Application of the Efros Theorem to the Function Represented by the Inverse Laplace Transform of S−Μ Exp(−Sν)

S S symmetry

Article Application of the Efros Theorem to the Function Represented by the Inverse Laplace Transform of s−µ exp(−sν)

Alexander Apelblat 1 and Francesco Mainardi 2,*

1 Department of Chemical Engineering, Ben Gurion University of the Negev, Beer Sheva 84105, Israel; [email protected] 2 Dipartimento di Fisica e Astronomia, Università di Bologna, Via Irnerio 46, I-40126 Bologna, Italy * Correspondence: [email protected]

Abstract: Using a special case of the Efros theorem which was derived by Wlodarski, and operational calculus, it was possible to derive many infinite integrals, finite integrals and integral identities for the function represented by the inverse Laplace transform. The integral identities are mainly in terms of convolution integrals with the Mittag–Leffler and Volterra functions. The integrands of determined integrals include elementary functions (power, exponential, logarithmic, trigonometric and hyperbolic functions) and the error functions, the Mittag–Leffler functions and the Volterra functions. Some properties of the inverse Laplace transform of s−µexp(−sν) with µ ≥ 0 and 0 < ν < 1 are presented.

Keywords: efros theorem; inverse laplace transforms; wright functions; Mittag–Leffler functions; volterra functions; modified bessel functions; finite; infinite and convolution integrals

MSC: 26A33; 33C10; 33E12; 34A25; 44A20

Citation: Apelblat, A.; Mainardi, F. 1. Introduction Application of the Efros Theorem to the Function Represented by the Inversions of the Laplace transforms of exponential functions Inverse Laplace Transform of Z c + i∞ −µ (− ν) − 1 1 st s exp s . Symmetry 2021, 13, 354. fν,µ(t) = L {F(s)} = e F(s) ds ; c > 0, 2πi c − i∞ https://doi.org/10.3390/sym13020354 ν (1) e−s F(s) = µ ; 0 < ν < 1 ; µ ≥ 0, Academic Editor: Paolo Emilio Ricci s Received: 31 January 2021 were during the 1945–1970 period in a focus of attention of a number of well-known Accepted: 14 February 2021 mathematicians like Humbert [1], Pollard [2], Wlodarski [3], Mikusinski [4–7], Wintner [8], Published: 22 February 2021 Ragab [9] and Stankoviˇc[10]. In the case µ = 0, Mikusinski was able to obtain the inverse Laplace transform in terms of integral representations [7] Publisher’s Note: MDPI stays neu- tral with regard to jurisdictional clai- Z ∞ − st − sν ms in published maps and institutio- L{ fν,0(t)} = e fν,0(t) dt = e ; 0 < ν < 1 ; t > 0, 0 nal afﬁliations. Z ∞ 1 − ut − uν cos(πν) ν fν,0(t) = e e sin[u sin(πν)] du, (2) π 0 ∞ 2 Z ν πν − u cos( 2 ) ν πν fν,0(t) = e cos[u cos( 2 )] cos(ut) du, π 0 Copyright: c 2021 by the authors. Licensee MDPI, Basel, Switzerland. and also as the ﬁnite trigonometric integral This article is an open access article Z π distributed under the terms and con- ν − ξ fν,0(t) = ξe du ; 0 < ν < 1 ; t > 0 ditions of the Creative Commons At- π(1 − ν)t 0 ( − ) (3) tribution (CC BY) license (https:// 1 sin(νu) ν/ 1 ν sin[(1 − ν)u] ξ = . creativecommons.org/licenses/by/ tν/(1 − ν) sin u sin u 4.0/).

Symmetry 2021, 13, 354. https://doi.org/10.3390/sym13020354 https://www.mdpi.com/journal/symmetry Symmetry 2021, 13, 354 2 of 15

It was established that the functions fν,0(t) can be expressed in terms of exponential and parabolic cylinder functions when ν = 1/2 [9,11] and by help of the Airy functions and their ﬁrst derivatives for ν = 1/3 and ν = 2/3 [9,12]. For ν = 1/4, the solution was deduced by Barkai [13] in 2001 using Mathematica as a sum of three generalized hypergeometric functions, but the numerical result was uncertain, presumably for a bag in the computing program. In 2010–2012 Gorska and Penson [14,15] were able to represent fν,0(t) in terms of Mejer G functions. Earlier, in 1958 Ragab [9] expressed fν,0(t) in terms of MacRobert E functions for ν = 1/n. In 1952 Wlodarski [3] showed that when the generalized product Efros theorem [16] is applied to the Laplace transform of fν,µ(t) which is given in (1), then it is possible to derive the following formula

