Higher Order Thorin–Bernstein Functions

Abstract

We investigate subclasses of generalized Bernstein functions related to complete Bernstein and Thorin–Bernstein functions. Representations in terms of incomplete beta and gamma as well as hypergeometric functions are presented. Several special cases and examples are discussed.

1 Introduction

In this paper subclasses of the so-called generalized Bernstein functions of positive order are investigated. A non-negative function f defined on $(0,\infty )$ having derivatives of all orders is a generalized Bernstein function of order $\lambda >0$ if $f'(x)x^{1-\lambda }$ is a completely monotonic function. This is equivalent to f admitting an integral representation of the form

$$\begin{aligned} f(x)=ax^{\lambda }+b+\int _0^{\infty }\gamma (\lambda ,xt)\frac{d\mu (t)}{t^{\lambda }}, \end{aligned}$$

(1)

where a and b are non-negative numbers, and $\mu$ is a positive measure on $(0,\infty )$ making the integral converge for all $x>0$. Here, $\gamma$ denotes the incomplete gamma function

$$\begin{aligned} \gamma (\lambda ,t)=\int _0^te^{-u}u^{\lambda -1}\, du. \end{aligned}$$

The convergence of the integral in (1) is equivalent to

$$\begin{aligned} \int _0^{\infty }\frac{d\mu (t)}{(t+1)^{\lambda }}<\infty . \end{aligned}$$

The class of generalized Bernstein functions of order $\lambda$ is denoted by $\mathcal B_{\lambda }$ and was studied in [12].

A function g is called a generalized Stieltjes function of order $\lambda$ if there exist a positive measure $\nu$ on $[0,\infty )$ and a non-negative constant c such that

$$\begin{aligned} g(x)=\int _0^{\infty }\frac{d\nu (t)}{(x+t)^{\lambda }}+c, \end{aligned}$$

for $x>0$. The class of these functions, denoted by $\mathcal S_{\lambda }$, can also be characterized in terms of the Laplace transform ${\mathcal {L}}$: $g\in {\mathcal {S}}_{\lambda }$ if and only if

$$\begin{aligned} g(x)=\frac{1}{\Gamma (\lambda )}{\mathcal {L}}(t^{\lambda -1}\mathcal L(\nu )(t))(x)+c =\frac{1}{\Gamma (\lambda )}\int _0^{\infty }e^{-xt}t^{\lambda -1}\mathcal L(\nu )(t)\, dt+c. \end{aligned}$$

This relation will be used throughout the paper. Several properties of gene- ralized Stieltjes functions are given in [13] and [14]. (For $\lambda =1$ the class is the class of Stieltjes functions, denoted by ${\mathcal {S}}$.)

The classes $\mathcal B_{\lambda }$ and ${\mathcal {S}}_{\lambda }$ are closely related: in fact, the map $\Phi :\mathcal B_{\lambda }\rightarrow {\mathcal {S}}_{\lambda }$ defined as

$$\begin{aligned}\Phi (f)(x)=x{\mathcal {L}}(f)(x)\end{aligned}$$

is a bijection. See [12, Theorem 3.1].

Let us also recall the definition of a completely monotonic function of order $\alpha$. A function $f:(0,\infty )\rightarrow [0,\infty )$ is completely monotonic of order $\alpha$ if $x^{\alpha }f(x)$ is completely monotonic. This class was introduced and characterized in [11]. Let us remark, that $\alpha$ in this definition can be any real number. The class of completely monotonic functions of order $\alpha$ is denoted by ${\mathcal {C}}_{\alpha }$.

The functions studied in this paper correspond to putting some extra conditions on the measure $\mu$ in the representation (1).

Definition 1.1

We say that a function $f:(0,\infty )\rightarrow {\mathbb {R}}$ is a higher order Thorin–Bernstein function, if there exist $\lambda >0$ and $\alpha <\lambda +1$ such that

$$\begin{aligned} f(x)=a x^{\lambda }+b+\int _0^{\infty }\gamma (\lambda ,xt)\varphi (t)\, dt, \end{aligned}$$

(2)

where a and b are non-negative numbers, and $\varphi$ is a completely monotonic function of order $\alpha$.

For given numbers $\lambda$ and $\alpha$ the class of functions satisfying (2) is denoted by $\mathcal T_{\lambda ,\alpha }$, and they are called $(\lambda ,\alpha )$-Thorin–Bernstein functions.

We notice that the integral in (2) converges exactly when $t^\lambda \varphi (t)/(t+1)^\lambda$ is integrable on $(0,\infty )$.

For given $\lambda >0$ it follows that $\mathcal T_{\lambda ,\alpha _2}\subseteq {\mathcal {T}}_{\lambda ,\alpha _1}$ when $\alpha _1\le \alpha _2<\lambda +1$. When $\alpha =0$ the class ${\mathcal {T}}_{\lambda ,\alpha }$ reduces to the class of generalized complete Bernstein functions of order $\lambda$, and when $\alpha =1$ to the class of generalized Thorin–Bernstein functions of order $\lambda$. These classes were studied in [12].

The higher order Thorin–Bernstein functions are closely related to the incomplete Beta function B, defined for $x\in [0,1)$, $a>0$, and $b\in {\mathbb {R}}$ as

$$\begin{aligned} B(a,b;x)=\int _0^xt^{a-1}(1-t)^{b-1}\, dt. \end{aligned}$$

In fact, letting $\varphi (t)=t^{-\alpha }e^{-ct}=t^{-\alpha }\mathcal L(\epsilon _c)(t)$, for $c>0$, where $\epsilon _c$ denotes the point mass at c, it follows that

$$\begin{aligned} \int _0^{\infty }\gamma (\lambda ,xt)\varphi (t)\, dt=\Gamma (\lambda +1-\alpha )c^{\alpha -1}B\left( \lambda ,1-\alpha ;\frac{x}{x+c}\right) , \end{aligned}$$

which is a special case of Proposition 2.10. See also Example 2.12. Moreover, the incomplete Beta function is the main building block of the classes of higher order Thorin–Bernstein functions. See Corollary 3.2.

Bondesson introduced in [4, p. 150] the classes $\mathcal T_{1,\beta }$ and they also appear in [16, Chapter 8]. The higher order Thorin–Bernstein functions defined above play similar roles but in the context of generalized Bernstein functions of positive order $\lambda$.

