1 Introduction

A function \(f: (a, b)\subset \mathbb {R}\longrightarrow \mathbb {R} \) is completely monotonic if it is infinitely differentiable and

$$\begin{aligned} (-1)^n f^{(n)}(x)\ge 0 \end{aligned}$$

for all \(x\in (a, b)\) and \(n \in \mathbb {N}\). 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 introduced by Bernstein. Bernstein himself, and later Widder independently, discovered that a necessary and sufficient condition for f to be completely monotonic on \((0, \infty )\) is that

$$\begin{aligned} f(x)=\mathcal {L}(\mu )(x)=\int e^{-xt}\,\textrm{d}\mu (t), \end{aligned}$$

where \(\mu \) is a positive measure on \([0, \infty )\) 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, it is easy to see that the absolute value of the derivatives of digamma function (the polygamma functions), \(\psi ^{(n)}=(\Gamma '/\Gamma )^{(n)}\), are completely monotonic functions on \((0, \infty )\). Indeed,

$$\begin{aligned} (-1)^{n+1}\psi ^{(n)}(x)=\mathcal {L} \left( \frac{t^n}{1-e^{-t}} \right) (x) \end{aligned}$$

for all \(x\in (0, \infty )\) and \(n\in \mathbb {N}\setminus \{0\}\). (The digamma function \(\psi \) and its absolute value are neither completely monotonic on \(x\in (0, \infty )\).) In [4], Clark and Ismail introduced the functions

$$\begin{aligned} F_m(x)=x^m \psi (x), \quad G_m(x)=-x^m \psi (x). \end{aligned}$$

These authors proved that \(F_m^{(m+1)}\) is completely monotonic on \((0, \infty )\) for \(m\in \mathbb {N} \setminus \{0\}\) [4, Theorem 1.2] and that \(G_m^{(m)}\) is completely monotonic on \((0, \infty )\) for \(m=1, 2,\ldots , 16\) [4, Theorem 1.3]. Afterwards, they wrote “We believe Theorem 1.3 \([G_m^{(m)}\)is completely monotonic on \((0, \infty )]\)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 \(m\ge m_0\), the function \(G_m^{(m)}\) is not completely monotonic. From this and the relation [4, (2.4)]

$$\begin{aligned} G^{(m)}_m(x)= \mathcal {L}\left( t^m \left( \frac{t^m}{1-e^{-t}}\right) ^{(m)}\right) (x), \end{aligned}$$

it follows that the following conjecture of Clark and Ismail [4, Conjecture 1.4] is false:

Conjecture

The inequality

$$\begin{aligned} \left( \frac{x^n}{1-e^{-x}}\right) ^{(n)}>0 \end{aligned}$$
(1)

holds for all \(x\in (0, \infty )\) and \(n\in \mathbb {N}\).

Since (1) holds for \(n=1, 2,\ldots , 16\), Clark and Ismail proved that \(G_n^{(n)}\) is (strictly) completely monotonic on \((0, \infty )\) for these values of n. Regardless of the fact that the conjecture is not true, the inequality (1) is of interest in its own right. It remains an open problem to determine the smallest positive number a (respectively, positive integer \(n_0\)) such that (1) remains positive for all \(x\in (a, \infty )\) and \(n\in \mathbb {N}\) (respectively, \(x\in (0, \infty )\) and \(n>n_0\) with \(n\in \mathbb {N}\)). This problem was placed in [2, Section 4]. In [1, Theorem 2.1],Footnote 1 Al-Musallam and Bustoz proved that (1) holds for all \(x\in (2 \log 2, \infty )\) and \(n\in \mathbb {N}\). This was also proved independently in [2, p. 112] using the same idea: an inequality proved by Szegő [8, Theorem 17a, p. 168]. Our main theorem, which improves the result in [1], reads as follows:

Theorem

The inequality (1) holds for all \(x\in (\log 2, \infty )\) and \(n\in \mathbb {N}\).

As in [1, Theorem 3.1], the next result follows from the theorem above. The details are left to the reader.

Corollary

The inequality

$$\begin{aligned} \left( \frac{x^{n+\alpha }}{1-e^{-x}}\right) ^{(n)}>0 \end{aligned}$$

holds for all \(\alpha \in (0, \infty )\), \(x\in (\log 2, \infty )\), and \(n\in \mathbb {N}\).

Let us give now an application of our main result.

Example

It is easy to obtain from [5, (1), p. 11], for \(n\in \mathbb {N}{\setminus }\{0\}\), the power series

$$\begin{aligned} \left( \frac{x^{n}}{1-e^{-x}}\right) ^{(n)}=\frac{n!}{2}+\sum _{j=2}^{\infty } \frac{(j+n-1)!}{(j-1)!} \frac{B_j}{j!} x^{j-1} \end{aligned}$$

valid in the disk \(|x|<2\pi \) which extends to the nearest singularities \(x=\pm 2 \pi i\) 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:

$$\begin{aligned} S_n&=\frac{(n+0)!}{0!\,1!} B_0+\frac{(n+1)!}{1!\,2!} B_2+\frac{(n+3)!}{3!\,4!} B_4+\frac{(n+5)!}{4! \, 5!} B_6+\cdots \\&=n!+\sum _{j=1}^{\infty } \frac{(2j+n-1)!}{(2j-1)!} \frac{B_{2j}}{(2j)!}. \end{aligned}$$

(\(Recall \,\, that B_0=1.\)) Note that

$$\begin{aligned} \left. \left( \frac{x^{n}}{1-e^{-x}}\right) ^{(n)}\right| _{x=1}=S_n-\frac{n!}{2}. \end{aligned}$$

Since \(\log 2\approx 0.693147<1<2\pi \), our main result gives

$$\begin{aligned} S_n>\frac{n!}{2}, \end{aligned}$$

\(n\in \mathbb {N}\setminus \{0\}\). It is worth pointing out that from the results obtained in [1, 2], it is not possible to conclude this because \(2 \log 2 \approx 1.38629>1\). Now it only remains to check that \(S_n\) converges, which follows from

In [2], the relation of the function given in (1) with a function of Hardy and Littlewood was extensively explored.

