Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 445–460

SOME CLASSES OF COMPLETELY MONOTONIC FUNCTIONS

Horst Alzer and Christian Berg

Morsbacher Str. 10, D-51545 Waldbröl, Germany alzer@wmax03.mathematik.uni-wuerzburg.de

University of Copenhagen, Department of Mathematics Universitetsparken 5, DK-2100 Copenhagen, Denmark; berg@math.ku.dk

Abstract. We prove: (i) Let

F_{n}(x) = P_{n}(x)[e − (1 + \frac{1}{x} )^{x}] and G_{n}(x) = P_{n}(x)[(1 + \frac{1}{x} )^{x+1} − e],

where P_{n}(x) = x^{n} + ∑_{ν=0}^{n−1} c_{ν}x^{ν} is a polynomial of degree n ≥ 1 with real coefficients. F_{n} is completely monotonic if and only if n = 1 and c_{0} ≥ 11/12; and G_{n} is completely monotonic if and only if n = 1 and c_{0} ≥ 1/12.

(ii) The functions x ↦ e − (1 + 1/x)^{x} and x ↦ (1 + 1/x)^{x+1} − e are Stieltjes transforms and in particular they are completely monotonic.

(iii) Let a > 0 and b be real numbers. The function x ↦ (1 + a/x)^{x+b} − e^{a} is completely monotonic if and only if a ≤ 2b.

Part (i) extends and complements a recently published result of Sándor and Debnath, while part (iii) generalizes a theorem of Schur.

1. Introduction

A function f: (0, ∞) → R is said to be completely monotonic, if f has derivatives of all orders and satisfies

(−1)^{n}f^{(n)}(x) ≥ 0 for all x > 0 and n = 0, 1, 2, ….

This definition was introduced in 1921 by F. Hausdorff [23], who called such functions ‘total monoton’. Bernstein’s theorem, cf. [35, p. 161], states that f is completely monotonic if and only if

f(x) = ∫_{0}^{∞} e^{−xt} dμ(t),

where μ is a nonnegative measure on [0, ∞) such that the integral converges for all x > 0.

2000 Mathematics Subject Classification: Primary 26A48, 26D07; Secondary 30E20.

Completely monotonic functions have remarkable applications in different branches. For instance, they play a role in potential theory [8], probability theory [13], [20], [26], physics [16], numerical and asymptotic analysis [22], [36], and combinatorics [4]. A detailed collection of the most important properties of completely monotonic functions can be found in [35, Chapter IV], and in an abstract setting in [7].

In the recent past, various authors showed that numerous functions, which are defined in terms of gamma, polygamma, and other special functions, are completely monotonic and used this fact to derive many interesting new inequalities; cf. [2], [3], [6], [9], [10], [14], [15], [24], [25], [27], [28], [31]. An elegant majorization theorem for completely monotonic functions is given in [21]. In [28] the author provides a counterpart of the classical theorem of Bohr–Mollerup by characterizing the gamma function via the notion of complete monotonicity. Hankel’s determinant inequality for completely monotonic functions is proved in [35, p. 167], and in [19] it is shown that in connection with an interpolation problem there exists a close relation between completely monotonic functions and completely monotonic sequences.

Several related classes of functions are studied in [21], [30], [34], [35]. Historical remarks on this subject can be found in [11].

This work has been motivated by an article of J. Sándor and L. Debnath [33], who proved that the function

f(x) = (x + 1)[e − (1 + \frac{1}{x})^{x}]

is decreasing and convex on (0, ∞). It is tempting to ask for an extension of this result: is f completely monotonic? We prove that the answer is ‘yes’. Actually, we investigate a more general problem. We are looking for real polynomials P_{n}(x) = x^{n} + ∑_{ν=0}^{n−1} c_{ν}x^{ν} (n ∈ N) such that

(1.1)  x ↦ P_{n}(x)[e − (1 + \frac{1}{x})^{x}]

is completely monotonic. In view of the well-known inequalities

(1 + \frac{1}{x})^{x} < e < (1 + \frac{1}{x})^{x+1} (x > 0),

it is natural also to study the monotonicity behaviour of

(1.2)  x ↦ P_{n}(x)[(1 + \frac{1}{x})^{x+1} − e]

and its derivatives. In Section 3 we present all P_{n} such that the functions defined in (1.1) and (1.2), respectively, are completely monotonic. Moreover, we prove that the functions x ↦ e − (1 + 1/x)^{x} and x ↦ (1 + 1/x)^{x+1} − e are Stieltjes transforms, which implies that they are completely monotonic.

