# On Some Complete Monotonicity of Functions Related to Generalizedk−Gamma Function

Authors: Hesham Moustafa; Hanan Almuashi; Mansour Mahmoud
Journal: Journal of Mathematics
 (2021)

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## Abstract

In this paper, we presented two completely monotonic functions involving the generalizedk−gamma function Γkx and its logarithmic derivative ψkx, and established some upper and lower bounds for Γkx in terms of ψkx.


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## Body

## 1. Introduction

The ordinary gamma function is given by the following equation [1]:(1)Γu=limn⟶∞n!nu−1uu+1u+2,…,u+n−1,u>0,which was discovered by Euler when he generalized the factorial function to noninteger values. Lots of mathematicians studied the gamma function because of its great importance; for complete studies of the gamma function, please refer to [2, 3]. The digamma function is the logarithmic derivative of the gamma function and is given by [4]:(2)ψu=−γ−1u+∑i=1∞1i−1u+i,u>0,where γ=limn⟶∞∑s=1n1/s−logn is Euler–Mascheroni’s constant. A function Lx defined on an interval I is said to be completely monotonic if it possesses derivatives Lnu for all n∈ℕ such that(3)−1nLnu≥0,u∈I;n∈ℕ.Completely monotonic functions have remarkable applications in different fields such as in the theory of special functions, numerical and asymptotic analysis, probability, and physics. Some important properties of these functions were collected in [5], and for more information about this topic, we refer the reader to [6, 7].In 2007, Alzer and Batir [8] proved that the function(4)Mαu=lnΓu−ulnu+u−ln2π+12ψu+α,α≥0,is completely monotonic on 0,∞ if and only if α≥1/3, and they also proved the function −Mαu is completely monotonic on 0,∞ if and only if α=0. As a consequence, they introduced the following sharp bounds for Γ function in terms of ψ function:(5)exp−12ψu+c<Γu2πuue−u<exp−12ψu+d,u>0,with the best possible constants c=1/3 and d=0. In order to refine inequality (5), Batir [9] modified the function M0u and proved that the function(6)Tδu=M0u+16u−δ,which is completely monotonic on 0,∞ if and only if δ≥1/4, and the function −Tδu is completely monotonic on 0,∞ if and only if δ≤0. As a consequence, he deduced the following refinement of the inequality (5):(7)exp−12ψu−16u−a<Γu2πuue−u<exp−12ψu−16u−b,u>0,with the best possible constants a=1/4 and b=0.In many contexts, such as the combinatorics of creation, the annihilation operators, and the perturbative computation of Feynam integrals, the following symbol appears [10–12]:(8)un,k=uu+ku+2k,…,u+n−1k,uandk>0;n∈ℕ,which is a generalization of the ordinary Pochhammer symbol un=Γu+n/Γu, when k=1. Diaz and Pariguan [13] were motivated by the importance of un,k, and they called it “the Pochhammer k− symbol,” and they introduced the k−gamma function by(9)Γku=limn⟶∞n!knnku/k−1un,k=∫0∞vu−1e−vk/kdv,uandk>0,which satisfies the functional equation:(10)Γku+k=uΓku,Γkk=1,u,k>0.As special cases,Γ1u is the ordinary gamma function and the case k=2 is of particular interest since Γ2u=∫0∞vu−1e−v2/2dv is the Gaussian integral. The k−gamma and ordinary gamma functions are related by the relation(11)Γku=ku/k−1Γuk,uandk>0.Thek-analogue of the digamma function is given by [14]:(12)ψku=lnk−γk−1u+∑n=1∞1nk−1nk+u,uandk>0,and it satisfies the following relations for uandk>0 and m=0,1,2,…,:(13)ψkmu+k=−1mm!um+1+ψkmu,ψkk=lnk−γk.In 2018, Nantomah et al. [15] presented the following integral representations:(14)ψku=∫0∞2e−t−e−ktkt−e−ut1−e−ktdt,uandk>0,(15)ψkmu=−1m+1∫0∞tme−ut1−e−ktdt,uandk>0;m∈ℕ.Ege and Yildirim [16] obtained an inequality involving ψk′u+k and ψk″u+k which can be written, by using (13), as(16)−uψk″u−1u2−1ku<ψk′u−1ku<−uψk″u−12u2−1ku,kandu>0.In 2020, Yildirim [17] presented some monotonicity properties for ψkm functions, and he deduced the following inequalities:(17)12u<1klnu−ψku<12u+k12u2,uandk>0,(18)12u2+k6u3−k330u5<ψk′u−1ku<12u2+k6u3,uandk>0,(19)1u3<−ψk″u−1ku2<1u3+k2u4,uandk>0.For more information aboutΓk and ψk functions, we refer the reader to [18–20] and the related references therein.Nantomah et al. [21] introduced the following two-parameter deformation of the gamma and digamma functions for uandk>0 and p∈ℕ:(20)Γp,ku=p+1!kp+1pku/k−1uu+ku+2k,…,u+pk=∫0pvu−11−vkpkpdv,ψp,ku=1klnpk−∑i=0p1u+ik=1klnpk−∫0∞1−e−kp+1t1−e−kte−utdt.Nantomah et al. [22] introduced some complete monotonicity properties for Γp,k, and they deduced the following inequalities:(21)0≤1klnpkuu+pk+k−ψp,ku≤1u−1u+pk+k,uandk>0;p∈ℕ,(22)0≤ψp,k′u−1k1u−1u+pk+k≤1u2−1u+pk+k2,uandk>0;p∈ℕ.Motivated by these results, we will present ak−analogue of the functions Mαu and Tδu and study their monotonicity. As a consequence, we deduce some inequalities for Γk and ψkm functions.


