In this paper, a function involving the divided difference of the

psi function is proved to be completely monotonic, a class of inequalities involving

sum are founded, and an equivalent relation between the complete

monotonicity and one of the class of inequalities is established

Qi, Feng

English

Victoria University Eprints Repository

A COMPLETELY MONOTONIC FUNCTION INVOLVINGDIVIDED DIFFERENCE OF PSI FUNCTION AND ANEQUIVALENT INEQUALITY INVOLVING SUMFENG QIAbstract. In this paper, a function involving the divided difference of thepsi function is proved to be completely monotonic, a class of inequalities in-volving sum are founded, and an equivalent relation between the completemonotonicity and one of the class of inequalities is established.1. IntroductionIt is well known that the classical Euler’s gamma function Γ(x) is one of the mostimportant special functions and has much extensive applications in many branches,for example, statistics, physics, engineering, and other mathematical sciences. Thelogarithmic derivative of Γ(x), denoted by ψ(x) = Γ′(x)/Γ(x), is called psi ordigamma function, and the derivatives ψ(i)(x) for i ∈ N of the psi function ψ(x)are known as the polygamma or multigamma functions.Recall [10] that a function f is said to be completely monotonic on an intervalI if f has derivatives of all orders on I and 0 ≤ (−1)nf (n)(x) < ∞ for x ∈ I andn ≥ 0, and that a positive function f is called logarithmically completely mono-tonic on an interval I if f has derivatives of all orders on I and its logarithm ln fsatisfies 0 ≤ (−1)k[ln f(x)](k) < ∞ for all k ∈ N on I. The set of the completelymonotonic functions on I is denoted by C[I] and the set of the logarithmically com-pletely monotonic functions on I is denoted by L[I]. It is proved in [3, 10, 13, 15]that L[I] ⊂ C[I]. The well known Bernstein’s Theorem [16, p. 161] states thatf ∈ C[(0,∞)] if and only if f(x) = ∫∞0e−xs dµ(s), where µ is a nonnegative mea-sure on [0,∞) such that the integral converges for all x > 0. This expresses thatf ∈ C[(0,∞)] is a Laplace transform of the measure µ. In [3, Theorem 1.1] and[8, 13] it is pointed out that the logarithmically completely monotonic functions on(0,∞) can be characterized as the infinitely divisible completely monotonic func-tions. In the recent past, numerous functions, which are defined in terms of gamma,polygamma, and other special functions, are proved to be (logarithmically) com-pletely monotonic and this fact is used to derive many interesting new inequalities(See, for examples, [1, 3, 4, 5, 6, 8, 10, 13, 14, 15] and the references therein).2000 Mathematics Subject Classification. Primary 33B15, 26D15; Secondary 26A48, 26A51.Key words and phrases. completely monotonic function, psi function, polygamma function,inequality, sum, equivalence.The author was supported in part by the Science Foundation of Project for Fostering InnovationTalents at Universities of Henan Province, China.This paper was typeset using AMS-LATEX.12 F. QIThe Kershaw’s inequality [9] states that(x+s2)1−s<Γ(x+ 1)Γ(x+ s)<(x− 12+√s+14)1−s(1)for 0 < s < 1 and x ≥ 1, which improved the corresponding result in [7]. In [4, 6,11, 14], in order to obtain the best upper and lower bounds for double inequality(1), the monotonicity and convexity properties of the function[Γ(x+ t)Γ(x+ s)]1/(t−s)− x (2)in x ∈ (−α,∞) is verified, where s and t are nonnegative numbers and α =min{s, t}. In [4, 6, 11, 14], in order to