A theorem of I. Schur [29, pp. 30 and 186] states that the sequence n ↦ (1 + 1/n)^{n+b} (n = 1, 2, …) is decreasing if and only if b ≥ \frac{1}{2} . This result leads to the following question: for which real parameters a ≥ 0 and b is the function

x ↦ (1 + \frac{a}{x})^{x+b} − e^{a}

completely monotonic? In Section 3 we answer this question.

2. Lemmas

In order to prove our main results we need some lemmas, which we collect in this section. In our first lemma we present an integral representation, which plays a crucial role in the proof of the theorems given in Section 3.

Lemma 1. Let

(2.1)  f(x) = (x + 1)[e − (1 + \frac{1}{x})^{x}]  (x > 0).

Then we have

(2.2)  f(x) = \frac{e}{2} + \frac{1}{π} ∫_{0}^{1} \frac{s^{s}(1 − s)^{1−s} sin (πs)}{x + s} ds.

Proof. In the Addenda and Problems in [1, p. 127], it is stated that if a function f is holomorphic in the cut plane 𝒜 = C\(−∞, 0], and satisfies Im f(z) ≤ 0 for Im z > 0 and f(x) ≥ 0 for x > 0, then f is a Stieltjes transform, that is, it has the representation

f(x) = a + ∫_{0}^{∞} \frac{dμ(t)}{x + t} (x > 0),

where a ≥ 0 and μ is a nonnegative measure on [0, ∞) with ∫_{0}^{∞} dμ(t)/(1+t) < ∞. A proof is written out in [5]. The constant a is given by a = lim_{x→∞} f(x), and μ is the limit in the vague topology of measures

dμ(t) = lim_{y→0^{+}} −\frac{1}{π} Im f(−t + iy) dt,

i.e.

∫ φ dμ = lim_{y→0^{+}} −\frac{1}{π} ∫ Im f(−t + iy)φ(t) dt,

for all continuous functions φ on R with compact support. Let

f(z) = (1 + z)[e − exp(zLog (1 + 1/z))] (z ∈ 𝒜) ,

where Log denotes the principal branch of the logarithm. We have

f(1/z) = e(1 + 1/z)[1 − exp(\frac{Log (1 + z)}{z} − 1)],

so that the series representation

\frac{Log (1 + z)}{z} − 1 = −\frac{z}{2} + \frac{z^{2}}{3} − + ⋯ (|z| < 1)

implies

(2.3)  lim_{|z|→∞} f(z) = \frac{e}{2}  (z ∈ 𝒜) .

To prove that the harmonic function Im f satisfies Im f(z) ≤ 0 for Im z > 0, we use the maximum principle for subharmonic functions, cf. [18, p. 20], and show that the lim sup of Im f at all boundary points including infinity is less than or equal to 0. From (2.3) we conclude that this is true at infinity.

Let t ∈ R and z ∈ C with Im z > 0. If z tends to t, then

f(z) → \{ \begin{matrix}f(t), & if t > 0, \\ e − 1, & if t = 0, \\ (1 + t)[e − exp(t log \frac{1 + t}{|t|} − iπt)], & if −1 < t < 0, \\ 1, & if t = −1, \\ (1 + t)[e − exp(t log \frac{1 + t}{t})], & if t < −1. \end{matrix}

In particular, if y tends to 0^{+} , then we obtain for t ∈ R:

−\frac{1}{π} Im f(−t + iy) → \{ \begin{matrix}0, & if t ≤ 0 or t ≥ 1, \\ \frac{1}{π} (1 − t) exp(−t log (\frac{1 − t}{t})) sin(πt), & if 0 < t < 1. \end{matrix}

The limit above is uniform for t in compact subsets of the real axis, and therefore the (continuous) density on the right is the vague limit of the densities on the left. The proof of Lemma 1 is complete.

Remarks. (1) To obtain formula (2.2) we can alternatively use the residue theorem applied to the function z ↦ f(z)/(z − z_{0}) (z_{0} ∈ 𝒜), where the path of integration is a simple closed contour given by two circles |z| = ε and |z| = R (where ε < |z_{0}| < R) modified by inserting two horizontal segments with x ≤ 0, y = ±ε from the small circle to the big one and removing the left part of the small circle and a small portion of the big circle close to −R. If we let ε tend to 0 and then R tend to ∞, then we get (2.2).

