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Completely Monotonic Degrees for a Difference Between the Logarithmic and Psi Functions Feng Qi, Ai-Qi Liu

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Completely Monotonic Degrees for a Difference Between the Logarithmic and Psi Functions Feng Qi, Ai-Qi Liu Completely monotonic degrees for a difference between the logarithmic and psi functions Feng Qi, Ai-Qi Liu To cite this version: Feng Qi, Ai-Qi Liu. Completely monotonic degrees for a difference between the logarithmic and psi functions: A difference between logarithmic and psi functions. Journal of Computational and Applied Mathematics, Elsevier, 2019, 361, pp.366–371. 10.1016/j.cam.2019.05.001. hal-01728682v2 HAL Id: hal-01728682 https://hal.archives-ouvertes.fr/hal-01728682v2 Submitted on 19 Jan 2020 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. COMPLETELY MONOTONIC DEGREES FOR A DIFFERENCE BETWEEN THE LOGARITHMIC AND PSI FUNCTIONS FENG QI AND AI-QI LIU Abstract. In the paper, the authors firstly present a concise proof for com- plete monotonicity of a function involving a difference between the logarith- mic and psi functions, secondly compute completely monotonic degree of the above-mentioned function, and finally pose several conjectures on completely monotonic degrees of remainders and their derivatives for the asymptotic for- mula of the logarithm of the classical Euler gamma function. 1. Preliminaries Recall from [16, Chapter XIII], [31, Chapter 1], and [32, Chapter IV] that a function f is said to be completely monotonic on an interval I if f has derivatives of all orders on I and n (n) (−1) f (x) ≥ 0; x 2 I; n 2 f0g [ N: Theorem 12b in [32, p. 161] states that a necessary and sufficient condition for f(x) to be completely monotonic on (0; 1) is that Z 1 (1) f(x) = e−xt d µ(t); x 2 (0; 1); 0 where µ(t) is non-decreasing and the above integral converges for x 2 (0; 1). In other words or simply speaking, a function is completely monotonic on (0; 1) if and only if it is a Laplace transform of a non-negative measure. In [2, pp. 374{375, Theorem 1] and [8, Theorem 1], it was verified that the function α (2) x [ln x − (x)]; α 2 R Γ0(z) is completely monotonic on (0; 1) if and only if α ≤ 1, where (z) = Γ(z) is called the psi function and Γ(z) denotes the classical Euler gamma function which can be defined [1, 8, 18, 30] by Z 1 Γ(z) = tz−1e−t d t; <(z) > 0 0 2010 Mathematics Subject Classification. Primary 26A48; Secondary 33B15, 44A10. Key words and phrases. complete monotonicity; completely monotonic degree; difference; psi function; logarithmic function; concise proof; gamma function; asymptotic formula; remainder; conjecture. Please cite this article as \Feng Qi and Ai-Qi Liu, Completely monotonic degrees for a difference between the logarithmic and psi functions, Journal of Computational and Applied Mathematics 361 (2019), 366{371; available online at https://doi.org/10.1016/j.cam.2019.05.001." 1 2 F. QI AND A.-Q. LIU or by n!nz Γ(z) = lim n ; z 2 n f0; −1; −2;::: g: n!1 Q C k=0(z + k) Let f(x) be a completely monotonic function on (0; 1) and denote f(1) = α limx!1 f(x) ≥ 0. When the function x [f(x) − f(1)] is completely monotonic on (0; 1) if and only if 0 ≤ α ≤ r 2 R, we say that the number r is the completely monotonic degree of f(x) with respect to x 2 (0; 1); if the function xα[f(x)−f(1)] is completely monotonic on (0; 1) for all α 2 R, then we say that the completely monotonic degree of f(x) with respect to x 2 (0; 1) is 1. For convenience, a x notation degcm[f(x)] was designed to denote the completely monotonic degree r of f(x) with respect to x 2 (0; 1). For more information on completely monotonic degree and its properties, please refer to the papers [4, 10, 12, 13, 14, 15, 19, 20, 23, 24, 25, 26, 28, 29] and closely related references therein. The necessary and sufficient condition α ≤ 1 for the function (2) to be completely monotonic on (0; 1) means that the completely monotonic degree of the completely monotonic function ln x − (x) on (0; 1) is x degcm[ln x − (x)] = 1: In [30, Theorem 1.7], it was proved that the function x x2[ (x) − ln x] + 2 1 is strictly decreasing and convex on (0; 1) and, as x ! 