Qi’s conjectures on completely monotonic degrees of remainders of asymptotic formulas of di- and trigamma functions

Several conjectures posed by Qi on completely monotonic degrees of remainders for the asymptotic formulas of the digamma and trigamma functions are proved.

first introduced by the Swiss mathematician Leonhard Euler, is one of the most important functions in mathematical analysis. It often appears in asymptotic series, hypergeometric series, Riemann zeta function, number theory, and so on. It is well known that the digamma function or the psi function is defined by ψ(x) = Γ (x)/Γ (x), and ψ (i) (x) (i ≥ 1) is called the polygamma functions. In particular, ψ (x) is called the trigamma function. In [1], an interesting fact was verified that the function f (x) = x α [log(x)ψ(x)] is strictly completely monotonic on (0, ∞) if and only if α ≤ 1, which improved the results due to Anderson et al. [2]. A new proof was given by [5]. The above result implies that deg x cm [log(x)ψ(x)] = 1. It was proved in [27] that the function x 2 [ψ(x) -log(x)] + x/2 is strictly decreasing and convex on (0, ∞) and, as x → ∞, tends to -1/12. Then the result was improved by [3], and it was proved that the function Φ(x) = x 2 [ψ(x) -log(x)] + x/2 + 1/12 is completely monotonic on (0, ∞). A concise proof of the complete monotonicity of the function Φ(x) was presented by Qi and Liu [22]. Meanwhile, they proved that Furthermore, Qi and Liu [22] guessed that where B i denotes the ith Bernoulli number defined by the generating function In fact, R n (x) (n ≥ 0) was proved to be completely monotonic on (0, ∞) in [1] and [7]. In [11], the completely monotonic degree of the function R n (x) for n ≥ 0 on (0, ∞) was proved to be at least n. This means that the functions (-1) m [R n (x)] (m) for m, n ≥ 0 are completely monotonic on (0, ∞). Qi and Liu [22] asked a question: What are the completely monotonic degrees of the completely monotonic functions (-1) m [R n (x)] (m) for m, n ≥ 0 on (0, ∞)? Several conjectures posed by Qi can be recited as follows: (1) The completely monotonic degrees of R n (x) for n ≥ 0 with respect to x ∈ (0, ∞) satisfy and deg x cm R n (x) = 2(n -1), n ≥ 2; (2) The completely monotonic degrees of -R n (x) for n ≥ 0 with respect to x ∈ (0, ∞) satisfy and deg x cm -R n (x) = 2n -1, n ≥ 2; (3) The completely monotonic degrees of (-1) m R (m) n (x) for m ≥ 2 and n ≥ 0 with respect to x ∈ (0, ∞) satisfy and deg x cm (-1) m R n (x) (m) = m + 2(n -1), n ≥ 2.
In this paper, following the method due to Koumandos and Pedersen [11], we prove that and deg x cm -R 1 (x) = 2. It is worth noting that almost at the same time that we submitted our paper, Qi and Mahmoud [23,24] independently confirmed (1.9) and (1.10). Meanwhile, motivated by (1.7) and (1.8), which can also be verified as in [15], Qi and Mahmoud [23,24] corrected and modified two conjectures stated in (3) as follows: The completely monotonic degrees of (-

A lemma
A necessary and sufficient condition for complete monotonicity was given by Bernstein's theorem (see Theorem 12b in [29, p. 161]), which states that f is completely monotonic where μ is nondecreasing and the integral converges for 0 < x < ∞. Bernstein's theorem was extended by Koumandos and Pedersen [11]. The theorem of Koumandos and Pedersen is a key and useful tool in our paper. Now, we state it as a lemma.
where the integral converges for all x > 0 and p is r The theorem of Koumandos and Pedersen contains a simple characterization of completely monotonic functions like x r f (x). It is very helpful for us to evaluate the completely monotonic degrees for certain functions.

Proofs of the conjectures
Now, we present our theorems and proofs.
Remark 2 In fact, in reference [5] and closely related references therein, it was proved that the function x[log(x)ψ(x)] is completely monotonic on (0, ∞) and that This implies that the function is also completely monotonic on (0, ∞) and that it tends to 0 as x → ∞. Hence, the function [log(x)ψ(x)] -1/(2x) is completely monotonic and its completely monotonic degree is 1 at least. This result can be applied to prove the above theorem.
So we have that p (t) is a Radon measure on [0, ∞). According to Lemma 1, we conclude that x 2 φ(x) is completely monotonic on (0, ∞), which implies that For the proof of deg x cm [-R 1 (x)] ≤ 2, see [22].
Remark 3 In reference [3], it was proved that the function is completely monotonic on (0, ∞) and that it tends to 0 as x → ∞. This implies that the completely monotonic degree of the function ψ(x) -log(x) + 1 2x + 1 12x 2 is 2 at least. This is also a part of the proof of Theorem 2 in reference [22] by Qi and Liu. Our method, with the help of the Koumandos-Pedersen theorem, is the same as that of Qi and Liu, a method of integration-by-part, essentially.
Remark 4 Almost at the same time that we submitted our paper, Qi and Mahmoud [23,24] independently proved Theorems 3 and 4.
Remark 5 In the proof of Theorems 1 and 4, Koumandos and Pedersen's theorem is applied with r = 2, in the proof of Theorem 2 with r = 3, and in the proof of Theorem 3 with r = 1. However, Koumandos and Pedersen's theorem cannot be applied with an r one unity larger in Theorems 1, 2, 3, and 4. Let us take the proof of Theorem 1 as an example. In the proof of Theorem 1, we got p (t) = e t (e t -1) 3 (t -2)e t + t + 2 , and proved p (t) > 0 for t > 0. So we obtained deg x cm [(-1) 2 R 0 (x)] ≥ 2. In fact, we can further obtain p (t) = -e t (e t -1) 4 (t -3)e 2t + 4te t + t + 3 .
According to the proof of Theorem 2, we can prove p (t) < 0 for t > 0 and get that p (t) is not increasing on [0, ∞). Thus, the function p (t) is not a Radon measure on [0, ∞). It means that the function p(t) in Theorem 1 does not satisfy the conditions of Lemma 1 with r = 3. In other words, we cannot get that x 3 R 0 (x) is completely monotonic on (0, ∞) and deg x cm [(-1) 2 R 0 (x)] ≥ 3. Similar discussion can be carried out in the proofs of the other three theorems.