

This expansion is valid only for0 ≤x ≤ 1 whenn ≥ 2 and is valid for0 <x < 1 whenn = 1.

The Fourier series of the Euler polynomials may also be calculated. Defining the functionsCν(x)=k=0cos((2k+1)πx)(2k+1)νSν(x)=k=0sin((2k+1)πx)(2k+1)ν{\displaystyle {\begin{aligned}C_{\nu }(x)&=\sum _{k=0}^{\infty }{\frac {\cos((2k+1)\pi x)}{(2k+1)^{\nu }}}\\[3mu]S_{\nu }(x)&=\sum _{k=0}^{\infty }{\frac {\sin((2k+1)\pi x)}{(2k+1)^{\nu }}}\end{aligned}}} forν>1{\displaystyle \nu >1} , the Euler polynomial has the Fourier seriesC2n(x)=(1)n4(2n1)!π2nE2n1(x)S2n+1(x)=(1)n4(2n)!π2n+1E2n(x).{\displaystyle {\begin{aligned}C_{2n}(x)&={\frac {\left(-1\right)^{n}}{4(2n-1)!}}\pi ^{2n}E_{2n-1}(x)\\[1ex]S_{2n+1}(x)&={\frac {\left(-1\right)^{n}}{4(2n)!}}\pi ^{2n+1}E_{2n}(x).\end{aligned}}} Note that theCν{\displaystyle C_{\nu }}  andSν{\displaystyle S_{\nu }}  are odd and even, respectively:Cν(x)=Cν(1x)Sν(x)=Sν(1x).{\displaystyle {\begin{aligned}C_{\nu }(x)&=-C_{\nu }(1-x)\\S_{\nu }(x)&=S_{\nu }(1-x).\end{aligned}}} 

They are related to theLegendre chi functionχν{\displaystyle \chi _{\nu }}  asCν(x)=Reχν(eix)Sν(x)=Imχν(eix).{\displaystyle {\begin{aligned}C_{\nu }(x)&=\operatorname {Re} \chi _{\nu }(e^{ix})\\S_{\nu }(x)&=\operatorname {Im} \chi _{\nu }(e^{ix}).\end{aligned}}} 

Inversion

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The Bernoulli and Euler polynomials may be inverted to express themonomial in terms of the polynomials.

Specifically, evidently from the above section onintegral operators, it follows thatxn=1n+1k=0n(n+1k)Bk(x){\displaystyle x^{n}={\frac {1}{n+1}}\sum _{k=0}^{n}{n+1 \choose k}B_{k}(x)} andxn=En(x)+12k=0n1(nk)Ek(x).{\displaystyle x^{n}=E_{n}(x)+{\frac {1}{2}}\sum _{k=0}^{n-1}{n \choose k}E_{k}(x).} 

Relation to falling factorial

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The Bernoulli polynomials may be expanded in terms of thefalling factorial(x)k{\displaystyle (x)_{k}}  asBn+1(x)=Bn+1+k=0nn+1k+1{nk}(x)k+1{\displaystyle B_{n+1}(x)=B_{n+1}+\sum _{k=0}^{n}{\frac {n+1}{k+1}}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}(x)_{k+1}} whereBn=Bn(0){\displaystyle B_{n}=B_{n}(0)}  and{nk}=S(n,k){\displaystyle \left\{{\begin{matrix}n\\k\end{matrix}}\right\}=S(n,k)} denotes theStirling number of the second kind. The above may be inverted to express the falling factorial in terms of the Bernoulli polynomials:(x)n+1=k=0nn+1k+1[nk](Bk+1(x)Bk+1){\displaystyle (x)_{n+1}=\sum _{k=0}^{n}{\frac {n+1}{k+1}}\left[{\begin{matrix}n\\k\end{matrix}}\right]\left(B_{k+1}(x)-B_{k+1}\right)} where[nk]=s(n,k){\displaystyle \left[{\begin{matrix}n\\k\end{matrix}}\right]=s(n,k)} denotes theStirling number of the first kind.

Multiplication theorems

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Themultiplication theorems were given byJoseph Ludwig Raabe in 1851:

For a natural numberm≥1,Bn(mx)=mn1k=0m1Bn(x+km){\displaystyle B_{n}(mx)=m^{n-1}\sum _{k=0}^{m-1}B_{n}{\left(x+{\frac {k}{m}}\right)}} En(mx)=mnk=0m1(1)kEn(x+km) for odd mEn(mx)=2n+1mnk=0m1(1)kBn+1(x+km) for even m{\displaystyle {\begin{aligned}E_{n}(mx)&=m^{n}\sum _{k=0}^{m-1}\left(-1\right)^{k}E_{n}{\left(x+{\frac {k}{m}}\right)}&{\text{ for odd }}m\\[1ex]E_{n}(mx)&={\frac {-2}{n+1}}m^{n}\sum _{k=0}^{m-1}\left(-1\right)^{k}B_{n+1}{\left(x+{\frac {k}{m}}\right)}&{\text{ for even }}m\end{aligned}}} 

Integrals

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Two definite integrals relating the Bernoulli and Euler polynomials to the Bernoulli and Euler numbers are:[5]

Another integral formula states[6]

with the special case fory=0{\displaystyle y=0} 

Periodic Bernoulli polynomials

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Aperiodic Bernoulli polynomialPn(x) is a Bernoulli polynomial evaluated at thefractional part of the argumentx. These functions are used to provide theremainder term in theEuler–Maclaurin formula relating sums to integrals. The first polynomial is asawtooth function.

Strictly these functions are not polynomials at all and more properly should be termed the periodic Bernoulli functions, andP0(x) is not even a function, being the derivative of a sawtooth and so aDirac comb.

The following properties are of interest, valid for allx{\displaystyle x} :

See also

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References

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  1. ^Hurtado Benavides, Miguel Ángel. (2020). De las sumas de potencias a las sucesiones de Appell y su caracterización a través de funcionales. [Tesis de maestría]. Universidad Sergio Arboleda.https://repository.usergioarboleda.edu.co/handle/11232/174
  2. ^Sergio A. Carrillo; Miguel Hurtado. Appell and Sheffer sequences: on their characterizations through functionals and examples. Comptes Rendus. Mathématique, Tome 359 (2021) no. 2, pp. 205-217. doi : 10.5802/crmath.172.https://comptes-rendus.academie-sciences.fr/mathematique/articles/10.5802/crmath.172/
  3. ^Lehmer, D.H. (1940). "On the maxima and minima of Bernoulli polynomials".American Mathematical Monthly.47 (8):533–538.doi:10.1080/00029890.1940.11991015.
  4. ^Zhi-Wei Sun; Hao Pan (2006). "Identities concerning Bernoulli and Euler polynomials".Acta Arithmetica.125 (1):21–39.arXiv:math/0409035.Bibcode:2006AcAri.125...21S.doi:10.4064/aa125-1-3.S2CID 10841415.
  5. ^Takashi Agoh & Karl Dilcher (2011)."Integrals of products of Bernoulli polynomials".Journal of Mathematical Analysis and Applications.381:10–16.doi:10.1016/j.jmaa.2011.03.061.
  6. ^Elaissaoui, Lahoucine & Guennoun, Zine El Abidine (2017). "Evaluation of log-tangent integrals by series involving ζ(2n+1)".Integral Transforms and Special Functions.28 (6):460–475.arXiv:1611.01274.doi:10.1080/10652469.2017.1312366.S2CID 119132354.

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