## 2. Preliminary

Using relation (11), we have(23)ψkx=1klnk+1kψxk,xandk>0,ψkmx=1km+1ψmxk,xandk>0;m∈ℕ.Also, using the following asymptotic formula ofΓ function [4](24)lnΓy∼y−12lny−y+ln2π+∑i=1∞B2i2i2i−1y2i−1,y⟶∞,with relation (11), we conclude for k>0 that(25)lnΓkx∼ln2πk+xk−12lnx−xk+∑i=1∞k2i−1B2i2i2i−1x2i−1,x⟶∞,(26)ψkx∼1klnx−12x−∑i=1∞k2i−1B2i2ix2i,x⟶∞,and for s∈ℕ,(27)ψksx∼−1s−1s−1!kxs−−1ss!2xs+1+−1s+1∑i=1∞s+2i−1!k2i−1B2i2i!x2i+s,x⟶∞.In the following, we will use the following important result [23].Corollary 1.
LetS be a real-valued function defined on x>x0; x0∈ℝ with Sx tends to zero as x⟶∞. Then, for r∈ℝ+, Sx>0, if Sx+r<Sx for all x>x0 and Sx<0, if Sx+r>Sx for all x>x0.
Now, we will prove the following auxiliary results.Lemma 1.
Fork>0, we have(28)ψk′x+k3>1kx,∀x>k3,ψk″x+k3<2xkψk″x+2kx2+2k2x,∀x≥83k100,ψk′″x+k3>2kx3,∀x≥76k100.Proof.
Consider the functionFkx=ψk′x+k/3−1/kx, and using (13), we get(29)Fkx+k−Fkx=−1x+k/32+1xx+k=−k3x−kxx+k3x+k2<0,∀x>k3,and using asymptotic expansion (27), we have limx⟶∞Fkx=0. Then, Corollary 1 gives us that Fkx>0 for all x>k/3. Now, we let Hkx=−k/2ψk″x+k/3+xψk″x+1/x2+1/kx and use (13) to get(30)Hkx+k−Hkx=kψk″x+k+k5+10k4x+10k3x2+9k2x3+27kx4+27x5x2k+x2k+3x3,≑Nkx,Nkx+k−Nkx=k2hkx−0.83kx2k+x32k+x2k+3x34k+3x3,where(31)1018hkx=507528366752760409361k9+159366156531359440200300k8x+10476615622621545898964k7x2+1061604508702355386252k6x3+10819056153366876366k5x4+1010140271547017402k4x5+1012654834692196k3x6+10141892006748k2x7+10163091689kx8+10182187x9>0,x≥0.
Then,Nkx+k−Nkx>0 for all x≥0.83k with limx⟶∞Nkx=0; hence, we have Nkx<0 for all x≥0.83k. It follows that Hkx+k−Hkx<0 for all x≥0.83k with limx⟶∞Hkx=0. Then, Hkx>0 for all x≥0.83k. Finally, consider the function Lkx=k/2ψk′″x+k/3−1/x3 and use (13) to obtain(32)Lkx+k−Lkx=−k2fkx−0.76kx3k+x3k+3x4,where(33)9765625fkx=13183088k5+2823864000k4x+10116487500k3x2+13331953125k2x3+7593750000kx4+1582031250x5>0,for x>0. Then, Lkx+k−Lkx<0 with limx⟶∞Lkx=0. Hence, Lkx>0 for all x≥0.76k.
Using relation (12) ofψk, we conclude the following corollary which will be useful in the next part.Corollary 2.
Letx∈0,∞ and k>0. Then, the following limits are valid:(34)limx⟶0xm+1ψkmx=−1m+1m!,m=0,1,2,…,and(35)limx⟶0xmψkmx+kb=0,m∈ℕandb>0.


