Logarithmically completely monotonic functions relating to the gamma function

J. Math. Anal. Appl. 321 (2006) 405–411 www.elsevier.com/locate/jmaa

Logarithmically completely monotonic functions relating to the gamma function ✩ Chao-Ping Chen a,∗ , Feng Qi b a College of Mathematics and Informatics, Henan Polytechnic University, Jiaozuo City, Henan 454010, China b Research Institute of Mathematical Sciences, Henan Polytechnic University, Jiaozuo City, Henan 454010, China

Received 14 April 2005 Available online 13 September 2005 Submitted by H.M. Srivastava

Abstract In this paper, the logarithmically complete monotonicity of the function ex Γ (x + β)/x x+β−α in (0, ∞) for α ∈ R and β  0 is considered and the corresponding result by G.D. Anderson, R.W. Barnard, K.C. Richards, M.K. Vamanamurthy and M. Vuorinen is generalized. As applications of these results, some inequalities between identric mean and ratio of two gamma functions by J.D. Keˇcki´c and P.M. Vasi´c are extended. © 2005 Elsevier Inc. All rights reserved. Keywords: Gamma function; Logarithmically completely monotonic function; Mean value; Inequality

1. Introduction A function f is said to be completely monotonic on an interval I if f has derivatives of all orders on I and (−1)n f (n) (x)  0

(1)

for x ∈ I and n  0. Let C denote the set of completely monotonic functions. ✩

The authors were supported in part by the Science Foundation of the Project for Fostering Innovation Talents at Universities of Henan Province, China. * Corresponding author. E-mail addresses: [email protected], [email protected] (C.-P. Chen), [email protected], [email protected] (F. Qi). URL: http://rgmia.vu.edu.au/qi.html (F. Qi). 0022-247X/$ – see front matter © 2005 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2005.08.056

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A positive function f is said to be logarithmically completely monotonic on an interval I if its logarithm ln f satisfies  (k) 0 (2) (−1)k ln f (x) for k ∈ N on I . Let L on (0, ∞) stand for the set of logarithmically completely monotonic functions. A function f on (0, ∞) is called a Stieltjes transform if it can be written in the form ∞ dμ(s) , (3) f (x) = a + s +x 0

where a is a nonnegative number and μ a nonnegative measure on [0, ∞) satisfying ∞ 1 dμ(s) < ∞. 1+s 0

The set of Stieltjes transforms is denoted by S. The notion “logarithmically completely monotonic function” was posed explicitly in [25] and published formally in [24] and a much useful and meaningful relation L ⊂ C between the completely monotonic functions and the logarithmically completely monotonic functions was proved in [24,25]. Motivated by the papers [25,26], among other things, it is proved in [9] that S \ {0} ⊂ L ⊂ C. An anonymous referee of this paper points out that the relation L ⊂ C is an old result and can be found in [12]. The class of logarithmically completely monotonic functions can be characterized as the infinitely divisible completely monotonic functions which are established by Horn in [17, Theorem 4.4] and restated in [9, Theorem 1.1]. There have been a lot of literature about the (logarithmically) completely monotonic functions, for example, [5,6,8–11,15,16,19, 24–26,29] and the references therein. is well known that the classical Euler gamma function Γ (x) is defined by Γ (z) =  ∞It z−1 t e−t dt for Re z > 0, which is one of the most important special functions [1,14,27, 0 28] and has much extensive applications in many branches, for example, statistics, physics, engineering, and other mathematical sciences, and the logarithmic derivative of Γ (x), denoted by ψ(x) = Γ  (x)/Γ (x) = [ln Γ (x)] , is called psi or digamma function. There exists a very extensive literature on these functions. In particular, inequalities, monotonicity and complete monotonicity properties for these functions have been published. Please refer to the papers [2–4] and the references therein. In [7] it is proved√that the function x 1/2−x ex Γ (x) is decreasing and logarithmically convex from (0, ∞) onto ( 2π, ∞) and the function x 1−x ex Γ (x) is increasing and logarithmically concave from (0, ∞) onto (1, ∞). Let α ∈ R and β  0 be real numbers, define ex Γ (x + β) (4) x x+β−α in (0, ∞). In this article, as a generalization of some results of [7], we are about to consider the logarithmically complete monotonicity property of the function fα,β (x) in (0, ∞). Our main results are as follows. fα,β (x) =

Theorem 1. The function fα,β (x) defined by (4) is logarithmically completely monotonic in (0, ∞) if 2α  1  β.

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Theorem 2. Let α be a real number. The function fα,1 (x) is logarithmically completely monotonic in (0, ∞) if and only if 2α  1. So is the function [fα,1 (x)]−1 if and only if α  1. Remark 1. From the monotonicity of the functions f1/2,1 (x) and [f1,1 (x)]−1 , we conclude the double inequality bb−1 a−b Γ (b) bb−1/2 a−b < e < e Γ (a) a a−1/2 a a−1

(5)

for b > a > 0. This extends the range of a result by J.D. Keˇcki´c and P.M. Vasi´c in [18], which says that inequality (5) holds for b > a  1. See also [2, p. 342] and [22]. Remark 2. Denote by   1 bb 1/(b−a) I (a, b) = e aa

(6)

for positive numbers a and b with a = b the so-called identric mean [13,23]. It was proved in [2] that the inequality  r  s b−a b Γ (b)  b Γ (b) < I (a, b) (7) < a Γ (a) a Γ (a) is valid for all real numbers b > a  1 if and only if r  1/2 and s  γ , where γ = 0.577 . . . stands for the Euler–Mascheroni constant. From inequality (5), it is not difficult to obtain an analogous of inequality (7) as follows:  1/2 b Γ (b) b Γ (b) < I (a, b)b−a < , b > a > 0. (8) a Γ (a) a Γ (a) 2. Proofs of the theorems Proof of Theorem 1. It is clear that ln fα,β (x) = x + ln Γ (x + β) − (x + β − α) ln x,   α−β . ln fα,β (x) = ψ(x + β) − ln x + x Using the representations [21, p. 16] for x > 0, ∞ ψ

