THREE CLASSES OF LOGARITHMICALLY COMPLETELY MONOTONIC FUNCTIONS INVOLVING GAMMA AND PSI

FUNCTIONS

FENG QI

Abstract. By a simple approach, two classes of functions involving Euler’s gamma function and originating from certain problems of traffic flow are proved to be logarithmically completely monotonic and a class of functions involving the psi function is showed to be completely monotonic.

1. Introduction

Recall [12, 22, 23] that a function f is said to be completely monotonic on an interval I if f has derivatives of all orders on I and 0 ≤ (−1)^{n}f^{(n)}(x) < ∞ for x ∈ I and n ≥ 0. The set of the completely monotonic functions on I is denoted by C[I]. The well known Bernstein’s Theorem [23, p. 161] states that f ∈ C[(0, ∞)] if and only if f(x) = ∫_{0}^{∞} e^{−xs} dμ(s), where μ is a nonnegative measure on [0, ∞) such that the integral converges for all x > 0. This expresses that f ∈ C[(0, ∞)] if and only if f is a Laplace transform of the measure μ.

Recall [1, 6, 12, 15, 17, 19, 20] also that a positive function f is called logarithmically completely monotonic on an interval I if f has derivatives of all orders on I and its logarithm ln f satisfies 0 ≤ (−1)^{k}[ln f(x)]^{(k)} < ∞ for all k ∈ ℕ on I. The set of the logarithmically completely monotonic functions on I is denoted by L[I]. In [2, Theorem 1.1] and [6, 20] it is pointed out that the logarithmically completely monotonic functions on (0, ∞) can be characterized as the infinitely divisible completely monotonic functions studied by Horn in [7, Theorem 4.4].

It was proved in [2, 12, 19, 20, 22] that L[I] ⊂ C[I], but not conversely. Stimulated by the papers [17, 19], among other things, it was further revealed in [2] that S \ {0} ⊂ L[(0, ∞)] ⊂ C[(0, ∞)], where S denotes the set of Stieltjes transforms.

The Kershaw’s inequality in [8] states that the double inequality

(x + \frac{s}{2})^{1−s} < \frac{Γ(x + 1)}{Γ(x + s)} < (x − \frac{1}{2} + \sqrt{s + \frac{1}{4}} )^{1−s}  (1)

holds for 0 < s < 1 and x ≥ 1, where Γ denotes the classical Euler’s gamma function and ψ = \frac{Γ′}{Γ} , the logarithmic derivative of Γ, the psi function. If taking s = \frac{1}{2} in (1), then, for x > 1,

2000 Mathematics Subject Classification. Primary 33B15, 26A48, 26A51; Secondary 26D20.

Key words and phrases. logarithmically completely monotonic function, completely monotonic function, gamma function, psi function, inequality.

The author was supported in part by the Science Foundation of the Project for Fostering Innovation Talents at Universities of Henan Province, China.

This paper was typeset using AMS-L^{A}TEX.

\sqrt{x + \frac{1}{4}} < \frac{Γ(x + 1)}{Γ(x + 1/2)} < \sqrt{x + \frac{\sqrt{3} − 1}{2}} .  (2)

Let s and t be nonnegative numbers and α = min{s, t}. For x ∈ (−α, ∞), define

z_{s,t}(x) = \begin{cases}[\frac{Γ(x + t)}{Γ(x + s)}]^{1/(t−s)} − x, & s ≠ t, \\ e^{ψ(x+s)} − x, & s = t. \end{cases}  (3)

In order to establish the best bounds in Kershaw’s inequality (1), among other things, the papers [3, 5, 13, 18] established the following monotonicity and convexity property of z_{s,t}(x): The function z_{s,t}(x) is either convex and decreasing for |t − s| < 1 or concave and increasing for |t − s| > 1. This result was further generalized in the papers [10, 11].

In [4, p. 123] and [9], while studying certain problems of traffic flow, a double inequality below was obtained for n ∈ ℕ:

2Γ(n + \frac{1}{2}) ≤ Γ(\frac{1}{2})Γ(n + 1) ≤ 2^{n}Γ(n + \frac{1}{2}),  (4)

In [21], inequality (4) was extended and refined for x > 0 as

\sqrt{x} ≤ \frac{Γ(x + 1)}{Γ(x + 1/2)} ≤ \sqrt{x + \frac{1}{2}}.  (5)

It is clear that the double inequality (5) is weaker than (2).

