On the positivity of a certain function related with the Digamma function

It is proved that $$\left(\frac{x^n}{1-e^{-x}}\right)^{(n)}>0$$ for all $x\in (\log 2, \infty)$ and $n\in \mathbb{N}$, which improves the result of [Al-Musallam and Bustoz in Ramanujan J. 11 (2006) 399-402].


Introduction
(−1) n f (n) (x) ≥ 0 for all x ∈ (a, b) and n ∈ N. A function f (−x) is called absolutely monotonic on (−b, −a) if and only if f (x) is completely monotonic on (a, b).Absolutely monotonic functions were pioneeringly introduced by Bernstein.Bernstein himself, and later Widder independently, discovered that a necessary and sufficient condition for f to be completely monotonic on (0, ∞) is that where µ is a positive measure on [0, ∞) and the integral converges for all positive x. (These and other classical results on absolutely/completely monotonic functions can be found in [8, Chapter IV] and [3].)As it was remarked in [2], by Bernstein's theorem, its is easy to see that the absolute value of the digamma function, ψ = Γ ′ /Γ, and the absolute value of its derivatives (the polygamma functions) are completely monotonic functions on (0, ∞).Indeed, for all x ∈ (0, ∞) and n ∈ N.
In [4] Clark and Ismail introduced the functions They proved that F m is completely monotonic on (0, ∞)] is true for all m [...]".However, Alzer, Berg, and Koumandos [2, Theorem 1.1] proved that there exists an integer m 0 such that for all m ≥ m 0 the function G (m) m is not completely monotonic.From this and the relation [4, (2.4)] it follows that the following conjecture of Clark and Ismail [4, Conjecture 1.4] is false: Conjecture.
By showing that (2) holds for n = 1, 2, . . ., 16 (and using (1)), Clark and Ismail proved that G (n) n is (strictly) completely monotonic on (0, ∞) for these values of n.Regardless of the fact that the conjecture is not true, the inequality ( 2 As in [1, Theorem 3.1], now using the above theorem, the next result follows.(The details are left to the reader.)

Corollary.
x Example.It is easy to obtain from [5, (1), p. 11], for n ∈ N \ {0}, the power series valid in the disk |x| < 2π which extends to the nearest singularities x = ±2πi of x/(e x − 1).(The coefficients B j are the Bernoulli numbers.The odd Bernoulli numbers are all zero after the first, but it is a highly complex task to determine the even Bernoulli numbers.)Let us imagine that we are questioned about the sign of the following sum: Since log 2 ≈ 0.693147 < 1 < 2π, our main results gives It is worth pointing out that from the results obtained in [1,2], it is not possible to conclude this because 2 log 2 ≈ 1.38629 > 1.Now it only remains to check that S n converges, which follows from In [2] the relation of the function given in (2) with a function of Hardy and Littlewood was extensively explored.

Proof of the theorem
Set If c > 0 is arbitrary and fixed, the series converges uniformly on [c, ∞).We then write f n in the form Recall that [6, (5), p. 188] n!L n (x) = e x (d n /dx n ) (e −x x n ), L n being the Laguerre polynomial of degree n, and so

on [c, ∞).
There is a well-known connection between the Laguerre and Hermite polynomials due to Feldheim [6, (33), p. 195 H n being the Hermite polynomial of degree n.Write From the above expressions, we have where g j (t) = e −t 2 e −jx H 2 n (t) cos( 2j x t) for all x ∈ [c, ∞).(Recall that it is not true that uniform convergence is sufficient to allow the interchanging of the sum and integral when the integral is over an infinity interval.)However, the function g j is integrable and  Consequently, we can interchange the sum and integral to obtain Thus g(x) ≥ 0 if and only if x > log 2. This completes the proof.

e
−t 2 H 2 n (t) dt ≤ √ π2 n n! e −j c < ∞,and the integrability of g j is guaranteed.Moreover,