Necessary and sufficient conditions for a class of functions and their reciprocals to be logarithmically completely monotonic

Abstract

We prove that the function F _α,β (x) = x^αΓ^β(x)/Γ(βx) is strictly logarithmically completely monotonic on (0, ∞) if and only if (α, β) ∈ {(α, β) : β > 0, β ≥ 2α + 1, β ≥ α + 1}\{(α, β) : α = 0, β = 1} and that [F _α,β (x)]^-1 is strictly logarithmically completely monotonic on (0, ∞) if and only if (α, β) ∈ {(α, β ) : β > 0, β ≤ 2α + 1, β ≤ α + 1}\{(α, β ) : α = 0, β = 1}.

2010 Mathematics Subject Classification: 33B15; 26A48.

1 Introduction

For real and positive values of x the Euler gamma function Γ and its logarithmic derivative ψ, the so-called digamma functions are defined by

$$\Gamma \left(x\right)={\int }_0^{\infty }{t}^{x-1}{e}^{-t}dt,$$

(1.1)

$$\psi \left(x\right)=\frac{{\Gamma }^{\prime }\left(x\right)}{\Gamma \left(x\right)}=-\gamma +{\int }_0^{\infty }\frac{e^{-t}-e^{-xt}}{1-e^{-t}}dt,$$

(1.2)

where γ = 0.5772 ··· is the Euler's constant.

For extension of these functions to complex variable and for basic properties see [1]. Over the last half century, many authors have established inequalities and monotonicity for these functions [2–22].

We know that a real-valued function f : I → ℝ is said to be completely monotonic on I if f has derivatives of all orders on I and

$${\left(-1\right)}^n{f}^{\left(n\right)}\left(x\right)\ge 0$$

(1.3)

for all x ∈ I and n ≥ 0. Moreover, f is said to be strictly completely monotonic if inequalities (1.3) are strict.

We also know that a positive real-valued function f : I → (0, ∞) is said to be logarithmically completely monotonic on I if f has derivatives of all orders on I and its logarithm log f satisfies

$${\left(-1\right)}^k{\left[logf\left(x\right)\right]}^{\left(k\right)}\ge 0$$

(1.4)

for all x ∈ I and k ∈ ℕ. Moreover, f is said to be strictly logarithmically completely monotonic if inequalities (1.4) are strict.

Recently, the completely monotonic or logarithmically completely monotonic functions have been the subject of intensive research. In particular, many complete monotonicity and logarithmically complete monotonicity properties related to the gamma function, psi function, and polygamma function can be found in the literature [17, 18, 23–37]. In 1997, Merkle [38] proved that $F\left(x\right)=\frac{\Gamma \left(2x\right)}{{\Gamma }^2\left(x\right)}$ is strictly log-concave on (0, ∞). Later, Chen [39] showed that ${\left[F\left(x\right)\right]}^{-1}=\frac{{\Gamma }^2\left(x\right)}{\Gamma \left(2x\right)}$ is strictly logarithmically completely monotonic on (0, ∞). In [40], Li and Chen proved that $F_{\beta }\left(x\right)=\frac{{\Gamma }^{\beta }\left(x\right)}{\Gamma \left(\beta x\right)}$ is strictly logarithmically completely monotonic on (0, ∞) for β > 1, and that [F _β (x)]^-1 is strictly logarithmically completely monotonic on (0, ∞) for 0 < β < 1. The purpose of this article is to generalize Li and Chen's result. Our main result is as follows.

Theorem 1.1 Let α ∈ ℝ, β > 0 and F _α,β (x) = x^α Γ^β(x)/Γ(βx), then

• (1) F _α,β (x) is strictly logarithmically completely monotonic on (0, ∞) if and only if (α, β) ∈ {(α, β) : β > 0, β ≥ 2α + 1, β ≥ α + 1}\{(α, β) : α = 0, β = 1};

• (2) [F _α,β (x)]^-1 is strictly logarithmically completely monotonic on (0, ∞) if and only if (α, β) ∈ {(α, β) : β > 0, β ≤ 2α + 1, β ≤ α + 1}\{(α, β) : α = 0, β = 1}.

2 Lemma

In order to prove our Theorem 1.1, we need a lemma which we present in this section.

