COMMUNICATIONS ON doi:10.3934/cpaa.2009.8.1975 PURE AND APPLIED ANALYSIS Volume 8, Number 6, November 2009 pp. 1975–1989
COMPLETELY MONOTONIC FUNCTIONS INVOLVING DIVIDED DIFFERENCES OF THE DI- AND TRI-GAMMA FUNCTIONS AND SOME APPLICATIONS
Feng Qi College of Mathematics and Information Science, Henan Normal University, Xinxiang City, Henan Province, 453007, CHINA Bai-Ni Guo School of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, CHINA
(Communicated by Carmen Chicone)
Abstract. In the present paper, we establish the complete monotonicity of two functions involving divided differences of the digamma function ψ and the trigamma function ψ′. Applying these monotonicity, we provide an alternative proof for the monotonicity and convexity of a function derived from bound- ing the ratio of two gamma functions, procure the logarithmically completely monotonic property of a function involving the ratio of two gamma functions, and obtain new bounds for the ratio of two gamma functions.
1. Introduction and main results. 1.1. Recall [45, Chapter XIII] and [85, Chapter IV] that a function f(x) is said to be completely monotonic on an interval I if f(x) has derivatives of all orders on I and ( 1)nf (n)(x) 0 (1) − ≥ for x I and n 0. The∈ well-known≥ Bernstein’s Theorem [85, p. 160, Theorem 12a] states that a function f(x) on[0, ) is completely monotonic if and only if there exists a bounded and non-decreasing∞ function α(t) such that ∞ − f(x)= e xt dα(t) (2) Z0 converges for x [0, ). This says that a completely monotonic function f(x) on [0, ) is a Laplace∈ transform∞ of the measure α. It∞ is known that if a nonconstant function f(x) is completely monotonic then the strict inequality in (1) holds. This has been established in [27, p. 98]. See also the simplified proof of this result given in [82] and related sentences in [71, p. 82].
2000 Mathematics Subject Classification. Primary: 26A48, 26A51, 33B15; Secondary: 26D20, 65R10. Key words and phrases. Completely monotonic function, logarithmically completely monotonic function, monotonicity, convexity, divided difference, digamma function, trigamma function, ratio of two gamma functions. The first author was partially supported by the China Scholarship Council.
1975 1976 FENG QI AND BAI-NI GUO
1.2. Recall also [9, 61, 67] that a positive function f(x) is said to be logarithmically completely monotonic on an interval I if f(x) has derivatives of all orders on I and ( 1)n[ln f(x)](n) 0 (3) − ≥ for all x I and n N. It was∈ proved by∈ different approaches explicitly in [15, 61, 67] that a logarith- mically completely monotonic function must be completely monotonic, but not conversely. It was pointed out in [15, 31, 71] that the logarithmically completely monotonic functions on [0, ) are those completely monotonic functions on [0, ) for which the representing∞ measure α in (2) is infinitely divisible in the convolu-∞ tion sense: For each n N there exists a positive measure ν on [0, ) with n-th convolution power equal∈ to α. ∞
1.3. It is well-known that the classical Euler’s gamma function is defined by ∞ − − Γ(z)= tz 1e t dt (4) Z0 ′ for z > 0. The logarithmic derivative of Γ(z), denoted by ψ(z)= Γ (z) , is called the ℜ Γ(z) psi or digamma function, and ψ(k)(z) for k N are called the polygamma functions. In particular, ψ′(x) and ψ′′(x) for x> 0 are∈ called the trigamma and tetragamma functions, see [16, p. 71]. These functions are some of the most important special functions and have much extensive applications in many branches, for example, statistics, physics, engineering, and other mathematical sciences.