(2) The integral representation (2.2) implies that f is completely monotonic. This extends the monotonicity result of Sándor and Debnath mentioned in Section 1.

Lemma 2. Let

(2.4)  g(s) = \frac{1}{π} s^{s}(1 − s)^{1−s} sin (πs)  (0 ≤ s ≤ 1).

Then we have

(2.5)  \begin{aligned}∫_{0}^{1} g(s) ds & = \frac{e}{24}, \\ ∫_{0}^{1} sg(s) ds & = \frac{e}{48},  and \\ ∫_{0}^{1} \frac{1}{s}g(s) ds & = ∫_{0}^{1} \frac{1}{1 − s}g(s) ds = \frac{e}{2} − 1. \end{aligned}

Proof. Let f be the function given in (2.1). We set x = 1/y and obtain

(2.6)  x[f(x) − \frac{e}{2}] = \frac{ey/2 + e − (1 + y)^{1+1/y}}{y^{2}} .

The rule of l’Hospital yields

(2.7)  lim_{y→0} \frac{ey/2 + e − (1 + y)^{1+1/y}}{y^{2}} = \frac{e}{24}.

From (2.2), (2.6), and (2.7) we get

∫_{0}^{1} g(s) ds = lim_{x→∞} ∫_{0}^{1} \frac{x}{x + s}g(s) ds = lim_{x→∞} x[f(x) − \frac{e}{2}] = \frac{e}{24}.

Using g(s) = g(1 − s) we obtain

\begin{aligned}∫_{0}^{1} sg(s) ds & = ∫_{0}^{1/2} sg(s) ds + ∫_{1/2}^{1} sg(s) ds = ∫_{0}^{1/2} sg(s) ds + ∫_{0}^{1/2}(1 − s)g(s) ds \\ & = ∫_{0}^{1/2} g(s) ds = \frac{1}{2} ∫_{0}^{1} g(s) ds = \frac{e}{48}. \end{aligned}

450 Horst Alzer and Christian Berg Applying Lemma 1 we get

∫_{0}^{1} \frac{1}{1 − s}g(s) ds = ∫_{0}^{1} \frac{1}{1 − s}g(1 − s) ds = ∫_{0}^{1} \frac{1}{s}g(s) ds = lim_{x→0^{+}} f(x) − \frac{e}{2} = \frac{e}{2} − 1.

Lemma 3. Let

(2.8)  Δ_{1}(x) = \frac{e − (1 + 1/x)^{x}}{(1 + 1/x)^{x}[log (1 + 1/x) − 1/(x + 1)]} − x

and

(2.9)  Δ_{2}(x) = \frac{e − (1 + 1/x)^{x+1}}{(1 + 1/x)^{x+1}[log (1 + 1/x) − 1/x]} − x.

Then

(2.10) lim_{x→∞} Δ_{1}(x) = \frac{11}{12} and lim_{x→∞} Δ_{2}(x) = \frac{1}{12}.

Proof. We have

Δ_{1}(x) = \frac{b_{1}(x) + c_{1}(x)}{a_{1}(x)} and Δ_{2}(x) = \frac{b_{2}(x) + c_{2}(x)}{a_{2}(x)},

where

\begin{aligned}a_{1}(x) & = x^{2}[log (1 + \frac{1}{x}) − \frac{1}{x + 1}], \\ b_{1}(x) & = x[\frac{1}{2} − a_{1}(x)], \\ c_{1}(x) & = x^{2}[\frac{e}{(1 + 1/x)^{x}} − 1] − \frac{x}{2}, \end{aligned}

and

\begin{aligned}a_{2}(x) & = x^{2}[log (1 + \frac{1}{x}) − \frac{1}{x}], \\ b_{2}(x) & = −x[\frac{1}{2} + a_{2}(x)], \\ c_{2}(x) & = x^{2}[\frac{e}{(1 + 1/x)^{x+1}} − 1] + \frac{x}{2}. \end{aligned}

The limit relations

lim_{x→∞} a_{1}(x) = \frac{1}{2}, lim_{x→∞} b_{1}(x) = \frac{2}{3}, lim_{x→∞} c_{1}(x) = −\frac{5}{24},

and

lim_{x→∞} a_{2}(x) = −\frac{1}{2}, lim_{x→∞} b_{2}(x) = −\frac{1}{3}, lim_{x→∞} c_{2}(x) = \frac{7}{24}

lead to (2.10).

