# Some Notes of the Fractional Integral and Derivatives and Its Applications

Authors: Fen Qin; Fuyi Xu; Huizeng Qin
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, the definitions of the fractional derivative and integrals are given by the neutrix limit. Not only is it consistent with the classical results but also the representations of the fractional derivative and integrals are obtained forx−a−k, x−a−klnpx−a, x−a−kfx, and x−a−klnpx−afx, where k,p=1,2,…,fx is the analytic function.


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## Body

## 1. Introduction

Fractional calculus has been used to model physics and engineering processes widely since standard mathematical models of integer-order derivatives, including nonlinear models. In fact, in many fields such as mechanics, electronics, chemistry, biology, economics, notably control theory, and signal and image processing, fractional calculus has been playing more and more important roles in recent years [1–5]. There are several definitions for fractional derivatives and integrals.The Riemann–Liouville fractional integral and derivatives [6] are defined as.(1)JxαaRLfx=1Γα∫axx−sα−1fsds,Reα>0,and(2)DxαaRLfx=1Γm−αdmdxm∫axx−sm−α−1fsds,m−1<Reα<m,dmdxmfx,α=m,for β>−1 and x>a, respectively. The Grünwald–Letnikov fractional derivative [7] is defined as(3)DxαaGLfx=limh⟶0h−α∑j=0x−a/h−1jαjfx−jh,where x is the integer part of x,αj=Γα+1/j!Γ1+α−j. It has been proved that this definition is equivalent to(4)DxαaGLfx=∑j=0m−1fjax−aj−αΓj+1−α+1Γm−α∫axx−sm−α−1fmsds,dmdxmfx,α=m,=∑j=0m−1fjax−aj−αΓj+1−α+Jxm−αaRLfmx,dmdxmfx,α=m,where m−1<Reα<m,fx∈Cma,x. The Caputo fractional derivative, from [8], is defined as(5)DxαaCfx=1Γm−α∫axx−sm−α−1fmsds,m−1<Reα<m,dmdxmfx,α=m,=Jxm−αaRLfmx,m−1<Reα<m,dmdxmfx,α=m,where fx∈Cma,x. The fractional derivative defined above is called the left fractional integral and derivative. It is completely analogous to introducing the corresponding right fractional derivative. For example, the right Riemann–Liouville fractional derivatives is defined as(6)DbαxRLfx=−1mΓm−αdmdxm∫xbs−xm−α−1fsds,m−1<α<m,dmdxmfx,α=m.From the definitions (4) and (5), we have DxαaCfx=DxαaGLfx when fka=0,k=0,1,…,m−1.For the fractional integral and derivative of the power functionx−aβ, there are(7)JxαaRLx−aβ=Γβ+1Γβ+1+αx−aβ+α,and(8)DxαaRLx−aβ=DxαaCx−aβ=DxαaGLx−aβ=Jx−αaRLx−aβ=Γβ+1x−aβ−αΓβ+1−α,for Reα>0,Reβ>−1 and x>a. When β=−1,−2,…, (7) and (8) are clearly not valid. However,(9)Dnx−a−m=−m−n+1nx−a−m−n=−1nΓn+mΓmx−a−m−n,where the symbol αn represents the Pochhammer symbol, i.e.,(10)αn=αα+1⋯α+n−1=Γα+nΓα,and it requires a proper definition of the fractional derivative of x−a−k,k=1,2,…. To remedy this problem, the following definition is given by Mauro Bologna [9]:(11)Dxα0RLx−m=−1αΓα+mΓmx−m−α.It is clear that (8) and (11) there is no essential connection, to be more reasonable definition. In fact, if (11) is true, then we have by (7)(12)JxαaRLDxα0RLx−m=−1αΓα+mΓmJxαaRLx−α−m=−1αΓα+mΓmΓ−m−α+1Γ−m+1x−a−m=0,which is impossible.Therefore, the traditional definition of fractional derivatives forDαx−a−kk=1,2,… is unreasonable. In this paper, our goal is to give a reasonable definition and representations of fractional derivatives for x−a−k, x−a−klnpx−a, x−a−kfx, x−a−klnpx−afx, where k,p=1,2,…,fx is an analytic function. The properties of the fractional derivatives of the correlation function and the solution of the correlation fractional differential equation are discussed. For simplicity, we take.(13)m=Reα+1,Reα>0,α≠0,1,2,…,α,α=0,1,2,…,0,Reα<0.We use neutrix limit [9–16] to define the frational derivatives for (1)–(6).Definition 1.
1 For complex numberα, let(14)JxαaRLfx=1Γαℕ.limε⟶0+0∫a+εxx−sα−1fsds,Reα>0,(15)DxαaRLfx=1Γm−αdmdxmℕ.limε⟶0+0∫a+εxx−sm−α−1fsds,or(16)DxαaRLfx=dmdxmJxm−αaRLfx,(17)DxαaGLfx=ℕ⋅limε⟶0+0∑k=0m−1fka+εx−a−εk−αΓk+1−α,Jxm−αaRLfmx,and(18)DxαaCfx=1Γm−αℕ.limε⟶0+0∫a+εxx−sm−α−1fmsds=Jxm−αaRLfmx.
For convenience of writing, letJxαafx=JxαaRLfx,Dxαafx=DxαaRLfx.Remark 1.
IfReα<0,m=0, then Dxαafx=DxαaGLfx=DxαaCfx=Jx−αafx.
The so-called neutrix limit is defined as follows.Definition 2.
If(19)Fx,ε=ε−λ∑k=0NAkxεk+ε−μ∑k=0Mεk∑l=1LBk,lxlnlε+Cx+oε,for N,M,L=0,1,2,3,…,N<λ,M≤μ, then(20)ℕ.limε⟶0+0Fx,ε=Cx.
For example, by(21)x−sα−1s−aβ=x−aα−1∑l=0∞1−αll!s−al+βx−al,for Reα>0, we have(22)∫a+εxx−sα−1s−aβds=x−aα−1∫a+εx∑l=0∞1−αll!s−al+βx−alds.
By exchanging the order of integration and summing, we have(23)∫a+εxx−sα−1s−aβds=∑l=0∞1−αlx−aα+βl!l+β+1−∑l=0−β−11−αlεl+β+1x−aα−1−ll!l+β+1+oε,β≠−1,−2,…,∑l=0,l≠k−1∞1−αlx−aα−kl!l−k+1+1−αk−1x−aα−klnx−ak−1!−∑l=0k−21−αlεl+1−kx−aα−1−ll!l−k+1−1−αk−1x−aα−kk−1!lnε+oε,β=−k,k=1,2,….
From Definitions2 and (23), we obtain that(24)ℕ.limε⟶0+0∫a+εxx−sα−1s−aβds=x−aα+β∑l=0∞1−αll!l+β+1,for β≠−1,−2,… and(25)lℕ⋅limε⟶0+0∫a+εxx−sα−1s−a−kds=x−aα−k∑l=0,l≠k−1∞1−αll!l−k+1+1−αk−1lnx−ak−1!,for k=1,2,….
It should be pointed out that we can consider the fractional order derivatives ofx−aβfx,x−aβlnpx−afx based on definitions (14)–(18), where fx is the analytic function and fa≠0, and present the following two examples to illustrate our conclusions.