1, tends to − 12 . In [3, Theorem 1], it was discovered that the function x 1 (3) Φ(x) = x2[ (x) − ln x] + + 2 12 is completely monotonic on (0; 1). In the second section of this paper, motivated by the second proof of [8, Theo- rem 1] and with the aid of some conclusions in [33], we will present a concise proof, which is simpler and shorter than the corresponding proof in [3, Theorem 1], for complete monotonicity of the function Φ(x) on (0; 1). In the third section of this paper, we will compute the completely monotonic degree of the completely monotonic function Φ(x) 1 1 (4) φ(x) = = (x) − ln x + + x2 2x 12x2 on (0; 1). In the fourth section of this paper, we will pose conjectures on completely mono- tonic degrees of remainders for the asymptotic formula of ln Γ(x) and for polygamma functions (k)(x) for k ≥ 0. 2. Complete monotonicity of the function Φ(x) The first main result in this paper can be stated as the following theorem. Theorem 1. The function Φ(x) defined in (3) is completely monotonic on (0; 1), with the limits 1 (5) lim Φ(x) = and lim Φ(x) = 0: x!0+ 12 x!1 A DIFFERENCE BETWEEN LOGARITHMIC AND PSI FUNCTIONS 3 Proof. In [18, p. 140, 5.9.13], it was listed that Z 1 1 1 −tz (z) = ln z + − −t e d t: 0 t 1 − e By integration by parts, this formula can be reformulated as Z 1 1 1 1 −xt (x) − ln x + = − t e d t x 0 t e − 1 Z 1 Z 1 1 −xt t=1 0 −xt 1 1 0 −xt = − h(t)e t=0 − h (t)e d t = + h (t)e d t; x 0 2x x 0 where the function 81 1 > − ; t 6= 0 < t et − 1 h(t) = 1 > ; t = 0 :2 is convex on (0; 1), with the limits 1 (6) lim h0(t) = − and lim h0(t) = 0: t!0+ 12 t!1 For detailed information on the function h(t), please refer to [5, 6, 7, 21] and closely related references therein. Consequently, integrating by parts again yields 1 Z 1 1 Z 1 Z 1 Φ(x) = + x h0(t)e−xt d t = − h0(t) de−xt = h00(t)e−xt d t: 12 0 12 0 0 Accordingly, by virtue of [32, p. 161, Theorem 12b] mentioned at the beginning of this paper and with the help of the convexity for the function h(t) on (0; 1), we immediately see that the function Φ(x) is completely monotonic on (0; 1), while, in light of the limits in (6), we readily derive the limits in (5). The proof of Theorem 1 is complete. 3. Completely monotonic degree of the function φ(x) The second main result in this paper can be stated as the following theorem. Theorem 2. The completely monotonic degree of the completely monotonic func- tion φ(x) defined in (4) on (0; 1) is x (7) degcm[φ(x)] = 2: Proof. Since the relation Φ(x) = x2φ(x) and the function φ(x) is completely mono- tonic on (0; 1), see [22, Theorem 1], then Theorem 1 in [3] and Theorem 1 in this paper mean x (8) degcm[φ(x)] ≥ 2: On the other hand side, if xαφ(x) is completely monotonic on (0; 1), then its first derivative is not positive, that is, αxα−1φ(x) + xαφ0(x) ≤ 0 which can be rearranged as 0 3 0 1 1 1 xφ (x) x (x + 1) − + 2 − 3 α ≤ − = − x 2x 6x ! 2 2 x 1 φ(x) x [ (x) − ln x] + 2 + 12 4 F. QI AND A.-Q. LIU + 0 0 1 for x ! 0 , where we used the recursion relation (x + 1) = (x) − x2 and the first limit in (5). This means that x (9) degcm[φ(x)] ≤ 2: Combining (8) with (9) concludes that the completely monotonic degree of the function φ(x) on (0; 1) is 2. The proof of Theorem 2 is complete. 4. Conjectures The third main result in this paper is to pose conjectures on completely mono- tonic degrees of remainders for the asymptotic formula of ln Γ(x) and on completely monotonic degrees of derivatives of these remainders. In [22, Theorem 1], among other things, the function φ(x) and three functions 1 1 1 1 1 1 1 1 ln x − − (x); 0(x) − − − + ; + + − 0(x) 2x x 2x2 6x3 30x5 x 2x2 6x3 were proved to be completely monotonic on (0; 1). What are the completely mono- tonic degrees of these three functions on (0; 1)? We guess that 1 (10) degx ln x − − (x) = 1; cm 2x 1 1 1 (11) degx + + − 0(x) = 2; cm x 2x2 6x3 1 1 1 1 (12) degx 0(x) − − − + = 4: cm x 2x2 6x3 30x5 In [2, Theorem 8] and [11, Theorem 2], the functions " n # 1 1 X B2k 1 (13) R (x) = (−1)n ln Γ(x)− x− ln x+x− ln(2π)− n 2 2 (2k − 1)2k x2k−1 k=1 for n ≥ 0 were proved to be completely monotonic on (0; 1).
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