Some classes of completely monotonic functions 451

A proof for the following lemma can be found in [26].

Lemma 4. Let h: [0, ∞) → (0, 1] be a continuous function. If h is completely monotonic on (0, ∞), then we get for all x, y ≥ 0:

h(x)h(y) ≤ h(x + y).

The next lemma is given in [12, p. 83].

Lemma 5. Let f and g be functions such that f(g(x)) is defined for x > 0. If f and g′ are completely monotonic, then x ↦ f(g(x)) is also completely monotonic.

Our final lemma is well known and is stated only for easy reference.

Lemma 6. The sum and the product of completely monotonic functions are also completely monotonic.

3. Main results We are now in a position to determine all real polynomials P_{n} such that the

functions given in (1.1) and (1.2), respectively, are completely monotonic.

Theorem 1. Let P_{n}(x) = x^{n} + ∑_{ν=0}^{n−1} c_{ν}x^{ν} be a polynomial of degree n ≥ 1 with real coefficients. The function

x ↦ P_{n}(x)[e − (1 + \frac{1}{x})^{x}]

is completely monotonic if and only if n = 1 and c_{0} ≥ 11/12.

Proof. Let

F_{n}(x) = P_{n}(x)[e − (1 + \frac{1}{x})^{x}] (x > 0).

There exists a number x_{0} > 0 such that P_{n} is positive on (x_{0}, ∞). If F_{n} is completely monotonic, then we have

F′_{n}(x) = P′_{n}(x)[e − (1 + \frac{1}{x})^{x}] − P_{n}(x)(1 + \frac{1}{x})^{x}[log(1 + \frac{1}{x}) − \frac{1}{x + 1}] ≤ 0,

and for x > x_{0} this is equivalent to

(3.1)  x\frac{P′_{n}(x)}{P_{n}(x)} ≤ (1 + \frac{Δ_{1}(x)}{x})^{−1},

where Δ_{1} is defined in (2.8). Applying Lemma 3 we get

n = lim_{x→∞} x\frac{P′_{n}(x)}{P_{n}(x)} ≤ lim_{x→∞}(1 + \frac{Δ_{1}(x)}{x})^{−1} = 1.

This implies n = 1. Hence, P_{n}(x) = x + c_{0} , so that (3.1) yields

(3.2)  Δ_{1}(x) ≤ c_{0}  for all sufficiently large x.

From (2.10) and (3.2) we conclude that c_{0} ≥ 11/12.

Let

(3.3)  f_{c}(x) = (x + c)[e − (1 + \frac{1}{x})^{x}].

We have

f_{c}(x) = f_{11/12}(x) + \frac{c − 11/12}{x + 11/12}f_{11/12}(x).

If c ≥ 11/12, then x ↦ (c − 11/12)/(x + 11/12) is completely monotonic. Applying Lemma 6 we get: if f_{11/12} is completely monotonic, then the same is true for f_{c} with c ≥ 11/12.

We prove now that f_{11/12} is completely monotonic. Let g be defined in (2.4). Applying

(3.4) \frac{1}{(x + s)^{n}} = \frac{1}{(n − 1)!} ∫_{0}^{∞} e^{−xt}e^{−st}t^{n−1} dt  (x > 0, s ≥ 0, n = 1, 2, …)

with n = 1 we obtain

(3.5)  ∫_{0}^{1} \frac{g(s)}{x + s} ds = ∫_{0}^{∞} e^{−xt}h(t) dt,

where

(3.6)  h(t) = ∫_{0}^{1} e^{−ts}g(s) ds.

From (2.2) and (3.5) we get

(3.7)  f_{1}(x) = \frac{e}{2} + ∫_{0}^{∞} e^{−xt}h(t) dt.

