# Certain Properties of the Modified Degenerate Gamma Function

Authors: Kwara Nantomah
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we prove some inequalities satisfied by the modified degenerate gamma function which was recently introduced. The tools employed include Holder’s inequality, mean value theorem, Hermite–Hadamard’s inequality, and Young’s inequality. By some parameter variations, the established results reduce to the corresponding results for the classical gamma function.


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## Body

## 1. Introduction

In recent times, degenerate special functions and polynomials have been a subject of intense discussion. See, for example, [1–5] and the related references therein.In 2017, Kim and Kim [6] introduced the degenerate gamma function as(1)Γλx=∫0∞tx−11+λt−1/λdt,where λ∈0,∞ and 0<ℜx<1/λ. This was motivated by the degenerate exponential function which is defined as [6](2)eλt=1+λt1/λ,where λ∈0,∞. It is clear that limλ⟶0eλt=et and limλ⟶0Γλx=Γx, where Γx is the classical gamma function.In 2018, Kim et al. [7] introduced the modified degenerate gamma function which is defined as(3)Γλx=∫0∞tx−11+λ−1/λdt,where λ∈0,1 and ℜx>0. This definition is equivalent to(4)Γkx=∫0∞tx−11+1k−ktdt,where 1<k<∞ and ℜx>0. Here, limk⟶∞Γkx=Γx. The modified degenerate gamma function (3) satisfies the following properties [7]:(5)Γλ1=λln1+λ,(6)Γλx+1=λxln1+λΓλx,(7)Γλm+1=λm+1m!ln1+λm+1,m∈ℕ.Derivatives of the modified degenerate gamma function are given as(8)Γλrx=∫0∞lntrtx−11+λ−t/λdt,where r∈ℕ0.In a recent work, He et al. [8] introduced the modified degenerate digamma function which is defined as(9)ψλx=ddxlnΓλx=Γλ′xΓλxand has the following representations among others:(10)ψλx=−γ+lnλln1+λ+∑k=0∞x−1k+1k+x=−γ+lnλln1+λ−1x+∑k=1∞xkk+x=−γ+lnλln1+λ+∫011−tx−11−tdt,where γ is the Euler–Mascheroni constant. It also satisfies the following basic properties:(11)ψλ1=−γ+lnλln1+λ,ψλx+1=ψλx+1x,and similarly, it is clear that limλ⟶0ψλx=ψx, where ψx is the classical digamma function. For further properties of the function ψλx, one may refer to [8].In this paper, we continue to investigate the modified degenerate gamma function. Precisely, we prove some inequalities satisfied by this generalized function. The techniques we employed are analytical in nature.