Observe that inequality (4) can be rearranged for n > 1 as

1 ≤ [\frac{Γ(1/2)Γ(n + 1)}{2Γ(n + 1/2)}]^{1/(n−1)} ≤ 2.  (6)

Hinted by this, the following function g(x) was defined in [14] for x ∈ (−\frac{1}{2}, ∞):

g(x) = \begin{cases}[\frac{Γ(1/2)Γ(x + 1)}{2Γ(x + 1/2)}]^{1/(x−1)}, & x ≠ 1, \\ exp[1 − γ − ψ(\frac{3}{2})], & x = 1, \end{cases}  (7)

where γ = 0.57721566 ⋯ is Euler-Mascheroni’s constant, and, among other things, it was proved in [14] that the function g(x) is logarithmically complete monotonic in (−\frac{1}{2}, ∞): g(x) ∈ L[(−\frac{1}{2}, ∞)] with lim_{x→−1/2} g(x) = ∞ and lim_{x→∞} g(x) = 1. As consequences of this result, it is deduced that

2Γ(x + \frac{1}{2}) ≤ Γ(\frac{1}{2})Γ(x + 1) ≤ 2Γ(x + \frac{1}{2}) exp\{(x − 1)[1 − γ − ψ(\frac{3}{2})]\} (8)

for x ∈ [1, ∞) and

2Γ(x + \frac{1}{2}) ≤ Γ(\frac{1}{2})Γ(x + 1) ≤ 2^{x}Γ(x + \frac{1}{2})  (9)

THREE CLASSES OF LOGARITHMICALLY COMPLETELY MONOTONIC FUNCTIONS 3

for x ∈ (0, ∞). It was remarked in [14] that inequalities (8) and (9) extend (4) and (6), the right hand side inequality of (8) refines the right hand side inequality of (4) and (6), and the right hand side inequalities in (2) and (5) and the following inequality

\frac{Γ(x + 1)}{Γ(x + 1/2)} ≤ \frac{2}{Γ(1/2)} exp\{(x − 1)[1 − γ − ψ(\frac{3}{2})]\}  (10)

for x ∈ [1, ∞), which is deduced from the right hand side inequality of (8), are not included with each other respectively.

Now rewrite inequality (4) or (6) for n > 1 as

1 ≤ [\frac{Γ(1 + 1/2)}{Γ(1 + 1)} · \frac{Γ(n + 1)}{Γ(n + 1/2)}]^{1/(n−1)} ≤ 2.  (11)

The definition (7) of g(x) and inequality (11) motivate us to introduce a new function h_{β}(x) as follows: Let s and t be nonnegative numbers with s ≠ t, α = min{s, t} and β > −α. For x ∈ (−α, ∞), define

h_{β}(x) = \begin{cases}[\frac{Γ(β + t)}{Γ(β + s)} · \frac{Γ(x + s)}{Γ(x + t)}]^{1/(x−β)}, & x ≠ β, \\ exp[ψ(β + s) − ψ(β + t)], & x = β. \end{cases}  (12)

The first aim of this paper is to consider the logarithmically completely monotonic property of h_{β}(x) by a simple approach. Our first main result is the following Theorem 1.

Theorem 1. (1) If s > t, then h_{β}(x) ∈ L[(−α, ∞)] with

lim_{x→−α} h_{β}(x) = ∞  and  lim_{x→∞} h_{β}(x) = 1.  (13)
(2) If s < t, then [h_{β}(x)]^{−1} ∈ L[(−α, ∞)] with
lim_{x→−α} h_{β}(x) = 0  and  lim_{x→∞} h_{β}(x) = 1.  (14)

Remark 1. It is noted that taking s = 1, t = α = \frac{1}{2} and β = 1 in Theorem 1 can deduce one of the results obtained in [14], the logarithmically complete monotonicity of the function g(x) defined by (7).

Since h_{β}(x) is decreasing (or increasing) for s > t (or s < t), h_{β}(β) = exp[ψ(β + s) − ψ(β + t)] and lim_{x→∞} h_{β}(x) = 1, then the following double inequality (15), as a direct consequence of Theorem 1, is established easily.

Corollary 1. Let s and t be nonnegative numbers, α = min{s, t} and β > −α. If s > t, inequality

\frac{Γ(β + s)}{Γ(β + t)} ≤ \frac{Γ(x + s)}{Γ(x + t)} ≤ \frac{Γ(β + s)}{Γ(β + t)} exp\{(x − β)[ψ(β + s)] − ψ(β + t)\}  (15)

holds for x ∈ [β, ∞). If s < t, inequality (15) reverses.

Remark 2. If taking β = 1, s = 1 and t = \frac{1}{2} in (15), then inequality (8) is deduced. So, it can be said that inequality (15) is a generalization of (8).

In [16], it was showed that the function ln x − \frac{1}{2x} − ψ(x) ∈ C[(0, ∞)]. In order to prove Theorem 3 below, this result need to be generalized. Our second main result is the following Theorem 2.