In this paper we shall relate these functions to, among other things, gene-ralized Stieltjes functions and find representations in terms of hypergeometric functions. Proposition 2.10 gives an equivalent integral representation of the functions $f\in {\mathcal {T}}_{\lambda ,\alpha }$ in terms of the measure representing the completely monotonic function $t^{\alpha }\varphi (t)$ as a Laplace transform. This leads to a characterization in Theorem 2.14 of the class $\mathcal T_{\lambda ,\alpha }$ in terms of generalized Stieltjes functions of positive order: $f\in \mathcal T_{\lambda ,\alpha }\ \Leftrightarrow \ x^{1-\lambda }f'(x)\in {\mathcal {S}}_{\lambda +1-\alpha }$.

Based on the above result we prove that any generalized Bernstein function of order $\lambda >0$ is the pointwise limit of a sequence of functions from $\cup _{\alpha <\lambda +1}\mathcal T_{\lambda ,\alpha }$. This result generalizes the corresponding result for Bernstein functions, obtained by Bondesson.

The motivation for studying these classes comes from specific examples including the incomplete Beta function. The incomplete Beta function plays, as the incomplete Gamma functions, an extensive role in probability and mathematical statistics. It also appears in e.g. Monte Carlo sampling in statistical mechanics (see [9, 10]). For additional applications we refer to [5] and the references given therein.

In this paper all measures are supposed to be positive Radon measures.

2 Fundamental Results

As mentioned above the classes ${\mathcal {T}}_{\lambda ,\alpha }$ for fixed $\lambda >0$ are nested. Their intersection $\cap _{\alpha <\lambda +1}{\mathcal {T}}_{\lambda ,\alpha }$ turns out to be the functions $\{ax^{\lambda }+b\, |\, a,b\ge 0\}$. This can be seen from Theorem 2.14 as follows: if f belongs to the intersection then $x^{1-\lambda }f'(x)$ belongs to $\cap _{\beta >0}{\mathcal {S}}_{\beta }$. This intersection, however, is equal to the constant functions (see [17]).

The class ${\mathcal {T}}_{\lambda ,\alpha }$ is closed under pointwise limits:

Proposition 2.1

If $\{f_n\}$ is a sequence from $\mathcal T_{\lambda ,\alpha }$ that converges pointwise to $f:(0,\infty )\rightarrow {\mathbb {R}}$ then $f\in \mathcal T_{\lambda ,\alpha }$.

Proof

Since $\mathcal T_{\lambda ,\alpha }$ is a subclass of $\mathcal B_{\lambda }$ and $\mathcal B_{\lambda }$ is closed under pointwise limits then $f\in \mathcal B_{\lambda }$ and furthermore, $f_n'\rightarrow f'$. See [12, Proposition 2.4]. This entails $x^{1-\lambda }f_n'(x)\rightarrow x^{1-\lambda }f'(x)$. From Theorem 2.14 we also have $x^{1-\lambda }f_n'(x)\in \mathcal S_{\lambda +1-\alpha }$. Since the Stieltjes classes are also closed under pointwise limits (see [8, Theorem 10]), $x^{1-\lambda }f'(x)\in {\mathcal {S}}_{\lambda +1-\alpha }$. Using again Theorem 2.14 we see that $f\in \mathcal T_{\lambda ,\alpha }$. $\square$

Since $\mathcal B_{\lambda }$ is closed under pointwise limits the limit of any pointwise convergent sequence of functions from $\cup _{\alpha <\lambda +1}{\mathcal {T}}_{\lambda ,\alpha }$ belongs to $\mathcal B_{\lambda }$. Conversely, we have the following proposition.

Proposition 2.2

Let $f\in \mathcal B_{\lambda }$. Then there is a sequence from $\cup _{\alpha <\lambda +1}{\mathcal {T}}_{\lambda ,\alpha }$ that converges pointwise to f.

The proof of Proposition 2.2 relies on the fact that any completely monotonic function is the pointwise limit of a sequence of generalized Stieltjes functions of positive order. See [8]. For the reader’s convenience and in order to have a basis for proving Proposition 2.2 we describe briefly a slightly different approach for proving this result from [8].

If f is completely monotonic then by Bernstein’s theorem,

$$\begin{aligned} f(x)=\int _0^{\infty }e^{-xt}\, d\sigma (t) \end{aligned}$$

for some positive measure $\sigma$ on $[0,\infty )$. Then take $\sigma _n=\sigma |_{[0,n]}$ and let

$$\begin{aligned} f_n(x)=\int _0^{\infty }e^{-xt}\, d\sigma _n(t)= \int _0^{n}e^{-xt}\, d\sigma (t). \end{aligned}$$

These functions converge pointwise to f (and even uniformly on $[r,\infty )$ for any $r>0$) and they are bounded by $\sigma ([0,n])$.

Next define

$$\begin{aligned} h_k(s)=(1+s/k)^{-k}-e^{-s}, \ s\ge 0. \end{aligned}$$

It can be shown that $\{h_k\}$ is a non-negative and decreasing sequence of functions and furthermore that

$$\begin{aligned} \sup \{h_k(s)\, | \, s\ge 0\}\rightarrow 0 \end{aligned}$$

as $k\rightarrow \infty$. (In fact, one may argue that $h_k$ has a unique maximum on $[0,\infty )$ and that the maximal value does not exceed e/k.)

Now, given a completely monotonic function f we construct $\{f_n\}$ as above and then we choose, for any n so large that $\sigma ([0,n])>0$, $k_n$ such that

$$\begin{aligned} \sup \{h_{k_n}(s)\, | \, s\ge 0\}\le \frac{1}{n\sigma ([0,n])}. \end{aligned}$$

Next we put $g_n(x)=\int _0^n(1+tx/k_n)^{-k_n}\, d\sigma (t)$ and notice that $g_n\in {\mathcal {S}}_{k_n}$ and also that $g_n-f_n$ converges uniformly to 0, indeed:

$$\begin{aligned} |f_n(x)-g_n(x)|\le \int _0^nh_{k_n}(tx)\, d\sigma (t)\le \frac{1}{n}. \end{aligned}$$

This gives

$$\begin{aligned} g_n(x)=(g_n(x)-f_n(x))+f_n(x)\rightarrow f(x) \end{aligned}$$

as $n\rightarrow \infty$.