## 3. Two Completely Monotonic Functions InvolvingΓkx and ψkx Functions

Theorem 1.
The function(36)Mλ,kx=lnΓkx−ln2πk−xklnx+xk+k2ψkx+λk,λ≥0;k>0,is completely monotonic on 0,∞ if and only if λ≥1/3, and the function −Mλ,kx is completely monotonic on 0,∞ if and only if λ=0.Proof.
Using relation (15), we get(37)Mλ,k″x=ψk′x−1kx+k2ψk″x+λk=∫0∞e−xtkekt−1Φtdt,where(38)Φt=1−ekt+ktekt−kt22e1−λkt.
Letλ≥1/3, then we obtain(39)Φt≥1−ekt+ktekt−kt22e2/3kt=∑r=2∞∑l=2rrl2r−lktr+2r+1r+2!3r>0.
Consequently,Mλ,k″x is completely monotonic on 0,∞; hence, Mλ,k′x is increasing on 0,∞. Using asymptotic expansions (26) and (27), we have limx⟶∞Mλ,k′x=0, and then, Mλ,k′x<0. Hence, Mλ,kx is decreasing on 0,∞. Using asymptotic expansions (25) and (26), we have limx⟶∞Mλ,kx=0; hence, Mλ,kx>0. Therefore, Mλ,kx is completely monotonic on 0,∞. Conversely, if Mλ,kx is completely monotonic, then we get(40)xkMλ,kx=xklnΓkx2π/kxx/k−1/2e−x/k−12lnx+k2ψkx+λk>0,xandk>0.
From (25), we have(41)limx⟶∞xklnΓkx2π/kxx/k−1/2e−x/k=112.
Using asymptotic expansion (26), we get limx⟶∞x/kkψkx+λk−lnx=λ−1/2. Then, from (40), we conclude that 1/12+1/2λ−1/2≥0; hence, λ≥1/3. Now, for λ=0, we have(42)M0,k″x=∫0∞e−xtkekt−1∑r=1∞−rr+12ktr+2r+2!dt,and therefore, −M0,k″x is completely monotonic on 0,∞. Thus, M0,k′x is decreasing on 0,∞ with limx⟶∞M0,k′x=0; then, M0,p′x>0. Thus, M0,kx is increasing on 0,∞ with limx⟶∞M0,kx=0; hence, M0,kx<0. Then, −M0,kx is completely monotonic on 0,∞. Conversely, we assume that −Mλ,kx (with λ>0) is completely monotonic on 0,∞; then, Mλ,kx is negative on 0,∞. But this contradicts that limx⟶0Mλ,kx=∞; hence, λ=0.Theorem 2.
The function(43)Uβ,kx=lnΓkx−ln2πk−xklnx+xk+k2ψkx+k6x−βk,xandk>0,is completely monotonic on 0,∞ if and only if β≥1/4. Also, −Uβ,kx is completely monotonic on 0,∞ if and only if β≤0.Proof.
Using relation (15), we get(44)Uβ,k″x=ψk′x−1kx+k2ψk″x+k3x−βk3=∫0∞e−xtkekt−1φtdt,where(45)φt=1−ekt+ktekt−kt22ekt+kt26eβktekt−1.
Letβ≥1/4, then we obtain(46)φt≥11kt52880+7kt62304+∑r=5∞frr+2!ktr+2>0,t>0,where(47)fr=r+16r54r−3+14r7r+2∑l=2rrl4l>0,r≥5.
Then,Uβ,k″x is completely monotonic on 0,∞. Thus, Uβ,k″x is an increasing function on 0,∞; using asymptotic expansions (26) and (27), we get that limx⟶∞Uβ,k′x=0. Then, Uβ,kx is a decreasing function with limx⟶∞Uβ,kx=0; hence, Uβ,kx>0. Thus, we deduce that Uβ,kx is a completely monotonic function on 0,∞. Conversely, if Uβ,kx is a completely monotonic function, then it is positive and we obtain(48)β≥limx⟶∞xk+16lnΓkx−ln2π/k−x/klnx+x/k+k/2ψkx,k>0.
Using asymptotic expansions (25) and (26), we have(49)lnΓkx−ln2πk−xklnx+xk+k2ψkx=−k4x+k24x2+O1x3.
Then,(50)β≥limx⟶∞xk−4x2k4x+k=14.
Now, forβ≤0, then(51)φt≤1−ekt+ktekt−kt23ekt−kt26=−13∑r=2∞r2−1ktr+2r+2!<0,and hence, −Uβ,k″x is a completely monotonic function on 0,∞ for β≤0. Now, Uβ,k′x is a decreasing function on 0,∞ with limx⟶∞Uβ,k′x=0, and then, Uβ,p′x>0. Thus, Uβ,kx is an increasing function on 0,∞ with limx⟶∞Uβ,kx=0; hence, Uβ,kx<0. Therefore, we deduce that −Uβ,kx is a completely monotonic function on 0,∞ for β≤0. Conversely, if −Uβ,kx is a completely monotonic function, then Uβ,kx<0 and(52)β≤limx⟶0xk+16lnΓkx−6ln2π/k−6x/klnx+6x/k+3kψkx,k>0.
Using functional relations (10) and (13), we get β≤limx⟶0x/Akx=0, where(53)Akx=6xlnΓkx+k−6xln2πk−6x+x2klnx+6x2k+3kxψkx+k−3k.


