In this paper, we introduce new functions Yρ,rν(x) as a generalization of the Krätzel function. We investigate recurrence relations, Mellin transform, fractional derivatives, and integral of the function Yρ,rν(x). We show that the function Yρ,rν(x) is the solution of differential equations of fractional order.

1. Introduction

The Krätzel function is defined for x>0 by the integral(1)Zρνx=∫0∞tν-1e-tρ-x/tdt,where ρ∈R and ν∈C, such that R(ν)<0 for ρ≤0 (cf. [1]). For ρ≥1 the function (1) was introduced by Krätzel as a kernel of the integral transform as follows:(2)Kνρfx=∫0∞Zρνxtftdtx>0.The Krätzel function Zρνx is related to the modified Bessel function of the second kind Kν by the relationship(3)Z1νx=2xν/2Kν2x.The generalized Krätzel function Dρ,rν,αx is given in [2, 3] by the following relation:(4)Dρ,rν,αx=∫0∞tν-11+aα-1tρ1/α-1e-xt-rdt,where ρ∈R, r∈R+, ν∈C, and α>1. Kilbas and Kumar considered the special case for r=1 in [2], calculated fractional derivatives and fractional integrals of Dρ,1ν,α(x), and obtained a representation using Wright hypergeometric functions. On the other hand the general case of (1) is given in [2, (54), p. 845].

We consider the generalized Krätzel function Yρ,rν(x) defined by the integral(5)Yρ,rνx=∫0∞tν-1e-tρ-xt-rdt,for x>0, ρ∈R, r∈R+, and ν∈C. The function Yρ,rνx is a generalization of the Krätzel function Zρνx since(6)limr→1Yρ,rνx=Zρνx.If a=1 in (4), then (7)limα→1Dρ,rν,αx=Yρ,rνx.

We give some definitions and inequalities that will be needed. The Turán type inequalities (8)fnx·fn+2x-fn-1x2≥0,n=0,1,2,…are important and well known in many fields of mathematics (cf. [4]). A function f(x) is completely monotonic on 0,∞, if f has derivatives of all orders and satisfies the inequality(9)-1mfmx≥0for all x>0 and m∈N (cf. [5, Section IV]). A function f(x) is said to be log-convex on 0,∞, if(10)fαx1+1-αx2≤fx1αfx21-αfor all x1,x2>0 and α∈0,1 (cf. [5, p. 167]).

Let p,q∈R such that p>1 and 1/p+1/q=1. If f and g are real valued functions defined on a closed interval and fp, gq are integrable in this interval, then we have (11)∫abftgtdt≤∫abftpdt1/p∫abgtqdt1/q.The following inequality is due to Mitrinović et al. (cf. [6, p. 239]). Let f and g be two functions which are integrable and monotonic in the same sense on a,b and p is a positive and integrable function on the same interval, then the following inequality holds true:(12)∫abptftdt∫abptgtdt≤∫abptdt∫abptftgtdt,if and only if one of the functions f and g reduces to a constant.

The Mellin transform of the function f is defined by(13)Mfx;s=∫0∞xs-1fxdxwhen Mf(x);s exists. The Mellin transform of the generalized Krätzel function (5) is given by Kilbas and Kumar in [2].

The Laplace transform of the function f is defined by(14)Lfx;s=∫0∞e-sxfxdxprovided that the integral on the right-hand side exists.

The Liouville fractional integral is defined by(15)I-αfx=1Γα∫x∞t-xα-1ftdtand its derivatives I-α and D-α are(16)D-αfx=-ddxRα+1I-1-α+Rαfx=1Γ1-α+Rα∫x∞t-x-α+Rαftdt,where x>0, α∈C, and R(α)>0 (cf. [7, Section 5.1]).

We introduce new operators(17)Lλν≔-rxD-λ+1+λr-νD-λ,(18)Tλν≔r2x2D-2λ+2+rx2ν-3λr-rD-2λ+1+ν-rλν-2rλD-2λ,where ν∈C and λ>0.