L{g(t)} = G(s) ; 0 < ν < 1 ; µ ≥ 0, ν Z ∞ −1n G(s ) o − 1 L sµ = g(u)L {F(s)} du = 0 (4) Z ∞ n − usν o Z ∞ − 1 e g(u)L sµ du = g(u) fν,µ(u) du. 0 0 It was established in our recent paper [17] that for µ = 0 and µ = 1 − ν this func- tional expression can be written in terms of speciﬁc Wright functions Wν,µ(−t), sometimes referred to as the Mainardi functions

Fν(t) = W− ν,0(− t) ; 0 < ν < 1, (5) Mν(t) = W− ν,1 − ν(− t),

in the following way

ν Z ∞ −1n G(s ) o u du νL 1 − ν = g(u) Fν tν u ; 0 < ν < 1, s 0 ν Z ∞ (6) ν −1n G(s ) o u t L 1 − ν = g(u) Mν tν du. s 0

This follows from the fact that the functions Fν(t) and Mν(t) satisfy

ν 1 λ 1 λ e− λ s L Fν = L Mν = , λν tν tν tν s1 − ν (7) 0 < ν < 1 ; λ > 0.

It was also illustrated by us in [17], that in many cases, by using standard tables of the Laplace transforms [18–21], the left hand side Laplace transforms in (6) can be inverted and numerous infinite integrals, finite integrals and integral identities for the functions Fν(t) and Mν(t) can be derived. Let us recall that the Wright functions [22,23], considered initially as a some kind generalization of the Bessel functions, are defined as an entire functions of the argument z ∈ C and parameters λ > −1 and µ ∈ C by

∞ zk W (z) = ; λ > −1, µ ∈ . (8) λ,µ ∑ ( + ) C k=0 k! Γ λk µ

It is usual to distinguish them in two kinds, the ﬁrst kind with λ ≥ 0 and the second kind with λ = −ν and ν ∈ (0, 1), see, e.g., [11]. Symmetry 2021, 13, 354 3 of 15

Restricting our attention to positive argument t > 0, the Mainardi functions turn out to be particular Wright functions of the second kind expressed by the following series

∞ (− t)k 1 ∞ (−1)k + 1tk F (t) = = Γ(ν k + 1) sin(π ν k) , ν ∑ (− ) ∑ k = 1 k! Γ ν k π k = 1 k! ∞ (− t)k 1 ∞ (− t)k − 1 M (t) = = Γ(ν k) sin(π ν k) , (9) ν ∑ (− ( + ) + ) ∑ ( − ) k = 0 k! Γ ν k 1 1 π k = 1 k 1 !

Fν(t) = ν t Mν(t) .

The interest in the Wright functions of the second kind comes from the fact that they play an important role in solution of the linear partial differential equations of fractional order which describe a wide spectrum of phenomena including probability distributions, anomalous diffusion and diffusive waves [24–33]. In Section2, of this paper, by using the complex inversion formula in (1) and the Bromwich contour, the integral representation of fν,µ(t) with 0 < ν < 1 and 0 ≤ µ < 1 is derived and some basic properties of this inverse transform are established. The cases µ ≥ 1 can be dealt as well, see AppendixA for details. The next two Sections3 and4 are devoted to evaluation of infinite, finite and convolution integrals by using the Efros theorem in the Wlodarski form. The integrands of these integrals or integral identities include the elementary functions (power, exponential, logarithmic, trigonometric and hyperbolic functions) and the special functions (the error function, Mittag–Leffler functions and the Volterra functions). The last section provides concluding remarks. In derivations of these integrals direct and inverse Laplace transforms which are taken from tables of transforms [18–21] are always presented in mathematical expressions. All mathematical operations and manipulations with elementary and special functions, integrals and transforms are formal and their validity is assured by considering the restrictions usually imposed in the operational calculus.

2. Integral Representations of the Inverse Laplace Transform of s−µ exp(−sν) Before inﬁnite integrals in (6) will be evaluated, it is of interest to derive the function fν,µ(t) by performing the complex integration from (1). In investigated case, the branch point of the integrand exists and is located at the origin s = 0 and therefore the equivalent Bromwich contour is plotted in Figure1. The closed contour of integration ABCDEFA includes the line AB, the arcs BC and FA of a circle of radius R → ∞ with center at origin, the arc DE of a circle of radius r → 0 with center at origin, and two parallel lines CD and EF. The the cut along the negative axis ensures that F(s) is a single-valued function. However, according to the Cauchy lemma the integrals along the arcs BC and FA vanish as R → ∞.

Figure 1. The equivalent Bromwich contour. Symmetry 2021, 13, 354 4 of 15

As a consequence there are only three contributions coming from integrals on the CD and EF lines and from the small circle round origin, DE, so we have

ν e−s F(s) = ; 0 < ν < 1 ; µ ≥ 0, sµ Z ∞ Z π (10) 1 −ut − πi πi 1 treiθ iθ iθ fν,µ(t) = e F(ue ) − F(ue ) du + lim e F(re ) re dr. 2πi 0 2πi r→0 − π

However, the last trigonometric integral vanishes for µ < 1 and therefore the ﬁnal result of the complex integration from (10) is

Z ∞ −ut − uν cos(πν) 1 e ν fν,µ(t) = µ sin[u sin(πν) + πµ] du, π 0 u (11) t > 0 ; 0 < ν < 1 ; µ < 1.