### 3.1. Some Inequalities forΓkx,ψkx, and ψkmx Functions

As a consequence of the completely monotonicity properties of the two functionsMλ,kx and Uβ,kx, we conclude the following results.Corollary 3.
Foraandb≥0, the following inequality holds:(54)exp−k2ψkx+ka<Γkx2π/kxx/ke−x/k<exp−k2ψkx+kb,xandk>0,with the best possible constants a=1/3 and b=0.Proof.
The left-hand side of inequality (54) is equivalent x/kMa,kx>0 which leads to a≥1/3 as mentioned in the proof of Theorem 1. Using the increasing property of ψkx on 0,∞, we get e−k/2ψkx+k/3≥e−k/2ψkx+ka for a≥1/3, and then, a=1/3 is the best possible constant in (54). Also, Theorem 1 gives the right-hand side of inequality (54) for b=0. If there exists b>0 such that the right-hand side of (54) is valid for x∈0,∞, then we would have(55)limx⟶0Γkx<2πkexp−k2ψkkb,which contradicts that limx⟶0Γkx=∞. Hence, b=0 is the best possible constant in (54).Remark 1.
If we putk=1 in inequality (54), then we get inequality (5).Corollary 4.
Forcandd≥0, the following inequality holds:(56)k2ψk′x+kc<1klnx−ψkx<k2ψk′x+kd,xandk>0,with the best possible constants c=1/3 and d=0.Proof.
The left-hand side of inequality (56) is equivalent that(57)x2kMc,k′x=x2kSkx+x22kkψk′x+kc−1x−k6x2<0,where Skx=ψkx−1/klnx+1/2x+k/12x2. Using asymptotic expansions (26) and (27), we have limx⟶∞x2/kSkx=0 and(58)limx⟶∞x22kkψk′x+kc−1x−k6x2=limx⟶∞x22kk2x2−2kcx−c2k26x2x+ck2−ckxx+ck+O1x3=12−c+13.
Hence, we conclude from (57) that 1/2−c+1/3≤0, and then, c≥1/3. Using the decreasing property of ψk′x on 0,∞, we get that c=1/3 is the best possible constant in (56). Also, Theorem 1 gives the right-hand side of inequality (56) for d=0. If there exists d>0 such that the right-hand side of (56) is valid for x∈0,∞, then we would have(59)−limx⟶0xψkx<k2limx⟶0xψk′x+kd.
From (34) and (35), we have limx⟶0xψkx=−1 and limx⟶0xψk′x+kd=0 which contradict with (59). Hence, the best possible constant in (56) is d=0.Remark 2.
The lower bound of (56) improves the lower bound of (21), when p⟶∞, for all x>0.Remark 3.
From Lemma1, we conclude that the lower bound of (56) improves the lower bound of (17) for all x>k/3.Corollary 5.
Forgandh≥0, xandk>0, and m=2,3,4,…, the following inequality holds:(60)−−1mk2ψkmx+gk<−1mψkm−1x−m−2!kxm−1<−−1mk2ψkmx+hk,with the best possible constants g=1/3 and h=0.Proof.
Form=2,3,…, the left-hand side of (60) is equivalent that(61)−1mxm+1kMc,kmx=−1mxm+1kVkx+xm+12k−1mkψkmx+gk+m−1!xm+km!6xm+1>0,where(62)−1mVkx=−1mψkm−1x−m−2!kxm−1−m−1!2xm−km!12xm+1.