Proof of Proposition 2.2

Let $f\in \mathcal B_{\lambda }$. Then $x^{1-\lambda }f'(x)$ is completely monotonic, and thus of the form

$$\begin{aligned} x^{1-\lambda }f'(x)=\int _0^{\infty }e^{-xt}\, d \sigma (t). \end{aligned}$$

Next let $f_n(x)=\int _0^{n}e^{-xt}\, d\sigma (t)$ and construct $g_n\in {\mathcal {S}}_{k_n}$ as above such that

$$\begin{aligned} \sup \{|f_n(x)-g_n(x)|\,|\, x\ge 0\}\rightarrow 0,\ \text {as}\ n\rightarrow \infty . \end{aligned}$$

Now let $x>0$ be given. Then $\int _0^xt^{\lambda -1}(f_n(t)-g_n(t))\, dt\rightarrow 0$, as $n\rightarrow \infty$. Indeed, the integrand tends pointwise to 0 and

$$\begin{aligned} t^{\lambda -1}\sup _n\sup _{s\ge 0}|f_n(s)-g_n(s)| \end{aligned}$$

is an integrable majorant so that Lebesgue’s theorem on dominated convergence can be applied.

Furthermore,

$$\begin{aligned} \int _0^xt^{\lambda -1}g_n(t)\, dt=\int _0^xt^{\lambda -1}(g_n(t)-f_n(t))\, dt+\int _0^xt^{\lambda -1}f_n(t)\, dt. \end{aligned}$$

The first term tends to 0, and the second term tends, by monotone convergence, to

$$\begin{aligned} \int _0^xt^{\lambda -1}\int _0^{\infty }e^{-ts}\, d\sigma (s)\, dt=\int _0^xf'(t)\, dt=f(x)-f(0^+). \end{aligned}$$

Thus, defining

$$\begin{aligned} F_n(x)=\int _0^xt^{\lambda -1}g_n(t)\, dt, \end{aligned}$$

then $F_n(x)\rightarrow f(x)-f(0^+)$ as $n\rightarrow \infty$ and $x^{1-\lambda }F_n'(x)=g_n(x) \in {\mathcal {S}}_{k_n}$. From Theorem 2.14 it now follows that $F_n\in \mathcal T_{\lambda ,\lambda +1-k_n}$. This completes the proof. $\square$

By inspection of the construction preceeding the proof of Proposition 2.2 one sees that any bounded completely monotonic function is the uniform limit of a sequence of bounded generalized Stieltjes functions. This observation can be used to obtain the next corollary, which refines [12, Proposition 3.5].

Corollary 2.3

For any finite Borel measure $\mu$ on $[0,\infty )$ there exists a sequence of bounded generalized Bernstein functions $b_n$ of positive order and a sequence of non-negative numbers $c_n$ such that

$$\begin{aligned} c_n\epsilon _0+b_n'(x)\, dx\rightarrow \mu \ \text {weakly on}\ [0,\infty ). \end{aligned}$$

Proof

Given $\mu$ we notice that $f={\mathcal {L}}(\mu )$ is a bounded completely monotonic function. As remarked above there is a sequence $\{f_n\}$ where $f_n\in {\mathcal {S}}_{\lambda _n}$ is bounded such that $f_n\rightarrow f$ uniformly on $[0,\infty )$ as $n\rightarrow \infty$. The function $f_n$ can be represented as

$$\begin{aligned} f_n(x)=c_n+\int _0^{\infty }\frac{d\mu _n(t)}{(x+t)^{\lambda _n}}=\mathcal L\left( \sigma _n\right) (x), \end{aligned}$$

where

$$\begin{aligned} \sigma _n=c_n\epsilon _0+\frac{t^{\lambda _n-1}}{\Gamma (\lambda _n)}\mathcal L(\mu _n)(t)dt. \end{aligned}$$

Boundedness of $f_n$ means that $\sigma _n$ is a finite measure. Since ${\mathcal {L}}(\sigma _n)\rightarrow {\mathcal {L}}(\mu )$ pointwise on $[0,\infty )$ it follows that $\sigma _n\rightarrow \mu$ vaguely and $\sigma _n([0,\infty ))\rightarrow \mu ([0,\infty ))$, and therefore $\sigma _n\rightarrow \mu$ weakly. Defining

$$\begin{aligned} b_n(x)=\int _0^{x}\frac{t^{\lambda _n-1}}{\Gamma (\lambda _n)}\mathcal L(\mu _n)(t)dt, \end{aligned}$$

we see that $b_n\in {\mathcal {B}}_{\lambda _n}$, that $b_n$ bounded (indeed $b_n(x)\le \sigma _n([0,\infty ))$) and that $c_n\epsilon _0+b_n'(x)\, dx=\sigma _n$. This proves the assertion. $\square$

It should be noted that bounded generalized Bernstein functions have been characterized in [12, Proposition 3.9].

Remark 2.4

If $f\in \mathcal B_{\lambda }$ is represented by the triple $(a,b,\mu )$ and if $f_n\in {\mathcal {T}}_{\lambda ,\alpha _n}$, represented by the triple $(a_n,b_n,\mu _n)$, converges pointwise to f then

$$\begin{aligned} x^{\lambda -\alpha _n}{\mathcal {L}}(\mu _n)(x)\, dx\rightarrow \mu \ \text {vaguely on}\ x>0. \end{aligned}$$

Indeed, we have according to [12, Proposition 2.4] that

$$\begin{aligned} x^{1-\lambda }f_n'(x)\rightarrow x^{1-\lambda }f'(x)={\mathcal {L}}(\lambda a\epsilon _0+\mu )(x),\ \text {as}\ n\rightarrow \infty . \end{aligned}$$

Also, using Theorem 2.14

$$\begin{aligned} x^{1-\lambda }f_n'(x)&=\lambda a_n+\Gamma (\lambda +1-\alpha _n)\int _0^{\infty }\frac{d\mu _n(s)}{(s+x)^{\lambda +1-\alpha _n}}\\&={\mathcal {L}}(\lambda a_n \epsilon _0+s^{\lambda -\alpha _n}{\mathcal {L}}(\mu _n)(s))(x) \end{aligned}$$

and this gives

$$\begin{aligned} \lambda a_n \epsilon _0+s^{\lambda -\alpha _n}{\mathcal {L}}(\mu _n)(s)\,ds\rightarrow \lambda a\epsilon _0+\mu \end{aligned}$$

vaguely on $[0,\infty )$. See [2, Proposition 9.5]. Restriction to the open half line establishes the result.

Next we develop the theory aiming, among other things, at the proof of Theorem 2.14. The first lemma is a simple consequence of Fubini’s theorem.