The same integral representation has been derived in 1970 by Stankovic˙ [10] for the Wright function, but on the negative values of the argument t and which is presented here in our notation

ν 1 1 Z ∞ e−ut −u cos(πν) µ − 1 − = [ ν ( ) + ] t Wν,µ ν µ sin u sin πν πµ du, t π 0 u (12) t > 0 ; 0 < ν < 1 ; µ < 1,

and comparing (11) with (12) we have that the inverse exponential functions can be expressed in terms of the Wright functions, in agreement with the survey analysis by Mainardi and Consiglio [33] where also plots are presented,

1 f (t) = tµ − 1W − ; 0 < ν < 1 ; µ ≥ 0, . (13) ν,µ ν,µ tν

In particular, for µ = 1 − ν the Wright functions are reduced to the Mainardi functions Fν(t) and Mν(t) [11]

1 1 1 1 f − (t) = F = M ; 0 < ν < 1. (14) ν,1 ν ν ν tν tν ν tν

For ν = 1/2, µ = 0 and ν = µ = 1/2, the integrals in (11) become the Laplace transforms of trigonometric functions [18–21]

Z ∞ − 1/4t 1 −ut 1/2 e f1/2,0(t) = e sin(u ) du = √ , π 0 2 π t3/2 Z ∞ −ut Z ∞ −ut − 1/4t (15) 1 e 1/2 π 1 e 1/2 e f1/2,1/2(t) = sin(u + ) du = cos(u ) du = √ . π 0 u1/2 2 π 0 u1/2 π

Since [11,12] 1 1 2 F1/3 = K1/3 √ , (16) t1/3 3πt1/2 27t it follows from (11) and (14) that

1/3 " √ ! √ !# Z ∞ −ut − u /2 √ 1 e 3 1/3 3 1/3 f1/3,2/3(t) = 3 cos u − sin u du = 2π 0 u2/3 2 2 r (17) 1 λ 2 K1/3 √ . π t 27t Symmetry 2021, 13, 354 5 of 15

Differentiation of the integral (11) with respect to the argument t gives

ν dn f (t) (−1)n Z ∞ e−ut −u cos(πν) ν,µ = ( ) = [ ν ( ) + ] n fν,µ − n t µ − n sin u sin πν πµ du, dt π 0 u (18) t > 0 ; 0 < ν < 1 ; µ < 1 ; n = 0, 1, 2, 3, . . .,

and using rules of the operational calculus we have the initial and ﬁnal values of the function from [33]

1 − µ − sν fν,µ(t → +0) = lim [sF(s)] = s e = 0 ; 0 < ν < 1 ; µ < 1, s → ∞ (19) 1 − µ −sν fν,µ(t → ∞) = lim [sF(s)] = s e = 0. s → 0

This is in an agreement with the ﬁnding of Pollard [2] that for µ = 0, the inverse Laplace transform fν,0 is positive almost everywhere, but for µ < 0 Stankovic˙ [10] postu- lated that in a some interval this function is negative and has at least one zero. Expanding the exponent in F(s) into series it is possible to obtain the behaviour of the function for large values of the argument t

( ν ) e− s 1 sν s2ν L− 1{F(s)} = L− 1 = L− 1 1 − + − ... . sµ sµ 1! 2! (20) 1 1 1 1 fν,µ(t → ∞) ∼ − + − ... t1 − µ Γ(µ) Γ(µ − ν)tν 2Γ(µ − 2ν)t2ν

The integral of the fν,µ can be derived directly from

Z t ( − sν ) − 1 e fν,µ(u) du = L + = fν,µ + 1(t). (21) 0 s1 µ

In order to obtain the recurrence relations the following operational rule can be applied

( −sν ) −sν − sν d d e e e L t f − (t) = − [F(s)] = − = (µ − 1) + ν , (22) ν,µ 1 ds ds sµ − 1 sµ sµ − ν

and the inverse transforms in (22) are

t fν,µ − 1(t) = (µ − 1) fν,µ(t) + ν fν,µ − ν(t). (23)

However, from (18) with n = 1, we have

d fν,µ(t) = f − (t), (24) dt ν,µ 1 and therefore d fν,µ(t) t = (µ − 1) f (t) + ν f − (t). (25) dt ν,µ ν,µ ν