## 2. Lemmas

In this section, we shall present some important lemmas that will be frequently used and their proof may be found in [15, 16].Lemma 1.
For Beta functionBx,y, there is(26)Bx,−m=B−m,x=∑k=0,k≠m∞1−xkk!k−m=−1mΓxHm−γ−ψx−mm!Γx−m,for x≠0,−1,−2,…,m=0,1,2,…,(27)Bn,−m=B−m,n=∑l=0,l≠mn−1n−1l−1ll−m=t1,n=1,2,…,m,m=1,2,…,t2,n=m+1,m+2,…,m=1,2,…,where t1=−1nn−1!n−m!/m!,t2=−1mn−1!Hm−Hn−m−1/m!n−m−1!,Hn=∑l=1n1/l and(28)B−n,−m=−∑l=0m−1n+ll1m−l−∑l=0n−1m+ll1n−l,for m,n=1,2,….Lemma 2.
For the complex numberβ and p=1,2,…, there is(29)dmdxmx−aβlnpx−a=x−aβ−m∑l=0mplam,l1+β−mlnp−lx−a,where am,lx=dl/dxlxm,pl=0,l=p+1,p+2,…. In particular,(30)dmdxmx−aβlnx−a=x−aβ−m1+β−mmlnx−a+∑j=1m1β−m+j.Lemma 3.
Forβ≠−1,−2,…,n=1,2,…, one has(31)Bαpβ=Hpψα,βBαβ,B0,pα,β=Hpψα,βBα,β,where Bq,px,y=dp+q/dxqdypBx,y,Bαβ=Γβ/Γβ+α,Bαpβ=dp/dxpBαx|x=β.(32)Hpψα,β=∑l=0p−1Hp,lψα,β,H1,0ψα,β=ψβ−ψβ+α,Hn,0ψα,β=−1nn−1!ζn,β−ζn,β+α,Hn,lψα,β=ψβ−ψβ+αHn−1,l−1ψα,β+n−1!∑i=1n−2−1n−iζn−i,β−ζn−i,β+αHi,l−1ψα,βi!,l=1,2,…,n,n=1,2,…,p,Here, ψx is the digamma function defined by(33)ψx=ddxlnΓx=−γ−1x+∑l=1∞1l−1l+x,and ζp,z is the Hurwitz zeta function defined by(34)ζp,z=∑l=0∞1l+zp,ζp,0=ζp,1=ζp,where ζq is the Riemann zeta function.Lemma 4.
For complex numberα,ββ≠−1,−2,…,n=0,1,2,…, there is(35)∑l=0∞1−αll!l+βn+1=−1nB0,nα,βn!.Lemma 5.
For complex numberβ, there is.(36)∫a+εxs−aβlnps−ads=p!x−aβ+1∑j=0p−1jlnp−jx−aβ+1j+1p−j!−p!εβ+1∑j=0p−1jlnp−jεβ+1j+1p−j!,for β≠−1 and(37)∫a+εxs−a−1lnps−ads=lnp+1x−a−lnp+1εp+1.Lemma 6.
Forn,p,k=1,2,…, there are(38)x−a−n−1p+1∑l=1n+1p+1lan+1,l−nlnp+1−lx−a=lnpx−ax−an,and(39)x−a−n−kp!∑j=0p−1jp−j!1−kj+1∑l=0n+1p−jlan+1,l1−n−klnp−j−lx−a=x−a−klnpx−an.Lemma 7.
For the complex numberα,ββ≠−1,−2,…,p=1,2,…, there is(40)1Γm−α∑l=0mplam,ltt=1+β−m∑j=0p−lp−ljB0,ju,v|u=m−αv=β−m+1lnp−l−jx−a=1Γ−α∑j=0ppjB0,j−α,β+1lnp−jx−a.The numerical calculation shows that (40) is always true, but its theory is a little harder to prove.