Lemma 2.5

For any non-negative Borel measurable function g and any positive measure $\mu$ we have

$$\begin{aligned} \int _0^{\infty }g(t){\mathcal {L}}(\mu )(t)\, dt=\int _0^{\infty }{\mathcal {L}}(g)(s)\, d\mu (s). \end{aligned}$$

The next result is about Laplace transforms and convolution measures. For the reader’s convenience let us mention that the convolution measure $\mu *\nu$ of two measures $\mu$ and $\nu$ is given as the image measure of the product measure under the map $\tau (x,t)=x+t$.

Lemma 2.6

Let $\mu$ and $\nu$ be two measures on $[0,\infty )$ and let $\beta >0$. Then

$$\begin{aligned} \int _0^{\infty }t^{\beta -1}{\mathcal {L}}(\mu )(t){\mathcal {L}}(\nu )(t)\, dt=\Gamma (\beta )\int _0^{\infty }\frac{d(\mu *\nu )(s)}{s^{\beta }}. \end{aligned}$$

Proof

This follows from Lemma 2.5 and the fact that ${\mathcal {L}}(t^{\beta -1})(s)=\Gamma (\beta )/s^{\beta }$. $\square$

Remark 2.7

Because of positivity of the function and measure in Lemma 2.6 interchanging the order of integration is permitted. It may of course happen that the integrals equal $\infty$. For values of t near 0, ${\mathcal {L}}(\mu )(t){\mathcal {L}}(\nu )(t)$ is in general bounded from below by some positive constant. Hence, in order for the integral on the left hand side to be convergent, $\beta$ must be positive.

Corollary 2.8

Let $\lambda >0$ and $\alpha <\lambda +1$. For $f\in \mathcal B_{\lambda }$ and any measure $\mu$ on $[0,\infty )$ the following relation holds:

$$\begin{aligned} \int _0^{\infty }f(t)t^{-\alpha }{\mathcal {L}} (\mu )(t)\, dt&=\Gamma (\lambda +1-\alpha )\int _0^{\infty }\frac{d(\omega *\mu )(t)}{t^{\lambda +1-\alpha }}\\&=\Gamma (\lambda +1-\alpha )\int _0^{\infty }\int _0^{\infty }\frac{d\omega (u)}{(s+u)^{\lambda +1-\alpha }}\, d\mu (s), \end{aligned}$$

where $\omega$ is the measure in the Bernstein representation of the completely monotonic function $t^{-\lambda }f(t)$.

Proof

We know from [12, Corollary 2.1] that the function $t^{-\lambda }f(t)$ is completely monotonic, and hence is of the form $t^{-\lambda }f(t)={\mathcal {L}}(\omega )(t)$, for some positive measure $\omega$ on $[0,\infty )$. This gives, using Lemma 2.6,

$$\begin{aligned} \int _0^{\infty }f(t)t^{-\alpha }{\mathcal {L}} (\mu )(t)\, dt&= \int _0^{\infty }t^{\lambda -\alpha }{\mathcal {L}} (\omega )(t){\mathcal {L}}(\mu )(t)\, dt\\&= \Gamma (\lambda +1-\alpha )\int _0^{\infty }\frac{d (\omega *\mu )(s)}{s^{\lambda +1-\alpha }}. \end{aligned}$$

This proves the result. $\square$

Taking $\mu$ to be the point mass $\epsilon _x$ at x the corollary yields the following.

Corollary 2.9

Let $f\in \mathcal B_{\lambda }$ and suppose that $\alpha <\lambda +1$. Then

$$\begin{aligned} {\mathcal {L}}(t^{-\alpha }f(t))(x)=\Gamma (\lambda +1-\alpha )\int _0^{\infty }\frac{d\omega (s)}{(x+s)^{\lambda +1-\alpha }}, \end{aligned}$$

where ${\mathcal {L}}(\omega )(t)=t^{-\lambda }f(t)$. In particular, the Laplace transform maps the class $t^{-\alpha }\mathcal B_{\lambda }$ into ${\mathcal {S}}_{\lambda +1-\alpha }$.

Next, let us show some consequences of these results. The incomplete gamma function is, due to the relation (1), a fundamental building block in constructing functions in $\mathcal B_{\lambda }$.

Proposition 2.10

The following relation holds for any positive measure $\mu$, $\alpha <\lambda +1$ and $x>0$

$$\begin{aligned} \int _0^{\infty }\gamma (\lambda ,xt)t^{-\alpha }{\mathcal {L}} (\mu )(t)\, dt&=\Gamma (\lambda +1-\alpha )\int _0^{\infty }\int _0^{x}\frac{u^{\lambda -1}du}{(s+u)^{\lambda +1-\alpha }}\, d\mu (s)\\&=\Gamma (\lambda +1-\alpha )\int _0^{\infty }B\left( \lambda ,1-\alpha ; \frac{x}{x+s}\right) \, \frac{d\mu (s)}{s^{1-\alpha }}. \end{aligned}$$

Proof

Since the function $f(t)=\gamma (\lambda ,xt)$ belongs to $\mathcal B_{\lambda }$ and

$$\begin{aligned} \gamma (\lambda ,xt)t^{-\lambda }=\int _0^xe^{-tu}u^{\lambda -1}\, du, \end{aligned}$$

the corresponding measure in Corollary 2.9 is $d\omega (u)=\mathbbm {1}_{[0,x]}(u)u^{\lambda -1}du$. Applying now Corollary 2.8 we obtain

$$\begin{aligned} \int _0^{\infty }\gamma (\lambda ,xt)t^{-\alpha }{\mathcal {L}} (\mu )(t)\, dt=\Gamma (\lambda +1-\alpha )\int _0^{\infty }\int _0^{x}\frac{u^{\lambda -1}du}{(s+u)^{\lambda +1-\alpha }}\, d\mu (s). \end{aligned}$$

Since

$$\begin{aligned} \int _0^{x}\frac{u^{\lambda -1}du}{(s+u)^{\lambda +1-\alpha }}=\frac{1}{s^{1-\alpha }}B\left( \lambda ,1-\alpha ; \frac{x}{x+s}\right) \end{aligned}$$

(which follows by the change of variable $v=u/(u+s)$) for $\alpha <\lambda +1$, the proof is complete. $\square$

Letting $\alpha =0$ and $\mu$ be the point mass at s in Proposition 2.10 we record the following elementary relation (see also [6, 6.451.1]):

Corollary 2.11

$$\begin{aligned} \int _0^{\infty }e^{-xt}\gamma (\lambda ,ts)\, dt=\frac{\Gamma (\lambda )}{x}\frac{s^{\lambda }}{(x+s)^{\lambda }}. \end{aligned}$$

Let us mention a couple of examples: Again letting $\mu$ be the point mass at s in Proposition 2.10 we obtain the assertion in the motivating example mentioned in the introduction.