3. Integrals of the Inverse Laplace Transform of s−µ exp(sν) with Elementary Functions In the ﬁrst example the Wlodarski integral formula (4) is applied to the power function g(t) = tλ,

g(t) = tλ ; λ > 0, Γ(λ + 1) Γ(λ + 1) G(s) = ; G(sν) = , sλ + 1 s(λ + 1)ν (26) 1 Γ(λ + 1) Γ(λ + 1) Γ(λ + 1) t(λ + 1)ν +µ − 1 L−1 · = L−1 = , sµ s(λ + 1)ν s(λ + 1)ν + µ Γ[(λ + 1)ν + µ)] Symmetry 2021, 13, 354 6 of 15

and therefore we have

Z ∞ (λ + 1)ν +µ − 1 λ Γ(λ + 1) t u fν,µ(u) du = ; 0 < ν < 1 ; µ < 1 ; λ > 0. (27) 0 Γ[(λ + 1)ν + µ)]

The above results can be extended to functions g(t) which are deﬁned at ﬁnite intervals and to step functions jumping at integral values of variable t. Let start with

1 ; 0 < t < λ g(t) = . 0 ; t > λ ν 1 − e− λs 1 − e− λs G(s) = ; G(sν) = , s sν ( ν ) ( 1/ν ν ) 1 1 − e− λs 1 λ(µ/ν + 1)e− (λ s) (28) L−1 · = L−1 − = sµ sν sν + µ (λ1/νs)ν + µ ν + µ − 1 t [(µ − 1)/ν + 1] t − λ fν,ν + µ , Γ(ν + µ) λ1/ν

which leads to the ﬁnite integral

Z λ ν + µ − 1 t [(µ − 1)/ν + 1] t fν,µ(u) du = − λ fν,ν + µ( 1/ν ), 0 Γ(ν + µ) λ (29) 0 < ν < 1 ; µ < 1 ; λ > 0.

In the next example, from

1 − t ; 0 < t < 1 g(t) = , 0 ; t > 1 ν e−s + s − 1 e−s + sν − 1 G(s) = ; G(sν) = , s2 s2ν ( ν ) ( ν ) 1 e−s + sν − 1 1 1 e− s (30) L−1 · = L−1 − + = sµ s2ν s2ν + µ sν + µ s2ν + µ t2ν + µ − 1 tν + µ − 1 − + f + (t), Γ(2ν + µ) Γ(ν + µ) ν,2ν µ

we have Z 1 t2ν + µ − 1 tν + µ − 1 (1 − u) fν,µ(u) du = − + fν,2ν + µ(t). (31) 0 Γ(2ν + µ) Γ(ν + µ) Using   0 ; 0 < t < λ g(t) = (t − λ)µ , ; t > λ  Γ(µ + 1) ν e−λs e−λs G(s) = ; G(sν) = , sµ + 1 s(µ + 1)ν (32) ( ν ) ( 1/ν ν ) 1 e−λs λ[(1/ν + 1)µ + 1]e− (λ s) L−1 · = L−1 = sµ s(µ + 1)ν (λ1/νs)(µ + 1)ν + µ = λ[(1/ν + 1)µ − 1/ν + 1] f ( t ), ν,ν + νµ +µ λ1/ν it is possible to derive

Z ∞ µ [(1/ν + 1)µ − 1/ν + 1] t (t − λ) fν,µ(u) du = λ fν,ν + νµ +µ . (33) λ λ1/ν Symmetry 2021, 13, 354 7 of 15

In the next example the case of exponential function is considered:

g(t) = e− αt ; α > 0, 1 1 G(s) = ; G(sν) = , s + α sν + α (34) ν − 1 ν + µ −2 −1 1 1 −1 1 s t ν L · = L · = ? Eν(− αt ). sµ sν + α sν +µ −1 sν + α Γ(ν + µ − 1)

From (4) and (34) the convolution integral with the Mittag–Lefﬂer function

∞ tk E (t) = , ν ∑ Γ(kν + 1) k = 0 (35) sν − 1 L{E (−αtν)} = , ν sν + α is derived Z ∞ ν + µ −2 − αu t ν e fν,µ(u)du = ? Eν(− αt ), 0 Γ(ν + µ − 1) (36) 0 < ν < 1 ; µ < 1 ; α > 0 . By changing variable of integration x = t (cosθ)2, all convolution integrals can be expressed as in terms of ﬁnite trigonometric integrals. The shifted increasing and decreasing exponential functions are considered in the following two examples. From

0 ; 0 < t < λ ; 0 < ν < 1 g(t) = , 1 − e− (t − λ) ; t > λ ν e−λs e−λs G(s) = ; G(sν) = , s(s + 1) sν(sν + 1) ( ν ) ( 1/ν ν ) (37) 1 e−λs sν − 1 λ(2ν + µ − 1)/νe− (λ s) L−1 · = L−1 · = sµ sν(sν + 1) (sν + 1) (λ1/νs)(2ν + µ − 1) λ(ν + µ − 1)/νE (− tν) ∗ f ( t ), ν ν,2ν + µ − 1 λ1/ν we have the convolution of two functions. Z ∞ h − (u − λ)i (ν + µ − 1)/ν ν t 1 − e fν,µ(u) du = λ Eν(− t ) ? fν,2ν + µ − 1( ). (38) λ λ1/ν Similarly from