Lemma 2.1 Let α ∈ ℝ, β ∈ (0, 1) ∪ (1, ∞) and

$$h\left(t\right)=-\alpha e^{-\left(\beta +1\right)t}+\left(\alpha +1\right)e^{-\beta t}+\left(\alpha -\beta \right)e^{-t}+\beta -\alpha -1.$$

Then the following statements are true:

• (1) If β ≤ α + 1 and β ≤ 2α + 1, then h(t) < 0 for t ∈ (0, ∞);

• (2) If α + 1 < β < 2α + 1, then there exists λ ₁ ∈ (0, ∞) such that h(t) < 0 for t ∈ (0, λ ₁) and h(t) > 0 for t ∈ (λ ₁, ∞);

• (3) If β ≥ α + 1 and β ≥ 2α + 1, then h(t) > 0 for t ∈ (0, ∞);

• (4) If 2α + 1 < β < α + 1, then there exists λ ₂ ∈ (0, ∞) such that h(t) > 0 for t ∈ (0, λ ₂) and h(t) < 0 for t ∈ (λ ₂, ∞).

Proof Let h₁(t) = e^{(β + 1)t}h'(t) and $h_2\left(t\right)=e^{-t}{h}_1^{\prime }\left(t\right)$. Then simple computations lead to

$$h\left(0\right)=0,$$

(2.1)

$$\begin{array}{lll}\hfill h^{\prime }\left(t\right)& =\alpha \left(\beta +1\right)e^{-\left(\beta +1\right)t}-\beta \left(\alpha +1\right)e^{-\beta t}-\left(\alpha -\beta \right)e^{-t},& \hfill \\ \hfill h_1\left(0\right)& =h^{\prime }\left(0\right)=0,& \hfill \\ \hfill \end{array}$$

(2.2)

$$\begin{array}{lll}\hfill h_1\left(t\right)& =\alpha \left(\beta +1\right)-\beta \left(\alpha +1\right)e^t-\left(\alpha -\beta \right)e^{\beta t},& \hfill \\ \hfill {h^{\prime }}_1\left(t\right)& =-\beta \left(\alpha +1\right)e^t-\beta \left(\alpha -\beta \right)e^{\beta t},& \hfill \\ \hfill \end{array}$$

(2.3)

$$h_2\left(0\right)=h_1^{\prime }\left(0\right)=\beta \left(\beta -2\alpha -1\right),$$

(2.4)

$$h_2\left(t\right)=-\beta \left(\alpha +1\right)-\beta \left(\alpha -\beta \right)e^{\left(\beta -1\right)t}$$

(2.5)

and

$$h_2^{\prime }\left(t\right)=\beta \left(\beta -1\right)\left(\beta -\alpha \right)e^{\left(\beta -1\right)t}.$$

(2.6)

• (1) If β ≤ α + 1 and β ≤ 2α + 1, then we divide the proof into four cases.

Case 1 If 0 < β < 1 and α < β ≤ 2α + 1, then from (2.4) and (2.6) we clearly see that

$$h_2\left(0\right)\le 0,$$

(2.7)

$$h_2^{\prime }\left(t\right)<0.$$

(2.8)

Therefore, h(t) < 0 for t ∈ (0, ∞), which follows from (2.7) and (2.8) together with (2.1) and (2.2).

Case 2 If 0 < β < 1 and β ≤ α, then (2.5) and (2.6) lead to

$$\underset{t\to +\infty }{lim}{h}_2\left(t\right)=-\beta \left(\alpha +1\right)<0,$$

(2.9)

$$h_2^{\prime }\left(t\right)\ge 0.$$

(2.10)

Therefore, h(t) < 0 for t ∈ (0, ∞), which follows from (2.9) and (2.10) together with (2.1) and (2.2).

Case 3 If 1 < β ≤ α, then (2.4) and (2.6) lead to

$$h_2\left(0\right)<0,$$

(2.11)

$$h_2^{\prime }\left(t\right)\le 0.$$

(2.12)

From equations (2.1) and (2.2) together with inequalities (2.11) and (2.12), we clearly see that h(t) < 0 for t ∈ (0, ∞).