A standard source in the theory of fractional calculus is the book [8]. For applications of fractional calculus to science and engineering, we refer the reader to the articles [9–11].

In this paper, we investigate the properties of the functions Yρ,rνx and prove their composition of Yρ,rνx with fractional integral and derivatives I-αf(x), D-αf(x) given by (15) and (16) (cf. [2, 6, 12, 13]). In Section 3, we show that Yρ,rνx is the solution of differential equations of fractional order.

2. The Main Theorems

In this section, we will give some properties of generalized Krätzel functions Yρ,rν.

Lemma 1.

Let ρ∈Rρ≠0, r∈R+, ν∈C, R(s)>0 be such that Rν+rs>0 when ρ>0 and Rν+rs<0 when ρ<0. The Mellin transform of the function Yρ,rνx is given by(19)MYρ,rν;s=1ρΓsΓν+rsρx>0.

Proof.

Using (13) and (5), we have (20)MYρ,rν;s=∫0∞xs-1∫0∞tν-1e-tρ-xt-rdtdx.Changing the order of integration and using the substitution of xt-r=u, we have (21)MYρ,rν;s=∫0∞tν-1e-tρ∫0∞xs-1e-xt-rdxdt=Γs∫0∞tν+rs-1e-tρdt.Making the change of variable the integral tρ=z, and using the known formula (1) from [14, p. 145], we find that(22)MYρ,rν;s=Γsρ∫0∞zν+rs/ρ-1e-zdz=ΓsρΓν+rsρ,when ρ>0 and(23)MYρ,rν;s=Γsρ∫0∞zν+rs/ρ-1e-zdz=-ΓsρΓν+rsρ,when ρ<0.

Theorem 2.

We have the following relationship for the function Yρ,rνx:(24)Yρ,rνx=L1rt-ν/r-1e-t-ρ/r;x,where ρ∈R, r∈R+, ν∈C, and x>0.

Proof.

Using (5) and making the change of t-r=u, we obtain (25)Yρ,rνx=∫0∞tν-1e-tρ-xt-rdt=∫0∞1ru-ν/r-1e-u-ρ/re-xudu.Now the assertion (24) follows from the definition (14) of the Laplace transform.

Using the known formula (29) from [14, p. 146], we find that(26)Yρ,ρνx=1ρLt-ν/r-1e-t-ρ/r;x=2ρxν/2ρKν/ρ2x,for ρ=1:(27)Y1,1νx=Z1νx=2xν/2Kν2x,Y1/r,1/rνx=2rxν/2Kν2x.

Theorem 3.

If ρ∈R, r∈R+, ν∈C and x>0, then the following assertions are true:

The function Yρ,rνx satisfies the recurrence relation (28)νYρ,rνx=ρYρ,rν+ρx-rxYρ,rν-rx.

The function x→Yρ,rνx is completely monotonic on (0,∞).

Proof.

(a) The above recurrence relation could be verified by using integration by parts as follows: (29)Yρ,rνx=1νtνe-tρ-xt-r0∞-∫0∞1νtν-ρtρ-1+rxt-r-1e-tρ-xt-rdt=ρν∫0∞tν+ρ-1e-tρ-xt-rdt-rxν∫0∞tν-r-1e-tρ-xt-rdt=ρνYρ,rν+ρx-rxνYρ,rν-rx.

(b) From Bernstein-Widder theorem (see Theorem 1, [5, p. 145]), the function Yρ,rνx is completely monotonic on (0,∞) for all x>0. This could be verified directly as follows:(30)dndxnYρ,rνx=-1n-1Yρ,rν-nrx>0n=0,1,2,…,which follows via mathematical induction from (5) provided that ρ∈R, r∈R+, ν∈C and x>0. From Bernstein-Widder theorem, generalized forms of Krätzel function are completely monotonic on 0,∞ for all x>0. Due to (30), the functions are completely monotonic on 0,∞ for all x>0.