prove the monotonicity and convexity of(2), the functionδs,t(x) =ψ(x+ t)− ψ(x+ s)t− s −2x+ s+ t+ 12(x+ s)(x+ t), s 6= tψ′(x+ s)− 1x+ s− 12(x+ s)2, s = t(3)in x ∈ (−α,∞), involving the divided difference of the psi function, is derived.Recently, the positivity of the function δ0,0(x) = ψ′(x) − 1/x − 1/2x2 was provedin [2, 5, 12] respectively.The first aim of this paper is to prove the complete monotonicity property of thefunction δs,t(x).Theorem 1. Let s and t be nonnegative numbers and α = min{s, t}. Then thefunctions δs,t(x) for |t− s| < 1 and −δs,t(x) for |t− s| > 1 are completely mono-tonic in x ∈ (−α,∞).The second aim of this paper is to found a class of inequalities involving sum.Theorem 2. Let k be a nonnegative integer and θ > 0 a constant.If a > 0 and b > 0, thenk∑i=01(a+ θ)i+1(b+ θ)k−i+1+k∑i=01ai+1bk−i+1> 2k∑i=01(a+ θ)i+1bk−i+1(4)holds for b− a > −θ and reveres for b− a < −θ.If a < −θ and b < −θ, then inequalities2k∑i=01(a+ θ)i+1(b+ θ)2k−i+1+2k∑i=01ai+1b2k−i+1> 22k∑i=01(a+ θ)i+1b2k−i+1(5)and2k+1∑i=01(a+ θ)i+1(b+ θ)2k−i+2+2k+1∑i=01ai+1b2k−i+2< 22k+1∑i=01(a+ θ)i+1b2k−i+2(6)hold for b− a > −θ and reverse for b− a < −θ.If −θ < a < 0 and −θ < b < 0, then inequality (5) holds and inequality (6) isvalid for a+ b+ θ > 0 and is reversed for a+ b+ θ < 0.If a < −θ and b > 0, then inequality (5) holds and inequality (6) is valid fora+ b+ θ > 0 and is reversed for a+ b+ θ < 0.A COMPLETELY MONOTONIC FUNCTION AND AN EQUIVALENT INEQUALITY 3If a > 0 and b < −θ, then inequality (5) is reversed and inequality (6) holds fora+ b+ θ < 0 and reverses for a+ b+ θ > 0.If b = a− θ, then inequalities (4), (5) and (6) become equalities.Remark 1. The following shadows in Figure 1 describes the domains on the planeaOb such that Theorem 2 being valid.-a−θ O(−θ,−θ) r6b−θ````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````Figure 1.Remark 2. Hinted by Theorem 2, the following open problem is posed: Let k be anonnegative integer and θ a positive constant, discuss the validity of the inequality(a+ b)kak+1bk+1+(a+ b+ 2θ)k(a+ θ)k+1(b+ θ)k+1>2(a+ b+ θ)k(a+ θ)k+1bk+1(7)or its reverse for a, b 6∈ {0,−θ}.The final aim of this article is to establish an equivalent relationship betweenTheorem 1 and inequality (4) for positive numbers a and b.Theorem 3. Inequality (4) for positive numbers a and b is equivalent to Theorem 1.2. Proofs of theoremsThe basic tool of this paper is the following lemma.Lemma 1. Let f(x) be defined in an infinite interval I. If limx→∞ f(x) = 0 andf(x)− f(x+ ε) > 0 for any given ε > 0, then f(x) > 0 in I.Proof. By induction, for any x ∈ I, we havef(x) > f(x+ ε) > f(x+ 2ε) > · · · > f(x+ kε)→ 0as k →∞. The proof of Lemma 1 is complete. ¤4 F. QI2.1. Proof of Theorem 1.2.1.1. The case δs,s(x). We only need to prove the function δ0,0(x) = ψ′(x)−1/x−1/2x2 being completely monotonic in (0,∞).Successive differentiation of the function δ0,0(x) with respect to x > 0 yieldsδ(k)0,0 (x) = ψ(k+1)(x) +(−1)k+1k!xk+1+(−1)k+1(k + 1)!2xk+2(8)for nonnegative integer k.It is well known that the polygamma functions ψ(n)(x) can be expressed asψ(n)(x) = (−1)n+1∫ ∞0tn1− e−t e−xt dt (9)for n ∈ N, and1xr=1Γ(r)∫ ∞0tr−1e−xt dt (10)for x > 0 and r > 0.Applying formulas (9) and (10) in (8) yields(−1)kδ(k)0,0 (x) =∫ ∞0(t1− e−t − 1−t2)tke−xt dt > 0. (11)Thus, the function δ0,0(x) is completely monotonic in (0,∞).2.1.2. The case δs,t(x) for s 6= t. The function δs,t(x) can be rewritten asδs,t(x) =1t− s∫ tsψ′(x+u) du− 12[(1− 1t− s)1x+ t+(1+1t− s)1x+ s], (12)then, for nonnegative integer k,δ(k)s,t (x) =1t− s∫ tsψ(k+1)(x+ u) du− (−1)kk!2[(1− 1t− s)1(x+ t)k+1+(1 +1t− s)1(x+ s)k+1]. (13)Since limx→∞ δ(k)s,t (x) = 0 from (9), by Lemma 1, to show (−1)kδ(k)s,t (x) ≷ 0, it issufficient to verify (−1)k[δ(k)s,t (x)− δ(k)s,t (x+ 1)] ≷ 0.Taking the logarithm of the difference equation Γ(x+1) = xΓ(x) and consecutivedifferentiating yieldsψ(i−1)(x+ 1) = ψ(i−1)(x) +(−1)i−1(i− 1)!xi(14)for i ∈ N and x > 0.In the following, for our own convenience and simplicity, denote p = x + s > 0and q = x+ t > 0, when no confusion appears in the context.By using formulas (14) and (10), it is obtained that(−1)k[δ(k)s,t (x)− δ(k)s,t (x+ 1)]=(−1)kt− s∫ ts[ψ(k+1)(x+ u)− ψ(k+1)(x+ u+ 1)] du− k!2{[1− 1t− s][1qk+1− 1(q + 1)k+1]A COMPLETELY MONOTONIC FUNCTION AND AN EQUIVALENT INEQUALITY 5+[1 +1t− s][1pk+1− 1(p+ 1)k+1]}=(k + 1)!t− s∫ ts1(x+ u)k+2du− k!2{[1− 1t− s][1qk+1− 1(q + 1)k+1]+[1− 1t− s][1pk+1− 1(p+ 1)k+1]}=k!t− s[1pk+1− 1qk+1]− k!2{[1− 1t− s][1qk+1− 1(q + 1)k+1]+[1− 1t− s][1pk+1− 1(p+ 1)k+1]} (15)=k!2{(1 +1q − p)1(p+ 1)k+1+(1q − p − 1)1pk+1+(1− 1q − p)1(q + 1)k+1−(1 +1q − p)1qk+1}=12∫ ∞0zk[(1 +1q − p)e−(p+1)z +(1q − p − 1)e−pz+(1− 1q − p)e−(q+1)z −(1 +1q − p)e−qz]dz=12∫ ∞0zk+1(ez + 1)[e−(p+1)z + e−(q+1)z]×[1(q − p)ze(q−p)z − 1e(q−p)z + 1− 1zez − 1ez + 1]dz=12∫ ∞0zk+1(ez + 1)[e−(p+1)z + e−(q+1)z]×[1(q − p)z tanh(q − p)z2− 1ztanhz2]dz.Since the function [tanh(y/2)]/y is even in (−∞,∞) and decreasing in (0,∞), then1(q − p)z tanh(q − p)z2≷ 1ztanhz2is valid for |q − p| = |t− s| ≶ 1. This reveals that (−1)k[δ(k)s,t (x)− δ(k)s,t (x+ 1)] ≷ 0for |t− s| ≶ 1, which implies (−1)kδ(k)s,t (x) ≷ 0 for |t− s| ≶ 1.In conclusion, the function δs,t(x) is completely monotonic if |t− s| < 1 and−δs,t(x) is completely monotonic if |t− s| > 1. The proof of Theorem 1 is complete.2.2. Proof of Theorem 2. For real numbers p, q 6∈ {0,−1}, denote the functionin the pair of braces in the ninth and tenth lines of (15) by Tk(p, q). ThenTk(p, q) =(1q − p − 1){[1pk+1− 1qk+1]+[1(p+ 1)k+1− 1(q + 1)k+1]}+ 2[1(p+ 1)k+1− 1qk+1]= [1− (q − p)][1(p+ 1)k+1(q + 1)k+1k∑i=0(q + 1)i(p+ 1)k−i6 F. QI+1pk+1qk+1k∑i=0qipk−i]+2(q − p− 1)(p+ 1)k+1qk+1k∑i=0qi(p+ 1)k−i= (p− q + 1){1(p+ 1)k+1(q + 1)k+1k∑i=0(p+ 1)k−i(q + 1)i(16)+1pk+1qk+1k∑i=0pk−iqi − 2(p+ 1)k+1qk+1k∑i=0(p+ 1)k−iqi}= [1− (q − p)]{k∑i=01(p+ 1)i+1(q + 1)k−i+1+k∑i=01pi+1qk−i+1− 2k∑i=01(p+ 1)i+1qk−i+1}.2.2.1. For p > 0 and q > 0, formula (15) tells us that Tk(p, q) ≷ 0 for |q − p| ≶ 1.Considering (16), it is concluded that if p > 0 and q > 0 then inequalityk∑i=01(p+ 1)i+1(q + 1)k−i+1+k∑i=01pi+1qk−i+1> 2k∑i=01(p+ 1)i+1qk−i+1(17)holds for q − p > −1 and reveres for q − p < −1.2.2.2. For p < −1 and q < −1, utilizing (10) revealsTk(p, q) = (−1)k+1{(1 +1q − p)1[−(p+ 1)]k+1 +(1q − p − 1)1(−p)k+1+(1− 1q − p)1[−(q + 1)]k+1 −(1 +1q − p)1(−q)k+1}=(−1)k+1k!∫ ∞0zk{e(p+1)z − epz + epz − eqzq − p+ e(q+1)z − eqz + e(p+1)z − e(q+1)zq − p}dz=(−1)k+1k!∫ ∞0zk+1epz(1 + ez)[1 + e(q−p)z]×[1zez − 1ez + 1− 1(q − p)ze(q−p)z − 1e(q−p)z + 1]dz=(−1)kk!∫ ∞0zk+1epz(1 + ez)[1 + e(q−p)z]×[1(q − p)z tanh(q − p)z2− 1ztanhz2]dz.Thus, (−1)kTk(p, q) ≷ 0 for |q − p| ≶ 1. Combing with (16), it is deduced that ifp < −1 and q < −1 then inequalities2k∑i=01(p+ 1)i+1(q + 1)2k−i+1+2k∑i=01pi+1q2k−i+1> 22k∑i=01(p+ 1)i+1q2k−i+1(18)A COMPLETELY MONOTONIC FUNCTION AND AN EQUIVALENT INEQUALITY 7and2k+1∑i=01(p+ 1)i+1(q + 1)2k−i+2+2k+1∑i=01pi+1q2k−i+2< 22k+1∑i=01(p+ 1)i+1q2k−i+2(19)hold for q − p > −1 and reverse for q − p < −1.2.2.3. For −1 < p < 0 and −1 < q < 0, by using (10), we haveTk(p, q) =(1 +1q − p)1(p+ 1)k+1+(1q − p − 1)(−1)k+1(−p)k+1+(1− 1q − p)1(q + 1)k+1−(1 +1q − p)(−1)k+1(−q)k+1=1k!