1 Introduction

Fractional differential equations have been an active research area during the past few decades and they occur in many applications of physics and engineering. The Mittag-Leffler function appears as the solution of fractional order differential equations and fractional order integral equations. Some applications of the Mittag-Leffler function are as follows: studies of the kinetic equation, the telegraph equation [1], random walks, Levy flights, superdiffuse transport, and complex systems. Besides this, the Mittag-Leffler function appears in the solution of certain boundary value problems involving fractional integro-differential equations of Volterra type [2]. It has applications in applied problems, such as fluid flow, rheology, diffusive transport akin to diffusion, electric networks, probability, and statistical distribution theory. Various properties of the Mittag-Leffler functions were presented and surveyed in [3]. Furthermore, a different variant of the Mittag-Leffler function has been investigated in [4].

Let us start with giving the historical background of the Mittag-Leffler functions. The function E α (z),

E α (z)= k = 0 z k Γ ( α k + 1 ) ,
(1)

was defined and studied by Mittag-Leffler in the year 1903 in [57]. It is a direct generalization of the exponential series, since, for α=1, we have the exponential function. The function defined by

E α , β (z)= k = 0 z k Γ ( α k + β )
(2)

gives a generalization of equation (1). This generalization was studied by Wiman in 1905 [8, 9], Agarwal in 1953, and Humbert and Agarwal [10, 11] in 1953. Afterward, Prabhakar [12] introduced the generalized Mittag-Leffler function by

E β , γ δ (z):= n = 0 ( δ ) n Γ ( β n + γ ) z n n ! ,
(3)

where β,γ,δC with (β)>0. For δ=1, it reduces to the Mittag-Leffler function given in equation (2). Some of the properties of the generalized Mittag-Leffler function such as the Mellin transform, the inverse Mellin transform, and differentiation were given in [13]. On the other hand, monotony of the Mittag-Leffler function was given in [14].

In this paper, we extend the Mittag-Leffler function E α , β γ (z) in the following way. Since

E α , β γ (z)= k = 0 ( γ ) k Γ ( α k + β ) ( c ) k ( c ) k z k k ! ,

using the fact that

( γ ) k ( c ) k = B ( γ + k , c γ ) B ( γ , c γ ) ,

we extend the Mittag-Leffler function as follows:

E α , β ( γ ; c ) ( z ; p ) : = k = 0 B p ( γ + k , c γ ) B ( γ , c γ ) ( c ) k Γ ( α k + β ) z k k ! ( p 0 ; Re ( c ) > Re ( γ ) > 0 ) ,
(4)

where for B p (x,y) we have

B p (x,y)= 0 1 t x 1 ( 1 t ) y 1 e p t ( 1 t ) dt ( Re ( p ) > 0 , Re ( x ) > 0 , Re ( y ) > 0 ) ,
(5)

the extended Euler’s Beta function defined in [15] (see also [16]).

The organization of the paper is as follows: In Section 2, we give an integral representations of the extended Mittag-Leffler function in terms of Prabhakar’s Mittag-Leffler function and in terms of known elementary functions. The Mellin transform of the extended Mittag-Leffler function is obtained by means of the generalized Wright hypergeometric function [17]. In Section 3, we obtain fractional derivative representations of the extended Mittag-Leffler function and give some derivative formulas. In Section 4, we obtain the relationship between the extended Mittag-Leffler function and simple Laguerre polynomials and Whittaker’s functions.

2 Some properties of the extended Mittag-Leffler function

We begin with the following theorem, which gives the integral representation of the extended Mittag-Leffler function.

Theorem 1 (Integral representation)

For the extended Mittag-Leffler function, we have

E α , β ( γ ; c ) (z;p)= 1 B ( γ , c γ ) 0 1 t γ 1 ( 1 t ) c γ 1 e p t ( 1 t ) E α , β ( c ) (tz)dt,
(6)

where p0, Re(c)>Re(γ)>0, Re(α)>0, Re(β)>0.