## 3. Fractional Derivatives ofx−aβ and x−aβlnpx−a

Through the analytical continuation ofΓβ+1, we see that (7) and (8) still hold for α>0,β≠−1,−2,… and x>a. In this section, we shall consider the fractional order derivatives of x−aβ and x−aβlnpx−a for all complex β and p=1,2,…, and expect to get similar results according to the definition of neutrix limit. Our main results read as follows.Theorem 1.
According to definition (14)–(16), (7), and (8) still hold for the complex α,ββ≠−1,−2,−3,….Proof.
ForReα>0,β≠−1,−2,−3,…,x−a>0, using (21), Definitions 2 and (35) yields that(41)ℕ.limε⟶0+0∫a+εxx−sα−1s−aβds=x−aα+β∑l=0∞1−αll!l+β+1=Bα,β+1x−aα+β.
By (16) and (41), we have(42)Jxαax−aβ=Bα,β+1Γαx−sα+β=Γβ+1Γα+β+1x−aβ+α,which implies that (7) holds.
Employing (16) and (7), we have.(43)Dxαax−aβ=dmdxmJxm−αax−aβ=dmdxmΓβ+1Γm−α+β+1x−aβ+m−α=β−α+1mΓβ+1Γm−α+β+1x−aβ−α=Γβ+1Γβ−α+1x−aβ−α,which concludes that (8) holds.Corollary 1.
(1)
For the complexα,ββ≠−1,−2,−3,…, there is Dxαax−aβ=DxαaGLx−aβ=DxαaCLx−aβ=Γβ+1/Γβ+1−αx−aβ−α=Jx−αax−aβ.(2)
Dxαax−aα−n=0,n=1,2,….Proof.
(1)
ForReα<0, Remark 1. indicates that corollary naturally holds.ForReα>0, fx=x−aβ, β≠−1,−2,−3,⋯,x−a>0, we have(44)ℕ.limε⟶0+0∑j=0m−1fja+εx−a−εj−αΓj+1−α=ℕ.limε⟶0+0∑j=0m−1x−aj−α−1jβ−j+1jεβ−jΓj+1−α1−εx−aj−α=ℕ.limε⟶0+0∑j=0m−1x−aj−α−1jkjεβ−jΓj+1−α∑l=0∞1+α−jll!εlx−al=ℕ.limε⟶0+0∑l=0∞∑j=0m−1x−aj−l−α−1jkjεβ+l−jΓj+1−α1+α−jll!.Due toβ+l−j≠0 for l=0,1,2,…,j=0,1,2,…,m−1, we have by Definition 2.(45)N.limε⟶0+0∑j=0m−1fja+εx−a−εj−αΓj+1−α=0,which implies that(46)Jxm−αax−aβm=β−m+1mJxm−αax−aβ−m=β−m+1mΓβ−m+1Γβ−m+1+m−αx−aβ−m+m−α=Γβ+1Γβ+1−αx−aβ−α.Hence, the first part of the corollary holds.(2)
From (41) and Bm−α,α−n+1=Γm−αΓα−n+1/Γm−n+1,m≥n,0,m<n, we have(47)Jxm−αax−aα−n=Bm−α,α−n+1x−am−nΓm−αΓα−n+1x−am−nΓm−n+1,m≥n,0,m<n.Then,(48)Dxαax−aα−n=dmdxmJxm−αax−aα−n=dmdxmΓα−n+1x−am−nΓm−n+1,m≥ndmdxm0,m<n=0.Therefore, the second part of the corollary holds.Theorem 2.
For complexα,ββ≠−1,−2,−3,… and p=1,2,…, there is(49)Jxαxx−aβlnpx−a=x−aβ+αΓα∑j=0ppjB0,jα,β+1lnp−jx−a,or(50)Jxαxx−aβlnpx−a=Γβ+1x−aβ+αΓα+β+1∑j=0ppjHjψβ+1,α+β+1lnp−jx−a,for Reα>0 and(51)DxαaRLs−aβlnps−a=Jx−αaRLx−aβlnpx−a=x−aβ−αΓ−α∑j=0ppjB0,j−α,β+1lnp−jx−a,or(52)DxαaRLs−aβlnps−a=Γβ+1x−aβ−αΓβ+1−α∑j=0ppjHjψβ+1,β+1−αlnp−jx−a.Proof.
By (21) and exchanging the order of integration and then summing, we have(53)ℕ.limε⟶0+0∫a+εxx−sα−1s−aβlnps−ads=ℕ.limε⟶0+0∫a+εxx−aα−1∑l=0∞1−αll!s−al+βx−allnps−ads=ℕ.limε⟶0+0∑l=0∞1−αlx−aα−l−1l!∫a+εxs−al+βlnps−ads.
From (36), we have(54)∑l=0∞1−αlx−aα−l−1l!∫a+εxs−al+βlnps−ads=p!x−aα+β∑l=0∞1−αll!∑j=0p−1jl+β+1j+1p−j!lnp−jx−a−p!∑l=0∞1−αlx−aα−l−1l!∑j=0p−1jεl+β+1l+β+1j+1p−j!lnp−jε=p!x−aα+β∑j=0p−1jp−j!lnp−jx−a∑l=0∞1−αll!l+β+1j+1−p!∑l=0∞1−αlx−aα−l−1l!∑j=0p−1jεl+β+1l+β+1j+1p−j!lnp−jε.
By (16), (35), (53), (54), Lemma 3, and Definition 2, we get(55)Jxαax−aβlnpx−a=p!x−aα+βΓα∑j=0p−1jlnp−jx−ap−j!∑l=0∞1−αll!l+β+1j+1=x−aα+βΓα∑j=0ppjB0,jα,β+1lnp−jx−a=Γβ+1x−aβ+αΓα+β+1∑j=0ppjHjψβ+1,α+β+1lnp−jx−a,which implies that (49) and (50) hold.