0 ; 0 < t < λ ; 0 < ν < 1 g(t) = ; e− (t − λ) ; t > λ ν e−λs e−λs G(s) = ; G(sν) = ; (s + 1) (sν + 1) ( ν ) ( 1/ν ν ) (39) 1 e−λs sν − 1 λ(ν + µ − 1)/νe− (λ s) L−1 · = L−1 · = sµ (sν + 1) (sν + 1) (λ1/νs)(ν + µ − 1) λ(ν + µ − 2)/νE (− tν) ∗ f ( t ), ν ν,ν + µ − 1 λ1/ν it is possible to obtain

Z ∞ − (u − λ) (ν + µ − 2)/ν ν t e fν,µ(u) du = λ Eν(− t ) ? fν,ν + µ − 1( ). (40) λ λ1/ν Symmetry 2021, 13, 354 8 of 15

The logarithmic functions are the next group of elementary functions to be considered. In the simplest case from

g(t) = ln t ; C = eγ; ln(Cs) ln(Csν) G(s) = − ; G(sν) = − ; s sν 1 − ln(Csν) 1 −ν ln(Cs) + (ν − 1)γ L−1 · = L−1 · = (41) sµ sν sν + µ − 1 s (ν − 1)γtν + µ − 1 νtν + µ − 2 + ? ln t Γ(ν + µ ) Γ(ν + µ − 1)

where γ is the Euler constant, it follows that

Z ∞ (ν − 1)γtν + µ − 1 νtν + µ − 2 ln u fν,µ(u) du = + ? ln t . (42) 0 Γ(ν + µ ) Γ(ν + µ − 1)

In the more general case

g(t) = tλ − 1 ln t ; λ > 0 ; 0 < ν < 1; Γ(λ) Γ(λ)[ψ(λ) − ν ln s] G(s) = [ψ(λ) − ln s ] ; G(sν) = ; sλ sλν 1 [ψ(λ) − ν ln s] ψ(λ) + γν ν ln(Cs) (43) Γ(λ) L−1 · = Γ(λ) L−1 − · = sµ sλν s(λ ν + µ) s(λ ν + µ − 1) s Γ(λ)[ψ(λ) + γν] t(λ ν + µ −1) Γ(λ) νt(λ ν + µ −2) + ? ln tν, Γ(λ ν + µ) Γ(λ ν + µ − 1)

we have

Z ∞ (λ ν + µ −1) (λ ν + µ −2) λ − 1 Γ(λ)[ψ(λ) + γν] t Γ(λ) νt ν u ln u fν,µ(u) du = + ? ln t . (44) 0 Γ(λ ν + µ) Γ(λ ν + µ − 1)

Trigonometric and hyperbolic functions is the last groups of elementary functions to be considered. From:

g(t) = sin(λt) ; 0 < ν < 1; λ λ G(s) = ; G(sν) = ; s2 + λ2 s2ν + λ2 (45) 1 1 1 s2ν − 1 λt2ν + µ − 2 λL−1 · = λL−1 · = ? E (−λ2 t2ν), sµ s2ν + λ2 s2ν + µ − 1 s2ν + λ2 Γ(2ν + µ − 1) 2ν

the following integral identity is derived

Z ∞ 2ν + µ − 2 λt 2 2ν sin(λu) fν,µ(u) du = ? E2ν(−λ t ). (46) 0 Γ(2ν + µ − 1)

Similarly as in (45), for the hyperbolic sine function, the change is only in the sign

g(t) = sinh(λt) ; 0 < ν < 1; λ λ G(s) = ; G(sν) = ; s2 − λ2 s2ν − λ2 (47) ( 2ν − 1 ) 2ν + µ − 2 −1 1 1 −1 1 s λt 2 2ν λL · = λL · = ? E ν(λ t ), sµ s2ν − λ2 s2ν + µ − 1 s2ν − λ2 Γ(2ν + µ − 1) 2

and therefore we have

Z ∞ 2ν + µ − 2 λt 2 2ν sinh(λu) fν,µ(u) du = ? E2ν(λ t ). (48) 0 Γ(2ν + µ − 1) Symmetry 2021, 13, 354 9 of 15