Proof Using equation (5) in equation (4), we get

E α , β ( γ ; c ) (z;p)= k = 0 { 0 1 t γ + k 1 ( 1 t ) c γ 1 e p t ( 1 t ) d t } ( c ) k B ( γ , c γ ) z k Γ ( α k + β ) k ! .
(7)

Interchanging the order of summation and integration in equation (7), which is guaranteed under the assumptions given in the statement of the theorem, we get

E α , β ( γ ; c ) ( z ; p ) = 0 1 t γ 1 ( 1 t ) c γ 1 e p t ( 1 t ) k = 0 ( c ) k B ( γ , c γ ) ( t z ) k Γ ( α k + β ) k ! d t .
(8)

Using equation (3) in equation (8), we get the desired result. □

Corollary 2 Note that, taking t= u 1 + u in Theorem  1, we get

E α , β ( γ ; c ) ( z ; p ) = 1 B ( γ , c γ ) 0 u γ 1 ( u + 1 ) c e p ( 1 + u ) 2 u E α , β ( c ) ( u z 1 + u ) d u .
(9)

Corollary 3 Taking t= sin 2 θ in the Theorem  1, we get the following integral representation:

E α , β ( γ ; c ) ( z ; p ) = 1 B ( γ , c γ ) [ 2 0 π 2 sin 2 γ 1 θ cos 2 c 2 γ 1 θ e p sin 2 θ cos 2 θ ] × E α , β ( c ) ( z sin 2 θ ) d θ .
(10)

Now, using the definition of Prabhakar’s Mittag-Leffler’s function, Bayram and Kurulay obtained the recurrence formula [13]

E α , β ( c ) (tz)=β E α , β + 1 ( c ) (tz)+αz d d z E α , β + 1 ( c ) (tz).

Inserting the above recurrence relation into equation (6), we get the following recurrence relation for the extended Mittag-Leffler’s function.

Corollary 4 (Recurrence relation)

For the extended Mittag-Leffler function, we get

E α , β ( γ ; c ) (z;p)=β E α , β + 1 ( γ ; c ) (z;p)+αz d d z E α , β + 1 ( γ ; c ) (z;p),

where p0, Re(c)>Re(γ)>0, Re(α)>0, Re(β)>0.

In the next theorem, we give the Mellin transform of the extended Mittag-Leffler function in terms of the Wright generalized hypergeometric function. Note that the Wright generalized hypergeometric function is defined by [17]

Ψ q p ( z ) = Ψ q p [ ( a 1 , A 1 ) , ( a 2 , A 2 ) , , ( a p , A p ) ( b 1 , B 1 ) , ( b 2 , B 2 ) , , ( b p , B q ) , z ] = k = 0 j = 1 p Γ ( a j + A j k ) j = 1 q Γ ( b j + B j k ) z k k ! ,
(11)

where the coefficients A i (i=1,,p) and B j (j=1,,q) are positive real numbers such that

1+ j = 1 q B j i = 1 p A i 0.

Theorem 5 (Mellin transform)

The Mellin transform of the extended Mittag-Leffler function is given by

M { E α , β ( γ ; c ) ( z ; p ) ; s } = Γ ( s ) Γ ( c + s γ ) Γ ( γ ) Γ ( c γ ) 2 Ψ 2 [ ( c , 1 ) , ( β , γ ) , ( γ + s , 1 ) ( c + 2 s , 1 ) , z ] ( p 0 , Re ( c ) > Re ( γ ) > 0 , Re ( α ) > 0 , Re ( s ) > 0 , Re ( β ) > 0 ) ,
(12)

where Ψ 2 2 is the Wright generalized hypergeometric function.

Proof Taking the Mellin transform of the extended Mittag-Leffler function, we have

M { E α , β ( γ ; c ) ( z ; p ) ; s } = 0 p s 1 E α , β ( γ ; c ) (z;p)dp.
(13)

Using equation (6) in equation (13), we get

M { E α , β ( γ ; c ) ( z ; p ) ; s } = 1 B ( γ , c γ ) 0 p s 1 [ 0 1 t γ 1 ( 1 t ) c γ 1 e p t ( 1 t ) ] E α , β ( c ) ( t z ) d t d p .
(14)

Interchanging the order of integrals in equation (14), which is valid because of the conditions in the statement of the Theorem 5, we get

M { E α , β ( γ ; c ) ( z ; p ) ; s } = 1 B ( γ , c γ ) 0 1 [ t γ 1 ( 1 t ) c γ 1 E α , β ( c ) ( t z ) ] 0 p s 1 e p t ( 1 t ) d p d t .
(15)