It’s worth noting that if we use the derivative method with parameter, then(56)Jxαax−aβlnpx−a=dpdβpJxαax−aβ=dpdβpBα,β+1Γαx−aα+β=x−aα+βΓα∑j=0ppjB0,jα,β+1lnp−jx−a,which is exactly the same as the derivation above.
By (16) and (49), we have(57)Dxαax−aβlnpx−a=dmdxmJxm−αax−aβlnpx−a=dmdxmx−aβ+m−αΓm−α∑j=0ppjB0,jm−α,β+1lnp−jx−a=1Γm−α∑j=0ppjB0,jm−α,β+1dmdxmx−aβ+m−αlnp−jx−a.
In fact,Dxαas−aβlnps−a can be obtained from (29) and (57), but the expression is more complex. Here, we still use the following derivative method with parameter.(58)Dxαax−aβlnpx−a=dpdxpDxαax−aβ=dmdxmB−α,β+1Γ−αx−aβ−α=x−aβ−αΓ−α∑j=0ppjB0,j−α,β+1lnp−jx−a=Γβ+1x−aβ+αΓβ+1−α∑j=0ppjHjψβ+1,β+1−αlnp−jx−a,so, we get (51) and (52).Corollary 2.
For the complexα,ββ≠−1,−2,−3,…, there is.(59)Dxαas−aβlnps−a=DxαaGLs−aβlnps−a=DxαaCLs−aβlnps−a=Jx−αas−aβlnps−a.Proof.
Forfx=x−aβlnpx−a, by similar to the proof of Corollary 1, we find that (45) still holds. Hence, DxαaGLs−aβlnps−a=DxαaCLs−aβlnps−a.
By (29), we get(60)Jxm−αax−aβlnpx−am=∑l=0mplam,l1+β−mJxm−αax−aβ−mlnp−lx−a.
Moreover, from (49), we have(61)Jxm−αax−aβ−mlnp−lx−a=x−aβ−αΓm−α∑j=0p−lp−ljB0,jm−α,β−m+1lnp−l−jx−a.
Substituting (61) into (60), we get(62)DxαaCx−aβlnpx−a=Jxm−αax−aβlnpx−am=x−aβ−αΓm−α∑l=0mplam,ltt=1+β−m∑j=0p−lp−ljB0,ju,v|u=m−αv=β−m+1lnp−l−jx−a.
By (40) and (62), we have(63)DxαaCx−aβlnpx−a=x−aβ−αΓ−α∑j=0ppjB0,j−α,β+1lnp−jx−a=Dxαas−aβlnps−a.
This completes the proof of the corollary.Theorem 3.
For the complexReα>0 and k=1,2,3,…, there is(64)Jxαax−a−k=−1kx−aα−kγ+ψ1+α−k−Hk−1−lnx−ak−1!Γ1+α−k,for k−α≠1,2,… and(65)Dxαax−a−k=−1kx−a−α−kγ+ψ1−α−k−Hk−1−lnx−ak−1!Γ1−α−k,for α+k≠1,2,….Proof.
ForReα>0,k=1,2,3,…,x−a>0, by (21) and (26), we have(66)Jxαax−a−k=x−aα−kΓα∑l=0,l≠k−1∞1−αll!l−k+1+1−αk−1lnx−ak−1!=x−aα−kΓα−1k−1ΓαHk−1−γ−ψα+1−kk−1!Γα+1−k+1−αk−1lnx−ak−1!=−1kx−aα−kγ+ψα+1−k−Hk−1−lnx−ak−1!Γα+1−k,which implies that (64) holds.
By (16), (64), and (30), we get(67)Dxαax−a−k=dmdxmaJxm−αx−a−k=dmdxm−1kx−am−α−kγ+ψm−α+1−k−Hk−1−lnx−ak−1!Γm−α+1−k=−1k1−α−kmγ+ψm−α+1−kx−a−α−kk−1!Γm−α+1−k−−1kk−1!Γm−α+1−kdmdxmx−am−α−klnx−a=−1kγ+ψm−α+1−k−Hk−1x−a−α−kk−1!Γ1−α−k−−1kk−1!Γm−α+1−kdmdxmx−am−α−klnx−a=−1kx−a−α−kγ+ψm−α+1−k−Hk−1−lnx−a+H˜α+kmk−1!Γ1−α−k.where H˜αm=∑j=1m1/a−j. Using(68)ψm−α+1−k+H˜α+km=ψ1−α−k,and (67) yields that (65) holds.Remark 2.
Due to(69)limα⟶n−1kψ1−α−kk−1!Γ1−α−k=−1nkn,limα⟶n−1kx−a−α−kγ−Hk−1−lnx−ak−1!Γ1−α−k=0,for n=1,2,…, we have(70)limα⟶nDxαax−a−k=−1nknx−a−n−k=x−a−kn.
Under the definition given in this paper, whetherβ≠−1,−2,…, or β=−1,−2,…, there is always limα⟶nDxαax−aβ=dn/dxnx−aβ. So limα⟶nDxαax−aβ is just the usual derivative dn/dxnx−aβ, which implies that the definition in this paper is reasonable. For example, taking fx=x−a−k,α≠1,2,…, there is(71)ℕ.limε⟶0+0∑j=0m−1fja+εx−a−εj−αΓj+1−α=ℕ.limε⟶0+0∑j=0m−1x−aj−α−1jkjε−k−jΓj+1−α1−εx−aj−α=ℕ.limε⟶0+0∑j=0m−1x−aj−α−1jkjε−k−jΓj+1−α∑l=0∞1+α−j−lll!εlx−al=ℕ.limε⟶0+0∑j=0m−1x−aj−α−1jkjε−k−jΓj+1−α∑l=0k+j−11+α−j−lll!εlx−al+x−a−k−α∑j=0m−1−1jkjΓj+1−α1+α−2j−kk+jk+j!+oε=x−a−k−α∑j=0m−1−1jkjΓj+1−α1+α−2j−kk+jk+j!=−1kx−a−k−αk−1!∑j=0m−1j−αk+jk+jΓj+1−α.Corollary 3.