1.4. In order to show [6, Theorem 4.8]: the double inequality α β (n 1)! exp nψ(x) < ψ(n)(x) < (n 1)! exp nψ(x) , x> 0 (5) − x − − x − holds if and only if α n and β 0, it was established in the proof of [6, Theorem 4.8] that ≤ − ≥
′ ′′ p(x) [ψ (x)]2 + ψ (x) > , x> 0, (6) 900x4(x + 1)10 where p(x)=75x10 + 900x9 + 4840x8 + 15370x7 + 31865x6 + 45050x5 (7) + 44101x4 + 29700x3 + 13290x2 + 3600x + 450. The basic tools to present the inequality (6) are the following inequalities:
1 1 1 1 ′ 1 1 1 + + < ψ (x) < + + , x> 0 (8) x 2x2 6x3 − 30x5 x 2x2 6x3 and 1 1 1 ′′ 1 1 < ψ (x) < , x> 0. (9) − x2 − x3 − 2x4 −x2 − x3 The inequality (8) can be found in [30, p. 860, Theorem 4] and the inequality (9) is a special cases of [5, Theorem 9]. From (6), the inequality ′ ′′ [ψ (x)]2 + ψ (x) > 0 (10) COMPLETELY MONOTONIC FUNCTIONS AND SOME APPLICATIONS 1977 for x > 0 was deduced and used in the proof of [10, Theorem 2.1] to present a double inequality Γ(x) exp α eψ(x)ψ(x) eψ(x) +1 exp β eψ(x)ψ(x) eψ(x) +1 (11) − ≤ Γ(c) ≤ − 6eγ with α = 1 and β = π2 for x>c, where c =1.4616 is the only positive zero of the psi function ψ(x) on (0, ). · · · In order to prove [14, Theorem∞ 2.1] that the inequality π2 ψ(x) > ln γ ln e1/x 1 (12) 6 − − − holds for x 2, the inequality (10) was recovered in [14, Lemma 1.1] elegantly. In [66], the≥ inequality (10) was used to give a simple proof for the increasingly monotonic property of the function φ(x)= ψ(x) + ln e1/x 1 (13) − on (0, ). In [12∞, Remark 1.3], it was pointed out that the inequality (10) is a special case of the inequality n − 1+1/n ( 1)nψ(n+1)(x) < ( 1)n 1ψ(n)(x) (14) − n (n 1)! − − N for x> 0 and n . p In [7, Theorem∈ 4.3], the inequality (10) was applied to provide a sharp and generalized version of (11): For 0 ′ 1 1 ψ (x) (17) − x − 2x2 was proved respectively. In [8, 22, 34, 35, 46, 59, 60], the positivity of the function (17) was generalized step and step to the following conclusions: The function exΓ(x) (18) xx−α is logarithmically completely monotonic on (0, ) if and only if α 1 and so is its reciprocal if and only if α 1 ; The function ∞ ≥ ≤ 2 α ψ(x) ln x + (19) − x 1978 FENG QI AND BAI-NI GUO is completely monotonic on (0, ) if and only if α 1 and so is its negative if 1 ∞ ≥ and only if α 2 . In fact, it is not difficult to verify that these two conclusions are equivalent≤ to each other and they implies the complete monotonicity of the function (17). For more information on the origin, applications and generalizations of the function (18), please refer to [32, 33] and related references therein. 1.6. The first aim of this paper is to generalize the positivity of the functions ′ ′′ [ψ (x)]2 + ψ (x) (20) and (17), the logarithmically complete monotonicity of the function (18), and the complete monotonicity of the function (19) to the complete monotonicity of two functions involving the divided differences of the digamma function ψ(x) and the trigamma function ψ′(x). Our main results may be stated as follows. Theorem 1.1. Let s and t be real numbers and α = min s,t . Then