Qi, Feng

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A Class of Logarithmically Completely Monotonic Functions and Application to the Best Bounds in the Second Gautschi-Kershaw's Inequality

Victoria University Eprints Repository

A CLASS OF LOGARITHMICALLY COMPLETELY MONOTONICFUNCTIONS AND APPLICATION TO THE BEST BOUNDS INTHE SECOND GAUTSCHI-KERSHAW’S INEQUALITYFENG QIAbstract. In this article, the logarithmically complete monotonicity of thefunction [Γ(x+ b)/Γ(x+ a)]1/(a−b) exp[ψ(x + c)] are discussed, where a, b, care real numbers and Γ is the classical Euler’s gamma function. From this, thebest upper and lower bounds for Walls’ ratio Γ(x+ 1)/Γ(x+ s) are established,which refine the second Gautschi-Kershaw’s inequality.1. IntroductionIt is well known that the classical Euler’s gamma function Γ can be defined forx > 0 as Γ(x) =∫∞0e−ttx−1 d t. The digamma or psi function ψ is defined as thelogarithmic derivative of Γ and ψ(i) for i ∈ N are called polygamma functions.Recall [26] that a function f is said to be completely monotonic on an intervalI if f has derivatives of all orders on I and (−1)nf (n)(x) ≥ 0 for x ∈ I andn ≥ 0. Recall [2, 13, 20, 21, 22, 23] also that a function f is called logarithmicallycompletely monotonic on an interval I if f has derivatives of all orders on I and itslogarithm ln f satisfies 0 ≤ (−1)k[ln f(x)](k) < ∞ for all k ∈ N on I. For our ownconvenience, the sets of the completely monotonic functions and the logarithmicallycompletely monotonic functions on I are denoted respectively by C[I] and L[I]. In[2, 20, 21], it has been proved that L[I] ⊂ C[I]. For more information on the classesC[I] and L[I], please refer to [2, 13, 20, 21, 22, 23] and the references therein.The first and second Gautschi-Kershaw inequalities [5, 7, 9, 24] state that(x+s2)1−s<Γ(x+ 1)Γ(x+ s)<(x− 12+√s+14)1−s(1)andexp[(1− s)ψ(x+√s )] < Γ(x+ 1)Γ(x+ s)< exp[(1− s)ψ(x+s+ 12)](2)for s ∈ (0, 1) and x ≥ 1.2000 Mathematics Subject Classification. Primary 33B15, 65R10; Secondary 26A48, 26A51,26D20.Key words and phrases. logarithmically completely monotonic function, Gautschi-Kershaw’sinequality, gamma function, psi function, Wallis’s ratio.The author was supported in part by the Science Foundation of the Project for FosteringInnovation Talents at Universities of Henan Province, China.This paper was typeset using AMS-LATEX.12 F. QIIn [15], among other things, the increasing monotonicity of [Γ(s)/Γ(r)]1/(s−r) fors > 0 and r > 0 and inequalityexp [(s− r)ψ(s)] > Γ(s)Γ(r)> exp [(s− r)ψ(r)] (3)for s > r > 0 were obtained.Inequalities (1), (2) and (3) give the lower and upper bounds for the well knownWallis’ ratio Γ(x+ 1)/Γ(x+ s).In [3], it was proved that the functionsΓ(x+ 1)Γ(x+ s)(x+s2)s−1∈ C[(0,∞)] (4)andΓ(x+ s)Γ(x+ 1)exp[(1− s)ψ(x+s+ 12)]∈ C[(0,∞)] (5)for s ∈ (0, 1).Let s