Forα≠1,2,…,k=1,2,3,…, there is(72)DxαaCx−a−k=−1kx−a−α−kγ+ψ1−α−k−Hk+m−1−lnx−ak−1!Γ1−α−k=DxαaRLx−a−k+−1kHk−1−Hk+m−1x−a−α−kk−1!Γ1−α−k,and(73)DxαaGLx−a−k=−1kx−a−k−αk−1!∑j=0m−1j−αk+jk+jΓj+1−α+−1kx−a−α−kγ+ψ1−α−k−Hk+m−1−lnx−ak−1!Γ1−α−k=DxαaCx−a−k+−1kx−a−k−αk−1!∑j=0m−1j−αk+jk+jΓj+1−α=DxαaRLx−a−k+−1kx−a−k−αk−1!∑j=0m−1j−αk+jk+jΓj+1−α+Hk−1−Hk+m−1Γ1−α−k,By (72) and (73), we see that Dxαax−a−kDxαaGLx−a−k and DxαaCx−a−k are not equal for Reα>0.Remark 3.
Whenβ≠−1,−2,…, then dk−1/dxk−1Dxαax−aβ=dk−1/dxk−1DxαaRLx−aβ. However, when β=−1,−2,…, the relationship is broken. For example, by (64) and (30), we have(74)Dxαax−a−1=x−a−α−1lnx−aΓ−α−γ+ψ−αx−a−α−1Γ−α,dk−1dxk−1Dxαax−a−1=x−a−α−k1−α−kk−1Γ−αlnx−a+∑j=1k−11j−α−k−γ+ψ−αx−a−α−k1−α−kk−1Γ−α=x−a−α−kΓ1−α−klnx−a+∑j=1k−11j−α−k−γ−ψ−α=x−a−α−klnx−a−γ−ψ1−α−kΓ1−α−k.−1k−1k−1!Dxαax−a−k=Dxαax−a−1k−1=dk−1dxk−1DxαaRLx−a−1+Hk−1x−a−α−kΓ1−α−k,for α≠1,2,….
Obviously,dk−1/dxk−1Dxαax−a−1=Dxα+k−1ax−a−1≠DxαaRLx−a−1k−1.Theorem 4.
Forp,k=1,2,…, there is(75)Jxαax−a−klnpx−a=p!∑j=0p−1jCjα,klnp−jx−ap−j!+−1k−1lnp+1x−aΓα+1−kk−1!p+1x−aα−kfor Reα>0, where(76)Cjα,k=∑l=0,l≠k−1∞−1lΓα−ll!l−k+1j+1,and(77)Dxαax−a−klnpx−a=x−a−α−kp!∑j=0p−1jRjα,klnp−jx−ap−j!+−1k−1Qα,kx−a−α−klnp+1x−ap+1k−1!,with(78)Rjα,k=Cjm−α,k∑l=0mp−jlam,l1−α−kln−lx−a,Qα,k=1Γm−α+1−k∑l=0mp+1lam,l1−α−kln−lx−a.Proof.
By (21), we have(79)ℕ.limε⟶0+0∫a+εxx−sα−1s−a−klnps−ads=ℕ.limε⟶0+0∫a+εxx−aα−1∑l=0∞1−αll!s−al−kx−allnps−ads=ℕ.limε⟶0+0∑l=0∞1−αlx−aα−l−1l!∫a+εxs−al−klnps−ads.
From (36), (37), and (79), we get(80)Jxαax−a−klnpx−a=1Γαℕ.limε⟶0+0∫a+εxx−sα−1s−a−klnps−ads=p!x−aα−k∑l=0,l≠k−1∞1−αlΓαl!∑j=0p−1jlnp−jx−al−k+1j+1p−j!+1−αk−1x−aα−klnp+1x−aΓαk−1!p+1.
By1−αl/Γα=−1l/Γα−l and exchanging the sum order, we see that (75) holds.
Using (16) and (29) yields that(81)Dxαax−a−klnpx−a=dmdxmJxm−αax−a−klnpx−a=dmdxmp!∑j=0p−1jCjm−α,klnp−jx−ap−j!+−1k−1lnp+1x−aΓm−α+1−kk−1!p+1x−am−α−k=x−a−α−kp!∑j=0p−1jCjm−α,kp−j!∑l=0mp−jlam,l1−α−klnp−j−lx−a+−1k−1x−a−α−kΓm−α+1−kk−1!p+1∑l=0mp+1lam,l1−α−klnp+1−lx−a.
Hence, we conclude that (77) holds.Remark 4.
(1)
Forn=1,2,…, noting that(82)limα⟶n+0m−α=1,limα⟶n+0Cjm−α,k=11−kj+1,k=2,3,…,0,k=1,limα⟶n+0Γm−α+1−k=∞,k=2,3,…,1,k=1,and employing (38), (39), and (77), we have(83)limα⟶n+0Dxαax−a−1lnpx−a=dndxnx−a−1lnpx−a,limα⟶n+0Dxαas−a−klnps−a=dndxnx−a−klnpx−a.limα⟶n+0Dxαas−a−klnps−a is just the usual derivative dn/dxnx−a−klnpx−a,k=1,2,…, which implies that the definitions in this paper are reasonable.(2)
From (7), (8), (49), (51), (64), and (65), we can still get(84)JxαaDxαafx=DxαaJxαafx=fx,wherefx=x−a−k,x−a−klnpx−a, k,p=1,2,…. For example,(85)DxαaJxαax−a−k=Dxαa−1kx−aα−kγ+ψ1+α−k−Hk−1k−1!Γ1+α−k−−1kx−aα−klnx−ak−1!Γ1+α−k=0−−1kx−a−kB0,0−α,α−k+1lnx−a+B0,1−α,α−k+1k−1!Γ1+α−kΓ−α=x−a−k,whereB0,0−α,α−k+1=0 and −1kB0,1−α,α−k+1/k−1!Γ1+α−kΓ−α=−1 are used. Again, this shows that the definitions in this paper are reasonable.