Now taking u= p t ( 1 t ) in equation (15) and using the fact that Γ(s)= 0 u s 1 e u du, we get

M { E α , β ( γ ; c ) ( z ; p ) ; s } = Γ ( s ) B ( γ , c γ ) 0 1 t γ + s 1 ( 1 t ) c + s γ 1 E α , β ( c ) ( t z ) d t .
(16)

Using the definition of Prabhakar’s generalized Mittag-Leffler function E α , β ( c ) (tz) in equation (16), we get

M { E α , β ( γ ; c ) ( z ; p ) ; s } = Γ ( s ) B ( γ , c γ ) 0 1 t γ + s 1 ( 1 t ) c + s γ 1 k = 0 ( c ) k ( t z ) k Γ ( α k + β ) k ! d t .
(17)

Interchanging the order of summation and integration, which is valid for Re(c)>Re(γ)>0, Re(s)>0, Re(cγ+s)>0, Re(α)>0, Re(β)>0, we get

M { E α , β ( γ ; c ) ( z ; p ) ; s } = Γ ( s ) B ( γ , c γ ) k = 0 ( c ) k z k Γ ( α k + β ) k ! 0 1 t γ + k + s 1 ( 1 t ) c + s γ 1 d t .
(18)

Using the Beta function in equation (18), we have

M { E α , β ( γ ; c ) ( z ; p ) ; s } = Γ ( s ) Γ ( c + s γ ) B ( γ , c γ ) k = 0 ( c ) k z k Γ ( α k + β ) k ! Γ ( γ + k + s ) Γ ( γ + s ) Γ ( γ + s ) Γ ( c + k + 2 s ) .
(19)

Considering that ( c ) k = Γ ( c + k ) Γ ( c ) , B(γ,cγ)= Γ ( γ ) Γ ( c γ ) Γ ( c ) , and inserting equation (11) into equation (19), we get the result

M { E α , β ( γ ; c ) ( z ; p ) ; s } = Γ ( s ) Γ ( c + s γ ) B ( γ , c γ ) Γ ( c ) k = 0 z k Γ ( α k + β ) k ! Γ ( γ + k + s ) Γ ( c + k ) Γ ( c + k + 2 s ) = Γ ( s ) Γ ( c + s γ ) Γ ( γ ) Γ ( c γ ) 2 Ψ 2 [ ( c , 1 ) , ( β , α ) , ( γ + s , 1 ) ( c + 2 s , 1 ) , z ] .

 □

Corollary 6 Taking s=1 in Theorem  5, we get

0 E α , β ( γ ; c ) (z;p)dp= Γ ( c + 1 γ ) Γ ( γ ) Γ ( c γ ) 2 Ψ 2 [ ( c , 1 ) , ( β , α ) , ( γ + 1 , 1 ) ( c + 2 , 1 ) , z ] .

Corollary 7 Taking the inverse Mellin transform on both sides of equation (12), we get the elegant complex integral representation

E α , β ( γ ; c ) (z;p)= 1 2 π i Γ ( γ ) Γ ( c γ ) ν i ν + i Γ(s)Γ ( c + s γ ) 2 Ψ 2 [ ( c , 1 ) , ( β , α ) , ( γ + s , 1 ) ( c + 2 s , 1 ) , z ] p s ds,

where ν>0.

3 Derivative properties of the extended Mittag-Leffler function

The classical Riemann-Liouville fractional derivative of order μ is usually defined by

D z μ { f ( z ) } = 1 Γ ( μ ) 0 z f(t) ( z t ) μ 1 dt,Re(μ)<0,

where the integration path is a line from 0 to z in the complex t-plane. For the case m1<Re(μ)<m (m=1,2,3,), it is defined by

D z μ { f ( z ) } = d m d z m D z μ m { f ( z ) } = d m d z m { 1 Γ ( μ + m ) 0 z f ( t ) ( z t ) μ + m 1 d t } .

The extended Riemann-Liouville fractional derivative operator was defined by Özarslan and Özergin as follows.