Using (3.7), (3.4) with s = n = 1, and the convolution theorem for Laplace transforms we obtain

f_{11/12}(x) = \frac{x + 11/12}{x + 1}f_{1}(x) = \frac{e}{2} \frac{x + 11/12}{x + 1} + \frac{x + 11/12}{x + 1} ∫_{0}^{∞} e^{−xt}h(t) dt

= \frac{e}{2}[1 − \frac{1}{12} ∫_{0}^{∞} e^{−xt}e^{−t} dt] + [1 − \frac{1}{12} ∫_{0}^{∞} e^{−xt}e^{−t} dt] ∫_{0}^{∞} e^{−xt}h(t) dt

= \frac{e}{2} − \frac{e}{24} ∫_{0}^{∞} e^{−xt}e^{−t} dt + ∫_{0}^{∞} e^{−xt}h(t) dt (3.8)

− \frac{1}{12} ∫_{0}^{∞} e^{−xt}e^{−t} dt ∫_{0}^{∞} e^{−xt}h(t) dt

= \frac{e}{2} + ∫_{0}^{∞} e^{−xt}u(t) dt,

where

u(t) = h(t) − e^{−t}[\frac{e}{24} + \frac{1}{12} ∫_{0}^{t} e^{s}h(s) ds].

In order to prove that u is positive on (0, ∞) we set v(t) = e^{t}h(t) and w(t) = e^{t}u(t). Then we get

w(t) = v(t) − \frac{e}{24} − \frac{1}{12} ∫_{0}^{t} v(s) ds.

Let t > 0. Differentiation gives

w′(t) = v′(t) − \frac{1}{12}v(t) = e^{t}[\frac{11}{12}h(t) + h′(t)]

and

(3.9) \begin{aligned}w′(t)e^{−t} & = ∫_{0}^{11/12} e^{−ts}[11/12 − s]g(s) ds + ∫_{11/12}^{1} e^{−ts}[11/12 − s]g(s) ds \\ & ≥ e^{−(11/12)t} ∫_{0}^{11/12}[11/12 − s]g(s) ds \\ & + e^{−(11/12)t} ∫_{11/12}^{1}[11/12 − s]g(s) ds \\ & = e^{−(11/12)t} ∫_{0}^{1}[11/12 − s]g(s) ds. \end{aligned}

Applying (2.5) we obtain

∫_{0}^{1}[11/12 − s]g(s) ds = \frac{5e}{288},

so that (3.9) reveals that w′ is positive on (0, ∞). Hence, we get for t > 0:

w(t) > w(0) = u(0) = h(0) − \frac{e}{24} = ∫_{0}^{1} g(s) ds − \frac{e}{24} = 0.

This implies that w and u are positive on (0, ∞). From the integral representation (3.8) we conclude that f_{11/12} is completely monotonic.

Sándor has proved the weaker result that the function f_{c} in (3.3) is convex for c ≥ 11/12, cf. [32].

Our second result is a striking companion of Theorem 1.

Theorem 2. Let P_{n}(x) = x^{n} + ∑_{ν=0}^{n−1} c_{ν}x^{ν} be a polynomial of degree n ≥ 1 with real coefficients. The function

x ↦ P_{n}(x)[(1 + \frac{1}{x})^{x+1} − e]

is completely monotonic if and only if n = 1 and c_{0} ≥ 1/12.

Proof. Let

G_{n}(x) = P_{n}(x)[(1 + \frac{1}{x})^{x+1} − e] (x > 0).

There exists a number x_{1} > 0 such that P_{n} is positive on (x_{1}, ∞). If G_{n} is completely monotonic, then we have

(3.10) G′_{n}(x) = P′_{n}(x)[(1+\frac{1}{x})^{x+1}−e]+P_{n}(x)(1+\frac{1}{x})^{x+1}[log(1+\frac{1}{x})−\frac{1}{x}] ≤ 0.

For x > x_{1} the inequality (3.10) can be written as

(3.11)  x\frac{P′_{n}(x)}{P_{n}(x)} ≤ (1 + \frac{Δ_{2}(x)}{x})^{−1},

where Δ_{2} is given in (2.9). We let x tend to ∞ and conclude from (2.10) that n = 1. Hence, P_{n}(x) = x + c_{0} and (3.11) gives for all sufficiently large x:

Δ_{2}(x) ≤ c_{0}.

From (2.10) we obtain c_{0} ≥ 1/12.

It remains to show: if c ≥ 1/12, then

(3.12)  g_{c}(x) = (x + c)[(1 + \frac{1}{x})^{x+1} − e]

is completely monotonic. We have

g_{c}(x) = g_{1/12}(x) + \frac{c − 1/12}{x + 1/12}g_{1/12}(x).