## 4. Fractional Derivatives of Some Elementary Functions

In this section, we shall study the fractional order derivatives of some elementary functions. First of all, we give the following general theorem.Theorem 5.
Iffx is the analytic function, i.e.,(86)fx=∑l=0∞flax−all!,then the following formulas are true,(87)Dxαax−a−nfx=∑l=0n−1−1n−lflaγ+ψ1+l−α−n−Hn−l−1−lnx−al!n−l−1!Γ1−α−n+lx−aα+n−l+∑l=n∞flal−n!x−al−n−αl!Γl−n+1−α,(88)Dxαax−a−nlnpx−afx=1x−aα+n∑l=0n−1flax−all!p!∑j=0p−1jRjα,n−llnp−jx−ap−j!+−1k−1Qα,n−llnp+1x−ap+1n−l−1!+1Γ−αx−aα+n∑l=n∞flax−all!∑j=0ppjB0,j−α,l−n+1lnp−jx−a.In particular,(89)Dxαax−a−nebx−a=∑l=n∞bll−n!x−al−n−αl!Γl−n+1−α−∑l=0n−1−1n+lbllnx−a−γ−ψ1+l−α−n+Hn−l−1l!n−l−1!Γ1+l−α−nx−aα+n−l,Dxαax−a−nsinbx−a=∑l=n/2∞−1lb2l+1x−a2l+1−n−α2l+2−nnΓ2l+2−n−α+∑l=0n/2−1−1n+lb2l+1lnx−a−γ−ψ2−α−n+2l+Hn−2l−22l+1!n−2l−2!Γ2−α−n+2lx−aα+n−2l−1,Dxαax−a−ncosbx−a=∑l=n+1/2∞−1lb2lΓ2l−n+1x−a2l−n−α2l!Γ2l−n+1−α−∑l=0n+1/2−1−1n+lb2llnx−a−γ−ψ1−α−n+2l+Hn−2l−12l!n−2l−1!Γ1−α−n+2lx−aα+n−2l,and(90)DxαaRLcotx−a=lnx−a−γ−ψ−αΓ−αx−aα+1+∑k=1∞−4kB2kx−a2k−1−α2kΓ2n−α,DxαaRLcscx−a=lnx−a−γ−ψ−αΓ−αx−aα+1−2∑k=1∞−1k22k−1−1B2kx−a2k−1−α2kΓ2k−α,DxαaRLcothx−a=lnx−a−γ−ψ−αΓ−αx−aα+1+∑k=1∞4kB2kx−a2k−1−α2nΓ2k−α,DxαaRLcschx−a=lnx−a−γ−ψ−αΓ−αx−aα+1−∑k=1∞22k−1−1B2kx−a2k−1−αkΓ2k−α,where we have used the following equations(91)cotx−a=1x−a+∑k=1∞−4kB2kx−a2k−12k!,cscx−a=1x−a−2∑k=1∞−1k22k−1−1B2kx−a2k−12k!,cothx−a=1x−a+∑k=1∞4kB2kx−a2k−12k!,cschx−a=1x−a−2∑k=1∞22k−1−1B2kx−a2k−12k!.For the fractional derivative, there is the following Leibniz derivative formula(92)Dαafxgx=∑k=0∞αkDα−kafxDkgxx∈a,b,where αk=Γα+1/k!Γα+1−k,Dαa is any of Dxαa,DxαaCL or DxαaGL, Dxα−kafx and Dkgx are continuous in a,b. Since Dαa1=x−a−α/Γ1−α=Γαx−a−αsinαπ/π, by (90), we have(93)Dαagx=Dαa1⋅gx=Γα+1∑k=0∞xk−αsinα−kπk!α−kπDkgx.Using (92), we obtain that Dxαax−a−n=DxαaGLx−a−n=DxαaCx−a−n, which is not consistent with Corollary 3 so that (92) does not hold for gx=x−a−n,x−a−nlnpx−a. Then, we cannot calculate Dxαax−a−nDxαaGLx−a−n,DxαaCx−a−n, by (92). Therefore, Dxαacotx−a (Dxαacscx−a, Dxαacschx−a, and Dxαacothx−a) cannot be completed by (92).Ifgx is the analytic function, the Leibniz derivative formula (92) is still valid for the Riemann–Liouville fractional integral and derivatives [5]. For example, if fx=x−a−n,gx is an analytic function in Theorem 5, the following formula still holds:(94)Dxαax−a−ngx=∑k=0∞αkDα−kax−a−nDkgx=−1nΓ1+αn−1!x−aα+n∑k=0∞γ+ψ1−α+k−n−Hn−1−lnx−ax−akgkxk!Γ1+α−kΓ1−α+k−n=−1n−1Γ1+αsinαππn−1!x−aα+n×∑k=0∞−1k1−α+k−nn−1γ+ψ1−α+k−n−Hn−1−lnx−ax−akgkxk!.By (94), we have(95)DxαaRLx−a−nebx−a=−1nΓα+1ebx−an−1!x−aα+n∑k=0∞bkγ+ψ1+k−α−n−Hn−1−lnx−ax−akk!Γ1+k−α−nΓα−k+1,DxαaRLx−a−nsinbx−a=−1nΓα+1n−1!x−aα+n∑k=0∞bkγ+ψ1+k−α−n−Hn−1−lnx−asinbx+kπ/2x−akk!Γ1+k−α−nΓα−k+1,DxαaRLx−a−ncosbx−a=−1nΓα+1n−1!x−aα+n∑k=0∞bkγ+ψ1+k−α−n−Hn−1−lnx−acosbx+kπ/2x−akk!Γ1+k−α−nΓα−k+1.or(96)DxαaRLx−a−nebx−a=−1n−1Γ1+αsinαπebx−aπn−1!x−aα+n×∑k=0∞−1k1−α+k−nn−1γ+ψ1−α+k−n−Hn−1−lnx−abkx−akk!,DxαaRLx−a−nsinbx−a=−1n−1Γ1+αsinαππn−1!x−aα+n×∑k=0∞−bx−ak1−α+k−nn−1γ+ψ1−α+k−n−Hn−1−lnx−asinbx+kπ/2k!,DxαaRLx−a−ncosbx−a=−1n−1Γ1+αsinαππn−1!x−aα+n×∑k=0∞−bx−ak1−α+k−nn−1γ+ψ1−α+k−n−Hn−1−lnx−acosbx+kπ/2k!.The numerical calculation shows that (95) and (96) are exactly same as (89), but (89) is the calculated fastest and (95) is the calculated slowest.Forcotx−a,cscx−a,cothx−a,cschx−a, we have by (94)(97)DxαaRLgx=DxαaRLx−a−1x−agx=Γ1+αsinαππx−aα+1∑k=0∞−1kγ+ψ−α+k−lnx−ax−akk!x−agxk=Γ1+αsinαππx−aα∑k=0∞−1kx−akgkxk!α−k,where gx is one of the cotx−a,cscx−a,cothx−a,cschx−a,. The numerical calculation shows that (97) is exactly same as (90), but (90) is calculated faster.Remark 5.
Based on the derivative formula of elementary function, (94) is more convenient to calculate Dxαax−a−ngx, but the calculation efficiency is not as fast as (87).