In the case of cosine function, from;

g(t) = cos(λt) ; 0 < ν < 1; s sν G(s) = ; G(sν) = ; s2 + λ2 s2ν + λ2 (49) ν ( 2ν − 1 ) ν + µ − 2 −1 1 s −1 1 s t 2 2ν L · = L · = ? E ν(−λ t ), sµ s2ν + λ2 sν + µ − 1 s2ν + λ2 Γ(ν + µ − 1) 2

we have Z ∞ ν + µ − 2 t 2 2ν cos(λu) fν,µ(u) du = ? E2ν(−λ t ). (50) 0 Γ(ν + µ − 1) Similarly as in (47) and (48)

g(t) = cosh(λt) ; 0 < ν < 1; s sν G(s) = ; G(sν) = ; s2 − λ2 s2ν − λ2 (51) ν ( 2ν − 1 ) ν + µ − 2 −1 1 s −1 1 s t 2 2ν L · = L · = ? E ν(λ t ), sµ s2ν − λ2 sν + µ − 1 s2ν − λ2 Γ(ν + µ − 1) 2

for the hyperbolic cosine function we have

Z ∞ ν + µ − 2 t 2 2ν cosh(λu) fν,µ(u) du = ? E2ν(λ t ). (52) 0 Γ(ν + µ − 1)

In the case of the product of trigonometric and hyperbolic sine functions the direct and inverse Laplace transforms are

g(t) = sin(λt) sinh(λt) ; 0 < ν < 1; 2λ2s 2λ2sν G(s) = ; G(sν) = ; s4 + 4λ4 s4ν + 4λ4 1 sν 1 s4ν − 1 (53) 2λ2L−1 · = 2λ2L−1 · = sµ s4ν + 4λ4 s3ν + µ − 1 (s4ν + 4λ4) 2λ2 [ (3ν + µ −2) ? (− 4 4ν)] Γ(3ν + µ −1) t E4ν 4λ t ,

and therefore

Z ∞ 2 2λ (3ν + µ −2) 4 4ν sin(λu) sinh(λu) fν,µ(u) du = [t ? E4ν(− 4λ t )]. (54) 0 Γ(3ν + µ − 1)

For the product of trigonometric and hyperbolic sine functions we have

g(t) = cos(λt) cosh(λt) ; 0 < ν < 1; s3 s3ν G(s) = ; G(sν) = ; 4 + 4 4ν + 4 (55) s 4λ ( s 4λ ) 1 s3ν 1 s4ν − 1 tν + µ − 2 L−1 · = L−1 · = ? E (− 4λ4t4ν), sµ s4 + 4λ4 sν + µ − 1 s4ν + 4λ4 Γ(ν + µ − 1) 4ν

and therefore

Z ∞ ν + µ − 2 t 4 4ν cos(λu)(cosh(λu) fν,µ(u) du = ? E4ν(− 4λ t ). (56) 0 Γ(ν + µ − 1) Symmetry 2021, 13, 354 10 of 15

4. Integrals of the Inverse Laplace Transform of s−µ exp(−sν) with the Mitag-Lefﬂer, Error and Volterra Functions The Laplace transform of the two parameter Mittag–Lefﬂer function is [11,34–36]

sα − β Ltβ − 1 E (±λtα) = , α,β sα ∓ λ ∞ zk (57) E (z) = , α,β ∑ ( + ) k = 0 Γ kα β

and this permits to obtain

β − 1 α g(t) = t Eα,β(±λt ) ; 0 < ν < 1; sα − β s(α − β)ν G(s) = ; G(sν) = ; sα ∓ λ sαν ∓ λ (58) ( (α − β)ν ) ( αν − (β ν +µ) ) 1 s −1 s (β ν +µ − 1) α · = L = t E + (±λt ), sµ sαν ∓ λ sαν ∓ λ αν,β ν µ

and Z ∞ β − 1 α (β ν +µ − 1) α u Eα,β(±λu ) fν,µ(u) du = t Eαν,β ν + µ(±λt ). (59) 0 Thus, in both sides of expressions (59) appear the Mittag–Lefﬂer functions. Evidently, for β = 1 they are reduced to the classical Mittag–Lefﬂer functions.

Z ∞ α ( ν +µ − 1) α Eα,1(±λu ) fν,µ(u) du = t Eαν, ν + µ(±λt ), (60) 0 and for β = α and β = α + 1 we have

Z ∞ α − 1 α (α ν +µ − 1) α u Eα,α(±λu ) fν,µ(u) du = t Eαν,α ν + µ(±λt ), 0 Z ∞ (61) α α [(α + 1) ν +µ − 1] α u Eα,α + 1(±λu ) fν,µ(u) du = t Eαν,(α + 1) ν + µ(±λt ). 0 If β = 1/2, in the integrands of (61) the Mittag–Lefﬂer functions are expressed then by the error functions [35,36]

z2 α E1/2(±z) = e [1 ± erf(z)] ; z = λu , 1 2 E (±z) = √ ± zez [1 ± erf(z)] , 1/2,1/2 z (62) ez E (z) = √ erf(z). 1/2,3/2 z

If β is positive integer, the Mittag–Lefﬂer functions are expressed by elementary functions [36]. In particular cases, with ν = 1/2, µ = 0. ν = µ = 1/2 and ν = 1/3, µ = 2/3 the explicit form of the inverse transforms of exponential functions is known (see (15) and (17)). The Laplace transform of the error function is