the function { } ψ(x + t) ψ(x + s) 2x + s + t +1 − , s = t t s − 2(x + s)(x + t) 6 δ (x)= (21) s,t − 1 1 ψ′(x + s) , s = t − x + s − 2(x + s)2 for t s < 1 and δs,t(x) for t s > 1 are completely monotonic in x ( α, ). | − | − | − | ∈ − ∞ Theorem 1.2. Let s and t be real numbers and α = min s,t . Then the function { } ψ(x + t) ψ(x + s) 2 ψ′(x + t) ψ′(x + s) − + − , s = t ∆s,t(x)= t s t s 6 (22) ′ −2 ′′ − [ψ (x + s)] + ψ (x + s), s = t for t s < 1 and ∆s,t(x) for t s > 1 are completely monotonic in x ( α, ). | − | − | − | ∈ − ∞ Remark 1.1. The paper [38] contains some results closely related to Theorem 1.1 and Theorem 1.2. Also, Corollary 3 in [38] contains a result which is stronger than the complete monotonicity of the function ∆0,0(x). 1.7. In[37], D. Kershaw proved the following double inequality − 1−s s 1 s Γ(x + 1) 1 1 x + < < x + s + (23) 2 Γ(x + s) − 2 4 r ! for 0 Theorem 1.3. The function zs,t(x) is either convex and decreasing for t s < 1 or concave and increasing for t s > 1. | − | | − | In [17, 72, 73], several alternative proofs for Theorem 1.3 were provided respec- tively by using the convolution theorem of Laplace transforms, some asymptotic formulas and integral representations of the gamma function Γ(x), the digamma function ψ(x) and the polygamma functions ψ(k)(x) for k N, and other analytic techniques. ∈ The second aim of this paper is to supply a new proof of Theorem 1.3 by em- ploying Theorem 1.1 and Theorem 1.2. Remark 1.2. In [83], the function z1,1/2(x) was proved to be strictly decreasing on ( 1 , ). Applying this result, the paper [2] showed that − 2 ∞ n + A Ω − n + B < n 1 (26) r 2π Ωn ≤ r 2π for n N with the best possible constants A = 1 and B = π 1, where ∈ 2 2 − πn/2 Ω = (27) n Γ(1 + n/2) denotes the volume of the unit ball in Rn. For more results on inequalities for Ωn, please refer to [2, 3, 68] and closely-related references therein. Remark 1.3. In [40], the monotonicity results in Theorem 1.3 were applied to obtain an inequality which generalizes the sharpened Wallis’ double inequality validated in [18, 19, 20, 21, 23, 24, 25, 62] and the references therein. In [39], Theorem 1.3 has also been applied to obtain sharp estimates for the remainders of certain series. Remark 1.4. In the literature, the double inequality (23) is called the first Kershaw’s inequality. The second Kershaw’s inequality [37] states that Γ(x + 1) s +1 exp (1 s)ψ x + √s < < exp (1 s)ψ x + (28) − Γ(x + s) − 2 for s (0, 1) and x 1. Along other directions, Kershaw’s two double inequalities have∈ been investigated≥ systematically and historically in [29, 47, 48, 49, 50, 51, 52, 53, 55, 56, 57, 58, 63, 64, 69, 70, 72, 73, 74, 75, 76] and related references therein. For detailed information on this topic, please refer to the expository article [54]. 1.8. The third aim of this paper is to derive the logarithmically complete mono- Γ(x+s) tonicity of a function involving the ratio Γ(x+t) and to acquire new bounds for the Γ(x+s) ratio Γ(x+t) by utilizing Theorem 1.1. Our conclusions can be recited as the following theorem. 