## 5. Applications to the Fractional Differential and Integral Equations

At last, as an application of the definitions of the fractional derivative and integrals given by the neutrix limit, this section is devoted to study the solution of fractional differential and integral equations.Theorem 6 (see [1]).
IfReα,Reμ>0, then the solution of fractional integral equation.(98)Nt+cαJα0Nt−N0tμ−1=0,t>0,is given by(99)Nt=N0Γμtμ−1Eα,μ−cαtα,where Eα,βt is the generalized Mittag-Leffer function and it is defined as(100)Eα,βt=∑k=0∞tkΓkα+β.Through the analytical continuation, whenμ<0,μ≠0,−1,−2,…, the above conclusion is also established. For μ=0,−1,−2,…, (99) is not a solution of the fractional integral equation (98), but we have the following result.Theorem 7.
IfReα>0,n=1,2,…, then the solution of the following fractional integral equation(101)Nx+cαJxαaNx−k−N0x−an=0,x>a≥0,is given by(102)Nx=N0x−an−N0−1ncαx−aαx−ann−1!Eα,1+α−nψ−cαx−aα+−1nN0lnx−a−γ+Hn−1cαx−aα−nn−1!Eα,1+α−n−cαx−aα,where(103)Eα,βψt=∑k=0∞ψkα+βtkΓkα+β.Proof.
Applying the operator−cαkJxkαa to both sides of (101), we have by (64)(104)−cαkJxkαaNt−−cαk+1Jxk+1αaNx=N0−cαkJxkαax−a−n=−1nN0n−1!x−anγ+ψ1+kα−n−Hn−1−lnx−a−cαx−aαkΓ1+kα−n,k=1,2,…,N0x−a−n,k=0.
Summing up the above expression with respect tok from 0 to ∞ gives rise to(105)Nx=N0x−an+−1nN0n−1!x−an∑k=1∞γ+ψ1+kα−n−Hn−1−lnx−a−cαx−aαkΓ1+kα−n=N0x−an−N0−1ncαx−aαx−ann−1!Eα,1+α−nψ−cαx−aα−−1nN0γ−lnx−a−Hn−1cαx−aαEα,1+α−nψ−cαx−aαn−1!x−an.
Hence, we deduce that (102).Theorem 8.
IfReα>0,n=1,2,…, then the solution of the following fractional differential equation(106)DxαaNx+cαNx−N0x−an=0,x>a≥0,Dxα−kaRLNa+=bk,k=1,2,…,m,is given by(107)Nx=N0−1nγ−Hn−1−lnx−aEα,1+α−n−cαx−aαn−1!x−an−α+N0−1nEα,1+α−nψ−cαx−aαn−1!x−an−α+∑j=1mbjx−aα−jEα,1+α−j−cαx−aα.Proof.
Applying the operatorJxαa to both sides of (106), using (64) and(108)JxαaDxαaNx=Nx−∑j=1mDxα−jaNa+x−aα−jΓα+1−j=Nx−∑j=1mbjx−aα−jΓα+1−j,we get the following integral equation.(109)Nx−gR,nx,α=−cαJxαaNx,x>a≥0,where(110)gR,nx,α=N0−1nx−aα−nγ+ψ1+α−n−Hn−1n−1!Γ1+α−n−N0−1nx−aα−nlnx−an−1!Γ1+α−n+∑j=1mbjx−aα−jΓα+1−j.
Applying the operator−cαkJxkαax−a−k to both sides of (109), and then using (7), (58), and Lemma 3 yield that(111)−cαkJxkαaNx−−cαk+1Jxk+1αaNx=Gn,kx,α,where(112)Gn,kx,α=−cαkJtkα0gnx,a=N0−1nγ+ψ1+α−n−Hn−1−cαkn−1!Γ1+α−nΓα−n+1Γα−n+1+kαx−aα−n+kα+∑j=1mbj−cαkΓα+1−jΓα−j+1Γα−j+1+kαx−aα−j+kα−N0−1n−cαkn−1!Γ1+α−nx−akα+α−nΓα−n+1Γα−n+1+kαlnx−a−N0−1n−cαkn−1!Γ1+α−nx−akα+α−nΓα−n+1ψ1+α−n−ψ1+kα+α−nΓα−n+1+kα=N0−1nγ+ψ1+kα+α−n−Hn−1−cαkx−aα−n+kαn−1!Γα−n+1+kα+∑j=1mbj−cαkx−aα−j+kαΓα−j+1+kα−N0−1n−cαkx−aα−n+kαlnx−an−1!Γα−n+1+kα,k=1,2,….
Summing up the above expression with respect tok from 0 to ∞ yields that(113)Nx=N0−1nx−aα−nγ+ψ1+α−n−Hn−1n−1!Γ1+α−n−N0−1nx−aα−nlnx−an−1!Γ1+α−n+∑j=1mbjx−aα−jΓα+1−j+∑k=1∞N0−1nγ+ψ1+kα+α−n−Hn−1−cαkx−aα−n+kαn−1!Γkα+α−n+1+∑k=1∞∑j=1mbj−cαkx−aα−j+kαΓα−j+1+kα−∑k=1∞N0−1n−cαkx−aα−n+kαlnx−an−1!Γα−n+1+kα=N0−1nγ−Hn−1−lnx−aEα,1+α−n−cαx−aαn−1!x−an−α+N0−1nEα,α+1−nψ−cαx−aαn−1!x−an−α+∑j=1mbjx−aα−jEα,1+α−j−cαx−aα.Theorem 9.
IfReα>0,n=1,2,…, then the solution of the following fractional differential equation(114)DxαaCNt+cαNx−N0x−an=0,x>a≥0,Nm−ka+0=bk,k=1,2,…,m,is given by(115)Nt=N0−1nγ−Hn−1−lnx−aEα,1+α−n−cαx−aαn−1!x−an−α+N0−1nEα,1−nψ−cαx−aαn−1!x−an−α+∑j=1mbjx−am−jEα,1+m−j−cαx−aα.Proof.
Applying the operatorJxαaRL to both sides of (114), using (64) imply that(116)JxαaDxαaCNx=JxαaJxm−αaNmx=JxmaNmx=1Γmℕ.limε⟶0+0∫a+εxx−sm−1Nmsds=−ℕ.limε⟶0+0∑j=1mx−a−εm−jΓm+1−jNm−ja+ε=−∑j=1mx−am−jΓm+1−jNm−ja+0=−∑j=1mbjx−am−jm−j!.thus, we get the following integral equation.(117)Nx+cαJxαaNx=gC,nx,α,x>a≥0,where(118)gC,nx,α=N0−1nx−aα−nγ+ψ1+α−n−Hn−1n−1!Γ1+α−n−N0−1nx−aα−nlnx−an−1!Γ1+α−n+∑j=1mbjx−am−jm−j!,then, according to Theorem 8, we get (115).Theorem 10.
IfReα>0,n=1,2,…, then the solution of the following fractional differential equation(119)DxαaNx+cαNx−N0lnpx−ax−an=0,x>a≥0,Dxα−kaRLNa+=bk,k=1,2,…,m,is given by(120)Nt=p!x−an−α∑j=0p−1jCjα,nlnp−jx−ap−j!+−1n−1lnp+1x−aΓα+1−nn−1!p+1x−an−α+p!Γα+1−nx−an−α∑j=0p−1jCjα,np−j!∑l=0p−jp−jlEα,n,l−cαx−aαlnp−j−lx−a+−1n−1n−1!p+1x−an−α∑j=0p+1p+1jEα,n,j−cαx−aαlnp+1−jx−a+∑j=1mbjx−aα−jEα,1+α−j−cαx−aα,where(121)Eα,n,lt=∑k=1∞Hlψα+1−n,kα+α+1−ntkΓkα+α+1−n.Proof.
Applying the operatorJxαa to both sides of (118), using (75) and (108), we get(122)Nx+cαJxαaNx=gR,n,px,α,x>a≥0,where(123)gR,n,px,α=p!∑j=0p−1jCjα,nx−aα−nlnp−jx−ap−j!+−1n−1x−aα−nlnp+1x−aΓα+1−nn−1!p+1+∑j=1mbjx−aα−jΓα+1−j,for Reα>0.
Exactly according to the discussion of Theorems8 and 9, we obtain(124)Nt=∑k=0∞−cαkJxkαagR,n,px,α=p!∑j=0p−1jCjα,nx−aα−nlnp−jx−ap−j!+−1n−1x−aα−nlnp+1x−aΓα+1−nn−1!p+1+∑j=1mbjx−aα−jΓα+1−j+p!∑j=0p−1jCjα,nx−aα−np−j!∑l=0p−jp−jllnp−j−lx−a∑k=1∞−cαx−aαkB0,lkα,α−n+1Γkα+−1n−1x−aα−nΓα+1−nn−1!p+1∑j=0p+1p+1jlnp+1−jx−a∑k=1∞−cαx−aαkB0,jkα,α−n+1Γkα+∑j=1mbjx−aα−j∑k=1∞−cαx−aαkΓα−j+1+kα.
By Lemma3 and (124) we obtain (120).
Let us substitute the solutions of (102), (107), (115), and (120) into their respective equations (101), (106), (114), and (119), and make their error curves (see (1)–(4) in Figure 1). Without loss of generality.
Here, letbk=0,k=1,2,…,m. In (1)–(4) of Figure 1, the figure above is the solution curve and the figure below is the error curve. The numerical results show that the solutions of the fractional differential and integral equations obtained above satisfy their respective equations. It can be seen from (2) and (3) of Figure 1 that although the expressions of (107) and (115) are different, their graphs are almost the same. This shows that equations (106) and (114) do not describe a fundamentally different problem.
Finally, we verify the correctness of the above results by numerical solutions of fractional differential equations. According to the method in literature [5], for (106) and (119), the following iterative algorithms are, respectively, as follows:(125)Nk=hαN0xk−an−cαNk−1−∑j=0kwjαNk−j,and(126)Nk=hαN0lnpa0+hk−axk−an−cαNk−1−∑j=0kwjαNk−j,for k=m,m+1,…, where(127)xk=a0+hk,Nk=Nxk,wjα=−1jαj,a0>a.
Without loss of generality, letN0=1,Nk=Nxk,k=1,2,…,3m, where Nx is the analytic solution. The analytical solutions of (107) or (106) and the numerical solution obtained by (125) are shown in (1) in Figure 2. The analytical solutions of (120) or (119) and the numerical solution obtained by (126) are shown in (2) in Figure 2.
As can be seen from Figure2, the analytical solution obtained in this paper is basically consistent with the numerical solution.Figure 1
The solution curves and the error curve of the equation.Figure 2
The analytical solution and the numerical solution of (106) and 119