Using asymptotic expansion (27), we have limx⟶∞−1mxm+1/kVkx=0 and(63)limx⟶∞xm+12k−1mkψkmx+gk+m−1!xm+km!6xm+1=limx⟶∞xm+12km−1!gk∑i=0m−1xm−1−ix+gkixmx+gkm+km!6xm+1−km!2x+gkm+1+O1xm+2,=m!2g−13.
Hence, we conclude from (61) that m!/2g−1/3≥0, and this leads to g≥1/3. Since ψk′x is strictly completely monotonic on 0,∞, then the function −1mψkmx is increasing on 0,∞ for m=0,1,2,…, and then, g=1/3 is the best possible constant in (60). Also, Theorem 1 gives the right-hand side of inequality (60) for h=0. If there exists h>0 such that the right-hand side of (60) is valid for x∈0,∞, then we would have(64)−1mlimx⟶0xmψkm−1x<−1m+1k2limx⟶0xmψkmx+hk,m=2,3,…,h>0.
From (34) and (35), we have limx⟶0xmψkm−1x=−1mm−1! and also limx⟶0xmψkmx+hk=0, and this contradicts with (64). Then, the best possible constant in (60) is h=0.Remark 4.
Form=2, the lower bound of (60) improves the lower bound of (22), when p⟶∞, for all x>0.Remark 5.
From Lemma1, we conclude that the lower bound of (60) for m=2 improves the lower bound of (16) for all x≥0.83k.Remark 6.
From Lemma1, we conclude that the lower bound of (60) for m=3 improves the lower bound of (19) for all x≥0.76k.Corollary 6.
The following inequality holds:(65)exp−k2ψkx−k6x−k/4<Γkx2π/kxx/ke−x/k<exp−k2ψkx−k6x,xandk>0.Proof.
The left-hand side of inequality (65) is equivalent U1/4,kx>0, and the right-hand side of inequality (65) is equivalent U0,kx<0 in Theorem 2.Remark 7.
If we letk=1 in inequality (65), then we obtain inequality (7).Remark 8.
The upper bound of inequality (65) improves the upper bound of inequality (54) for all x>0.Corollary 7.
The following inequality holds:(66)k2ψk′x−k6x−k/42<1klnx−ψkx<k2ψk′x−k6x2,xandk>0.Proof.
The left-hand side of inequality (66) is equivalent U1/4,k′x<0, and the right-hand side of inequality (66) is equivalent U0,k′x>0 in Theorem 2.Remark 9.
The upper bound of inequality (66) improves the upper bound of inequality (56) for all x>0.Corollary 8.
The following inequality holds:(67)−−1mk2ψkmx−km!6x−k/4m+1<−1mψkm−1x−m−2!kxm−1<−−1mk2ψkmx−km!6xm+1,xandk>0;m=2,3,…,.Proof.
The left-hand side of inequality (67) is equivalent −1mU1/4,kmx>0, and the right-hand side of inequality (67) is equivalent −1mU0,kmx<0 in Theorem 2.Remark 10.
The upper bound of inequality (67) improves the upper bound of inequality (60) for all x>0.