1980 FENG QI AND BAI-NI GUO Theorem 1.4. For real numbers s and t, define − (x + t)1/2(t−s)−1/2 Γ(x + t) 1/(t s) , s = t (x + s)1/2(t−s)+1/2 Γ(x + s) 6 Hs,t(x)= (29) 1 1 exp + ψ(t + x) , s = t t + x 2(t + x) on ( α, ), where α = min s,t . − ∞ { } 1. For t s > 1, the function Hs,t(x) is logarithmically completely monotonic; | − | −1 2. For t s < 1, the function [Hs,t(x)] is logarithmically completely mono- tonic;| − | 3. For t s =1, the function Hs,t(x) identically equals 1 on ( α, ); 4. If s| −t <| 1, the inequality − ∞ | − | − Γ(x + t) 1/(t s) (x + s)1/2(t−s)+1/2 < (30) Γ(x + s) (x + t)1/2(t−s)−1/2 holds on ( α, ); if s t > 1, the inequality (30) is reversed on ( α, ). 5. On the interval− ∞(ρ, |) −(| α, ), the inequality − ∞ ∞ ⊂ − ∞ − − Γ(x + t) 1/(t s) (ρ + t)1/2(t−s)−1/2 Γ(ρ + t) 1/(t s) (x + s)1/2(t−s)+1/2 > (31) Γ(x + s) (ρ + s)1/2(t−s)+1/2 Γ(ρ + s) (x + t)1/2(t−s)−1/2 holds for 0 < t s < 1; if t s > 1, the inequality (31) is reversed. | − | | − | Remark 1.5. The form of the bounds in (30) and (31) are different from not only those in (23) and (28) but also others collected in the survey paper [54]. Remark 1.6. Finally, we remark that the complete monotonicity and inequalities for special functions are classical topics. The results of this type have various applications, especially in mathematical inequality theory and asymptotic analysis. In recent years there has been an increased interest on the complete monotonicity and inequalities of functions involving the gamma and polygamma functions and a large number of interesting results have been obtained by several authors. Except those mentioned above, we refer [11, 13, 65, 86, 87] and closely-related references therein to readers. 2. Lemmas. In order to prove our theorems, the following lemmas are needed. Lemma 2.1 ([1]). For i N and x> 0, the polygamma functions ψ(i)(x) have the following integral expressions:∈ ∞ i t − ψ(i)(x) = ( 1)i+1 e xt dt, (32) − 1 e−t Z0 − i−1 − − ( 1) (i 1)! ψ(i 1)(x +1)= ψ(i 1)(x)+ − − . (33) xi Lemma 2.2 (Hermite-Hadamard’s inequality [77, 78, 79, 80, 81]). For f(x) being convex on [a,b], a + b b f(a)+ f(b) (b a)f < f(x) dx< (b a) . (34) − 2 − 2 Za COMPLETELY MONOTONIC FUNCTIONS AND SOME APPLICATIONS 1981 Lemma 2.3 ([36]). For 1 1 1 1 1 ′ 1 1 1 + + < ψ (x) < + + . (36) x 2x2 6x3 − 30x5 x 2x2 6x3 Proof. This follows directly from applying the identity (33) for i =2 to 1 1 1 ′ 1 1 1 < ψ (x + 1) < + (37) 2x2 − 6x3 x − 2x2 − 6x3 30x5 gained in [62, p. 305, Corollary 1]. Remark 2.2. The inequalities (8), (9), (36), (37), and the positivity of the func- tion (17) are special cases of [5, Theorem 9] and [44, Theorem 1]. The complete monotonicity of δ0,0(x) or the function (17) is also a special cases of [5, Theorem 8]. A short proof of these results and some additional comments are given in [41]. Lemma 2.5. If f(x) is a function defined in an infinite interval I such that f(x) f(x + ε) > 0 and lim f(x)= δ − x→∞ for x I and some given constant ε> 0, then f(x) >δ on I. ∈ Proof. By induction, for any x I, ∈ f(x) >f(x + ε) >f(x +2ε) > >f(x + kε) δ · · · → as k . The proof of Lemma 2.5 is complete. → ∞ 3. Proofs of theorems. Now we are in a position to prove our theorems. 