## 2. Results and Discussion

Theorem 1.
Fors∈0,1 and x>0, the inequality holds:(12)λxln1+λ1−s≤Γλx+1Γλx+s≤λx+sln1+λ1−s.Proof.
The case fors=1 is obvious. So, let s∈0,1 and x>0. Then, by applying Holder’s inequality for integrals, we have(13)Γλx+s=∫0∞tx+s−11+λ−t/λdt=∫0∞t1−sx−11+λ−t1−s/λtsx1+λ−st/λdt≤∫0∞tx−11+λ−t/λdt1−s∫0∞tx1+λ−t/λdts=Γλx1−sΓλx+1s,and by using (6), we obtain(14)Γλx+s≤λxln1+λsΓλx.
By replacings with 1−s in (14), followed by substituting x by x+s, we obtain(15)Γλx+1≤λx+sln1+λ1−sΓλx+s.
Now, combining (14) and (15), we obtain(16)ln1+λλx+s1−sΓλx+1≤Γλx+s≤λxln1+λsΓλx,and by using (6), we obtain the desired results (12).Remark 1.
Inequality (16) can also be rearranged as(17)xx+s1−s≤Γλx+sλx/ln1+λsΓλx≤1,which is the degenerate form of Wendel’s inequality (see (7) of [9]). Furthermore, by Squeezes theorem, (17) implies that(18)limx⟶∞Γλx+sλx/ln1+λsΓλx=1,which is the degenerate form of Wendel’s asymptotic relation (see (1) of [9]). The limit (18) also implies that(19)limx⟶∞λxln1+λr−sΓλx+sΓλx+r=1.Theorem 2.
For0<u≤v, the inequality holds:(20)expv−uψλu≤ΓλvΓλu≤expv−uψλv.Proof.
The case foru=v is trivial. So, consider the function lnΓλx on the interval 0<u<v. Then, by the mean value theorem, there exist a k∈u,v such that(21)lnΓλv−lnΓλuv−u=ψλk.
Sinceψλx is increasing, then(22)ψλu<ψλk<ψλv,which yields(23)v−uψλu<lnΓλvΓλu<v−uψλv,and by exponentiation, we obtain the desired result (20).Corollary 1.
Fors∈0,1 and x>0, the inequality holds:(24)exp1−sψλx+s≤Γλx+1Γλx+s≤exp1−sψλx+1.Proof.
Letv=x+1 and u=x+s in Theorem 2.Corollary 2.
Forx≥0, the inequality holds:(25)λln1+λexpxψλx+1≤Γλx+1≤λln1+λexpxψλ1.Proof.
Letv=x+1 and u=1 in Theorem 2.Remark 2.
Inequality (24) is the degenerate form of inequality (3.4) of [10].Theorem 3.
For0<u≤v, the inequality holds:(26)expv−uψλu+ψλv2≤ΓλvΓλu≤expv−uψλu+v2.Proof.
Let0<u≤v, and consider the function ψλx on the interval u,v. Since ψλx is concave, then by the classical Hermite–Hadamard inequality, we have(27)ψλu+ψλv2≤1v−u∫uvψλtdt≤ψλu+v2,which translates to(28)v−uψλu+ψλv2≤lnΓλvΓλu≤v−uψλu+v2,and by exponentiation, we obtain the desired result (26).Corollary 3.
Fors∈0,1 and x>0, the inequality holds:(29)exp1−sψλx+1+ψλx+s2≤Γλx+1Γλx+s≤exp1−sψλx+s+12.Proof.
Letv=x+1 and u=x+s in Theorem 3.Corollary 4.
Forx≥0, the inequality holds:(30)λln1+λexpx2ψλx+1+ψλ1≤Γλx+1≤λln1+λexpxψλx2+1.Proof.
Letv=x+1 and u=1 in Theorem 3.Remark 3.
Inequalities (26), (29), and (30) are, respectively, better than (20), (24), and (25)Remark 4.
By lettingλ⟶0, inequality (29) reduces to(31)exp1−sψx+1+ψx+s2≤Γx+1Γx+s≤exp1−sψx+s+12.
The upper bound of (31) coincides with the upper bound of inequality (1.2) in the work [11] which was obtained by a different procedure. However, the lower bound of (31) is better than the lower bound of inequality (1.2) in [11] since(32)ψx+1+ψx+s2≥ψx+1ψx+s≥ψx+s≥ψx+s.
This is by virtue of the arithmetic-geometric mean inequality and the monotonicity property ofψx.Theorem 4.
Forx>0 and s∈0,1, the inequality holds:(33)lnλ+1λ1−sx+ss−1≤Γλx+sΓλx+1≤lnλ+1λs1−sΓλsx+ss−1.Proof.
LetΔx=x+s1−sΓλx+s/Γλx+1 for x>0 and s∈0,1. Then,(34)limx⟶0Δx=lnλ+1λs1−sΓλs.
Also, inequality (17) implies that(35)limx⟶∞Δx=lnλ+1λ1−s.
Furthermore,(36)Δ′xΔx=1−sx+s+ψλx+s−ψλx+1≤0,which shows that Δx is decreasing. Hence, Δ∞≤Δx≤Δ0 which yields (33)Remark 5.
As a particular case, by lettings=1/2 and λ⟶0, we obtain(37)2πx+12<Γx+1Γx+1/2<x+12.
The upper bound of (37) agrees with the upper bound of (3) in [12]. Comparing the lower bounds of (37) and (3) in [12] reveals that the lower bound of (37) is stronger if 0<x<1/π−2 and is weaker if x>1/π−2.Theorem 5.