Definition 8 ([18])

The extended Riemann-Liouville fractional derivative is defined as

D z μ , p { f ( z ) } = 1 Γ ( μ ) 0 z f(t) ( z t ) μ 1 exp ( p z 2 t ( z t ) ) dt,Re(μ)<0,Re(p)>0
(20)

and for m1<Re(μ)<m (m=1,2,3,)

D z μ , p { f ( z ) } = d m d z m D z μ m { f ( z ) } = d m d z m { 1 Γ ( μ + m ) 0 z f ( t ) ( z t ) μ + m 1 exp ( p z 2 t ( z t ) ) d t } ,

where the path of integration is a line from 0 to z in the complex t-plane. For the case p=0, we obtain the classical Riemann-Liouville fractional derivative operator.

We begin by the following theorem.

Theorem 9 Let p0, Re(μ)>Re(λ)>0, Re(α)>0, Re(β)>0. Then

D z λ μ , p { z λ 1 E α , β ( c ) ( z ) } = z μ 1 B ( λ , c λ ) Γ ( μ λ ) E α , β ( λ ; μ ) (z;p).

Proof Replacing μ by λμ in the definition of the extended fractional derivative operator (20), we get

D z λ μ , p { z λ 1 E α , β ( c ) ( z ) } = 1 Γ ( μ λ ) 0 z t λ 1 E α , β ( c ) ( t ) ( z t ) λ + μ 1 exp { p z 2 t ( z t ) } d t = z λ + μ 1 Γ ( μ λ ) 0 z t λ 1 E α , β ( c ) ( t ) ( 1 t z ) λ + μ 1 exp { p z 2 t ( z t ) } d t .
(21)

Taking u= t z in equation (21), we get

D z λ μ , p { z λ 1 E α , β ( c ) ( z ) } = z μ 1 Γ ( μ λ ) 0 1 u λ 1 ( 1 u ) λ + μ 1 exp { p u ( 1 u ) } E α , β ( c ) ( u z ) d u .
(22)

Comparing this result with equation (6), we get

D z λ μ , p { z λ 1 E α , β ( c ) ( z ) } = z μ 1 B ( λ , c λ ) Γ ( μ λ ) E α , β ( λ ; μ ) (z;p).

Whence the result. □

In the following theorem, we give the derivative properties of the extended Mittag-Leffler function.

Theorem 10 For the extended Mittag-Leffler function, we have the following derivative formula:

d n d z n { E α , β ( γ ; c ) ( z ; p ) } = ( c ) n E α , β + n α ( γ + n ; c + n ) (z;p),nN.
(23)

Proof Taking the derivative with respect to z in equation (6), we get

d d z { E α , β ( γ ; c ) ( z ; p ) } =c E α , β + α ( γ + 1 ; c + 1 ) (z;p).
(24)

Again taking the derivative with respect to z in equation (24), we get

d 2 d z 2 { E α , β ( γ ; c ) ( z ; p ) } =c(c+1) E α , β + 2 α ( γ + 2 ; c + 2 ) (z;p).
(25)

Continuing the repetition of this procedure n times, we get the desired result. □

Theorem 11 For the extended Mittag-Leffler function, the following differentiation formula holds:

d n d z n { z β 1 E α , β ( γ ; c ) ( λ z α ; p ) } = z β n 1 E α , β n ( γ ; c ) ( λ z α ; p ) .

Proof In equation (23), replace z by λ z α and multiply z β 1 , then taking the z-derivative n times, we get the result. □

Theorem 12 For the extended Mittag-Leffler function, the following differentiation formula holds:

d n d p n { E α , β ( γ ; c ) ( z ; p ) } = ( 1 ) n Γ ( γ n ) Γ ( c γ n ) Γ ( c ) Γ ( c 2 n ) Γ ( γ ) Γ ( c γ ) E α , β ( γ n ; c 2 n ) (z;p).

Proof Taking the p-derivative n times in equation (6), we get the result. □

4 Relations between the extended Mittag-Leffler function with Laguerre polynomial and Whittaker function

In this section, we give a representation of the extended Mittag-Leffler function in terms of Laguerre polynomials and Whittaker’s function.

Theorem 13 For the extended Mittag-Leffler function, we have

exp ( 2 p ) E α , β ( γ ; c ) ( z ; p ) = 1 B ( γ , c γ ) m , n , k = 0 L m ( p ) L n ( p ) ( c ) k Γ ( α k + β ) k ! z k B ( m + k + γ + 1 , n + c γ + 1 ) ,

where Re(c)>Re(γ)>0, Re(α)>0, Re(β)>0.

