# Generalized Fractional-Order Bernoulli Functions via Riemann-Liouville Operator and Their Applications in the Evaluation of Dirichlet Series

Authors: Jorge Sanchez-Ortiz
Journal: Abstract and Applied Analysis
 (2018)

---

## Abstract

In this work, we define a new class of functions of the Bernoulli type using the Riemann-Liouville fractional integral operator and derive a generating function for these class generalized functions. Then, these functions are employed to derive formulas for certain Dirichlet series.


---

## Body

## 1. Introduction

The Bernoulli polynomials are defined by the generating function [1](1)∑n=0∞Bnxtnn!=textet-1t<2π.When x=0, Bn=Bn(0) are called Bernoulli numbers. The following property is well known:(2)Bnx=∑j=0nnjBn-jxj.Also, the Bernoulli polynomials are defined by the following Fourier series [2]:(3)Bnx=-n!2πin∑k=-∞∞k-ne2πikx0≤x<1.Various generalizations of the Bernoulli polynomials have been proposed. For example, Natalini [3] gave the following generalization: (4)E1,m+1text=∑n=0∞Bnm-1xtnn!,where Eα,β(t) is the two-parametric Mittag-Leffler function, so that, obviously, Bn(x):=Bn[0](x). Another generalization is given by Balanzario [4]:(5)Bnx=∫0xBn-1ydy+∫01y-1Bn-1ydy,where B0(x) is given and n≥1. In case B0(x)=1 for x∈[0,1), then Bn(x)·n! is the usual n-th Bernoulli polynomial. Balanzario and Sanchez [5] derive the following generating function for Bn(x) defined in (5):(6)∑k=0∞Bkxtk=textet-1a0-∫01B0′1-yety-1tdy+∫0xetx-yB0′ydy,where B0(x) is given and a0=∫01‍B0(x)dx; they used these generalized Bernoulli polynomials to derive formulas of certain Dirichlet series.Rahimkhani et al. [6] define the fractional-order Bernoulli functions, such as the functions obtained by changing the variable t to xα in (3), and applied these functions for solving the fractional Fredholem-Volterra integrodifferential equations.In the present paper, new functions called generalized fractional-order Bernoulli functions are defined by a generalization of (5) and obtain a generalization of the generating function (6). Also, given a generalization of the Fourier series (3), we use these functions to derive formulas for certain Dirichlet series and finally, some examples are shown.


## 2. Preliminaries

In this section, we give some basic definitions and properties of fractional calculus theory which are used in this work.Definition 1.
The Riemann-Liouville fractional integral of orderα∈R+ is defined by (7)Iαfx=1Γα∫0xftx-t1-αdt,where x>0 and Γ is the Gamma function.It can be directly verified that(8)Iαtβx=ΓβΓβ+αxβ+α-1,where α>0 and β>0.Definition 2.
The Caputo fractional derivative of orderα∈R+ is defined by (9)Dαfx=Ir-αDrfx,where D=d/dx, r=[α]+1 for α∉N0 and r=α for α∈N0.Now, whenα∈R+, the Caputo fractional differential operator Dα provides operation inverse to the Riemann-Liouville fractional integration operator Iα; the proof can be seen in [7].Lemma 3.
Letα∈R+ and f(x) a continuous function in the interval [0,1]. Then, (DαIαf)(x)=f(x).Now, we define the Laplace transform of a functionf(x) of a variable x∈R+ by (10)Lfxk=∫0∞e-kxfxdxk∈C,if the integral converges and its inverse by (11)L-1fkx=12πi∫γ-i∞γ+i∞ekxfkdk,with γ>σ, where σ is the abscissa of convergence.Under suitable conditions, the Laplace transform of the Caputo fractional derivativeDαf is given by [7](12)LDαfxk=kαLfxk-∑j=0r-1kα-j-1Djf0.Definition 4.
The two-parametric Mittag-Leffler function(13)Eα,βz=∑k=0∞zkΓαk+βα>0,β∈C,generalizes the classical Mittag-Leffler function (14)Eαz=∑k=0∞zkΓαk+1α>0.Using Definition4, we obtain the formulas (15)E1,mz=1zm-1ez-∑k=0m-2zkk!,Rexβ-1E1,βiλx=xβ-1E2,β-λx2,Imxβ-1E1,βiλx=λxβE2,1+β-λx2,where m∈N, β>0, and λ∈R. From above equations, we have(16)xmE2,1+m-λ2x2=1λmcos⁡mπ2-λx-∑k=0m-1λxkk!cosmπ2-kπ2.The following differentiation formula is an immediate consequence of Definition 4(17)ddzmzβ-1Eα,βzα=zβ-m-1Eα,β-mzαm∈N.Using Definition 4 and term-by-term integration, we arrive at(18)1Γα∫0zz-tμ-1Eα,βλtαtβ-1dt=zμ+β-1Eα,β+μλzα,where μ>0 and β>0. From (18) we obtain(19)IαtnαE2,1+nαλt2x=xn+1αE2,1+n+1αλx2,(20)Iαt1+nαE2,2+nαλt2x=x1+n+1αE2,2+n+1αλx2.It follow from the well-known discrete orthogonality relation(21)∑h=0m-1e2πihk/m=m,ifk≡0modm,0,ifk≢0modmand formula (18) that(22)Iαcosh⁡λtx=xαE2,α+1λx2,(23)Iαsenhλtλx=xα+1E2,α+2λx2α>0.Now, we state an important relation between the Laplace transform and Mittag-Leffler function; the proof can be seen in [8].Lemma 5.
The following formula is true:(24)∫0∞e-kxxβ-1Eα,βaxαdx=kα-βkα-a,where α>0, β>0, and k>|a|1/α.