Setting r→1 and using (28), the equation yields(31)Yρ,1νx=ρνYρ,1ν+ρx-xνYρ,1ν-1x.Then using (31) and (6), we obtain the relation(32)νZρνx=ρZρν+ρx-xZρν-1x,(cf. 2.1 of Theorem 1 from [12]).

Theorem 4.

Let ν1,ν2,ρ∈R, 0<λ<1, and x>0, then the following assertions hold true:

The function ν→Yρ,rνx is log-convex on R: (33)Yρ,rλν1+1-λν2,αx≤Yρ,rν1xλYρ,rν2x1-λ.

The function x→Yρ,rνx is log-convex on 0,∞: (34)Yρ,rνλx1+1-λx2≤Yρ,rνx1λYρ,rνx21-λ.

The function Yρ,rνx satisfies the following relation:(35)Yρ,rνtr=tνYr,ρ-νtρ.

Proof.

(a) Using (5) and (11), we obtain(36)Yρ,rλν1+1-λν2x=∫0∞tλν1+1-λν2-1e-tρ-xt-rdt=∫0∞tν1-1e-tρe-xt-rλtν2-1e-tρe-xt-r1-λdt≤∫0∞tν1-1e-tρ-xt-rdtλ∫0∞tν2-1e-tρ-xt-rdt1-λ=Yρ,rν1xλYρ,rν2x1-λ,where λ∈0,1, ν1,ν2,ρ∈R, α>1, and x>0. Thus, ν→Yρ,rνx is log-convex on R.

(b) The integrand in (5) is a log-linear convex function of x. By using (11), we have (37)Yρ,rνλx1+1-λx2=∫0∞tν-1e-tρe-λx1+1-λx2t-rdt=∫0∞tν-1e-tρe-x1t-rλtν-1e-tρe-x2t-r1-λdt≤∫0∞tν-1e-tρe-x1t-rdtλ∫0∞tν-1e-tρe-x2t-rdt1-λ=Yρ,rνx1λYρ,rνx21-λ,where λ∈0,1, ν,ρ∈R, r>0, and x1,x2>0. Thus, x→Yρ,rνx is log-convex on (0,∞).

(c) Again using (5), we conclude that (38)Yρ,rνx=∫0∞tν-1e-tρ-xt-rdt,t=x1/ru-1=∫0∞x1/ru-1ν-1e-x1/ru-1ρe-xx1/ru-1-rx1/ru-2du=xν/r∫0∞u-ν-1e-ur-xρ/ru-ρdu=xν/rYr,ρ-νxρ/ror for the change of x=tr, we obtain (35).

Moreover, since Yρ,rνx is log-convex on R, we have Turán type inequality(39)Yρ,rν1+ν2/2x2≤Yρ,rν1xYρ,rν2xfor ν1,ν2,ρ∈R, α>1, and x>0. Making the change of variable ν1=ν-2 and ν2=ν, the equation yields(40)fρν,αx=Yρ,rν-1x2-Yρ,rν-2xYρ,rνx≤0which is valid for ν,ρ∈R, α>1, and x>0.

Using (39) and making the change of variables ν1=ν-n-1 and ν2=ν-n+1, we have(41)Yρ,rν-nx2≤Yρ,rν-n-1xYρ,rν-n+1x.

Theorem 5.

If ν,ρ∈R, r∈R+ and x>0, then the following inequality holds true:(42)ΓνrYρ/r,1-1/rx≤xν-1/rΓ1rYρ,r-νx.

Proof.