Theorem 2. Let α ∈ ℝ. Then ψ(x) − ln x + \frac{α}{x} ∈ C[(0, ∞)] if and only if α ≥ 1 and ln x − \frac{α}{x} − ψ(x) ∈ C[(0, ∞)] if and only if α ≤ \frac{1}{2}.

For x ∈ (0, ∞), define

p(x) = \begin{cases}[\frac{x^{x}}{Γ(x + 1)}]^{1/(1−x)}, & x ≠ 1, \\ e^{−γ}, & x = 1. \end{cases}  (16)

In [14], among other things, the logarithmically complete monotonicity of p(x) was proved: p(x) ∈ L[(0, ∞)] with lim_{x→0+} p(x) = 1 and lim_{x→∞} p(x) = \frac{1}{e}. Motivated by inequality (11) and the definition of h_{β}(x) in (12), a more general function than p(x) can be introduced: For x ∈ (0, ∞) and α > 0, let

p_{α}(x) = \begin{cases}[\frac{Γ(α + 1)}{α^{α}} · \frac{x^{x}}{Γ(x + 1)}]^{1/(α−x)}, & x ≠ α, \\ \frac{exp[ψ(α + 1) − 1]}{α}, & x = α. \end{cases}  (17)

It is clear that p_{1}(x) = p(x).

The third aim of this paper is to show the logarithmically completely monotonicity of the function p_{α}(x) for any fixed α > 0 by a simple approach. Our third main result is the following Theorem 3.

Theorem 3. For any fixed α > 0, p_{α}(x) ∈ L[(0, ∞)] with

lim_{x→0+} p_{α}(x) = \frac{\sqrt[α]{Γ(α + 1)}}{α}  and  lim_{x→∞} p_{α}(x) = \frac{1}{e}.  (18)

2. Proofs of theorems

It is well-known (see [12, 15, 17, 18, 19, 20]) that, for x > 0 and ω > 0,

\frac{1}{x^{ω}} = \frac{1}{Γ(ω)} ∫_{0}^{∞} t^{ω−1}e^{−xt} dt,  (19)

and that, for k ∈ ℕ and x > 0,

ψ(x) = ln x + ∫_{0}^{∞}(\frac{1}{u} − \frac{1}{1 − e^{−u}} )e^{−xu} du,  (20)
ψ^{(k)}(x) = (−1)^{k+1} ∫_{0}^{∞} \frac{t^{k}e^{−xt}}{1 − e^{−t}} dt.  (21)

Proof of Theorem 1. Without loss of generality, assume s > t. For x ≠ β, taking logarithm of the function h_{β}(x) gives

ln h_{β}(x) = \frac{1}{x − β} [ln \frac{Γ(x + s)}{Γ(β + s)} − ln \frac{Γ(x + t)}{Γ(β + t)}]

= \frac{ln Γ(x + s) − ln Γ(β + s)}{x − β} − \frac{ln Γ(x + t) − ln Γ(β + t)}{x − β}

= \frac{1}{x − β} ∫_{β}^{x} ψ(u + s) du − \frac{1}{x − β} ∫_{β}^{x} ψ(u + t) du

= \frac{1}{x − β} ∫_{β}^{x}[ψ(u + s) − ψ(u + t)] du

= \frac{1}{x − β} ∫_{β}^{x} ∫_{t}^{s} ψ′(u + v) dv du

≜ \frac{1}{x − β} ∫_{β}^{x} Φ_{s,t}(u) du

= ∫_{0}^{1} Φ_{s,t}((x − β)u + β) du

and, by differentiating ln h_{β}(x) with respect to x,

[ln h_{β}(x)]^{(k)} = ∫_{0}^{1} u^{k}Φ_{s,t}^{(k)}((x − β)u + β) du  (22)

for k ∈ ℕ.

If x = β, formula (22) is also valid.

Formula (21) implies that ψ′ ∈ C[(0, ∞)] and Φ_{s,t}(u) ∈ C[(−t, ∞)]. This means that (−1)^{i}[Φ_{s,t}(u)]^{(i)} ≥ 0 holds in u ∈ (−t, ∞) for any nonnegative integer i. Thus,

(−1)^{k}[ln h_{β}(x)]^{(k)} = ∫_{0}^{1} u^{k}{(−1)^{k}Φ_{s,t}^{(k)}((x − β)u + β)} du ≥ 0  (23)

in (−t, ∞) for k ∈ ℕ. The proof of Theorem 1 is complete. □

Proof of Theorem 2. Formulas (19) and (20) imply that

ψ(x) − ln x + \frac{α}{x} = ∫_{0}^{∞}[\frac{(1 − u)e^{u} − 1}{u(e^{u} − 1)} + α]e^{−xu} du ≜ ∫_{0}^{∞}[θ(u) + α]e^{−xu} du,

θ′(u) = \frac{(u^{2} + 2)e^{u} − 1 − e^{2u}}{u^{2}(e^{u} − 1)^{2}} ≜ \frac{θ_{1}(u)}{u^{2}(e^{u} − 1)^{2}} ,

θ′_{1}(u) = [u^{2} + 2u + 2 − 2e^{u}]e^{u} ≜ e^{u}θ_{2}(u),

θ′_{2}(u) = 2(1 + u − e^{u}) < 0.

