In the paper, new upper bounds in the second Kershaw’s double

inequality and its generalizations involving the gamma, psi and polygamma

functions are established, some known results are refined

Guo, Senlin

Qi, Feng

English

Victoria University Eprints Repository

NEW UPPER BOUNDS IN THE SECOND KERSHAW’S DOUBLEINEQUALITY AND ITS GENERALIZATIONSFENG QI AND SENLIN GUOAbstract. In the paper, new upper bounds in the second Kershaw’s doubleinequality and its generalizations involving the gamma, psi and polygammafunctions are established, some known results are refined.1. IntroductionIt is well known thatG(a, b) =√ab , L(a, b) =b− aln b− ln a, (1)I(a, b) =1e(bbaa)1/(b−a), A(a, b) =a+ b2(2)for positive numbers a and b with a 6= b are called respectively the geometric mean,the logarithmic mean, the identric or exponential mean and the arithmetic meanand that inequalitiesG(a, b) < L(a, b) < I(a, b) < A(a, b) (3)are valid. See [4, 20] and the references therein.In [10], the following two double inequalities were established:(x+s2)1−s<Γ(x+ 1)Γ(x+ s)<(x− 12+√s+14)1−s, (4)exp[(1− s)ψ(x+√s )] < Γ(x+ 1)Γ(x+ s)< exp[(1− s)ψ(x+s+ 12)], (5)where 0 < s < 1, x ≥ 1, Γ stands for the classical Euler’s gamma function, ψdenotes the logarithmic derivative of Γ, and ψ(i) for i ∈ N are called the polygammaDate: Completed on 25 January 2007.2000 Mathematics Subject Classification. 26A48, 26A51, 26D20, 33B10, 33B15, 65R10.Key words and phrases. Kershaw’s double inequality, identric mean, exponential mean, gammafunction, psi function, polygamma function, inequality, upper bound, generalization.This paper was typeset using AMS-LATEX.12 F. QI AND S. GUOfunctions. Inequalities (4) and (5) are called in the literature the first and secondKershaw’s double inequality respectively.In [2, Theorem 2.7], the second Kershaw’s double inequality (5) was generalizedand extended as follows:−∣∣ψ(n+1)(A(x, y))∣∣ > ∣∣ψ(n)(x)∣∣− ∣∣ψ(n)(y)∣∣x− y > −∣∣ψ(n+1)(L−(n+2)(x, y))∣∣, (6)where x and y are positive numbers and n is a positive integer.Recently, the following generalization, extension and refinement of the secondKershaw’s double inequality (5) was obtained in [16]: For s, t ∈ R with s 6= t, thefunction [Γ(x+ s)Γ(x+ t)]1/(s−t) 1eψ(L(s,t;x))(7)is decreasing in x > −min{s, t}. In particular, for s, t ∈ R and x > −min{s, t}with s 6= t,eψ(L(s,t;x)) <[Γ(x+ s)Γ(x+ t)]1/(s−t)< eψ(A(s,t;x)), (8)where L(s, t;x) = L(x + s, x + t) and A(s, t;x) = A(x + s, x + t) for s, t ∈ R andx > −min{s, t} with s 6= t.There have been a lot of literature about Kershaw’s two double inequalities andtheir history, background, refinements, extensions, generalizations and applications.For more information, please refer to [5, 7, 8, 9, 11, 12, 13, 14, 15, 21, 22, 23, 24,26, 28, 31, 33] and the references therein.The aim of this paper is to refine the right hand side inequality in (8) and theleft hand side inequality in (6). These results refine, extend and generalize the righthand side inequality in the second Kershaw’s double inequality (5).The main results of this paper are the following theorems.Theorem 1. For s, t ∈ R with s 6= t and x > −min{s, t}, inequality[Γ(x+ s)Γ(x+ t)]1/(s−t)≤ eψ(I(s,t;x)) (9)holds, where I(s, t;x) = I(x+ s, x+ t).Theorem 2. For s, t ∈ R with s 6= t and x > −min{s, t}, inequality(−1)n[ψ(n−1)(x+ s)− ψ(n−1)(x+ t)]s− t ≤ (−1)nψ(n)(I(s, t;x)), (10)holds, where I(s, t;x) = I(x+ s, x+ t) and n ∈ N.NEW UPPER BOUNDS IN THE SECOND KERSHAW’S DOUBLE INEQUALITY 3In Section 2, a key and necessary lemma is presented. In Section 3, a simplemethod and polished techniques are employed to verify Theorem 1 and Theorem 2.In Section 4, some remarks on Theorem 1 and Theorem 2, the first and second Ker-shaw’s double inequalities (4) and (5), the (logarithmically) complete monotonicityof mean values and two conjectures are given.2. A lemmaIn order to prove our theorems, the following lemma is key and necessary.Lemma 1. For positive numbers a and b with a 6= b, inequality(−1)ib− a∫ baψ(i)(u) du ≤ (−1)iψ(i)(I(a, b)) (11)is valid for all nonnegative integer i, where I(a, b) stands for the exponential oridentric mean.Proof. It was obtained in [6] (see also [4, p. 274, Lemma 2]) that if g is strictlymonotonic, f is strictly increasing and f ◦ g−1 is convex (or concave, respectively)on an interval I, theng−1(1t− s∫ tsg(u) du)≤ f−1(1t− s∫ tsf(u) du)(12)holds (or reverses, respectively) for s, t ∈ I. It is easy to see that the functionsf(x) = (−1)iψ(i)(x) for i ≥ 0 and g(x) = lnx are strictly increasing and g−1(x) =ex. Direct computation givesg−1(1t− s∫ tsg(u) du)= I(s, t), (13)h(x) , f ◦ g−1(x) = (−1)iψ(i) (ex) (14)andh′′(x) = (−1)iex[ψ(i+1) (ex) + exψ(i+2) (ex)] = (−1)iu[ψ(i+1)(u) + uψ(i+2)(u)],where u = ex. It was proved in [32] that the function x∣∣ψ(i+1)(x)∣∣ − α∣∣ψ(i)(x)∣∣is completely monotonic in (0,∞) if and only if 0 ≤ α ≤ i ∈ N. This impliesh′′(x) ≤ 0 in (0,∞). Consequently, the function h(x) is concave for all nonnegativeintegers i ≥ 0. Thus, the conditions of the reversed inequality of (12) are satisfiedby taking f(x) = (−1)iψ(i)(x) and g(x) = lnx for i ≥ 0. This leads to (11). Theproof of Lemma 1 is complete. ¤4 F. QI AND S. GUO3. Proofs of theoremsNow we are in a position to prove simply and elegantly our main results statedin Section 1.Proof of Theorem 1. LetFs,t(x) =[Γ(x+ s)Γ(x+ t)]1/(t−s)eψ(I(s,t;x)) (15)for s, t ∈ R with s 6= t and x > −min{s, t}. Taking logarithm of Fs,t(x) and usingthe mean value theorem yieldslnFs,t(x) = ψ(I(s, t;x))− ln Γ(x+ s)− ln Γ(x+ t)s− t= ψ(I(s, t;x))− 1s− t∫ stψ(x+ u) du.Applying inequality (11) to i = 0 and both a = x + s and b = x + t shows thatlnFs,t(x) ≥ 0 and Fs,t(x) ≥ 1 which is equivalent to (9). The proof of Theorem 1is complete. ¤Proof of Theorem 2. Inequality (6) can be rewritten as(−1)n+1s− t∫ stψ(n+1)(x+ u) du ≤ (−1)n+1ψ(n+1)(I(s, t;x)),which follows from inequality (11). The proof of Theorem 2 is complete. ¤4. RemarksRemark 1. Since I(a, b) < A(a, b) for positive numbers a and b with a 6= b, in-equality (9) refines the right hand side inequalities in (5) and (8). This means thatTheorem 1 refines, extends and generalizes the right hand side inequality in thesecond Kershaw’s double inequality (5).By the same argument, it is easy to see that inequality (10) refines the left handside inequality in (6).Remark 2. The case of n = 1 in inequality (10) is not included in [2, Theorem 2.7].Remark 3. Recently, the following sufficient and necessary conditions are presentedin [28]: For real numbers a, b, c and ρ = min{a, b, c}, the functionHa,b,c(x) = (x+ c)b−aΓ(x+ a)Γ(x+ b)(16)NEW UPPER BOUNDS IN THE SECOND KERSHAW’S DOUBLE INEQUALITY 5is logarithmically completely monotonic in x ∈ (−ρ,∞) if and only if (a, b, c) ∈{(a, b, c) : (b − a)(1 − a − b + 2c) ≥ 0} ∩ {(a, b, c) : (b − a)(|a − b| − a − b + 2c) ≥0} \ {(a, b, c) : a = c + 1 = b + 1} \ {(a, b, c) : b = c + 1 = a + 1}, and thefunction Hb,a,c(x) is logarithmically completely monotonic in (−ρ,∞) if and only if(a, b, c) ∈ {(a, b, c) : (b−a)(1−a−b+2c) ≤ 0}∩{(a, b, c) : (b−a)(|a−b|−a−b+2c) ≤0} \ {(a, b, c) : b = c + 1 = a + 1} \ {(a, b, c) : a = c + 1 = b + 1}. From this, thebest bounds in the first Kershaw’s double inequality (4) can be deduced.Remark 4. Recall [25, 27, 29, 30] that a function f is said to be logarithmically com-pletely monotonic on an interval I if its logarithm ln f satisfies (−1)k[ln f(x)](k) ≥ 0for k ∈ N on I. Recall also [1] that if f (k)(x) for some nonnegative integer k is com-pletely monotonic