Let pt=e-xtr, ft=tν-1 and gt=e-t-ρ. The function ft is increasing on 0,∞ for ν≥1 and is decreasing for ν≤1. On the other hand, we observe that, for all ρ>0,(43)g′tgt=ρt-ρ-1>0. Thus, gt is increasing if and only if ρ>0. Moreover, making the change of tr=u and using the known formula (1) from [14, p. 137], we have(44)∫0∞ptdt=∫0∞e-xtrdt=1rΓ1rx-1/r.Making the change of t-r=u, we find(45)∫0∞ptftdt=∫0∞e-xtrtν-1dt=1rΓνrx-ν/r.Making the change of variable t=u-1/r and using (6), we have (46)∫0∞ptgtdt=∫0∞e-xtre-t-ρdt=1r∫0∞u-1/r-1e-uρ/re-x/udu=1rYρ/r,1-1/rx.Using (5) and making the change of variable t=u-1, we find (47)∫0∞ptftgtdt=∫0∞e-xtrtν-1e-t-ρdt=∫0∞u-ν-1e-uρ-xu-rdu=Yρ,r-νx.Finally, by using the relation (12), we obtain the inequality (42): (48)∫0∞ptftdt∫0∞ptgtdt≤∫0∞ptdt∫0∞ptftgtdt1rΓνrx-ν/r·1rYρ/r,1-1/rx≤1rΓ1rx-1/r·Yρ,r-νxxν-1/rΓ1rYρ,r-νx≥ΓνrYρ/r,1-1/rx.

If we choose r→1 in (42), then we have(49)xν-1Yρ,1-νx≥ΓνYρ,1-1x.As a result, we find the following inequality by using (6):(50)Zρ-νx≥x1-νΓνZρ-1x.

3. Differential Equations of Fractional Order

In this section, we show that Yρ,rνx is the solution of differential equations of fractional order.

Theorem 6.

If α,ν∈C, R(α)>0, and ρ>0, then the following identity holds true:(51)I-αYρ,rνx=Yρ,rν+rαx.

Proof.

Applying (15), (5), and relation (11) of [15, p. 202], we obtain (52)I-αYρ,rνx=1Γα∫x∞t-xα-1dt∫0∞uν-1e-uρ-tu-rdu=∫0∞uν-1e-uρ1Γα∫x∞t-xα-1e-tu-rdtdu=∫0∞uν-1e-uρI-αe-tu-rxdu=∫0∞uν-1e-uρe-xu-rurαdu=∫0∞uν+rα-1e-uρe-xu-rdu=Yρ,rν+rαx.

Theorem 7.

If α,ν∈C, R(α)>0, and ρ>0 then we have(53)D-αYρ,rνx=Yρ,rν-rαx.

Proof.

Using (16), (5), and (51), we obtain (54)D-αYρ,rνx=-ddxRα+1I-1-α+RαYρ,rνx=-ddxRα+1Yρ,rν+r1-α+Rαx=-ddxRα+1∫0∞tν+r1-α+Rα-1e-tρ-xt-rdt=∫0∞tν+r1-α+Rα-1e-tρ-ddxRα+1e-xt-rdt=∫0∞tν+r1-α+Rα-1e-tρ1trRα+1e-xt-rdt=∫0∞tν-αr-1e-tρe-xt-rdt=Yρ,rν-rαx.

Corollary 8.

If α,β, and ν∈C, R(α)>0, R(β)>0, and ρ>0, then we have(55)D-αI-βYρ,rνx=I-βD-αYρ,rνx=Yρ,rν+β-α+1-r1+Reα.

Theorem 9.

If ν∈C and ρ>0, then the following identity holds true:(56)LρνYρ,rνx=-ρYρ,rν+1-rρx.

Proof.

Applying (17) to (5), we get (57)LρνYρ,rνx=-rxD-ρ+1Yρ,rνx+ρr-νD-ρYρ,rνx=-rxYρ,rν-ρ+1rx+ρr-νYρ,rν-ρrx=-rx∫0∞tν-ρ+1r-1e-tρ-xt-rdt+ρr-ν∫0∞tν-ρr-1e-tρ-xt-rdt=-∫0∞tν-ρr-1rxtr+ν-ρre-tρ-xt-rdt.Using the formula (58)tν-ρre-xt-r′=tν-ρr-1ν-ρr+xrt-re-xt-rand applying the integration by parts, we find (59)LρνYρ,rνx=-∫0∞tν-ρre-xt-r′e-tρdt=-tν-ρre-xt-re-tρ0∞+∫0∞tν-ρre-xt-r-ρtρ-1e-tρdt=0-ρ∫0∞tν-ρr+ρ-1e-xt-re-tρdt=-ρYρ,rν+1-rρx.