Proof We start by recalling the useful identity used in [18]

exp ( p t ( 1 t ) ) =exp(2p) m , n = 0 L n (p) L m (p) t m + 1 ( 1 t ) n + 1 ;0<t<1.
(26)

Using equation (26) in equation (6), we get

E α , β ( γ ; c ) ( z ; p ) = 1 B ( γ , c γ ) 0 1 t γ 1 ( 1 t ) c γ 1 exp ( 2 p ) × m , n = 0 L n ( p ) L m ( p ) t m + 1 ( 1 t ) n + 1 E α , β c ( t z ) d t .
(27)

Now, taking into account the series expansion of Prabhakar’s generalized Mittag-Leffler’s function E α , β c (tz) in equation (27), we have

E α , β ( γ ; c ) ( z ; p ) = exp ( 2 p ) B ( γ , c γ ) 0 1 t γ 1 ( 1 t ) c γ 1 × m , n = 0 L n ( p ) L m ( p ) t m + 1 ( 1 t ) n + 1 k = 0 ( c ) k ( t z ) k Γ ( α k + β ) k ! d t = exp ( 2 p ) B ( γ , c γ ) 0 1 t γ 1 ( 1 t ) c γ 1 × m , n , k = 0 L n ( p ) L m ( p ) ( c ) k Γ ( α k + β ) k ! t m + k + 1 ( 1 t ) n + 1 z k d t .
(28)

Interchanging the order of integration and summation in equation (28), which can be done under the assumptions of the theorem, we have

E α , β ( γ ; c ) ( z ; p ) = exp ( 2 p ) B ( γ , c γ ) m , n , k = 0 L n ( p ) L m ( p ) ( c ) k Γ ( α k + β ) k ! z k × B ( m + k + γ + 1 , n + c γ + 1 ) .
(29)

Multiplying both sides of equation (29) by exp(2p), we get the result. □

In the following theorem, we give the extended Mittag-Leffler function in terms of Whittaker’s function.

Theorem 14 For the extended Mittag-Leffler function we have

exp ( 3 p 2 ) E α , β ( γ ; c ) (z;p)= Γ ( c γ + 1 ) B ( γ , c γ ) m , k = 0 L m ( p ) ( c ) k Γ ( α k + β ) k ! p m + k + γ 1 2 W γ 2 c m 1 2 , m + k + γ 2 (p).

Proof Considering the following equality:

exp ( p t ( 1 t ) ) =exp ( p 1 t ) exp ( p t ) ,

and using the generating function of the Laguerre polynomials, we get

exp ( p t ( 1 t ) ) =exp(p)exp ( p t ) (1t) m = 0 L m (p) t m .
(30)

Taking equation (30) into account in equation (6), we have

E α , β ( γ ; c ) ( z ; p ) = 1 B ( γ , c γ ) 0 1 t γ 1 ( 1 t ) c γ 1 e p t ( 1 t ) E α , β ( c ) ( t z ) d t = 1 B ( γ , c γ ) 0 1 t γ 1 ( 1 t ) c γ 1 exp ( p ) exp ( p t ) × ( 1 t ) m = 0 L m ( p ) t m E α , β ( c ) ( t z ) d t .
(31)

By use of Prabhakar’s generalized Mittag-Leffler function E α , β ( c ) (tz) in equation (31), we get

E α , β ( γ ; c ) ( z ; p ) = exp ( p ) B ( γ , c γ ) 0 1 t γ 1 ( 1 t ) c γ exp ( p t ) m = 0 L m ( p ) t m k = 0 ( c ) k t k z k Γ ( α k + β ) k ! d t .
(32)

Interchanging the order of summation and integration in equation (32), we get

E α , β ( γ ; c ) ( z ; p ) = exp ( p ) B ( γ , c γ ) m , k = 0 L m ( p ) ( c ) k z k Γ ( α k + β ) k ! 0 1 t m + k + γ 1 ( 1 t ) c γ exp ( p t ) d t .
(33)

Finally, using the following integral representation [19]:

0 1 t μ 1 ( 1 t ) ν 1 exp ( p t ) d t = Γ ( ν ) p μ 1 2 exp ( p 2 ) W 1 μ 2 ν 2 , μ 2 ( p ) ( Re ( ν ) > 0 , Re ( p ) > 0 ) ,

in equation (33), we get the result. □