Letu>1 and 1/u+1/v=1. Then,(38)Γλx+y≤Γλux1/uΓλvy1/v,holds for x>0 and y>0.Proof.
By Holder’s inequality for integrals, we have(39)Γλx+y=∫0∞tx+y−11+λ−t/λdt=∫0∞tx−1/u1+λ−t/λuty−1/v1+λ−t/λvdt≤∫0∞tux−11+λ−t/λdt1/u∫0∞tvy−11+λ−t/λdt1/v=Γλux1/uΓλvy1/u,which concludes the proof.Remark 6.
Applying Young’s inequality on the right-hand side of (38) reveals that(40)Γλx+y≤Γλuxu+Γλvyv.Theorem 6.
Letr1,r2∈2n:n∈ℕ0, k1>1, 1/k1+1/k2=1, and r1/k1+r2/k2∈ℕ0. Then, inequality holds for x>0 and y>0:(41)Γλr1/k1+r2/k2xk1+yk2≤Γλr1x1/k1Γλr2y1/k2,Proof.
By using (8) and Holder’s inequality, we have(42)Γλr1/k1+r2/k2xk1+yk2=∫0∞lntr1/k1+r2/k2tx/k1+y/k2−11+λ−t/λdt=∫0∞lntr1/k1tx−1/k11+λ−t/λk1lntr2/k2ty−1/k21+λ−t/λk2dt≤∫0∞lntr1tx−11+λ−t/λdt1/k1∫0∞lntr2ty−11+λ−t/λdt1/k2=Γλr1x1/k1Γλr2y1/k2,which concludes the proof.Remark 7.
Ifr1=r2=r, then (41) reduces to(43)Γλrxk1+yk2≤Γλrx1/k1Γλry1/k2,which implies that function (8) is log-convex for any even order derivative. Moreover, if r=0 in (43), we obtain(44)Γλxk1+yk2≤Γλx1/k1Γλy1/k2,which shows that the modified degenerate gamma function is log-convex.Remark 8.
Ifr1=r, r2=r+2, k1=k2=2, and x=y, then (41) reduces to the Turan-type inequality:(45)Γλr+1x2≤ΓλrxΓλr+2x.Remark 9.
Ifr1=s−1, r2=s+1, k1=k2=2, and x=y, then (35) reduces to the Turan-type inequality:(46)Γλsx2≤Γλs−1xΓλs+1x,where s∈2n+1:n∈ℕ0. This is the degenerate version of main result of [13].Theorem 7.
Forx≥1, the inequality holds:(47)Γλx≥λln1+λ−1+1x.Proof.
Letϕx=Γλx−1/x. Then,(48)ϕx+1−ϕx=Γλx+1−Γλx+1x−1x+1=Γλxλxln1+λ−1+1x−1x+1>0.
Thus,ϕx is increasing, and for x≥1, we have ϕx≥ϕ1 which completes the proof.Lemma 1.
The functionβx=xΓλ′x is increasing for all x>0.Proof.
By using (6) and (9) and the monotonicity property of ψλx, we have(49)βx=xΓλxψλx=lnλ+1λΓλx+1ψλx,and consequently, we obtain(50)λlnλ+1β′xΓλx+1=ψλx+1ψλx+ψλ′x>ψλ2x+ψλ′x>0,which completes the proof.Theorem 8.
The inequalities hold forx>0:(51)ΓλxΓλ1x≥λlnλ+12,(52)Γλx+Γλ1x≥2λlnλ+1.Proof.
By lettingk1=k2=2 and replacing x and y with 1+x and 1+1/x in (44), we obtain(53)lnΓλ1+x2+12x≤lnΓλ1+x2+lnΓλ1+1/x2.
Also, sincex+1/x≥2 for x>0, then 1+x/2+1/2x≥2. Now, let ϕx=ΓλxΓλ1/x for x>0. Then, by using (6), we have(54)ϕx=lnλ+1λ2Γλ1+xΓλ1+1x.
Next, by using (5), (53), and (54), we obtain(55)lnϕx≥2lnlnλ+1λ+2lnΓλ1+x2+12x≥2lnlnλ+1λ+2lnΓλ2=lnλlnλ+12,which gives (51). Next, let θx=Γλx+Γλ1/x for x>0. Then,(56)xθ′x=xΓλ′x−1xΓλ′1x.
It follows from Lemma1 that θx is increasing if x>1 and decreasing if 0<x<1. For both cases, we have θx>θ1=2λ/lnλ+1 which gives inequality (52).Remark 10.
Inequality (52) can be obtained from inequality (51) by applying the arithmetic-geometric mean inequality.Theorem 9.
Letr,s∈2n:n∈ℕ0 and r≥s. Then, inequality holds for x>0:(57)expΓλrx2≤expΓλr−sx⋅expΓλr+sx.Proof.
We adopt the technique of Mortici [14] to estimate the function(58)Γλr−sx+Γλr+sx2−Γλrx=12∫0∞lntr−stx−11+λ−t/λdt+12∫0∞lntr+stx−11+λ−t/λdt−∫0∞lntstx−11+λ−t/λdt=12∫0∞1lnts+lnts−2lntrtx−11+λ−t/λdt=12∫0∞1−lnts2lntr−stx−11+λ−t/λdt≥0.
Thus,(59)2Γλrx≤Γλr−sx+Γλr+sx,and by exponentiation, we arrive at (57).


## 3. Concluding Remarks

In this work, we have proved several inequalities satisfied by the modified degenerate gamma function which was recently introduced. Whenλ⟶0, the established results reduce to the corresponding results for the classical gamma function. It is our fervent hope that the present results will inspire further research on the modified degenerate gamma function.



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