Example 2.12

For $s>0$ the function

$$\begin{aligned} x\mapsto B\left( \lambda ,1-\alpha ;\frac{x}{x+s}\right) \end{aligned}$$

belongs to ${\mathcal {T}}_{\lambda ,\alpha }$.

Example 2.13

For a positive integer n and $c>0$ the function

$$\begin{aligned} g_{c,n}(x)=(-1)^{n-1}\left( \log \left( \frac{x+c}{c}\right) +\sum _{k=1}^{n-1}\frac{(-1)^kx^k}{k}c^{-k}\right) \end{aligned}$$

belongs to ${\mathcal {T}}_{n,n}$. (See Proposition 4.2 with $\mu =\epsilon _c$.) Notice that

$$\begin{aligned} g_{c,n}(x)=x^n/c^n\, \int _0^{\infty }e^{-xt/c}E_n(t)\, dt, \end{aligned}$$

where $E_n$ is the generalized exponential integral (see [5, 8.19.24] and the end of Sect. 5).

Also the function

$$\begin{aligned} x\mapsto \Gamma (\lambda )x^{\lambda }\int _0^{\infty }\int _{t_n}^{\infty }\cdots \int _{t_2}^{\infty }\int _{t_1}^{\infty }\frac{ds}{(x+s)^{\lambda }s}dt_1\cdots dt_{n-1}d\mu (t_n) \end{aligned}$$

(where $\mu$ is any positive measure making this multiple integral converge) belongs to ${\mathcal {T}}_{\lambda ,n}$. (See Proposition 4.4 with $\beta =0$.)

Theorem 2.14

For a function $f:(0,\infty )\rightarrow [0,\infty )$ we have

$$\begin{aligned} f\in \mathcal T_{\lambda ,\alpha }\ \Leftrightarrow \ x^{1-\lambda }f'(x)\in {\mathcal {S}}_{\lambda +1-\alpha }. \end{aligned}$$

Proof

Assume that $f\in \mathcal T_{\lambda ,\alpha }$. In Definition 1.1 the function $t^{\alpha }\varphi (t)$ is the Laplace transform of a positive measure $\mu$, and using Proposition 2.10, f can be written in the form

$$\begin{aligned} f(x)&=ax^{\lambda }+b+\int _0^{\infty }\gamma (\lambda ,xt)t^{-\alpha }{\mathcal {L}} (\mu )(t)\, dt\\&=ax^{\lambda }+b+\Gamma (\lambda +1-\alpha )\int _0^{\infty }\int _0^{x}\frac{u^{\lambda -1}du}{(s+u)^{\lambda +1-\alpha }}\, d\mu (s). \end{aligned}$$

This gives

$$\begin{aligned} f'(x)=\lambda ax^{\lambda -1}+\Gamma (\lambda +1-\alpha )x^{\lambda -1} \int _0^{\infty }\frac{d\mu (s)}{(s+x)^{\lambda +1-\alpha }}, \end{aligned}$$

showing that $x^{1-\lambda }f'(x)\in {\mathcal {S}}_{\lambda +1-\alpha }$.

Conversely, if $x^{1-\lambda }f'(x)\in {\mathcal {S}}_{\lambda +1-\alpha }$ then

$$\begin{aligned} f'(x)=x^{\lambda -1}\int _0^{\infty }e^{-xs}s^{\lambda -\alpha }\varphi (s)\, ds+cx^{\lambda -1}, \end{aligned}$$

where $\varphi$ is completely monotonic and $c\ge 0$. Since f is increasing, integration of this relation yields

$$\begin{aligned} f(x)-f(0+)&=\int _0^{\infty }\int _0^xt^{\lambda -1}e^{-ts}\, dts^{\lambda -\alpha }\varphi (s)ds+\frac{c}{\lambda }x^{\lambda }\\&=\int _0^{\infty }\gamma (\lambda ,sx)s^{-\alpha }\varphi (s)ds+\frac{c}{\lambda }x^{\lambda }. \end{aligned}$$

By definition, $s^{-\alpha }\varphi (s)$ is completely monotonic of order $\alpha$ and so $f\in {\mathcal {T}}_{\lambda ,\alpha }$. This completes the proof. $\square$

Remark 2.15

Theorem 2.14 yields

• (a) when $\alpha =1$: $f\in {\mathcal {T}}_{\lambda ,1}\Leftrightarrow x^{1-\lambda }f'(x)\in {\mathcal {S}}_{\lambda }$ (see also [12, Theorem 4.1]);

• (b) when $\alpha =\lambda$: $f\in {\mathcal {T}}_{\lambda ,\lambda } \Leftrightarrow x^{1-\lambda }f'(x)\in {\mathcal {S}}$;

• (c) when $\lambda =1$: $f\in {\mathcal {T}}_{1,\alpha } \Leftrightarrow f'\in {\mathcal {S}}_{2-\alpha }$. See also Remark 3.5.

3 Representation Via Hypergeometric Functions

In this section we prove that the $(\lambda ,\alpha )$-Thorin–Bernstein functions admit an integral representation in terms of the hypergeometric function ${}_2F_1$.

Theorem 3.1

A function f belongs to $\mathcal T_{\lambda ,\alpha }$ if and only if there are non-negative constants a and b and a positive measure $\mu$ such that

$$\begin{aligned} f(x)=ax^{\lambda }+b+\frac{\Gamma (\lambda +1-\alpha )x^{\lambda }}{\lambda }\int _0^{\infty }\frac{{}_2F_1(\alpha ,1; \lambda +1; -x/s)}{(s+x)^{\lambda -\alpha }}\frac{d\mu (s)}{s}. \end{aligned}$$

In the affirmative case, $\mu$ is the measure such that $t^{\alpha }\varphi (t)={\mathcal {L}}(\mu )(t)$, $\varphi$ being the function in Definition 1.1.

