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Partial solutions to several conjectures on completely monotonic degrees for remainders in asymptotic expansions of the digamma function

Motivated by several conjectures posed in the paper"F. Qi and A.-Q. Liu, Completely monotonic degrees for a difference between the logarithmic and psi functions, J. Comput. Appl. Math., vol. 361, pp. 366--371 (2019); available online at https://doi.org/10.1016/j.cam.2019.05.001", the authors compute several completely monotonic degrees of the remainders in the asymptotic expansions of the logarithm of the gamma function and in the asymptotic expansions of the logarithm of the digamma function.

in [39, p. 161] states that a necessary and sufficient condition for h to be completely monotonic on the infinite interval (0, ∞) is that h(t) = ∞ 0 e −ts d σ(s), s ∈ (0, ∞), (1.1) where σ(s) is non-decreasing and the above integral converges for s ∈ (0, ∞). In other words, a function is completely monotonic on (0, ∞) if and only if it is a Laplace transform of a non-negative measure.
Definition 1.1. Let h(t) be a completely monotonic function on (0, ∞) and let h(∞) = lim t→∞ h(t) ≥ 0. If the function t α [h(t) − h(∞)] is completely monotonic on (0, ∞) if and only if 0 ≤ α ≤ r ∈ R, then we say that h(t) is of completely monotonic degree r; if t α [h(t) − h(∞)] is completely monotonic on (0, ∞) for all α ∈ R, then we say that the completely monotonic degree of h(t) is ∞.
The function t α [h(t) − h(∞)] in Definition 1.1 can be essentially regarded as the ratio h(t)−h(∞) 1/t α between the completely monotonic function h(t) − h(∞) and the α power of the completely monotonic function 1 t on (0, ∞). This is the reason why we designed in [8] a notation deg t cm [h(t)] to denote the completely monotonic degree r of h(t) with respect to t ∈ (0, ∞). According to this idea, we can define the completely monotonic degree r of h(t) with respect to g(t), denoted by deg g(t) cm [h(t)], as the largest number α such that the ratio h(t)−h(∞) [g(t)−g(∞)] α between the completely monotonic function h(t) and the α power of the completely monotonic function g(t) on (0, ∞).
Proposition 1.2 in [29] can be modified as is a bounded and non-decreasing measure on (0, ∞). For more information on completely monotonic degrees and their properties, please refer to the papers [13,15,16,17,18,24,27] and closely related references therein.
Theorem 2.1. For n ≥ 0, the completely monotonic degrees of the remainder R n (t) with respect to t ∈ (0, ∞) satisfy the two equalities in (2.5).
For n ≥ 0, the completely monotonic degrees of the functions with respect to t ∈ (0, ∞) satisfy the two equalities in (2.7) and a double inequality 3. Proof of Theorem 2.1 Now we start out to prove our main results stated in Theorem 2.1.

Proofs of equalities in (2.5). From Binet's first formula
for t > 0 in [3, p. 28, Theorem 1.6.3] and [19, p. 11], it is easy to see that and The inequality (2.4) means that the completely monotonic degree of the remainder R n (t) for n ≥ 0 with respect to t ∈ (0, ∞) is at least n. In particular, we have If t θ R 0 (t) and t λ R 1 (t) were completely monotonic on (0, ∞), then their first derivatives should be nonpositive. As a result, using Combining those inequalities in (3.1) and (3.2) concludes those equalities in (2.5).

Proofs of equalities in
by the equation (2.1), we conclude the first conclusion in (2.7).
The equation (2.2) established in [30,Theorem 2] is equivalent to the second conclusion in (2.7).

Proof of the equality (2.8). Let
for n ≥ 0, where the empty sum is understood to be 0. Then, by virtue of the formulas [21, p. 140, 5.9.13], and for n ≥ 0. This means that The inequality (2.4) tells us that the remainder R n (t) for n ≥ 0 is completely monotonic in t ∈ (0, ∞). Then, by definition of completely monotonic functions, it is ready that the remainder −R ′ n (t) for n ≥ 0 is completely monotonic on (0, ∞).
Since the function v e v −1 − 1 + v 2 is even in v ∈ R, by virtue of the equation (2.3), it follows that for |v| < 2π and n ≥ 0. Accordingly, we have for 0 ≤ ℓ ≤ 2n or ℓ = 2m with m, n ≥ 0, where are the Stirling numbers of the second kind. Taking ϑ = θ = 1 in (3.5) leads to Making use of (3.7) and lim for ℓ, n ≥ 0 and t > 0. Consequently, by the limits (3.4) and (3.8), integrating by parts inductively 2n − 1 times in the equation (3.3) gives Using the relation for v > 0 and n ≥ 1. Using the relation Consequently, we have Hence, the function t 2n−1 [−R ′ n (t)] is completely monotonic on (0, ∞). This means that If the function t α [−R ′ n (t)] were completely monotonic on (0, ∞), then its first derivative is negative, hence This means that deg t cm [−R ′ n (t)] ≤ 2n, n ≥ 1.

Remarks
Finally we list several remarks on main results and Qi's conjectures in this paper. Remark 4.1. As in the derivation of (3.9), integrating by parts R ′ n (t) inductively 2n ≥ 2 times in (3.3) yields for n ≥ 1, where, by (3.6), for v > 0 and n ≥ 1. It is easy to see that for v ∈ (0, ∞). Therefore, the function K 2 (v) is positive on (0, ∞). By Theorem 12b in [39, p. 161] stated in (1.1), we conclude that the function This supplies an alternative and partial proof for the second equality in (2.7).
Directly proving these three inequalities should be difficult and these functions involving in these three inequalities are important in special functions and number theory. These are two reasons why we write this remark.   5.4. Data availability statements. Data sharing is not applicable to this article as no new data were created or analyzed in this study.