Case 4 If β > 1 and α < β ≤ α + 1, then we clearly see that

$$\underset{t\to +\infty }{lim}h\left(t\right)=\beta -\alpha -1\le 0.$$

(2.13)

From (2.3)-(2.6), we know that

$$\underset{t\to +\infty }{lim}{h}_1\left(t\right)=+\infty ,$$

(2.14)

$$h_2\left(0\right)<0,$$

(2.15)

$$\underset{t\to +\infty }{lim}{h}_2\left(t\right)=+\infty ,$$

(2.16)

$$h_2^{\prime }\left(t\right)>0.$$

(2.17)

From (2.15)-(2.17), we clearly see that there exists t₁> 0 such that h₂(t) < 0 for t ∈ (0, t₁) and h₂(t) > 0 for t ∈ (t₁, ∞). Hence, h₁(t) is strictly decreasing in [0, t₁] and strictly increasing in [t₁, ∞).

From (2.2) and (2.14) together with the monotonicity of h₁(t), we know that there exists t₂> 0 such that h₁(t) < 0 for t ∈ (0, t₂) and h₁(t) > 0 for t ∈ (t₂, ∞). Hence, h(t) is strictly decreasing in [0, t₂] and strictly increasing in [t₂, ∞).

Therefore, h(t) < 0 for t ∈ (0, ∞) follows from (2.1) and (2.13) together with the monotonicity of h(t).

(2) If α + 1 < β < 2α + 1, then we clearly see that

$$\underset{t\to +\infty }{lim}h\left(t\right)=\beta -\alpha -1>0$$

(2.18)

and (2.14)-(2.17) hold again. From the proof of Case 4 in Lemma 2.1(1), we know that there exists λ > 0 such that h(t) is strictly decreasing in [0, λ] and strictly increasing in [λ, ∞).

Therefore, Lemma 2.1(2) follows from (2.1) and (2.18) together with the monotonicity of h(t).

(3) If β ≥ α + 1 and β ≥ 2α + 1, then we divide the proof into three cases.

Case I If β > 1 and β ≥ 2α + 1, then

$$\beta >\alpha$$

(2.19)

and it follows from (2.4) that

$$h_2\left(0\right)\ge 0.$$

(2.20)

Equation (2.6) and inequality (2.19) lead to

$$h_2^{\prime }\left(t\right)>0.$$

(2.21)

Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1), (2.2), (2.20), and (2.21).

Case II If 0 < β < 1 and α ≤ -1, then from (2.5) and (2.6) we clearly see that

$$\underset{t\to +\infty }{lim}{h}_2\left(t\right)=-\beta \left(\alpha +1\right)\ge 0,$$

(2.22)

$$h_2^{\prime }\left(t\right)<0.$$

(2.23)

Inequalities (2.22) and (2.23) imply that

$$h_2\left(t\right)>0$$

(2.24)

for t ∈ (0, ∞).

Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1) and (2.2) together with (2.24).

Case III If 0 < α + 1 ≤ β < 1, then we clearly see that

$$\underset{t\to +\infty }{lim}h\left(t\right)=\beta -\alpha -1\ge 0.$$

(2.25)

It follows from (2.3)-(2.6) that

$$\underset{t\to +\infty }{lim}{h}_1\left(t\right)=-\infty ,$$

(2.26)

$$h_2\left(0\right)=\beta \left(\beta -2\alpha -1\right)>0,$$

(2.27)

$$\underset{t\to +\infty }{lim}{h}_2\left(t\right)=-\beta \left(\alpha +1\right)<0,$$

(2.28)

$$h_2^{\prime }\left(t\right)<0.$$

(2.29)

Inequalities (2.27)-(2.29) imply that there exists t₃> 0 such that h₂(t) > 0 for t ∈ (0, t₃) and h₂(t) < 0 for t ∈ (t₃, ∞). Hence, h₁(t) is strictly increasing in [0, t₃] and strictly decreasing in [t₃, ∞).

It follows from (2.2) and (2.26) together with the monotonicity of h₁(t) that there exists t₄> 0 such that h₁(t) > 0 for t ∈ (0, t₄) and h₁(t) < 0 for t ∈ (t₄, ∞). Hence, h(t) is strictly increasing in [0, t₄] and strictly decreasing in [t₄, ∞).

Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1) and (2.25) together with the monotonicity of h(t).

(4) If 2α + 1 < β < α + 1, then we clearly see that

$$\underset{t\to +\infty }{lim}h\left(t\right)=\beta -\alpha -1<0$$

(2.30)

and (2.26)-(2.29) hold again.