## 3.1. Some Inequalities forΓkx,ψkx, and ψkmx Functions

As a consequence of the completely monotonicity properties of the two functionsMλ,kx and Uβ,kx, we conclude the following results.Corollary 3.
Foraandb≥0, the following inequality holds:(54)exp−k2ψkx+ka<Γkx2π/kxx/ke−x/k<exp−k2ψkx+kb,xandk>0,with the best possible constants a=1/3 and b=0.Proof.
The left-hand side of inequality (54) is equivalent x/kMa,kx>0 which leads to a≥1/3 as mentioned in the proof of Theorem 1. Using the increasing property of ψkx on 0,∞, we get e−k/2ψkx+k/3≥e−k/2ψkx+ka for a≥1/3, and then, a=1/3 is the best possible constant in (54). Also, Theorem 1 gives the right-hand side of inequality (54) for b=0. If there exists b>0 such that the right-hand side of (54) is valid for x∈0,∞, then we would have(55)limx⟶0Γkx<2πkexp−k2ψkkb,which contradicts that limx⟶0Γkx=∞. Hence, b=0 is the best possible constant in (54).Remark 1.
If we putk=1 in inequality (54), then we get inequality (5).Corollary 4.
Forcandd≥0, the following inequality holds:(56)k2ψk′x+kc<1klnx−ψkx<k2ψk′x+kd,xandk>0,with the best possible constants c=1/3 and d=0.Proof.
The left-hand side of inequality (56) is equivalent that(57)x2kMc,k′x=x2kSkx+x22kkψk′x+kc−1x−k6x2<0,where Skx=ψkx−1/klnx+1/2x+k/12x2. Using asymptotic expansions (26) and (27), we have limx⟶∞x2/kSkx=0 and(58)limx⟶∞x22kkψk′x+kc−1x−k6x2=limx⟶∞x22kk2x2−2kcx−c2k26x2x+ck2−ckxx+ck+O1x3=12−c+13.
Hence, we conclude from (57) that 1/2−c+1/3≤0, and then, c≥1/3. Using the decreasing property of ψk′x on 0,∞, we get that c=1/3 is the best possible constant in (56). Also, Theorem 1 gives the right-hand side of inequality (56) for d=0. If there exists d>0 such that the right-hand side of (56) is valid for x∈0,∞, then we would have(59)−limx⟶0xψkx<k2limx⟶0xψk′x+kd.
From (34) and (35), we have limx⟶0xψkx=−1 and limx⟶0xψk′x+kd=0 which contradict with (59). Hence, the best possible constant in (56) is d=0.Remark 2.
The lower bound of (56) improves the lower bound of (21), when p⟶∞, for all x>0.Remark 3.
From Lemma1, we conclude that the lower bound of (56) improves the lower bound of (17) for all x>k/3.Corollary 5.
Forgandh≥0, xandk>0, and m=2,3,4,…, the following inequality holds:(60)−−1mk2ψkmx+gk<−1mψkm−1x−m−2!kxm−1<−−1mk2ψkmx+hk,with the best possible constants g=1/3 and h=0.Proof.
Form=2,3,…, the left-hand side of (60) is equivalent that(61)−1mxm+1kMc,kmx=−1mxm+1kVkx+xm+12k−1mkψkmx+gk+m−1!xm+km!6xm+1>0,where(62)−1mVkx=−1mψkm−1x−m−2!kxm−1−m−1!2xm−km!12xm+1.