and t be nonnegative numbers and α = min{s, t}. In [5, Theorem 5] and[24], the result (5) was generalized to[Γ(x+ s)Γ(x+ t)]1/(t−s)exp[ψ(x+s+ t2)]∈ L[(−α,∞)]. (6)In [14, 18], the monotonicity of the functionqα,β(t) =e−αt − e−βt1− e−t , t 6= 0β − α, t = 0(7)for real numbers α and β with (α, β) 6∈ {(0, 1), (1, 0)} and α 6= β was established,and then the paper [10] considered the logarithmically complete monotonicity ofthe more general functionHa,b,c(x) = (x+ c)b−aΓ(x+ a)Γ(x+ b)(8)for x ∈ (−ρ,∞), where a, b and c are real numbers and ρ = min{a, b, c}, andobtained the following conclusions:(1) Ha,b,c(x) ∈ L[(−ρ,∞)] if (a, b, c) ∈{a + b ≥ 1, c ≤ b < c + 12} ∪ {a > b ≥c+ 12}∪{2a+1 ≤ a+b ≤ 1, a < c}∪{b−1 ≤ a < b ≤ c}\{a = c+1, b = c}.(2) [Ha,b,c(x)]−1 ∈ L[(−ρ,∞)] if (a, b, c) ∈{(a, b, c) : a+ b ≥ 1, c ≤ a < c+ 12}∪ {(a, b, c) : b > a ≥ c + 12} ∪ {(a, b, c) : b < a ≤ c} ∪ {(a, b, c) : b + 1 ≤a, c ≤ a ≤ c+1}∪{(a, b, c) : b+c+1 ≤ a+b ≤ 1}\{(a, b, c) : a = c+1, b =c} \ {(a, b, c) : b = c+ 1, a = c}.These (logarithmically) complete monotonicity mentioned above can be appliedto acquire the best bounds in Gautschi-Kershaw’s inequalities (1) and (2). For moredetailed information, please refer to [4, 5, 6, 7, 8, 9, 10, 11, 12, 15, 16, 17, 19, 24, 25]and the references therein.The main aim of this article is to generalize the logarithmically complete mono-tonicity (6). The main result of this paper is the following Theorem 1.A CLASS OF LOGARITHMICALLY COMPLETELY MONOTONIC FUNCTIONS 3Theorem 1. Let a, b and c be real numbers and ρ = min{a, b, c}. DefineFa,b,c(x) =[Γ(x+ b)Γ(x+ a)]1/(a−b)exp[ψ(x+ c)], a 6= bexp[ψ(x+ c)− ψ(x+ a)], a = b 6= c(9)for x ∈ (−ρ,∞). Furthermore, let θ(t) be an implicit function defined by equationet − t = eθ(t) − θ(t) (10)in (−∞,∞). Then θ(t) is decreasing and tθ(t) < 0 for θ(t) 6= t, and(1) Fa,b,c(x) ∈ L[(−ρ,∞)] if (a, b, c) ∈ D1(a, b, c), whereD1(a, b, c) = {c ≥ a, c ≥ b} ∪ {c ≥ a, 0 ≥ c− b ≥ θ(c− a)}∪ {c ≤ a, c− b ≥ θ(c− a)} \ {a = b = c}; (11)(2) [Fa,b,c(x)]−1 ∈ L[(−ρ,∞)] if (a, b, c) ∈ D2(a, b, c), whereD2(a, b, c) = {c ≤ a, c ≤ b} ∪ {c ≥ a, c− b ≤ θ(c− a)}∪ {c ≤ a, 0 ≤ c− b ≤ θ(c− a)} \ {a = b = c}. (12)Remark 1. The numerical computation of θ(t) defined by (10) can be carried outby using the well known software Mathematica 5.2, for example, as follows:θ(0.5) = −0.599 · · · , θ(1) = −1.4937 · · · , θ(1.5) = −2.928 · · · ,θ(2) = −5.3844 · · · , θ(2.5) = −9.6824 · · · , θ(3) = −17.085 · · · ,θ(−0.5) = 0.42864 · · · , θ(−1) = 0.75078 · · · , θ(−1.5) = 1.0028 · · · ,θ(−2) = 1.2065 · · · , θ(−2.5) = 1.3756 · · · , θ(−3) = 1.5193 · · · .As an application of Theorem 1, the following inequalities are obtained.Theorem 2. Let D1(a, b, c) and D2(a, b, c) be defined by (11) and (12) respectively.If (a, b, c) ∈ D1(a, b, c), then[Γ(x+ b)Γ(x+ a)]1/(b−a)< exp[ψ(x+ c)] (13)for