## 3. Generalized Fractional-Order Bernoulli Functions

In this section, first we define a new set of fractional-order Bernoulli functions by means of the Riemann-Liouville fractional integration operator.Definition 6.
LetB0α(x) be a periodic function of period 1. We define the fractional-order Bernoulli functions by(25)Bnαx=IαBn-1αx-1αΓα∫01Bn-1αt1-t-αdt,where α>0 and n∈N.In the case,α=1, then Bn1(x) are the generalization of the Bernoulli polynomials defined in (5). For example, when B0α(x)=1 for 0≤x<1, the first two fractional-order Bernoulli functions are (26)B1αx=xαΓ1+α-1Γ2+α,B2αx=x2αΓ1+2α-xαΓ1+αΓ2+α+1Γ2+α2-1Γ2+2α.The functions defined (25) satisfy the following properties:(27)∫01Bnαxdx=0andDαBnαx=Bn-1αx.These assertions are followed by integrating (25) and Lemma 3, given that (Dαc)(x)=0, for c∈R.In the following theorem, we obtain a generating function for the fractional-order Bernoulli functions defined in (25).Theorem 7.
LetB0α(x) be a periodic function of period 1. Suppose that B0α(x) has a continuous derivative in the open interval (0,1). Let A0=∫01‍B0α(x)dx and {Bnα} be the sequence defined by (25). Then for α>0,(28)∑n=0∞Bnαxtn=EαtxαEα,2tA0-∫01yDB0α1-yEα,2tyαdy+∫0xDB0yEαtx-yαdy.Proof.
We proceed formally as in [9, Problem 9.785]. Consider the following fractional differential equation:(29)DαGx,t-tGx,t=DαB0αx,for a B0α(x) given function and (30)Gx,t=∑n=0∞Bnαxtn.Applying the Laplace transform to (29) and using (12), we obtain (31)LGx,tk=kαkα-tLB0αxk+1kα-t∑j=0m-1kα-j-1DjG0,t-DjB0α0,where m=[α]+1. Then, using the inverse Laplace transform in above equation, we arrive to the equation (32)Gx,t=L-1kαkα-tLB0αxkx+G0,t-B0α0L-1kα-1kα-tx.Therefore, by Lemma (25) and given that L[δ(x)](k)=1, we get (33)Gx,t=G0,t-B0α0Eαtxα+Eαtxα∗Dδx∗B0αx=G0,tEαtxα+∫0xEαtx-yαDB0αydy,where δ(x) is the Dirac delta function, and(34)f∗gx=∫0xfygx-ydyis the convolution of the functions f and g. Now, we integrate (33) from 0 to 1 with respect to x and by (27) and (18) we obtain (35)A0=G0,tEα,2t+∫01∫0xEαtx-yαDB0αydydx.Solving for G(0,t) and substituting in (33) we obtain our result.Observe that if we setα=1 and B0α(x)=1 for x∈[0,1) in Theorem 7, then we obtain the corresponding unification and generalization of the generating function (1) of the usual Bernoulli polynomials. In case α=1 in Theorem 7, we obtain the generating function (6).In the next theorem, we compute the fractional-order Bernoulli functions defined in (25) through the two-parametric Mittag-Leffler function.Theorem 8.
LetB0α(x) be a periodic function of period one and piecewise continuous in the open interval(0,1). Let A0 and {Bnα} be as in Theorem 7. Then for n≥1, (36)Bnαx=A0xnαΓ1+nα-1Γ2+nα+∑k=1n-1Bkα0xn-kαΓ1+n-kα-1Γ2+n-kα+2∑j=1∞AjxnαE2,1+nα-λj2x2-E2,2+nα-λj2+4π∑j=1∞jBjx1+nαE2,2+nα-λj2x2-E2,3+nα-λj2,where λj=2πj and Aj and Bj are the Fourier coefficients of B0α(x).Proof.
The proof is by mathematical induction onn. Since B0α(x) is piecewise continuous, then we can consider its Fourier series (37)B0αx=A0+2∑j=1∞Ajcos⁡λjx+Bjsin⁡λjx.Let n=1. Then by (22), (23), and (25) we obtain(38)B1αx=IαA0x+2∑j=1∞AjIαcos⁡λjtx+BjIαsin⁡λjtx-A0αΓα∫0111-t-αdt+2αΓα∑j=1∞∫01Ajcos⁡λjt+Bjsin⁡λjt1-t-αdt⁡=A0xαΓ1+α-1Γ2+α+2∑j=1∞AjxαE2,1+α-λj2x2-E2,2+α-λj2+2∑j=1∞λjBjx1+αE2,2+α-λj2x2-E2,3+α-λj2.Now, we assume the theorem true for a given n and we will prove that it is valid for n+1. From (25) (39)Bn+1αx=IαBnαx-1αΓα∫01Bnαt1-t-αdt,applying (8), (19), and (20) and the above equation we get the result.