Since x ↦ (c − 1/12)/(x + 1/12) (with c ≥ 1/12) is completely monotonic, we conclude from Lemma 6 that it suffices to prove that g_{1/12} is completely monotonic. Let f_{1} and h be the functions defined in (3.3) and (3.6), respectively. Using the identity

(3.13)  g_{1}(x) = (1 + \frac{1}{x})[e − f_{1}(x)],

(3.7), (3.4) with s = 0, n = 1, and the convolution theorem for Laplace transforms, we get

g_{1/12}(x) = \frac{x + 1/12}{x + 1}g_{1}(x) = (1 + \frac{1}{12x})[e − f_{1}(x)]

= (1 + \frac{1}{12x})[\frac{e}{2} − ∫_{0}^{∞} e^{−xt}h(t) dt]

= \frac{e}{2} + ∫_{0}^{∞} e^{−xt}σ(t) dt,

where

σ(t) = −h(t) + \frac{e}{24} − \frac{1}{12} ∫_{0}^{t} h(s) ds.

We have to show that σ is nonnegative on (0, ∞). Applying (2.5), (3.6), and

\begin{aligned}∫_{0}^{t} h(s) ds & = ∫_{0}^{t} ∫_{0}^{1} e^{−su}g(u) du ds = ∫_{0}^{1} g(u) ∫_{0}^{t} e^{−su} ds du \\ & = ∫_{0}^{1} \frac{1 − e^{−tu}}{u} g(u) du, \end{aligned}

we obtain the integral representation

σ(t) = ∫_{0}^{1}(1 − e^{−tu})[1 − \frac{1}{12u}]g(u) du.

Using (2.5) we get for t > 0:

σ(t) ≥ (1 − e^{−t/12}) ∫_{0}^{1/12}[1 − \frac{1}{12u}]g(u) du

+ (1 − e^{−t/12}) ∫_{1/12}^{1}[1 − \frac{1}{12u}]g(u) du

= (1 − e^{−t/12}) ∫_{0}^{1}[1 − \frac{1}{12u}]g(u) du = \frac{1}{12}(1 − e^{−t/12}) > 0.

This completes the proof of Theorem 2.

Remark. In analogy with Lemma 1 we obtain that g_{1} is a Stieltjes transform. Indeed, from (3.13), (2.2), and (2.5) we get

(x + 1)[(1 + \frac{1}{x})^{x+1} − e] = \frac{e}{2} + \frac{1}{x} + ∫_{0}^{1} \frac{(1 − s)g(s)}{s} \frac{ds}{x + s},

where g is given in (2.4).

Theorem 3. The functions p(x) = e−(1+1/x)^{x} and q(x) = (1+1/x)^{x+1} −e are Stieltjes transforms with the representations

p(x) = \frac{1}{x + 1} + ∫_{0}^{1} \frac{g(s)}{1 − s} \frac{ds}{x + s}, q(x) = \frac{1}{x} + ∫_{0}^{1} \frac{g(s)}{s} \frac{ds}{x + s},

and in particular they are completely monotonic. Here, g denotes the function defined in (2.4).

Proof. Using (2.2) and (2.5) we obtain

p(x) − \frac{1}{x + 1} = \frac{1}{x + 1}[∫_{0}^{1} \frac{g(s) ds}{1 − s} + ∫_{0}^{1} \frac{g(s) ds}{x + s}] = ∫_{0}^{1} \frac{g(s) ds}{(1 − s)(x + s)}.

Letting x tend to 0 we get

∫_{0}^{1} \frac{g(s) ds}{s(1 − s)} = e − 2,

which implies

\begin{aligned}q(x) − \frac{1}{x} & = \frac{e − 1}{x} − (1 + \frac{1}{x})p(x) = \frac{e − 2}{x} − (1 + \frac{1}{x}) ∫_{0}^{1} \frac{g(s) ds}{(1 − s)(x + s)} \\ & = ∫_{0}^{1} \frac{g(s) ds}{s(x + s)}. \end{aligned}

Remarks. (1) Using (3.3) and (3.12) we obtain the identities

p(x) = \frac{1}{x + 11/12}f_{11/12}(x)  and  q(x) = \frac{1}{x + 1/12}g_{1/12}(x).

Lemma 6 and Theorems 1 and 2 imply that p and q are completely monotonic.

(2) Applying Lemma 4 and Theorem 3 we get the following functional inequalties involving p:

0 < \frac{p(x)p(y)}{p(x + y)} ≤ e − 1 (x, y > 0).