x ∈ (−ρ,∞) and[Γ(x+ b)Γ(x+ a)]1/(b−a)≥[Γ(δ + b)Γ(δ + a)]1/(b−a)exp[ψ(x+ c)− ψ(δ + c)] (14)for x ∈ [δ,∞) are valid, where δ is a constant greater than −ρ.If (a, b, c) ∈ D2(a, b, c), inequalities (13) and (14) are reversed.As a direct consequence of Theorem 2, the best lower and upper bounds forWallis’s ratio Γ(x+ 1)/Γ(x+ s) are established below, which improve the secondGautschi-Kershaw’s inequality (2) and inequality (3).Theorem 3. Let θ(t) be defined by (10), p(t) = t− θ(t− 1) in (−∞,∞) and p−1stand for the inverse function of p. Then inequalitiesΓ(x+ 1)Γ(x+ s)< exp[(1− s)ψ(x+ p−1(s))] (15)4 F. QIfor x ∈ (−s,∞) andΓ(x+ 1)Γ(x+ s)≥ Γ(δ + 1)Γ(δ + s)exp[ψ(x+ p−1(s))− ψ(δ + p−1(s))] (16)for x ∈ (δ,∞) are valid for s ∈ (0, 1), where δ > −s and s ≤ p−1(s) ≤ 1.Remark 2. By the software Mathematica 5.2, some values of p(t) are calculatednumerically:p(0.1) = −0.59286 · · · , p(0.2) = −0.43195 · · · , p(0.3) = −0.26780 · · · ,p(0.4) = −0.10013 · · · , p(0.5) = 0.071355 · · · , p(0.6) = 0.24702 · · · ,p(0.7) = 0.42726 · · · , p(0.8) = 0.61249 · · · , p(0.9) = 0.80322 · · · .This shows that 0.5 < p−1(s) < 1 for s ∈ (0, 1) approximately.2. Proofs of theoremsProof of Theorem 1. It is well known [1] thatψ(x) =∫ ∞0(e−uu− e−xu1− e−u)du (17)for x > 0.For a = b, by (17), it is clear thatlnFa,a,c(x) = ψ(x+ c)− ψ(x+ a) =∫ ∞0e−au − e−cu1− e−u e−xu duand(−1)i[lnFa,a,c(x)](i) =∫ ∞0e−au − e−cu1− e−u uie−xu dufor i ∈ N. Therefore Fa,a,c(x) ∈ L[(−a,∞)] if a = b < c and [Fa,a,c(x)]−1 ∈L[(−c,∞)] if c < a = b.For b 6= a, taking the logarithm of the function Fa,b,c(x), differentiating andusing (17) yieldslnFa,b,c(x) = ψ(x+ c)− ln Γ(x+ b)− ln Γ(x+ a)b− a= ψ(x+ c)− 1b− a∫ baψ(x+ t) d t=∫ ∞0(1b− a∫ bae−tu d t− e−cu)e−xu1− e−u du=∫ ∞0[e(c−a)u − e(c−b)uu(b− a) − 1]e−(x+c)u1− e−u duand(−1)i[lnFa,b,c(x)](i) =∫ ∞0[e(c−a)u − e(c−b)uu(b− a) − 1]uie−(x+c)u1− e−u du,∫ ∞0[gc−a,c−b(u)− 1]uie−(x+c)u1− e−u dufor i ∈ N.A CLASS OF LOGARITHMICALLY COMPLETELY MONOTONIC FUNCTIONS 5Lethα,β(u) , gα,β(u)− 1 = [eαu − αu]− [eβu − βu](α− β)u (18)for u > 0 and (α, β) ∈ R2 with α 6= β. It is easy to see that the function et − tis decreasing in (−∞, 0) and increasing in (0,∞). See Figure 1. Consequently, ifα ≥ 0 and β ≥ 0, the function hα,β(u) is positive in (0,∞); if 0 ≥ α and 0 ≥ β,the function hα,β(u) is negative in u ∈ (0,∞). Let θ(t) be defined by (10). It ist1 t2Plot?Exp?t? ? t, ?t, ?9, 3??-8 -6 -4 -2 22.557.51012.51517.5Figure 1. The graph of et − t in (−9, 3) pictured by Mathematica 5.2apparent that tθ(t) < 0 for θ(t) 6= t. As a result, if α ≥ 0 and 0 ≥ β ≥ θ(α), thenhα,β(u) is positive; if α ≥ 0 and β ≤ θ(α), then the function hα,β(u) is negative in(0,∞); if α ≤ 0 and 0 ≤ β ≤ θ(α), then hα,β(u) is negative; if α ≤ 0 and β ≥ θ(α),then the function hα,β(u) is positive in (0,∞)From the positivity or negativity of the