A class of logarithmically completely monotonic functions

Applied Mathematics Letters 21 (2008) 1134–1141 www.elsevier.com/locate/aml

A class of logarithmically completely monotonic functions Senlin Guo a , H.M. Srivastava b,∗ a Department of Mathematics, University of Manitoba, Winnipeg, Manitoba R3T 2N2, Canada b Department of Mathematics and Statistics, University of Victoria, Victoria, British Columbia V8W 3R4, Canada

Received 28 October 2007; accepted 28 October 2007

Abstract The main object of this work is to give some conditions for a class of functions to be logarithmically completely monotonic. Our result is shown to be an extension of a result which was proven in the recent literature on this subject. c 2008 Elsevier Ltd. All rights reserved.

Keywords: Necessary and sufficient conditions; Logarithmically completely monotonic functions; Completely monotonic functions; Gamma functions; Psi (or digamma) functions; Polygamma functions

1. Introduction, definitions and the main results We begin by recalling from [12] that a function f is said to be completely monotonic on R+ := (0, ∞) if, for all n ∈ N0 := N ∪ {0}

(N := {1, 2, 3, · · ·}),

the following inequality holds true: 0 5 (−1)n f (n) (x) < ∞

(x ∈ R+ ).

The set of all completely monotonic functions on R+ is denoted here by C M(R+ ). S. Bernstein (see, for example, [12, Chapter IV, Section 12]) proved that a function f is completely monotonic on the interval R+ if and only if there exists an increasing function α(t) on [0, ∞) such that Z ∞ f (x) = e−xt dα(t). 0

A function f is said to be logarithmically completely monotonic (see [1,9]) on R+ if f is positive and, for all n ∈ N, 0 5 (−1)n [ln f (x)](n) < ∞

(x ∈ R+ ).

The class of all logarithmically completely monotonic functions on R+ is denoted here by LC M(R+ ). ∗ Corresponding author.

E-mail addresses: [email protected] (S. Guo), [email protected] (H.M. Srivastava). c 2008 Elsevier Ltd. All rights reserved. 0893-9659/$ - see front matter doi:10.1016/j.aml.2007.10.028

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It is known from [6,9] that LC M(R+ ) ⊂ C M(R+ ). The Euler gamma function 0(z) is defined, for R(z) > 0, by Z ∞
t z−1 e−t dt R(z) > 0 . 0(z) := 0

For some recent developments involving the gamma function and its incomplete versions γ (κ, z) and 0(κ, z), see the works by (for example) Srivastava [10] and Lin et al. [7]. The psi (or digamma) function ψ(z) is the logarithmic derivative of 0(z), that is, Z z 0 0 (z) ψ(t)dt, ψ(z) = or log 0(z) = 0(z) 1 and ψ (k) (z) (k ∈ N) are called the polygamma functions (see, for details, [8,11]). Let α be a real parameter and suppose that β is a non-negative parameter. Define a two-parameter function f α,β (x) by f α,β (x) :=

ex 0(x + β) x x+β−α

(x ∈ R+ ).

The following result is proved in [2]: Theorem A. For β = 1, if α 5 21 , then f α,β ∈ LC M(R+ ). Among other results, the following result is established in [5]: Theorem B. For β = 1, f α,β ∈ LC M(R+ ) if and only if α5

1 . 2

In this work, we give some sufficient conditions for f α,β ∈ LC M(R+ )

(0 < β < 1).

We also establish a necessary and sufficient condition for ! √ 1 3 + + 5β<∞ . f α,β ∈ LC M(R ) 2 6 We first state our main results as follows. h  Theorem 1. (1) For β ∈ 0, 12 , if   2 −4 2 α 5 β − e (1 − β) exp , 1−β then f α,β ∈ LC M(R+ ).