## 4. Evaluation by Certain Dirichlet Series

For the proof of the following theorems one proceeds as in Balanzario [10], using Theorem 8 and (16).Theorem 9.
Let{fj} be a sequence of complex numbers of period T, so that fj+T=fj for all j∈N. Let B0α(x), {Bnα}, and Aj be as in Theorem 8. Suppose nα is par and fT-j=fj for each j∈{1,2,…,T-1} and suppose nα is impar and fT-j=-fj for each j∈{1,2,…,T-1} and fT=0. Then(40)∑j=1∞Ajjnαfj-∑k=0nα-1γknα2πjkk!cos⁡π2nα-k=2πnα2∑m=1TβmnαBnαmT-A0Tnα∑m=1TβmnαmnαΓ1+nα-∑k=1n-1γn-kαnαΓ1+n-kαBkα0,where (41)βmnα=1T∑k=1Tfkcos⁡nαπ2-2πmkT,γknα=1Tk∑m=1Tmkβmnα.Theorem 10.
Assume the notation of Theorem10. If nα is impar and fT-j=fj for each j∈{1,2,…,T-1} and if nα is par and fT-j=-fj for each j∈{1,2,…,T-1} and fT=0, then(42)∑j=1∞Bjjαn-fj+∑k=0nαφknα2πjkk!sinπ2nα-k=2πnα2∑m=1TμmnαBnαmT-A0Tnα∑m=1TμmnαmnαΓ1+nα-∑k=1n-1φn-kαnαΓ1+n-kαBkα0,where (43)μmnα=1T∑k=1Tfksin⁡nαπ2-2πmkT,φknα=1Tk∑m=1Tmkμmnα.Finally, some examples are given.Example 1.
As the first example, we considerB0α(x) be of period one such that B0α(x)=x1/2 for 0≤x<1. Let α=1/2, n=2, T=8, λ1=λ5=-1, λ3=λ7=1, and λ2=λ4=λ6=λ8=0. Applying Theorem 9, we get (44)∑j=1∞Fs2jj5/2=π265-33,where Fs(z) is the Fresnel sine integral given by ∫0z‍sin⁡(πt2/2)dt.Example 2.
Here is another example of Theorem9. Let B0α(x) be of period one such that B0αx=xcos⁡(πx) for 0≤x<1. Let α=2, n=1, T=10, λ1=λ2=λ3=λ4=λ6=λ7=λ8=λ9=1, and λ5=λ10=0. Then (45)∑j=1∞1+4j2j21-4j22λj=-π525-5+25+5+π25027-25.Example 3.
LetB0α(x) be of period one such that B0α(x)=x1/2sin⁡(πx) for 0≤x<1. Let α=1/2, n=2, T=8, λ1=λ3=λ5=λ7=1, and λ2=λ4=λ8=0. By applying Theorem 10, we obtain(46)∑j=1∞Fs4j+22j+13/2-Fs4j-22j-13/2λj=π22+Fc2+π310-42​1F254;32,94;-π24+​1F254;32,94;-π216,where Fc(z) is the Fresnel cosine integral and ​1F2 is the hypergeometric function defined by (47)​1F2a;b,c;z=∑k=0∞akbkckzkk!,ak=Γa+kΓa.



---

*Source: 4875916-2018-12-12.xml*