Proof

In the first relation in Proposition 2.10 we perform the change of variable $v=u/x$. This gives us

$$\begin{aligned} f(x)-ax^{\lambda }-b&=\int _0^{\infty }\gamma (\lambda ,xt)t^{-\alpha }{\mathcal {L}} (\mu )(t)\, dt\\ \quad&=\Gamma (\lambda +1-\alpha )x^{\lambda }\int _0^{\infty }\frac{1}{s^{\lambda +1-\alpha }}\int _0^{1}\frac{v^{\lambda -1}dv}{(1+vx/s)^{\lambda +1-\alpha }}\, d\mu (s). \end{aligned}$$

Next, a combination of Euler’s integral representation of the ${}_2F_1$ and Euler’s transformation (see [1, Theorem 2.2.1 and Theorem 2.2.5]) yields

$$\begin{aligned} f(x)-ax^{\lambda }-b&=\frac{\Gamma (\lambda +1-\alpha )x^{\lambda }}{\lambda }\int _0^{\infty }\frac{{}_2F_1(\lambda +1-\alpha ,\lambda ;\lambda +1;-x/s)}{s^{\lambda +1-\alpha }}\, d\mu (s)\\&=\frac{\Gamma (\lambda +1-\alpha )x^{\lambda }}{\lambda }\int _0^{\infty }\frac{{}_2F_1(\alpha ,1;\lambda +1;-x/s)}{s^{\lambda +1-\alpha }(1+x/s)^{\lambda -\alpha }}\, d\mu (s). \end{aligned}$$

This proves the asserted formula. $\square$

We record the following equivalent characterizations, obtained by using Pfaff’s transformation ([1, Theorem 2.2.5]), Proposition 2.10 and the identity

$$\begin{aligned} B(c,d;z)=\frac{z^c}{c}\, {}_2F_1\left( c,1-d;c+1; z\right) , \end{aligned}$$

see [5, 8.17.7].

Corollary 3.2

The following statements are equivalent for a function $f:(0,\infty )\rightarrow [0,\infty )$.

• (a) $f\in \mathcal T_{\lambda ,\alpha }$,

• (b) f can be represented as

  $$\begin{aligned} f(x)=ax^{\lambda }+b+\frac{\Gamma (\lambda +1-\alpha )x^{\lambda }}{\lambda }\int _0^{\infty }\frac{{}_2F_1(\lambda , \alpha ; \lambda +1; x/(x+s))}{(s+x)^{\lambda }}\frac{d\mu (s)}{s^{1-\alpha }}, \end{aligned}$$

• (c) f can be represented as

  $$\begin{aligned} f(x)=ax^{\lambda }+b+\Gamma (\lambda +1-\alpha )\int _0^{\infty }B\left( \lambda ,1-\alpha ;\frac{x}{x+t}\right) \frac{d\mu (t)}{t^{1-\alpha }}. \end{aligned}$$

Remark 3.3

As was shown in Proposition 2.1, if $f_n\in \mathcal T_{\lambda ,\alpha }$ converges pointwise to f then f also belongs to $\mathcal T_{\lambda ,\alpha }$. Letting $f_n$ correspond to the triple $(a_n,b_n,\mu _n)$ in (c) of Corollary 3.2 and f to $(a,b,\mu )$, then $\mu _n\rightarrow \mu$ vaguely on $(0,\infty )$. To see this we notice

$$\begin{aligned} x^{1-\lambda }f_n'(x)&=\lambda a_n+\Gamma (\lambda +1-\alpha )\int _0^{\infty }\frac{d\mu _n(t)}{(x+t)^{\lambda +1-\alpha }}\quad \text {and}\\ x^{1-\lambda }f'(x)&=\lambda a+\Gamma (\lambda +1-\alpha )\int _0^{\infty }\frac{d\mu (t)}{(x+t)^{\lambda +1-\alpha }}, \end{aligned}$$

from which it follows that $\mu _n\rightarrow \mu$ vaguely (see [16, Theorem 2.2]).

Remark 3.4

Suppose $\lambda =\alpha =1$. Since ${}_2F_1(1,1;2;x)=-(1/x)\log (1-x)$ the function f in Corollary 3.2 takes the form

$$\begin{aligned} f(x)=ax+b+\int _0^{\infty }\log \left( \frac{x+t}{t}\right) \, d\mu (t), \end{aligned}$$

which is in accordance with the representation of ordinary Thorin–Bernstein functions. See [16, Theorem 8.2]. See also Proposition 4.2.

Remark 3.5

In the special case where $\lambda =1$ and $0<\alpha <1$ observe that for $-1<z<1$,

$$\begin{aligned} z\, {}_2F_1\left( 1, \alpha ; 2; z\right) =\int _0^z\frac{dt}{(1-t)^{\alpha }}. \end{aligned}$$

Substituting $z=x/(x+s)$ it follows that

$$\begin{aligned} \frac{x}{x+s}\, {}_2F_1\left( 1,\alpha ; 2; \frac{x}{x+s}\right) =\frac{1}{1-\alpha }-\frac{1}{1-\alpha }\left( \frac{s}{x+s}\right) ^{1-\alpha }. \end{aligned}$$

Therefore, Corollary 3.2(b) gives us that any $f\in {\mathcal {T}}_{1,\alpha }$ satisfies

$$\begin{aligned} f'(x)=a+\Gamma (2-\alpha )\int _0^{\infty }\frac{d\mu (s)}{(x+s)^{2-\alpha }}, \end{aligned}$$

which is [16, Proposition 8.10].

4 Integer Values of the Parameters

We find in Proposition 4.2 another form of the representation of functions from $\mathcal T_{\lambda ,\alpha }$ in the case where $\lambda =\alpha =n$, n being a positive integer. First, a lemma.