∫ ∞0zk(eqz + epz)[1q − pe(q−p)z − 1e(q−p)z + 1+ 1][e−(p+q+1)z + (−1)k]dz.If k is even, then Tk(p, q) > 0 and inequality (18) holds for all −1 < p < 0 and−1 < q < 0. If k is odd, then Tk(p, q) ≷ 0 for p+ q + 1 ≶ 0, and inequality (19) isvalid for p+ q + 1 > 0 and is reversed for p+ q + 1 < 0.2.2.4. For p < −1 and q > 0, using (10) showsTk(p, q) =(1 +1q − p)(−1)k+1[−(p+ 1)]k+1 +(1q − p − 1)(−1)k+1(−p)k+1+(1− 1q − p)1(q + 1)k+1−(1 +1q − p)1qk+1=1k!∫ ∞0zk(1 + ez)epz(ez − 1ez + 1+1q − p)[(−1)k+1 − e−(p+q+1)z]dz.If k is even, then Tk(p, q) < 0 and inequality (18) holds for all p < −1 and q > 0by (16). If k is odd, then Tk(p, q) ≷ 0 for p+ q + 1 ≷ 0, and inequality (19) holdsfor p+ q + 1 > 0 and reverses for p+ q + 1 < 0.2.2.5. For p > 0 and q < −1, by using (10), we obtainTk(p, q) =(1 +1q − p)1(p+ 1)k+1+(1q − p − 1)1pk+1+(1− 1q − p)(−1)k+1[−(q + 1)]k+1 −(1 +1q − p)(−1)k+1(−q)k+1=1k!∫ ∞0zkeqz(1 + ez)[ez − 1ez + 1− 1q − p][(−1)k+1 − e−(p+q+1)z]dz.If k is even, then Tk(p, q) < 0 and inequality (18) is reversed for all p > 0 andq < −1. If k is odd, Tk(p, q) ≶ 0 for p+ q + 1 ≶ 0, and then inequality (19) holdsfor p+ q + 1 < 0 and reverses for p+ q + 1 > 0.2.2.6. Letting θ > 0 and substituting p = a/θ and q = b/θ into (17) leads tok∑i=01(a/θ + 1)i+1(b/θ + 1)k−i+1+k∑i=01(a/θ)i+1(b/θ)k−i+1> 2k∑i=01(a/θ + 1)i+1(b/θ)k−i+1, (20)8 F. QIwhich is equivalent to (4).Similarly, inequalities (5) and (6) can be deduced by applying p = a/θ andq = b/θ into (18) and (19).2.2.7. Substituting q = p− 1 into the first two lines in (16) leads immediately toTk(p, p− 1) = 0, then inequalities (17), (18) and (19) become equalities. Hence, ifb = a − θ then inequalities (4), (5) and (6) also become equalities. The proof ofTheorem 2 is complete.2.3. Proof of Theorem 3. For a > 0 and b > 0, suppose inequality (4) holdsfor b − a > −θ and reveres for b − a < −θ. If assume a = pθ and b = qθ in (4),then simplifying yields that inequality (17) holds for q − p > −1 and reveres forq − p < −1, which is equivalent to Tk(p, q) ≷ 0 for |q − p| ≶ 1. This means that(−1)k[δ(k)s,t (x)− δ(k)s,t (x+ 1)] ≷ 0 for |t− s| ≶ 1, which implies (−1)kδ(k)s,t (x) ≷ 0 for|t− s| ≶ 1. Theorem 1 follows.Conversely, if Theorem 1 is valid, then (−1)kδ(k)s,t (x) ≷ 0 for |t− s| ≶ 1, whichimplies the function (−1)kδ(k−1)s,t (x) is increasing/decreasing for |t− s| ≶ 1, and(−1)kδ(k−1)s,t (x)− (−1)kδ(k−1)s,t (x+ 1) = (−1)k[δ(k−1)s,t (x)− δ(k−1)s,t (x+ 1)]≶ 0for |t− s| ≶ 1, which means that (−1)k[δ(k)s,t (x) − δ(k)s,t (x + 1)] ≷ 0 for |t− s| ≶ 1.Further, combining this with (15) and (16) leads to (17), and then (20). Inequality(4) is proved. The proof of Theorem 3 is complete.References[1] H. Alzer, Sharp inequalities for the digamma and polygamma functions, Forum Math. 16(2004), 181–221.[2] N. Batir, Some new inequalities for gamma and polygamma functions, J. Inequal. PureAppl. Math. 6 (2005), no. 4, Art. 103. Available online at http://jipam.vu.edu.au/article.php?sid=577. RGMIA Res. Rep. Coll. 7 (2004), no. 3, Art. 1. Available online at http://rgmia.vu.edu.au/v7n3.html.[3] C. Berg, Integral representation of some functions related to the gamma function, Mediterr.J. 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cari´ c, The best bounds in Gautschi’s inequality,