3.1. Proof of Theorem 1.1. 3.1.1. The case δs,s(x). This is equivalent to the complete monotonicity of the func- tion ′ 1 1 δ (x)= ψ (x) . (38) 0,0 − x − 2x2 This can be deduced easily from the logarithmically complete monotonicity of the function (18) or the complete monotonicity of the function (19). However, for consistency, we would like to give it a new proof as follows. Successive differentiation of the function δ0,0(x) with respect to x> 0 yields ( 1)k+1k! ( 1)k+1(k + 1)! δ(k)(x)= ψ(k+1)(x)+ − + − (39) 0,0 xk+1 2xk+2 for nonnegative integer k. Formula (32) means directly that lim δ(k)(x)=0. x→∞ 0,0 k (k) Hence, in virtue of Lemma 2.5, to prove ( 1) δ0,0 (x) > 0 on (0, ) for nonnegative integer k, it suffices to show − ∞ ( 1)k[δ(k)(x) δ(k)(x + 1)] > 0. (40) − 0,0 − 0,0 1982 FENG QI AND BAI-NI GUO From (33), it is concluded that ( 1)k[δ(k)(x) δ(k)(x + 1)] = ( 1)k[ψ(k+1)(x) ψ(k+1)(x + 1)] − 0,0 − 0,0 − − 1 1 (k + 1)! 1 1 k! − xk+1 − (x + 1)k+1 − 2 xk+2 − (x + 1)k+2 (k + 1)! 1 1 (k + 1)! 1 1 = k! xk+2 − xk+1 − (x + 1)k+1 − 2 xk+2 − (x + 1)k+2 (k + 1)! 1 1 1 1 = + k! 2 xk+2 (x + 1)k+2 − xk+1 − (x + 1)k+1 1/xk+2 +1/(x + 1)k+2 x+1 1 = (k + 1)! du . (41) 2 − uk+2 Zx Utilizing the inequality (34) in the line (41) deduces the inequality (40), as a result, the function δ (x) is completely monotonic on (0, ). 0,0 ∞ 3.1.2. The case δ (x) for s = t. The function δ (x) can be rewritten as s,t 6 s,t t ′ ψ (x + u) du 1 1 1 1 1 δ (x)= s 1 + 1+ , s,t t s − 2 − t s x + t t s x + s R − − − then, for nonnegative integer k, 1 t δ(k)(x)= ψ(k+1)(x + u) du s,t t s − Zs ( 1)kk! 1 1 1 1 − 1 + 1+ . − 2 − t s (x + t)k+1 t s (x + s)k+1 − − (k) k (k) Since limx→∞ δs,t (x)=0 from (32), by Lemma 2.5, to show ( 1) δs,t (x) ≷ 0, it is sufficient to verify − ( 1)k[δ(k)(x) δ(k)(x + 1)] = ( 1)k[δ (x) δ (x + 1)](k) ≷ 0. (42) − s,t − s,t − s,t − s,t Using the formula (33) and a standard argument reveals that 1 t ψ′(x + u) ψ′(x + u + 1) δ (x) δ (x +1)= − du s,t − s,t t s 2 − Zs 1 t ψ′(x + u + 1) ψ′(x + u + 2) + − du t s 2 − Zs t 1/(x + u + 1)3 du t 1/(x + u)3 du + s s t 2 − t 2 Rs 1/(x + u + 1) du Rs 1/(x + u) du 1 t 1 t 1 = R du +R du 2(t s) (x + u)2 (x + u + 1)2 − Zs Zs t 1/(x + u + 1)3 du t 1/(x + u)3 du + s s t 2 − t 2 Rs 1/(x + u + 1) du Rs 1/(x + u) du 1 1 1 = R +R 2 (x + s + 1)(x + t + 1) (x + s)(x + t) x +1+(s + t)/2 x + (s + t)/2 + (x + s + 1)(x + t + 1) − (x + s)(x + t) COMPLETELY MONOTONIC FUNCTIONS AND SOME APPLICATIONS 1983 1 (s t)2 = − − 2(x + s)(x + s + 1)(x + t)(x + t + 1) ≷ 0 if and only if 1 (s t)2 ≷ 0 which is equivalent to t s ≶ 1 immediately. Since 1 − − | − | x is completely monotonic in (0, ) and a product of finite completely monotonic functions is also completely monotonic,∞ then the function δ (x) δ (x + 1) s,t − s,t 1 (s t)2 − − is completely monotonic on ( α, ), that is, − ∞ (k) (k) [δ (x) δ (x + 1)](k) δ (x) δ (x + 1) ( 1)k s,t − s,t = ( 1)k s,t − s,t > 0 − 1 (s t)2 − 1 (s t)2 − − − − in x ( α, ). Hence, the inequality (42) holds true on ( α, ) for t s ≶ 1. The proof∈ − of∞ Theorem 1.1 is complete. − ∞ | − | 3.2. Proof of Theorem 1.2. 