Using asymptotic expansion (27), we have limx⟶∞−1mxm+1/kVkx=0 and(63)limx⟶∞xm+12k−1mkψkmx+gk+m−1!xm+km!6xm+1=limx⟶∞xm+12km−1!gk∑i=0m−1xm−1−ix+gkixmx+gkm+km!6xm+1−km!2x+gkm+1+O1xm+2,=m!2g−13.
Hence, we conclude from (61) that m!/2g−1/3≥0, and this leads to g≥1/3. Since ψk′x is strictly completely monotonic on 0,∞, then the function −1mψkmx is increasing on 0,∞ for m=0,1,2,…, and then, g=1/3 is the best possible constant in (60). Also, Theorem 1 gives the right-hand side of inequality (60) for h=0. If there exists h>0 such that the right-hand side of (60) is valid for x∈0,∞, then we would have(64)−1mlimx⟶0xmψkm−1x<−1m+1k2limx⟶0xmψkmx+hk,m=2,3,…,h>0.
From (34) and (35), we have limx⟶0xmψkm−1x=−1mm−1! and also limx⟶0xmψkmx+hk=0, and this contradicts with (64). Then, the best possible constant in (60) is h=0.Remark 4.
Form=2, the lower bound of (60) improves the lower bound of (22), when p⟶∞, for all x>0.Remark 5.
From Lemma1, we conclude that the lower bound of (60) for m=2 improves the lower bound of (16) for all x≥0.83k.Remark 6.
From Lemma1, we conclude that the lower bound of (60) for m=3 improves the lower bound of (19) for all x≥0.76k.Corollary 6.
The following inequality holds:(65)exp−k2ψkx−k6x−k/4<Γkx2π/kxx/ke−x/k<exp−k2ψkx−k6x,xandk>0.Proof.
The left-hand side of inequality (65) is equivalent U1/4,kx>0, and the right-hand side of inequality (65) is equivalent U0,kx<0 in Theorem 2.Remark 7.
If we letk=1 in inequality (65), then we obtain inequality (7).Remark 8.
The upper bound of inequality (65) improves the upper bound of inequality (54) for all x>0.Corollary 7.
The following inequality holds:(66)k2ψk′x−k6x−k/42<1klnx−ψkx<k2ψk′x−k6x2,xandk>0.Proof.
The left-hand side of inequality (66) is equivalent U1/4,k′x<0, and the right-hand side of inequality (66) is equivalent U0,k′x>0 in Theorem 2.Remark 9.
The upper bound of inequality (66) improves the upper bound of inequality (56) for all x>0.Corollary 8.
The following inequality holds:(67)−−1mk2ψkmx−km!6x−k/4m+1<−1mψkm−1x−m−2!kxm−1<−−1mk2ψkmx−km!6xm+1,xandk>0;m=2,3,…,.Proof.
The left-hand side of inequality (67) is equivalent −1mU1/4,kmx>0, and the right-hand side of inequality (67) is equivalent −1mU0,kmx<0 in Theorem 2.Remark 10.
The upper bound of inequality (67) improves the upper bound of inequality (60) for all x>0.



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