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h

i (2) For β ∈ 12 , 1 , if   1 2 α 5 min 3β − 3β + 1, , 2 then f α,β ∈ LC M(R+ ). Theorem 2. Let ! √ 1 3 β ∈ {0} ∪ + ,∞ . 2 6 "

Then f α,β ∈ LC M(R+ ) if and only if α5

1 . 2

The following theorems deal with the complete monotonicity property of the two-parameter function gα,β (x) defined by gα,β (x) := x + ln 0(x + β) − (x + β − α) ln x

(x ∈ R+ ),

which is related to the functions f α,β (x). Here, in the definition (1), we assume that α∈R

β = 0.

and

Theorem 3. Suppose that the function gα,β ∈ C M(R+ ). Then α = 12 . h  √ Theorem 4. Let β ∈ {0} ∪ 21 + 63 , ∞ . Then gα,β ∈ C M(R+ ) if and only if α = 12 . 2. Proofs of Theorems 1–4 Proof of Theorem 1. It is clear that ln f α,β (x) = x + ln 0(x + β) − (x + β − α) ln x and [ln f α,β (x)]0 = ψ(x + β) − ln x +

α−β . x

It is also known that (see [4, p. 884]) Z ∞ tn (n) n+1 ψ (x) = (−1) e−xt dt 1 − e−t 0

(n ∈ N; x ∈ R+ )

and 1 1 = n x 0(n)

∞

Z

t n−1 e−xt dt 0

(n ∈ N; x ∈ R+ ).

(1)

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Hence, for n ∈ N \ {1}, (−1)n [ln f α,β (x)](n) (n − 2)! (β − α)(n − 1)! + xn x n−1 Z ∞ Z ∞ n−1 t n−2 −xt −(x+β)t t n−1 e−xt dt t e dt + (β − α) e dt − 1 − e−t 0 0

= (−1)n ψ (n−1) (x + β) − ∞

Z = 0

∞

Z

δ(t)

= 0

t n−2 e−xt dt et − 1

(n ∈ N \ {1}),

(2) (3) (4)

where δ(t) := (β − α)t (et − 1) + te(1−β)t − et + 1. Direct computations show that δ 0 (t) = (β − α)tet + (1 − β)te(1−β)t + e(1−β)t + (β − α − 1)et + α − β, δ(0) = 0,

(5)

δ (0) = 0,

(6)

δ 00 (t) = et g(t),

(7)

0

and

where g(t) := 2β − 2α − 1 + (β − α)t + 2(1 − β)e−βt + (1 − β)2 te−βt .

(8)

It is easily verified that the function   2 −4 2 h(β) := β − e (1 − β) exp 1−β is strictly increasing and strictly concave from     1 1 β ∈ 0, onto h(β) ∈ −e−2 , . 2 4 From this fact and the conditions of Theorem 1, we thus obtain g(0) = 0.

(9)

From (8) we get g 0 (t) = β − α + (1 − β)e−βt [1 − 3β − β(1 − β)t], g (t) = 00

If α 5

3β 2

β(1 − β)e−βt [2(2β

− 1) + β(1 − β)t].

1 2

(11)

− 3β + 1, then

g 0 (0) = 3β 2 − 3β + 1 − α = 0. If

(10)

(12)

5 β 5 1, then, from (11), we get

g 00 (t) = 0.

(13)

By (12) and (13), we obtain g 0 (t) = 0

(t = 0).

(14)

By (9) and (14), we have g(t) = 0

(t = 0).

(15)

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Hence, under the conditions of Theorem 1(2), from (7) and (15), we find that δ 00 (t) = 0

(t = 0).

(16)

Combining (6) and (16), we have δ 0 (t) = 0

(t = 0),

from which and (5), we obtain δ(t) = 0

(t = 0),

(17)

In view of (4), we have proved that, under the conditions of Theorem 1(2), we have (−1)n [ln f α,β (x)](n) = 0

(x ∈ R+ ; n ∈ N \ {1}).

(18)

Now, if 0 < β < 21 , we set t0 :=

2(1 − 2β) . β(1 − β)

Then, from (11), it is clear that g 00 (t) < 0

for t < t0

g 00 (t) > 0

for t > t0 .

and

Therefore, we have g 0 (t) = g 0 (t0 )

(t = 0).

Simple computation shows that g (t0 ) = β − α − e 0

−4



2 (1 − β) exp 1−β 2



.

Hence, if α 5 β − e−4 (1 − β)2 exp



2 1−β

 and

0<β<

1 , 2

then g 0 (t) = 0

(t = 0).