From the proof of Case III in Lemma 2.1(3) we know that there exists μ > 0 such that h(t) is strictly increasing in [0, μ] and strictly decreasing in [μ, ∞).

Therefore, Lemma 2.1(4) follows from (2.1) and (2.30) together with the monotonicity of h(t).

3 Proof of Theorem 1.1

Proof of Theorem 1.1 Let E₁ = {(α, β) : 0 < β < 1, β ≥ α + 1}, E₂ = {(α, β) : β > 1, β ≥ 2α +1}, E₃ = {(α, β) : α < 0, β = 1}, E₄ = {(α, β) : α = 0, β = 1}, E₅ = {(α, β) : α + 1 < β < 2α + 1}, E₆ = {(α, β) : β > 0, 2α + 1 < β < α + 1}, E₇ = {(α, β) : 0 < β < 1, β ≤ 2α + 1}, E₈ = {(α, β) : β > 1, β ≤ α + 1} and E₉ = {(α, β) : α > 0, β = 1}. Then

$$\begin{array}{c}\left\{\left(\alpha ,\beta \right):\alpha \in ℝ,\beta >0\right\}={\cup }_{i=1}^9{E}_i,\\ \left\{\left(\alpha ,\beta \right):\beta >0,\beta \ge 2\alpha +1,\beta \ge \alpha +1\right\}\backslash \left\{\left(\alpha ,\beta \right):\alpha =0,\beta =1\right\}=E_1\cup E_2\cup E_3,\\ \left\{\left(\alpha ,\beta \right):\beta >0,\beta \le 2\alpha +1,\beta \le \alpha +1\right\}\backslash \left\{\left(\alpha ,\beta \right):\alpha =0,\beta =1\right\}=E_7\cup E_8\cup E_9.\end{array}$$

• (1) We divide the proof of Theorem 1.1(1) into five cases.

Case 1.1 (α, β) ∈ E₁ ∪ E₂. From (1.1), (1.2), and applying

$${\psi }^m(x)=(-{1)}^{m+1}{\displaystyle {\int }_0^{\infty }}\frac{t^m}{1-e^{-t}}{e}^{-xt}dt(x>\mathrm{0,}m=\mathrm{1,2,...}),$$

we obtain for n ≥ 1,

$$\begin{array}{l}{(-1)}^n{[\log F_{\alpha \mathrm{,}\beta }(x)]}^{(n)}\\ =(-{1)}^n\left[{(-1)}^{n-1}\frac{\alpha (n-1)!}{x^n}+\beta {\psi }^{(n-1)}(x)-{\beta }^n{\psi }^{(n-1)}(\beta x)\right]\\ =-\alpha {\displaystyle {\int }_0^{\infty }}{s}^{n-1}{e}^{-xs}ds+\beta {\displaystyle {\int }_0^{\infty }}\frac{s^{n-1}}{1-e^{-s}}{e}^{-xs}ds-{\beta }^n{\displaystyle {\int }_0^{\infty }}\frac{t^{n-1}}{1-e^{-t}}{e}^{-\beta xt}dt\\ =-\alpha {\beta }^n{\displaystyle {\int }_0^{\infty }}{t}^{n-1}{e}^{-\beta xt}dt+{\beta }^{n+1}{\displaystyle {\int }_0^{\infty }}\frac{t^{n-1}}{1-e^{-\beta t}}{e}^{-\beta xt}dt-{\beta }^n{\displaystyle {\int }_0^{\infty }}\frac{t^{n-1}}{1-e^{-t}}{e}^{-\beta xt}dt\\ ={\beta }^n{\displaystyle {\int }_0^{\infty }}\frac{t^{n-1}{e}^{-\beta xt}}{(1-e^{-t})(1-e^{-\beta t})}h(t)dt\mathrm{,}\end{array}$$

(3.1)

where

$$h\left(t\right)=-\alpha e^{-\left(\beta +1\right)t}+\left(\alpha +1\right)e^{-\beta t}+\left(\alpha -\beta \right)e^{-t}-\alpha +\beta -1.$$

(3.2)

Therefore, F _α,β (x) is strictly logarithmically completely monotonic on (0, ∞), which follows from (3.1) and (3.2) together with Lemma 2.1(3).