3.2.1. The case ∆s,s(x). This case is equivalent to the complete monotonicity of ′ 2 ′′ the function ∆0,0(x) = [ψ (x)] + ψ (x). By standard computation, it follows that k−1 k 1 − ∆(k)(x)=2 − ψ(i+1)(x)ψ(k i+1)(x)+ ψ(k+2)(x), 0,0 i i=0 X (k) for k 1. Formula (32) means clearly that lim →∞ ∆ (x) = 0 for nonnegative ≥ x 0,0 k (k) integer k. Hence, in order to show ( 1) ∆0,0(x) > 0 for nonnegative integer k, by Lemma 2.5, it is sufficient to prove that− ( 1)k[∆(k)(x) ∆(k)(x + 1)] = ( 1)k[∆ (x) ∆ (x + 1)](k) − 0,0 − 0,0 − 0,0 − 0,0 is positive for nonnegative integer k. This follows from combination of the following facts: From (33), it is obtained that 2 ′ 1 1 ∆ (x) ∆ (x +1)= ψ (x) ; 0,0 − 0,0 x2 − x − 2x2 2 The function x2 is completely monotonic on (0, ) clearly; The function δ0,0(x) is completely monotonic on (0, ), which was proved∞ in Theorem 1.1 just now; A product of finite completely monotonic∞ functions is also completely monotonic. k (k) 3.2.2. The case ∆s,t(x) for s = t. In order to show ( 1) ∆s,t (x) ≶ 0 for nonnega- tive integer k, by Lemma 2.5,6 it is sufficient to show− that ( 1)k[∆(k)(x) ∆(k)(x + 1)] = ( 1)k[∆ (x) ∆ (x + 1)](k) ≶ 0 (43) − s,t − s,t − s,t − s,t on ( α, ) for nonnegative integer k. By− using∞ the formula (33), it is obtained that ′ ′ ′′ ′′ ∆ (x) ∆ (x +1)= [Φ (x)]2 [Φ (x + 1)]2 +Φ (x) Φ (x + 1) s,t − s,t s,t − s,t s,t − s,t ′ ′ 2 Φ (x)+Φ (x + 1) t 1 t 1 = s,t s,t du du t s 2 (x + u)2 − (x + u)3 − " Zs Zs # 1984 FENG QI AND BAI-NI GUO t ′ ′ t 3 1 1 Φs,t(x)+Φs,t(x + 1) s 1/(x + u) du =2 2 du t t s s (x + u) 2 − 1/(x + u)2 du − Z " Rs # t ′ ′ t 3 1 ψ (x + u)+ ψ (x + u + 1) s 1/(xR + u) du =2 du t t s s 2 − 1/(x + u)2 du − Z Rs 1 t 1 1 t 1 du =2δ (x) R du . × t s (x + u)2 s,t t s (x + u)2 − Zs − Zs 1 t 1 It is clear that the function t−s s (x+u)2 du of x is completely monotonic on ( α, ). Combining with the fact that ( 1)kδ(k)(x) ≷ 0on( α, ) for t s ≶ 1 − ∞ R − s,t − ∞ | − | obtained in Theorem 1.1, the inequality (43) for t s ≶ 1 holds on ( α, ). The proof of Theorem 1.2 is complete. | − | − ∞ 3.3. Proof of Theorem 1.3. 3.3.1. Convexity and concavity in Theorem 1.3 for s = t. Standard differentiating and simplifying yields 6 ′ [z (x)+ x][ψ(x + t) ψ(x + s)] z (x)= s,t − 1 (44) s,t t s − − and 2 ′′ z (x)+ x [ψ(x + t) ψ(x + s)] ′ ′ z (x)= s,t − + [ψ (x + t) ψ (x + s)] s,t t s t s − − − t 2 t 1 ′ 1 ′′ = [z (x)+ x] ψ (x + u) du + ψ (x + u) du s,t t s t s ( − Zs − Zs ) , [zs,t(x)+ x]∆s,t(x). From Theorem 1.2, it is clear that ′′ zs,t(x) = ∆s,t(x) ≷ 0 zs,t(x)+ x if and only if t s ≶ 1. So, the convexity and concavity of the function zs,t(x) follows readily.| − | 3.3.2. Convexity in Theorem 1.3 for s = t. It is clear that ′ ′ z (x)= ψ (x + s)eψ(x+s) 1, s,s − ′′ ′ 2 ′′ ψ(x+s) zs,s(x) = [ψ (x + s)] + ψ (x + s)]e . ′′ The requirement zs,s(x) > 0 follows easily from Theorem 1.2 or [14, Lemma 1.1] ′ 2 ′′ which says that ∆0,0(x) = [ψ (x)] + ψ (x) > 0 for x> 0. 