It is clear that the function θ_{2}(u) is decreasing in (0, ∞). From θ_{2}(0) = 0, it follows that θ_{2}(u) < 0 and θ′_{1}(u) < 0 in (0, ∞) and that θ′_{1}(u) decreases in (0, ∞). Since θ′_{1}(0) = 0, then θ′_{1}(u) < 0 and θ_{1}(u) is decreasing in (0, ∞). From θ_{1}(0) = 0, it follows that θ_{1}(u) < 0 and θ′(u) < 0 in (0, ∞), and then the function θ(u) is decreasing in (0, ∞). L’Hôspital’s rule yields that lim_{u→0+} θ(u) = −\frac{1}{2} and lim_{u→∞} θ(u) = −1. Thus,

θ(u) + α \begin{cases}≥ 0, & α ≥ 1 \\ ≤ 0, & α ≤ \frac{1}{2} \end{cases}

on [0, ∞]. Since [ψ(x)−ln x+ \frac{α}{x}]^{(k)} = (−1)^{k} ∫_{0}^{∞}[θ(u)+α]u^{k}e^{−xu} du for nonnegative integer k, then ψ(x) − ln x + \frac{α}{x} ∈ C[(0, ∞)] if α ≥ 1 (or ln x − \frac{α}{x} − ψ(x) ∈ C[(0, ∞)] if α ≤ \frac{1}{2}).

If ψ(x)−ln x+\frac{α}{x} ∈ C[(0, ∞)] (or ln x−\frac{α}{x}−ψ(x) ∈ C[(0, ∞)]), then ψ(x)−ln x+\frac{α}{x} > 0 (or < 0), this leads to α > x[ln x − ψ(x)] (or α < x[ln x − ψ(x)]) in (0, ∞). Since lim_{x→0+}\{x[ln x − ψ(x)]\} = 1 (or lim_{x→∞}\{x[ln x − ψ(x)]\} = \frac{1}{2}), then α ≥ 1 (or α ≤ \frac{1}{2}). The proof of Theorem 2 is complete. □

Proof of Theorem 3. From the well known differences equation Γ(x + 1) = xΓ(x), it follows easily that

ψ(x + 1) − ψ(x) = \frac{1}{x}  (24)

for x > 0. For x ≠ α, taking logarithm of p_{α}(x) and using (24) gives

ln p_{α}(x) = \frac{1}{x − α}[ln \frac{Γ(x + 1)}{Γ(α + 1)} − \frac{x^{x}}{α^{α}} ]

= \frac{ln Γ(x + 1) − ln Γ(α + 1)}{x − α} − \frac{x ln x − α ln α}{x − α}

= \frac{1}{x − α} ∫_{α}^{x} ψ(u + 1) du − \frac{1}{x − α} ∫_{α}^{x}[1 + ln x] du

= \frac{1}{x − α} ∫_{α}^{x}[ψ(u) − ln u + \frac{1}{u} − 1] du

≜ \frac{1}{x − α} ∫_{α}^{x}[Ψ(u) − 1] du

= ∫_{0}^{1} Ψ((x − α)u + α) du − 1

and, by differentiating ln p_{α}(x) with respect to x,

[ln p_{α}(x)]^{(k)} = ∫_{0}^{1} u^{k}Ψ^{(k)}((x − α)u + α) du  (25)

for k ∈ ℕ.

If x = α, formula (25) is also valid. From Theorem 2, it follows that Ψ(u) ∈ C[(0, ∞)]. This implies (−1)^{i}Ψ^{(i)}(u) ≥ 0

for nonnegative integer i. As a result, for k ∈ ℕ,

(−1)^{k}[ln p_{α}(x)]^{(k)} = ∫_{0}^{1} u^{k}{(−1)^{k}Ψ^{(k)}((x − α)u + α)} du ≥ 0  (26)

holds in x ∈ (0, ∞). The proof of Theorem 3 is complete. □

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(F. Qi) College of Mathematics and Information Science, Henan Normal University, Xinxiang City, Henan Province, 453007, China; Research Institute of Mathematical Inequality Theory, Henan Polytechnic University, Jiaozuo City, Henan Province, 454010, China

E-mail address: qifeng@hpu.edu.cn, fengqi618@member.ams.org, qifeng618@hotmail.com,
qifeng618@msn.com, 316020821@qq.com
URL: http://rgmia.vu.edu.au/qi.html