Lemma 4.1

For $z\in (-1,1)\setminus \{0\}$ we have

$$\begin{aligned} \frac{{}_2F_1(\lambda ,\alpha ;\lambda +1;z)}{\lambda }=\sum _{k=0}^{\infty }\frac{1}{\lambda +k}\frac{(\alpha )_k}{k!}z^k= \frac{1}{z^{\lambda }}\int _0^{z/(1-z)}u^{\lambda -1}(1+u)^{\alpha -\lambda -1}\, du. \end{aligned}$$

Proof

The first equality holds by definition. Furthermore,

$$\begin{aligned} \sum _{k=0}^{\infty }\frac{1}{\lambda +k}\frac{(\alpha )_k}{k!}z^k&= \frac{1}{z^{\lambda }}\sum _{k=0}^{\infty }\frac{1}{\lambda +k}\frac{(\alpha )_k}{k!}z^{k+\lambda }\\&= \frac{1}{z^{\lambda }}\int _0^zt^{\lambda -1}\sum _{k=0}^{\infty }\frac{(\alpha )_k}{k!}t^{k}\, dt\\&= \frac{1}{z^{\lambda }}\int _0^zt^{\lambda -1}\frac{1}{(1-t)^{\alpha }}\, dt. \end{aligned}$$

The change of variable $u=t/(1-t)$ transforms this expression into

$$\begin{aligned} \frac{1}{z^{\lambda }}\int _0^{z/(1-z)}u^{\lambda -1}(1+u)^{\alpha -\lambda -1}\, du, \end{aligned}$$

and this proves the lemma. $\square$

Proposition 4.2

Suppose that n is a positive integer. A function f belongs to ${\mathcal {T}}_{n,n}$ if and only if

$$\begin{aligned} f(x)=ax^{n}+b+(-1)^{n-1}\int _0^{\infty }\left( \log \left( \frac{x+s}{s}\right) +\sum _{k=1}^{n-1}\frac{(-1)^kx^k}{ks^k}\right) \, s^{n-1}\, d\mu (s). \end{aligned}$$

Proof

Corollary 3.2 gives us

$$\begin{aligned} f(x)=ax^{n}+b+\frac{x^{n}}{n}\int _0^{\infty }\frac{{}_2F_1(n, n; n+1; x/(x+s))}{(s+x)^{n}}\frac{d\mu (s)}{s^{1-n}}, \end{aligned}$$

and Lemma 4.1 yields (for $x\ne 0$)

$$\begin{aligned} \frac{{}_2F_1(n, n; n+1; x/(x+s))}{n}=\frac{(s+x)^{n}}{x^n}\int _0^{x/s}\frac{u^{n-1}}{1+u}\, du. \end{aligned}$$

The integral in the right-hand side of this relation can be rewritten using Taylor’s formula with integral remainder:

$$\begin{aligned} \int _0^{x/s}\frac{u^{n-1}}{1+u}\, du=(-1)^{n-1}\left( \log (1+x/s)+\sum _{k=1}^{n-1}\frac{(-x/s)^k}{k}\right) . \end{aligned}$$

In this way the proposition is proved. $\square$

Remark 4.3

In the case where $\alpha =1$ and $\lambda$ is a positive rational number, the integrand in the representation in Corollary 3.2(b) can be summed. Indeed, writing $\lambda =n+p/q$, where n, p and q are non-negative integers and $p<q$ we have

$$\begin{aligned} \sum _{k=0}^{\infty }\frac{1}{k+\lambda }\left( \frac{x}{x+t}\right) ^{k+\lambda }&= \sum _{k=0}^{\infty }\frac{1}{k+p/q}\left( \frac{x}{x+t}\right) ^{k+p/q}\\&\quad - \sum _{k=0}^{n-1}\frac{1}{k+p/q}\left( \frac{x}{x+t}\right) ^{k+p/q}. \end{aligned}$$

The infinite sum on the right hand side can be rewritten as

$$\begin{aligned} \sum _{k=0}^{\infty }\frac{1}{k+p/q}\left( \frac{x}{x+t}\right) ^{k+p/q}=-\sum _{k=0}^{q-1}e^{-2\pi i kp/q}\log \left( 1-\left( \frac{x}{x+t}\right) ^{1/q}e^{2\pi i k/q}\right) . \end{aligned}$$

This formula can be found in [15]. We indicate how to prove a more general formula below. The representation thus takes the form

$$\begin{aligned} f(x)&=ax^{n+p/q}+b\\ {}&+ \Gamma (n+p/q)\int _0^{\infty }\Bigg \{ -\sum _{k=0}^{q-1}e^{-2k\pi ip/q}\log \left( 1-\left( \frac{x}{x+t}\right) ^{1/q}e^{2k\pi i/q}\right) \\&\quad -\left( \frac{x}{x+t}\right) ^{p/q}\sum _{k=0}^{n-1}\frac{1}{k+p/q}\left( \frac{x}{x+t}\right) ^k\Bigg \} \, d\mu (t). \end{aligned}$$

In particular, for $\lambda =n+1/2$, using $2\tanh ^{-1}(s)=\log ((1+s)/(1-s))$ for $|s|<1$,

$$\begin{aligned} f(x)&=ax^{n+1/2}+b+ 2\Gamma (n+1/2)\int _0^{\infty }\left\{ \tanh ^{-1}\left( \left( \frac{x}{x+t}\right) ^{1/2}\right) \right. \\&\quad -\left. \sum _{k=0}^{n-1}\frac{1}{2k+1}\left( \frac{x}{x+t}\right) ^{k+1/2}\right\} \, d\mu (t). \end{aligned}$$

The summation formula alluded to: Suppose that $g(z)=\sum _{n=1}^{\infty }a_nz^n$ converges for $|z|<1$. Then, interchanging the order of summation and using elementary properties of the roots of unity it follows that

$$\begin{aligned} \sum _{k=0}^{q-1}e^{-2\pi i kp/q}g\left( z^{1/q}e^{2\pi i k /q}\right) = qz^{p/q}\sum _{k=0}^{\infty }a_{p+kq}z^k. \end{aligned}$$

The formula needed above corresponds to $a_n=1/n$.

In the next proposition we examine in more detail the representation of functions in $\mathcal T_{\lambda ,\alpha }$ taking into account both the integer part and the fractional part of the order $\alpha$ of complete monotonicity of $\varphi$.

Proposition 4.4

Suppose that $\varphi$ is a completely monotonic function of order $\alpha =n+\beta$, $n\in {\mathbb {N}}$, $\beta \in [0,1)$. There exists a positive measure $\mu$ such that, with

$$\begin{aligned} \xi _n(t)=\frac{1}{(n-1)!}\int _0^t(t-u)^{n-1}\,d\mu (u), \end{aligned}$$

we have

$$\begin{aligned}&\int _0^{\infty }\gamma (\lambda ,xt)\varphi (t)\, dt\\&=\Gamma (\lambda -\beta +1)\int _0^{\infty }B\left( \lambda ,1-\beta ; \frac{x}{x+t}\right) \frac{\xi _n(t)}{t^{1-\beta }}\,dt\\&=\Gamma (\lambda -\beta +1)\int _0^{\infty }\int _{t_n}^{\infty }\cdots \int _{t_1}^{\infty }B\left( \lambda ,1-\beta ; \frac{x}{x+s}\right) \frac{ds}{s^{1-\beta }}dt_1\cdots d\mu (t_n). \end{aligned}$$

Proof

According to [11] we may write $\varphi$ as $\varphi (t)=t^{-\beta }{\mathcal {L}}(\xi _n)(t)$ where $\xi _n$ is the fractional integral of positive integer order

$$\begin{aligned} \xi _{n}(s)=\frac{1}{(n-1)!}\int _0^s(s-u)^{n-1}\, d\mu (u). \end{aligned}$$

The first equality now follows from Proposition 2.10.