Corollary 10.

If ν∈C and ρ>0, then the function Yρ,rν(x) is a solution of the differential equation of fractional order(60)rxD-ρ+1Yρ,rνx+ν-ρrD-ρYρ,rνx-ρYρ,rν+1-rρx=0.

Remark 11.

If ν∈C, and ρ=r=1, then the function Y1,1ν(x)=Z1ν(x) is a solution of the following differential equation: (61)xy′′+ν-1y′-y=0(cf. [13, (30), p. 20]).

Theorem 12.

If ν∈C and ρ>0, then the function Yρ,rνx is a solution of the differential equation of fractional order (62)TρνYρ,rνx+ρ2Yρ,rν-2r-1ρx-ρ2r-1Yρ,rν-2r-1ρx=0.

Proof.

Using (18), (5), and (53), we get(63)TρνYρ,rνx=∫0∞r2x2t2r+2ν-3ρr-rrxtrtν-2rρ-1e-xt-re-tρdt+∫0∞ν-rρν-2rρtν-2rρ-1e-xt-re-tρdt.If we take the derivative as the proof of Theorem 9, then we arrive at(64)ν-rρ+rxtrtν-2rρe-xt-r′=r2x2t2r+2ν-3ρr-rrxtr+ν-rρν-2rρtν-2rρ-1e-xt-r.Substituting (64), (16), into (63) and applying the integration by parts, we get(65)TρνYρ,rνx=∫0∞ν-rρ+rxtrtν-2rρe-xt-r′·e-tρdt=ν-rρ+rxtrtν-2rρe-xt-r·e-tρ0∞+ρ∫0∞ν-rρ+rxtrtν-2rρ+ρ-1e-xt-re-tρdt=ρ∫0∞ν-rρ+rxtrtν-2r-1ρ-1e-xt-re-tρdt.If we rewrite the expression in (65) relation as (66)ν-rρ+rxtr=ν-2r-1ρ+rxtr+r-1ρ,then we have(67)TρνYρ,rνx=ρ∫0∞ν-2r-1ρ+rxtrtν-2r-1ρ-1e-xt-re-tρdt+ρ2r-1∫0∞tν-2r-1ρ-1e-xt-re-tρdt.If we evaluate the integral on the right-hand side of relation (64) and apply the integration by parts, we arrive at (62) as follows: (68)TρνYρ,rνx=ρ∫0∞tν-2r-1ρe-xt-r′e-tρdt+ρ2r-1Yρ,rν-2r-1ρ=ρ·tν-2r-1ρe-xt-re-tρ0∞-ρ2∫0∞tν-2r-1ρ-1e-xt-re-tρdt+ρ2r-1Yρ,rν-2r-1ρ=-ρ2Yρ,rν-2r-1ρ+ρ2r-1Yρ,rν-2r-1ρ, where (69)tν-2r-1ρe-xt-r′=ν-2r-1ρ+rxtrtν-2r-1ρ-1e-xt-r.

Remark 13.

If ν∈C and ρ=1, then the function Y1,1ν(x)=Z1ν(x) is a solution of the differential equation of fourth order (70)x2yIV+2ν-4xy′′′+ν-1ν-2y′′+y=0(cf. [13, p. 21]).

4. Conclusion

Mejer’s G functions, which are generalization of hypergeometric functions, are Mellin-Barnes integrals. Generalized Krätzel functions, Yρ,rν(x) could be written in terms of H-functions, which are generalization of G-function, as a Mellin-Barnes integral. Furthermore, the integral transform with the kernel Yρ,rν(x) could be investigated.

Competing Interests

The authors declare that they have no conflict of interests regarding the publication of this paper.

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