(19)

If β = 0, (19) is trivial. We have thus proved that, under the conditions of Theorem 1(1), (19) holds true. Like in the arguments from (14) to (17), by combining (5) to (7), (9) and (19), we obtain δ(t) = 0

(t = 0),

from which, in conjunction with (4), we see that, for n = 2, (18) also holds true under the conditions of Theorem 1(1). By using the fact that (see [3, p. 47])   1 1 ψ(x) = ln x − +O (x → ∞), 2x x2 it is easily observed that     1 β 1 α−β 0 [ln f α,β (x)] = ln 1 + − + +O x 2(x + β) x x2

(x → ∞).

Thus, for any parameters α and β, we have lim [ln f α,β (x)]0 = 0.

x→∞

(20)

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(x ∈ R+ ).

(21)

By (18), we also have [ln f α,β (x)]00 = 0

From (20) and (21), we thus obtain [ln f α,β (x)]0 5 0

(x ∈ R+ ).

Consequently, for all n ∈ N, (18) holds true. The proof of Theorem 1 is evidently completed.



Proof of Theorem 2. From Theorem 1(2) of [5] and the following relationship: f α,0 = f α,1 , we see that the condition of Theorem 2 is necessary. It is proved in [2] that, for β = 1, if α 5 12 , then f α,β ∈ LC M(R+ ). From this fact and Theorem 1(2), together with f α,0 = f α,1 , we know that the condition of Theorem 2 is also sufficient. The proof of Theorem 2 is thus completed.  Proof of Theorem 3. First of all, we note that gα,β (x) = ln f α,β (x). If gα,β ∈ C M(R+ ), then f α,β ∈ LC M(R+ ). Hence, by Theorem 1(2) of [5] and noting that f α,0 = f α,1 , we obtain 1 . 2 On the other hand, since (see [3, p. 47])     1 ln(2π ) 1 +O ln x − x + ln 0(x + β) = x + β − 2 2 x α5

(22)

(x → ∞),

(23)

we have gα,β (x) = ln x(α − 1/2) + ln

√

  1 2π + O x

(x → ∞).

(24)

In view of the fact that gα,β (x) = 0, we find from (24) that   √ 1 O x ln 2π 1 − (x → ∞), α− =− 2 ln x ln x which readily yields    √ O x1 1 ln 2π  = 0, α − = lim − − 2 x→∞ ln x ln x that is, α=

1 . 2

(25)

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In view of (22) and (25), we immediately conclude that 1 . 2 This evidently completes the proof of Theorem 3. α=



Proof of Theorem 4. By Theorem 3, the condition of Theorem 4 is necessary. On the other hand, since gα,β (x) = ln f α,β (x), by Theorem 2, we obtain (n)

(−1)n gα,β (x) = 0

(x ∈ R+ ; n ∈ N).

(26)

In particular, we have 0 (x) 5 0 gα,β

(x ∈ R+ ),

which means that gα,β (x) is decreasing on R+ . By (23), for α = 21 , we obtain   √ 1 gα,β (x) = ln 2π + O (x → ∞). x Hence gα,β (x) = lim gα,β (x) = ln x→∞

√ 2π ,

(27)

from which we readily see that (26) is also valid for n = 0. The proof of Theorem 4 is thus completed.  3. Corollaries and consequences Several interesting corollaries and consequences of our main results (Theorems 1 to 4) can easily be deduced by appropriately specializing the parameters α and β. Thus, for example, we can deduce the following corollary as a consequence of Theorem 1. h i Corollary. (1) For β ∈ 14 , 12 , if α 5 β − 41 , then f α,β ∈ LC M(R+ ).  i (2) For β ∈ 21 , 34 , if α 5 β − 31 , then f α,β ∈ LC M(R+ ).  i (3) For β ∈ 43 , 1 , if α 5 β − 12 , then f α,β ∈ LC M(R+ ). Acknowledgements The present investigation was supported, in part, by the Natural Sciences and Engineering Research Council of Canada under Grant OGP0007353. References [1] R.D. Atanassov, U.V. Tsoukrovski, Some properties of a class of logarithmically completely monotonic functions, C. R. Acad. Bulgare Sci. 41 (1988) 21–23. [2] C.-P. Chen, F. Qi, Logarithmically completely monotonic functions relating to the gamma function, J. Math. Anal. Appl. 321 (2006) 405–411.

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