function hα,β(u), the logarithmicallycompletely monotonicity of the function Fa,b,c(x) is obtained:(1) If either c − a ≥ 0 and c − b ≥ 0 or c − a ≥ 0 and 0 ≥ c − b ≥ θ(c − a) orc− a ≤ 0 and c− b ≥ θ(c− a), then Fa,b,c(x) ∈ L[(−ρ,∞)];(2) If either c−a ≤ 0 and c−b ≤ 0 or c−a ≥ 0 and c−b ≤ θ(c−a) or c−a ≤ 0and 0 ≤ c− b ≤ θ(c− a), then [Fa,b,c(x)]−1 ∈ L[(−ρ,∞)].The proof of Theorem 1 is complete. ¤Proof of Theorem 2. For a and b being two constants, as x → ∞, the followingasymptotic formula is given in [1, p. 257 and p. 259]:xb−aΓ(x+ a)Γ(x+ b)= 1 +(a− b)(a+ b− 1)2x+O(1x2)= 1 +O(1x). (19)In [17], it was proved that ψ(x) − lnx + α/x ∈ C[(0,∞)] if and only if α ≥ 1 andlnx− α/x−ψ(x) ∈ C[(0,∞)] if and only if α ≤ 1/2. From this, it is deduced thatlnx− 1x< ψ(x) < lnx− 12x(20)in (0,∞). Utilization of (19) and (20) leads tolimx→∞Fa,b,c(x) = limx→∞{exp[ψ(x+ c)]x[1 +O(1x)]1/(b−a)}6 F. QI= limx→∞exp[ψ(x+ c)]x= 1.Hence, if (a, b, c) ∈ D1(a, b, c), then Fa,b,c(x) > 1 which is equivalent to inequality(13); if (a, b, c) ∈ D2(a, b, c), then Fa,b,c(x) < 1 which is equivalent to the reversedinequality of (13).Let δ be a constant greater than −ρ. If (a, b, c) ∈ D1(a, b, c), then Fa,b,c(x) ≤Fa,b,c(δ) in x ∈ [δ,∞), which is equivalent to inequality (14); if (a, b, c) ∈ D2(a, b, c),then Fa,b,c(x) ≥ Fa,b,c(δ) in x ∈ [δ,∞), which is equivalent to the reversed inequalityof (14). The proof of Theorem2 is complete. ¤Proof of Corollary 3. Taking a = 1 and b = s ∈ (0, 1) in (13) and (14) leads toΓ(x+ 1)Γ(x+ s)< exp[(1− s)ψ(x+ c)] (21)for x ∈ (−ρ,∞) andΓ(x+ 1)Γ(x+ s)≥ Γ(δ + 1)Γ(δ + s)exp{(1− s)[ψ(x+ c)− ψ(δ + c)]} (22)for x ∈ (δ,∞), where δ > −ρ, c ≤ 1 and c − s ≥ θ(c − 1) ≥ 0. Since θ(t) isstrictly decreasing in t ∈ (−∞,∞), then the function p(t) is strictly increasing int ∈ (−∞,∞). Thus, inequalities (21) and (22) validate for c ≥ p−1(s) and s ∈ (0, 1).Since ψ(x + c) is increasing and ψ(x + c) − ψ(δ + c) for x > δ is decreasing withrespect to c, inequalities (21) and (22) have a best upper bound and a best lowerbound exp[(1−s)ψ(x+p−1(s))] and [Γ(δ + b)/Γ(δ + a)]1/(b−a) exp[ψ(x+p−1(s))−ψ(δ + p−1(s))]respectively. The proof of Corollary 3 is complete. ¤References[1] M. Abramowitz and I. A. 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Qi) College of Mathematics and Information Science, Henan Normal University,Xinxiang City, Henan Province, 453007, China; Research Institute of Mathematical In-equality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010,ChinaE-mail address: qifeng@hpu.edu.cn, fengqi618@member.ams.org, qifeng618@hotmail.com,qifeng618@msn.com, 316020821@qq.comURL: http://rgmia.vu.edu.au/qi.html

A Class of Logarithmically Completely Monotonic Functions and Application to the Best Bounds in the Second Gautschi-Kershaw's Inequality

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