λ g(t) = erf ; 0 < ν < 1, 2t1/2 1/2 ν/2 1 − e− λs 1 − e− λs G(s) = ; G(sν) = , s sν ( ν/2 ) ( 2/ν ν/2 ) 1 1 − e− λs 1 λ2µ/νe− (λ s) (63) L−1 · = L−1 − = sµ sν sµ (λ2/νs)µ ! tµ − 1 tν/2 − λ2(µ − 1)/ν f , Γ(µ) ν/2,µ λ2/ν Symmetry 2021, 13, 354 11 of 15

which yields ! Z ∞ µ − 1 ν/2 λ t 2(µ − 1)/ν t erf fν,µ(u) du = − λ fν/2,µ . (64) 0 2u1/2 Γ(µ) λ2/ν

The Volterra functions are deﬁned by the following integrals [37]:

Z ∞ tu ν(t) = du, 0 Γ(u + 1) Z ∞ tu + α ν(t, α) = du, (65) 0 Γ(u + α + 1) Z ∞ uβtu + α µ(t, β, α) = du, 0 Γ(β + 1)Γ(u + α + 1)

and their Laplace transforms are

1 L{ν(λt)} = s ; λ > 0, s ln λ λα L{ν(λt, α)} = s , sα + 1 ln (66) λ λα L{µ((λt, β, α)} = . s β + 1 sα + 1 ln λ The logarithmic functions in the Laplace transforms permit to express the integrals of the Volterra functions with inverse Laplace transform of exponential function in terms of convolution integrals. From

g(t) = ν(λt) ; λ > 0, 1 1 G(s) = ; G(sν) = ; 0 < ν < 1, s sν s ln sν ln λ λ (67) 1 1 1 1 L−1 · = L−1 · = sµ sν ln[(s/λ1/ν)ν] sν + µ − 1 νs ln(s/λ1/ν) tν + µ − 2 ∗ ν(λ1/νt), νΓ(ν + µ − 1)

it follows that Z ∞ ν + µ − 2 t 1/ν ν(λu) fν,µ(u) du = ? ν(λ t). (68) 0 νΓ(ν + µ − 1) Similarly from

g(t) = ν(λt, ρ) ; λ, ρ > 0 ρ ρ λ ν λ G(s) = ; G(s ) = ν ; 0 < ν < 1, ρ + 1 s s s ln s(ρ + 1)ν ln λ λ ( ) ( ) (69) 1 1 1 1 λρL−1 · = λρL−1 · = sµ s(ρ + 1)ν ln[(s/λ1/ν)ν] sµ − 1 νs(ρ + 1)ν + 1 ln(s/λ1/ν) tµ − 2 ?, ν[λ1/νt, (ρ + 1)ν], Γ(µ − 1)νλ

we have Z ∞ µ − 2 t 1/ν ν(λu, ρ) fν,µ du = ? ν[λ t, (ρ + 1)ν]. (70) 0 Γ(µ − 1)νλ Symmetry 2021, 13, 354 12 of 15

Finally, the Laplace transform of the generalized Volterra function is

g(t) = µ(λt, ξ, ρ) ; λ, ρ, ξ > 0 ρ ρ λ ν λ G(s) = ; G(s ) = ν ; 0 < ν < 1, ρ + 1 s ξ + 1 s s [ln ] s(ρ + 1)ν[ln ]ξ + 1 λ λ 1 1 λρL−1 · = (71) sµ s(ρ + 1)ν{ln[(s/λ1/ν)ν]}ξ + 1 λρ 1 1 L−1 · = νξ + 1 sµ − 1 s(ρ + 1)ν + 1[ln(s/λ1/ν)]ξ + 1 tµ − 2 ? µ[λ1/νt, ξ, (ρ + 1)ν], Γ(µ − 1)λνξ + 1

and (65) yields

Z ∞ µ − 2 t 1/ν µ(λu, ξ, ρ) fν,µ du = + ? µ[λ t, ξ, (ρ + 1)ν]. (72) 0 Γ(µ − 1)λνξ 1

The number of similar convolution integrals can be signiﬁcantly enlarged if the Volterra functions are multiplied by tn with n = 1, 2, 3, ... , then their Laplace transforms should be differentiated n times. The result of such differentiations are linear combinations of these functions [34–36]. For example, with n = 1 and λ = 1 we have

d 1 ρ + 1 1 ( ) = L− 1 − = L− 1 + = tν t, ρ + ρ + ρ + ds sρ 1 ln s s 2 ln s s 2(ln s)2 (73) (ρ + 1) ν[t, (ρ + 1)ν] + µ[t, 1, (ρ + 1)ν].