Moreover, since lim_{x→0} p(x) = e − 1 and lim_{x→∞}(p(x))^{2}/p(2x) = 0, we conclude that both bounds are best possible.

(3) A holomorphic function f: C \ R → C is said to be a Pick function, if f(\overline{z}) = \overline{f(z)} and Im f(z) ≥ 0 for Im z > 0; cf. [17]. It follows immediately from the first part of Theorem 3 that p̃(z) = (1+1/z)^{z} is a Pick function. The function z ↦ (1+z)^{1/z} is holomorphic on C\(−∞, 0] and has a negative imaginary part in the upper half-plane, because p̃ is a Pick function. Like in the proof of Lemma 1 it can be seen by [1, p. 127] that it is a Stieltjes transform with the representation

(1 + x)^{1/x} = 1 + \frac{1}{x + 1} + \frac{1}{π} ∫_{1}^{∞} \frac{sin (π/t)}{(t − 1)^{1/t}} \frac{dt}{x + t} (x > 0).

This formula follows also from the representation of p in Theorem 3 by a change of variable.

(4) A function f: (0, ∞) → [0, ∞) is called a Bernstein function, if f has derivatives of all orders and f′ is completely monotonic; cf. [8]. Since p̃′ = −p′ we conclude that p̃ is a Bernstein function. Theorem 3 immediately implies that x ↦ (1 + 1/x)^{x+1} is a Stieltjes transform. Using that if f is a Stieltjes transform, then so is x ↦ 1/f(1/x), and if f is a nonzero Stieltjes transform, then 1/f is a Bernstein function, cf. [5], we obtain that x ↦ (1+x)^{1+1/x} is a Bernstein function.

Our next theorem extends in part the results on complete monotonicity given in Theorem 3 and provides a generalization of Schur’s monotonicity theorem mentioned in Section 1.

Theorem 4. Let a > 0 and b be real numbers. The function

x ↦ (1 + \frac{a}{x})^{x+b} − e^{a}

is completely monotonic if and only if a ≤ 2b.

Proof. Let

q_{a,b}(x) = (1 + \frac{a}{x})^{x+b} − e^{a}.

If q_{a,b} is completely monotonic, then we get

(3.14)  q′_{a,b}(x) = (1 + \frac{a}{x})^{x+b}[log(1 + \frac{a}{x}) − \frac{a(x + b)}{x(x + a)}] ≤ 0.

We set x = at. Then (3.14) yields

t[(t + 1) log (1 + 1/t) − 1] ≤ b/a (t > 0).

Letting t tend to ∞, we get 1/2 ≤ b/a.

Next, we define for x > 0 and b ≥ a/2 > 0:

y_{a,b}(x) = −(x + b) log (1 + a/x).

Differentiation gives

y′_{a,b}(x) = \frac{a(x + b)}{x(x + a)} − log (1 + \frac{a}{x}) and −y′′_{a,b}(x) = \frac{a}{(x + a)^{2}} [\frac{2b − a}{x} + \frac{ab}{x^{2}} ].

Applying (3.4) and Lemma 6 we obtain that −y′′_{a,b} is completely monotonic. Since lim_{x→∞} y′_{a,b}(x) = 0, it follows that y′_{a,b} is also completely monotonic. Using Lemma 5 (with g = y_{a,b} and f = 1/ exp) and the limit relation lim_{x→∞} q_{a,b}(x) = 0, we conclude that exp(−y_{a,b}) = q_{a,b} + e^{a} and q_{a,b} are completely monotonic.

Remark. It remains an open problem to determine all real parameters α > 0 and β such that

p_{α,β}(x) = e^{α} − (1 + \frac{α}{x})^{x+β}

is completely monotonic. Theorem 3 implies that p_{1,0} is completely monotonic. This result can be extended easily: if 0 < α ≤ 1, then p_{α,0} is completely monotonic. Let 0 < α ≤ 1 and z_{α}(x) = p_{α,0}(αx). Then we have

−z′_{α}(x) = α[e − p_{1,0}(x)]^{α−1}(−p′_{1,0}(x)).

Using Theorem 3, Lemma 5 (with g(x) = e − p_{1,0}(x) and f(x) = x^{α−1} ), and Lemma 6 we conclude that −z′_{α} is completely monotonic. Since lim_{x→∞} z_{α}(x) = 0, we obtain that z_{α} and p_{α,0} are also completely monotonic.

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Received 28 November 2001