## 6. Conclusions

As defined in this paper, the following function is consistent with traditional fractional integrals and derivatives:(128)x−aβfx,x−aβlnpx−afx,where fx is the analytic function, β≠−1,−2,…,p=1,2,…,fa≠0. The fractional integrals and derivatives of the following function can be obtained:(129)x−a−kfx,x−a−klnpx−afx,where fx is the analytic function, k=1,2,…,p=1,2,…,fa≠0. The following Leibniz derivative formula of the Riemann–Liouville fractional derivative still holds:(130)Dxαax−a−ngx=∑k=0∞αkDxα−kax−a−nDkgx=−1n−1Γ1+αsinαππn−1!x−aα+n×∑k=0∞−1k1−α+k−nn−1γ+ψ1−α+k−n−Hn−1−lnx−ax−akgkxk!,and the properties(131)Dxα+lax−a−kfx=dldxlDxαax−a−kfx,limα⟶nDxα+lax−a−kfx=dndxnx−a−kfx,limα⟶n+0Dxα+lax−a−klnpx−afx=dndxnx−a−klnpx−afx,Dxαax−aα−k=0,k=1,2,…,are true.Therefore, we present the definitions of the fractional derivative and integrals given by the neutrix limit in our paper. Based on it, the solutions to some of the differential and integral equations given in the previous section cannot be expressed by traditional methods, and the solutions to these differential and integral equations are also expressed by the definitions in this paper.



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