5. Conclusions By applying the Efros theorem in the form established by Wlodarski it was possible to derive a number of infinite integrals, finite integrals and integral identities with the function which represent the Laplace inverse transform of s−µ exp(−sν) with 0 < ν < 1 and 0 ≤ µ < 1. The extension to the cases µ ≥ 1 is dealt in AppendixA Derived by us integrals include in integrands elementary functions (power, exponential, logarithmic, trigonometric and hyperbolic functions) and the error functions, the Mittag–Leffler functions and the Volterra functions. Many results appear in form of the convolution integrals. Performing the inversion by the complex integration, it was possible to show that the inverse Laplace inverse transform, which means the original function, can be also expressed in terms of the Wright functions and for particular values of parameters by the Mainardi functions. Using rules of operational calculus some properties of the inverse Laplace transform were derived.

Author Contributions: These authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript. Funding: This research received no external funding. Acknowledgments: The research work of F.M. has been carried out in the framework of the activities of the National Group of Mathematical Physics (GNFM, INdAM). Both the authors would like to acknowledge the unknown reviewers for their constrictive comments. Conﬂicts of Interest: The authors declare no conﬂict of interest.

Appendix A. Additional Properties of the Inverse of the Laplace Transform of s−µ exp(−sν)

The restriction posed on the second parameter of the function fν,µ(t), i.e., µ < 1 (in the inverse of the Laplace transform s−µ exp(−sν)) can be removed by using rules of operational calculus. In the general case, for a non-negative parameter ρ ≥ 0, this Symmetry 2021, 13, 354 13 of 15

inverse transform can always be expressed as the convolution integral which includes the corresponding power function:

( − sν ) ( −sν ) λ − 1 − 1 e − 1 1 e t fν ρ(t) = L = L · = ? fν µ(t), , sρ sλ sµ Γ(λ) , (A1) ρ = λ + µ.

In the particular case µ = 1, it reduces to the simple integral, because 1/s expresses the integration operation:

Z t − 1 1 − sν fν,1(t) = L e = fν,0(u) du. (A2) s 0

The product of two identical or different Laplace inverse transforms can be derived in the form of convolution integrals from

( ν ) ( ν ) ( 1/ν ν ) − s ν − 2s − (2 s) − 1 e e− s − 1 e 2µ/ν − 1 e L · µ = L = 2 L , sµ s s2µ (21/νs)2µ ( ν ) ( ν ) ( 1/ν ν ) (A3) − s ν − 2s − (2 s) − 1 e e− s − 1 e (µ + ρ)/ν − 1 e L · ρ = L = 2 L , sµ s sµ + ρ (21/νs)(µ + ρ)

which gives

Z t (2µ − 1)/ν t fν,µ(t) ? fν,µ(t) = fν,µ(t − ξ) fν,µ(ξ)dξ = 2 fν,2µ 1/ν , 0 2 Z t (A4) (µ + ρ − 1)/ν t fν,µ(t) ? fν,ρ(t) = fν,µ(t − ξ) fν,ρ(ξ)dξ = 2 fν,µ + ρ 1/ν . 0 2 These results can be generalized to n-fold integrals when exists the factor 1/sn, with n = 1, 2, 3, . . .. In the case of the product of identical Laplace inverse transforms it is reduced to the following convolution integral

1 Z t (t − u)n − 1 L− 1 ( ) = ( ) n F s f u du. (A5) s 0 (n − 1)!

In similar way, if the inverse transform is in a more general form with 0 < ν < 1,

( ν ) ( ν ) e− αs e− βs L f (t); α = L ; L f (t); β = L ; α, β > 0, ν,µ sµ ν,ρ sµ ( ν ν ) ( ν ) e− αs e− βs e− (α + β)s L− 1 · = L− 1 = (A6) sµ sρ sµ + ρ ( 1/ν ν ) e−[(α + β) s] (α + β)(µ +ρ)/ν L− 1 , [(α + β)1/νs]µ + ρ

we have Z t fν,µ(t; α) ? fν,ρ(t; β) = fν,µ(t − ξ; α) fν,ρ(ξ; β)dξ = 0 (A7) (α + β)(µ + ρ − 1)/ν f t ; (α + β) . ν,µ + ρ (α+β)1/ν Symmetry 2021, 13, 354 14 of 15

The differentiation of Laplace inverse transforms with respect to the parameters ν and µ gives

− ν − ν ( ) ν s s − ν ∂ fν,µ t s e ln s e ln(sC) ln Ce s γ L = − = − · + − ; C = e , ∂ν sµ sµ − ν − 1 s sµ ν − sν (A8) ∂ fν,µ(t) µe L = − , ∂µ sµ + 1

which yields the following inverses of (A8)

∂ fν,µ(t) = ln t ? f − − (t) + γ f − (t), ∂ν ν,µ ν 1 ν,µ ν Z t (A9) n ∂ fν,µ(t) o L ∂µ = −µ fν,µ + 1(t) = −µ fν,µ(ξ) dξ, 0 where γ denotes the Euler constant.

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