on an interval I, but f (k−1)(x) is not completely monotonic onI, then f(x) is called a completely monotonic function of k-th order on an intervalI. It has been proved in [3, 17, 25, 27] that a logarithmically completely monotonicfunction on an interval I must be completely monotonic on I. The logarithmicallycompletely monotonic functions have close relationships with both the completelymonotonic functions and Stieltjes transforms. For detailed information, please referto [3, 17, 18, 30, 34] and the references therein.In [19], it was proved that the logarithmic mean L(s, t;x) is a completely mono-tonic function of first order in x > −min{s, t} for s, t ∈ R with s 6= t. By standardargument, it is easy to verify that the reciprocal of the identric mean I(s, t;x) islogarithmically completely monotonic in x > −min{s, t} for s, t ∈ R with s 6= t.Remark 5. Formula (13) gives an integral representation of the exponential oridentric mean I(s, t) for positive numbers s and t as follows:I(s, t) = exp(1t− s∫ tslnudu)(17)which is not found in the book [4]. From this, it is easy to obtain that the identricmean I(s, t;x) for s, t ∈ R and x > −min{s, t} is a completely monotonic functionof first order.Remark 6. It is conjectured that the functions defined by (7) and (15) are loga-rithmically completely monotonic in x > −min{s, t} for s, t ∈ R with s 6= t.6 F. QI AND S. GUORemark 7. Let p ∈ R and a and b be positive numbers with a 6= b. The generalizedlogarithmic mean of order p of a and b is defined [4, p. 385] byLp(a, b) =[bp+1 − ap+1(p+ 1)(b− a)]1/p, p 6= −1, 0;b− aln b− ln a, p = −1;1e(bbaa)1/(b−a), p = 0.(18)Let Lp(s, t;x) = Lp(x + s, x + t) for s, t ∈ R and x > −min{s, t} with s 6= t. It isnatural to ask for the best number p such that the functioneψ(Lp(s,t;x))[Γ(s+ x)Γ(t+ x)] 1t−s(19)is either decreasing, or increasing, or (logarithmically) completely monotonic in x.References[1] R. D. Atanassov and U. V. Tsoukrovski, Some properties of a class of logarithmically com-pletely monotonic functions, C. R. Acad. Bulgare Sci. 41 (1988), no. 2, 21–23.[2] N. Batir, On some properties of digamma and polygamma functions, J. Math. Anal. Appl.(2006), http://dx.doi.org/10.1016/j.jmaa.2006.05.065.[3] C. Berg, Integral representation of some functions related to the gamma function, Mediterr.J. Math. 1 (2004), no. 4, 433–439.[4] P. S. Bullen, Handbook of Means and Their Inequalities, Mathematics and its Applications,Volume 560, Kluwer Academic Publishers, Dordrecht/Boston/London, 2003.[5] J. Bustoz and M. E. H. Ismail, On gamma function inequalities, Math. Comp. 47 (1986),659–667.[6] G. T. Cargo, Comparable means and generalized convexity, J. Math. Anal. 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Qi) Research Institute of Mathematical Inequality Theory, Henan PolytechnicUniversity, Jiaozuo City, Henan Province, 454010, ChinaE-mail address: qifeng@hpu.edu.cn, fengqi618@member.ams.org, qifeng618@hotmail.com,qifeng618@msn.com, qifeng618@qq.com, qifeng618@gmail.comURL: http://rgmia.vu.edu.au/qi.html(S. Guo) Department of Mathematics, University of Manitoba, Winnipeg, MB, R3T2N2, CanadaE-mail address: sguo@hotmail.com

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New Upper Bounds in the Second Kershaw's Double Inequality and its Generalizations

Victoria University Institutional Repository (VUIR)

In the paper, new upper bounds in the second Kershaw’s double\ud
inequality and its generalizations involving the gamma, psi and polygamma\ud
functions are established, some known results are refined

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New Upper Bounds in the Second Kershaw's Double Inequality and its Generalizations

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