Case 1.2 (α, β) ∈ E₃. Then we clearly see that

$${(-1)}^n{[log{F}_{\alpha \mathrm{,}\beta }(x)]}^{(n)}=(-{1)}^n\frac{\alpha (n-1)!(-{1)}^{n-1}}{x^n}=-\frac{\alpha (n-1)!}{x^n}>0$$

(3.3)

for all x > 0.

Therefore, F _α,β (x) is strictly logarithmically completely monotonic on (0, ∞), which follows from (3.3).

Case 1.3 (α,β) ∈ E₄. Then F _α,β (x) = 1 and

$${\left(-1\right)}^n{\left[log{F}_{\alpha ,\beta }\left(x\right)\right]}^{\left(n\right)}=0.$$

(3.4)

Therefore, F _α,β (x) is not strictly logarithmically completely monotonic on (0, ∞), which follows from (3.4).

Case 1.4 (α, β) ∈ E₅∪E₆∪E₇∪E₈. Then F _α,β (x) is not strictly logarithmically completely monotonic on (0, ∞), which follows from Lemmas 2.1(2), 2.1(4), 2.1(1), and equations (3.1) and (3.2).

Case 1.5 (α, β) ∈ E₉. Then

$${\left(-1\right)}^n{\left[log{F}_{\alpha ,\beta }\left(x\right)\right]}^{\left(n\right)}=-\frac{\alpha \left(n-1\right)!}{x^n}<0$$

(3.5)

for all x > 0.

Therefore, F _α,β (x) is not strictly logarithmically completely monotonic on (0, ∞), which follows from (3.5).

• (2) We divide the proof of Theorem 1.1(2) into five cases.

Case 2.1 (α, β) ∈ E₇ ∪ E₈. Then from (3.1) we get

$$\begin{array}{c}{\left(-1\right)}^n{\left\{log{\left[F_{\alpha ,\beta }\left(x\right)\right]}^{-1}\right\}}^{\left(n\right)}\\ =-{\beta }^n{\int }_0^{\infty }\frac{t^{n-1}{e}^{-\beta xt}}{\left(1-e^{-t}\right)\left(1-e^{-\beta t}\right)}h\left(t\right)dt,\end{array}$$

(3.6)

where

$$h\left(t\right)=-\alpha e^{-\left(\beta +1\right)t}+\left(\alpha +1\right)e^{-\beta t}+\left(\alpha -\beta \right)e^{-t}-\alpha +\beta -1.$$

(3.7)

Therefore, [F _α,β (x)]^-1 is strictly logarithmically completely monotonic on (0, ∞), which follows from (3.6) and (3.7) together with Lemma 2.1(1).

Case 2.2 (α, β) ∈ E₉. Then

$${\left(-1\right)}^n{\left\{log{\left[F_{\alpha ,\beta }\left(x\right)\right]}^{-1}\right\}}^{\left(n\right)}=\frac{\alpha \left(n-1\right)!}{x^n}>0$$

(3.8)

for all x > 0.

Therefore, [F _α,β (x)]^-1 is strictly logarithmically completely monotonic on (0, ∞), which follows from (3.8).

Case 2.3 (α, β) ∈ E₄. Then [F _α,β (x)]^-1 = 1 and

$${\left(-1\right)}^n{\left\{log{\left[F_{\alpha ,\beta }\left(x\right)\right]}^{-1}\right\}}^{\left(n\right)}=0.$$

(3.9)

Therefore, [F _α,β (x)]^-1 is not strictly logarithmically completely monotonic on (0, ∞), which follows from (3.9).

Case 2.4 (α, β) ∈ E₁ ∪ E₂ ∪ E₅ ∪ E₆. Then [F _α,β (x)]^-1 is not strictly logarithmically completely monotonic on (0, ∞), which follows from equations (3.6) and (3.7) and Lemmas 2.1(3), 2.1(2) and 2.1(4).

Case 2.5 (α, β) ∈ E₃. Then

$${\left(-1\right)}^n{\left\{log{\left[F_{\alpha ,\beta }\left(x\right)\right]}^{-1}\right\}}^{\left(n\right)}=\frac{\alpha \left(n-1\right)!}{x^n}<0$$

(3.10)

for all x > 0.

Therefore, [F _α,β (x)]^-1 is not strictly logarithmically completely monotonic on (0, ∞), which follows from (3.10). □