3.3.3. Monotonicity in Theorem 1.3. From an inequality ′ eψ(x)ψ (x) < 1 (45) ′ for x > 0 obtained in [14, Lemma 1.1], it is deduced easily that zs,s(x) < 0, and then zs,s(x) is decreasing. From the convexity and concavity of zs,t(x), it is deduced immediately that if ′ t s > 1 the first derivative zs,t(x) is decreasing and, if 0 < t s < 1 the function | ′− | | − | zs,t(x) is increasing. Therefore, in order to obtain monotonicity of zs,t(x), it is ′ sufficient to verify limx→∞ zs,t(x)=0. COMPLETELY MONOTONIC FUNCTIONS AND SOME APPLICATIONS 1985 The inequality (36) means ′ lim x[ψ(x + t) ψ(x + s)] = (t s) lim xψ (x + ξ) x→∞ − − x→∞ 1 1 1 (t s) lim x + + = t s ≤ − x→∞ x + ξ 2(x + ξ)2 6(x + ξ)3 − and ′ lim x[ψ(x + t) ψ(x + s)] = (t s) lim xψ (x + ξ) x→∞ − − x→∞ 1 1 1 1 (t s) lim x + + = t s, ≥ − x→∞ x + ξ 2(x + ξ)2 6(x + ξ)3 − 30(x + ξ)5 − where ξ is between s and t. As a result, lim x[ψ(x + t) ψ(x + s)] = t s. (46) x→∞ − − Applying a = x + s and b = x + t into the inequality (35) and rearranging leads to x+t−1 1/(s−t) 1/(s−t) (1 + t/x) − Γ(x + t) e< xs t (1 + s/x)x+s−1 Γ(x + s) − (1 + t/x)x+t−1/2 1/(s t) < e. (1 + s/x)x+s−1/2 For any constant β R, it is clear that ∈ x+t−β 1/(s−t) x+t−β 1/(s−t) (1 + t/x) limx→∞(1 + t/x) 1 lim − = − = . x→∞ (1 + s/x)x+s β lim →∞(1 + s/x)x+s β e x Thus, 1/(s−t) − Γ(x + t) lim xs t =1. (47) x→∞ Γ(x + s) Rearranging (44) and utilizing (46) and (47) reveals 1/(t−s) ′ Γ(x + t) ψ(x + t) ψ(x + s) lim zs,t(x) + 1 = lim − x→∞ x→∞ Γ(x + s) t s − 1/(t−s) − Γ(x + t) x[ψ(x + t) ψ(x + s)] = lim xs t − x→∞ Γ(x + s) t s − 1/(t−s) − Γ(x + t) lim →∞ x[ψ(x + t) ψ(x + s)] = lim xs t x − x→∞ Γ(x + s) t s − =0. The proof of Theorem 1.3 is complete. 3.4. Proof of Theorem 1.4. The logarithmically completely monotonic prop- ′ erties of the function Hs,t(x) follow directly from [ln Hs,t(x)] = δs,t(x) and the definition of the logarithmically completely monotonic functions. It was established in [84] that the classical asymptotic relation Γ(x + s) lim = 1 (48) x→∞ xsΓ(x) 1986 FENG QI AND BAI-NI GUO holds for real s and x. This implies that 1/2(t−s) 1/(t−s) (x + t) x − Γ(x + t) H (x)= xs t 1 (49) s,t (x + s)1/2(t−s) · Γ(x + s) → (x + s)(x + t) as x . Since the function Hps,t(x) for 0 < t s < 1 is increasing on ( α, ), employing→ ∞ the limit (49) concludes that the inequality| − | − ∞ − (x + t)1/2(t−s)−1/2 Γ(x + t) 1/(t s) < 1 (50) (x + s)1/2(t−s)+1/2 Γ(x + s) holds on ( α, ). Similarly, the function Hs,t(x) for t s > 1 is decreasing on ( α, ), which− ∞ implies that the reversed version of the| − inequality| (50) is valid. Hence,− ∞ the inequality (30) is proved. The inequality (31) comes from the fact that the inequality Hs,t(ρ) < Hs,t(x) on (ρ, ) holds if 0 < t s < 1 or reverses if t s 1. The proof of Theorem 1.4 is complete.∞ | − | | − |≥ Acknowledgements. The authors express many thanks to the anonymous referees for valuable suggestions and comments. REFERENCES [1] M. Abramowitz and I. A. 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