(m)

(x) = (−1)

m+1 0

tm e−xt dt, 1 − e−t

and m! x m+1

∞ =

t m e−xt dt,

0

it is obtained that for n  2,

m = 0, 1, 2, . . . ,

m = 1, 2, . . . ,

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 (n) (n − 2)! (β − α)(n − 1)! (−1)n ln fα,β (x) = (−1)n ψ (n−1) (x + β) − n−1 + xn x ∞ n−1 ∞ ∞ t −(x+β)t n−2 −xt = e dt − t e dt + (β − α) t n−1 e−xt dt 1 − e−t 0

0

∞ =

δα,β (t) 0

t n−2 e−xt eβt − e(β−1)t

0

dt,

(9)

where    δα,β (t) = t + (β − α)t − 1 eβt − e(β−1)t   ∞      t k 1 2 −α t + . k(β − α) β k−1 − (β − 1)k−1 − β k − (β − 1)k = 2 k! k=3

It is easy to see that for 2α  1 = β,   ∞    tk 1 δα,1 (t) = − α t2 + > 0. k(1 − α) − 1 2 k!

(10)

k=3

Now we are in a position to prove δα,β (t) > 0 for 2α  1 < β. It is sufficient to show that   k(β − α) β k−1 − (β − 1)k−1 > β k − (β − 1)k for k  3 and 2α  1 < β, which is equivalent to Jk (β − 1, β) =

(k − 1)[β k − (β − 1)k ] < (k − 1)(β − α), k[β k−1 − (β − 1)k−1 ]

where Jr (a, b) denotes the one-parameter mean [13,20,23] of two positive numbers a and b, which is defined by ⎧ r+1 r+1 r(b −a ) ⎪ ⎪ ⎨ (r+1)(br −a r ) , r = 0, −1, r = 0, Jr (a, b) = L(a, b), (11) ⎪ ⎪ ⎩ [G(a,b)]2 , r = −1, L(a,b)

for a = b and Jr (a, a) = a. Since Jr (a, b) is strictly increasing in r when a = b and lim Jr (a, b) = min{a, b},

r→−∞

lim Jr (a, b) = max{a, b},

r→∞

then the sequence {Jk (β − 1, β)}k∈N is strictly increasing and limk→∞ Jk (β − 1, β) = β, therefore, it is enough to prove that β  (k − 1)(β − α) for k  3 and 2α  1 < β, which implies 2α  β. Hence, δα,β (t) > 0 in (0, ∞) for 2α  1  β, this means by considering (9) that when n  2,  (n) (−1)n ln fα,β (x) >0 in (0, ∞) for 2α  1 < β. From the asymptotic expansion in [1, p. 259],   1 1 +O 2 ψ(x) = ln x − 2x x

(12)

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as x → ∞, it is concluded that     ln fα,β (x) < lim ln fα,β (x) x→∞      1 β 1 α−β = lim ln 1 + − + +O 2 x→∞ x 2(x + β) x x = 0, hence, for 2α  1  β and n ∈ N,  (n) >0 (−1)n ln fα,β (x) in (0, ∞). The proof of Theorem 1 is complete.

2

Remark 3. If ck = [β k − (β − 1)k ]/k, then Jk (β − 1, β) = ck /ck−1 . Actually {ck }k∈N is a Stieltβ jes moment sequence, namely, ck = β−1 x k dx x , and any Stieltjes moment sequence satisfies ck2  ck−1 ck+1 just by the Cauchy–Schwarz inequality. This means that {Jk (β − 1, β)}k∈N is increasing. Therefore, there is no need to refer to a complicated one-parameter mean in order to reveal the monotonicity of the sequence {Jk (β − 1, β)}k∈N . Proof of Theorem 2. By Theorem 1, it is obvious that the function fα,1 (x) is logarithmically completely monotonic in (0, ∞) for 2α  1. Conversely, if fα,1 (x) is logarithmically completely monotonic in (0, ∞), then for all real x > 0,   α ln fα,1 (x) = ψ(x) − ln x + < 0, x which is equivalent to   1 α < x ln x − ψ(x) → 2 as x → ∞, which follows from utilizing (12). Now we show that the function [fα,1 (x)]−1 is logarithmically completely monotonic in (0, ∞) for α  1. It is easy to see from (10) that δα,1 (t) < 0 for α  1, so from (9) it is deduced that  (n) 1 (−1)n ln >0 fα,1 (x) in (0, ∞) for α  1 and n  2. Apparently,   1 α ln = − + ln x − ψ(x) fα,1 (x) x is increasing, thus,     1 1 ln < lim ln x→∞ fα,1 (x) fα,1 (x)      α 1 − 2α 1 = 0. +O 2 = lim − + ln x − ψ(x) = lim x→∞ x→∞ x 2x x Hence, for α  1 and n ∈ N,  (n) 1 n (−1) ln >0 fα,1 (x)

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holds in (0, ∞). Finally, assume that fα,11(x) is logarithmically completely monotonic in (0, ∞). By definition, this gives us that for all real x > 0,   1 α ln = − + ln x − ψ(x) < 0, fα,1 (x) x which implies   α > x ln x − ψ(x) → 1 as x tends to 0. The proof of Theorem 2 is complete.

2

Acknowledgments The authors express heartily thanks to two anonymous referees for their valuable comments on this article.

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