2 Proof of the theorem

Set

$$\begin{aligned} f_n(x)=\frac{\textrm{d}^n}{\textrm{d} x^n}\left( \frac{x^n}{1-e^{-x}}\right) . \end{aligned}$$

If \(c>0\) is arbitrary and fixed, the series

$$\begin{aligned} \frac{1}{1-e^{-x}}=\sum _{j=0}^\infty e^{-j x} \end{aligned}$$

converges uniformly on \([c, \infty )\). We then write \(f_n\) in the form

$$\begin{aligned} f_n(x)=\sum _{j=0}^\infty \displaystyle \frac{\textrm{d}^n}{\textrm{d} x^n}\left( e^{-j x} x^n\right) . \end{aligned}$$

Recall that [6, (5), p. 188] \(n! L_n(x)=e^x (\textrm{d}^n/\textrm{d}x^n)\left( e^{-x} x^n\right) \), \(L_n\) being the Laguerre polynomial of degree n, and so

$$\begin{aligned} n! e^{-jx}L_n(jx)=\frac{\textrm{d}^n}{\textrm{d} x^n}\left( e^{-j x} x^n\right) . \end{aligned}$$

Hence,

$$\begin{aligned} f_n(x)=n! \sum _{j=0}^\infty L_n(jx) e^{-jx} \end{aligned}$$

on \([c, \infty )\). There is a well-known connection between the Laguerre and Hermite polynomials due to Feldheim [6, (33), p. 195]:

$$\begin{aligned} \int _0^\infty e^{-t^2} H_n^2(t) \cos (2^{1/2} y\,t)\,\textrm{d}t=\sqrt{\pi } 2^{n-1} n! L_n(y^2), \end{aligned}$$

\(H_n\) being the Hermite polynomial of degree n. Write

$$\begin{aligned} y^2=j\,x. \end{aligned}$$

From the above expressions, we have

$$\begin{aligned} f_n(x)=\frac{1}{\displaystyle \sqrt{\pi }\, 2^{n-1} } \sum _{j=0}^\infty \int _0^\infty g_j(t)\, \textrm{d}t, \end{aligned}$$

where

$$\begin{aligned} g_j(t)=e^{-t^2} e^{-jx} H_n^2(t) \cos (\sqrt{2 j\, x} \,t) \end{aligned}$$

for all \(x\in [c, \infty )\). Recall that it it is not true that uniform convergence is sufficient to allow interchange of the sum and integral when the integral is over an infinite interval. However, we claim that the function \(g_j\) is integrable and

$$\begin{aligned} \sum _{j=0}^\infty \int _0^\infty |g_j(t)|\,\textrm{d}t<0, \end{aligned}$$

for all \(x\in [c, \infty )\). (These conditions allow the interchanging of the above sum and integral, see, for instance, [7, Corollary 17.4.7].) Indeed, since

$$\begin{aligned} |g_j(t)|<e^{-t^2} e^{-j\,x} H_n^2(t), \end{aligned}$$

we see at once that

$$\begin{aligned} \int _0^\infty |g_j(t)|\,\textrm{d}t&< e^{-j\,x}\, \int _0^\infty e^{-t^2} H_n^2(t)\,\textrm{d}t\\&< e^{-j\,x}\, \int _{-\infty }^\infty e^{-t^2} H_n^2(t)\,\textrm{d}t\le \sqrt{\pi } 2^n n!\, e^{-j\,c}<\infty , \end{aligned}$$

and the integrability of \(g_j\) is guaranteed. Moreover,

$$\begin{aligned} \sum _{j=0}^\infty \int _0^\infty |g_j(t)|\,\textrm{d}t&<\sqrt{\pi }\, 2^n\, n!\,\sum _{j=0}^\infty e^{-j\,x}\\&\le \sqrt{\pi }\, 2^n\,n!\,\displaystyle \frac{e^c}{e^{c}-1}<\infty . \end{aligned}$$

Consequently, we can interchange the sum and integral to obtain

$$\begin{aligned} \sqrt{\pi } 2^{n-1} f_n(x)=\int _0^\infty e^{-t^2} H_n^2(t) \sum _{j=0}^\infty e^{-jx} \cos (\sqrt{2 j\, x} \,t)\,\textrm{d}t. \end{aligned}$$

Finally, note that

$$\begin{aligned} \sum _{j=0}^\infty e^{-jx} \cos (\sqrt{2 j\, x} \,t)>1- \sum _{j=1}^\infty e^{-jx} =1-\frac{1}{e^{x}-1}=g(x). \end{aligned}$$

Thus \(g(x)\ge 0\) if and only if \(x>\log 2\). This completes the proof.