Certain logarithmically N-alternating monotonic functions involving gamma and q-gamma functions,

Completely Monotonic and Related Functions,

Completely monotonic functions involving the gamma and q-gamma functions,

Integral representation of some functions related to the gamma function,

Monotonicity and convexity for the gamma function,

On some properties of the gamma function,

Sharp inequalities for the digamma and polygamma functions,

Some completely monotonic functions involving polygamma functions and an application,

Some completely monotonic functions involving the gamma and polygamma functions,

Some elementary inequalities relating to the gamma and incomplete gamma function,

Some extensions of W. Gautschi’s inequalities for the gamma function,

Some new inequalities for gamma and polygamma functions,

The best bounds in Gautschi-Kershaw inequalities,

The best bounds in Kershaw’s inequality and two completely monotonic functions,

The Laplace Transform,

A Completely Monotonic Function Involving Divided Difference of Psi Function and an Equivalent Inequality Involving Sum

Victoria University Institutional Repository (VUIR)

In this paper, a function involving the divided difference of the\ud
psi function is proved to be completely monotonic, a class of inequalities involving\ud
sum are founded, and an equivalent relation between the complete\ud
monotonicity and one of the class of inequalities is established

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A Completely Monotonic Function Involving Divided Difference of Psi Function and an Equivalent Inequality Involving Sum

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