As it is well known, we have

$$\begin{aligned} \frac{1}{(n-1)!}\int _0^t(t-u)^{n-1}\, d\mu (u)=\int _0^t\int _0^{t_1}\cdots \int _0^{t_{n-1}}d\mu (t_n)dt_{n-1}\cdots dt_1 \end{aligned}$$

and this gives the second equality. $\square$

Remark 4.5

When $n=0$ the contents of Proposition 4.4 are described in Proposition 2.10.

Let us complete this section by relating our results with earlier results for generalized complete Bernstein functions and generalized Thorin–Bernstein functions presented in [12].

Remark 4.6

Proposition 4.4 extends two results from [12]:

• (a) A function f belongs to ${\mathcal {T}}_{\lambda ,0}$ if and only if

  $$\begin{aligned} f(x)=ax^\lambda + b+ \Gamma (\lambda )\,\int _0^{\infty }\left( \frac{x}{x+s}\right) ^{\lambda }\,\frac{d\mu (s)}{s}. \end{aligned}$$

  (See also [12, Proposition 3.12].)

• (b) A function f belongs to ${\mathcal {T}}_{\lambda ,1}$ if and only if

  $$\begin{aligned} f(x)=ax^\lambda + b+ \Gamma (\lambda )\,\int _0^{\infty }\left( \frac{x}{x+s}\right) ^{\lambda }\,\frac{h(s)}{s}\, ds, \end{aligned}$$

  where $h=\xi _1$ is non-negative and increasing. (See also [12, Proposition 4.1].)

5 Additional Results and Comments

For non-negative $\alpha$, it is easily seen that $(-1)^n\varphi ^{(n)}$ is completely monotonic of order $\alpha$ if $\varphi$ is completely monotonic of order $\alpha$. Indeed, this follows by writing $\varphi (t)=t^{-\alpha }{\mathcal {L}}(\sigma )(t)$ and using Leibniz’ formula:

$$\begin{aligned} t^{\alpha }(-1)^n\varphi ^{(n)}(t) =\sum _{k=0}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) (\alpha )_k\frac{1}{t^k}\mathcal L(s^{n-k}d\sigma (s))(t). \end{aligned}$$

Denoting by $m_r$ the measure on $(0,\infty )$ having density $s^{r-1}/\Gamma (r)$ w.r.t. Lebesgue measure (for $r>0$) we notice that ${\mathcal {L}}(m_r)(t)=t^{-r}$. Thus the relation above can be written as

$$\begin{aligned} t^{\alpha }(-1)^n\varphi ^{(n)}(t)={\mathcal {L}} \left( s^nd\sigma (s)+\sum _{k=1}^n\left( {\begin{array}{c}n\\ k\end{array}}\right) (\alpha )_k (m_{k}*s^{n-k}d\sigma (s))\right) (t). \end{aligned}$$

Definition 5.1

For $\alpha \in [0,\lambda )$ and a non-negative integer n, $\mathcal T_{\lambda ,\alpha }^{(n)}$ is the subclass of $\mathcal T_{\lambda ,\alpha }$ consisting of the functions of the form

$$\begin{aligned} f(x)=a x^{\lambda }+b+ \int _0^{\infty }\gamma (\lambda ,xt)(-1)^n\varphi ^{(n)}(t)\, dt, \end{aligned}$$

(3)

where a and b are non-negative numbers, and $\varphi$ is a completely monotonic function of order $\alpha$.

(It should be noted that the condition $\alpha <\lambda$ appears because that e.g. $-\varphi '(t)$ dominates $t^{-\alpha -1}$ for t near 0.)

Proposition 5.2

Suppose that $n\ge 1$. If f belongs to $\mathcal T_{\lambda ,\alpha }^{(n)}$ with the representation (3) then $f(x)/x^{\lambda }$ is in ${\mathcal {S}}_{\lambda -\alpha }$. It has the representation

$$\begin{aligned} f(x)/x^\lambda =a +bx^{-\lambda }+\Gamma (\lambda -\alpha )\int _0^{\infty }\frac{d\sigma _n(t)}{(x+t)^{\lambda -\alpha }}, \end{aligned}$$

(4)

where

$$\begin{aligned} \sigma _n=s^{n-1}d\sigma (s)+\sum _{k=1}^{n-1}\left( {\begin{array}{c}n-1\\ k\end{array}}\right) (\alpha )_k\, m_k*(s^{n-1-k}d\sigma (s)), \end{aligned}$$

(5)

$\sigma$ being the positive measure such that $t^{\alpha }\varphi (t)={\mathcal {L}}(\sigma )(t)$.

Conversely, if $a,b\ge 0$ and $\sigma$ is any positive measure for which the Laplace transform converges then defining the measure $\sigma _n$ by (5), f given by (4) belongs to $\mathcal T_{\lambda ,\alpha }^{(n)}$. The corresponding function $\varphi$ in (3) is given by $\varphi (t)=t^{-\alpha }\mathcal L(\sigma )(t)$ and the measure $\sigma _n$ is related to $\varphi$ in the following way: $(-1)^{n-1}\varphi ^{(n-1)}(t)=t^{-\alpha }\mathcal L(\sigma _{n})(t)$.

Proof

When $\varphi$ is a completely monotonic function of order $\alpha \ge 0$ we write $t^{\alpha }\varphi (t)=\mathcal L(\sigma )(t)$, $\varphi (t)={\mathcal {L}}(\mu )(t)$, and notice that $(-1)^n\varphi ^{(n)}(t)= {\mathcal {L}}(s^nd\mu (s))(t)$. This gives, using Lemma 2.5 and Corollary 2.11,

$$\begin{aligned} \int _0^{\infty }\gamma (\lambda ,xt)(-1)^n\varphi ^{(n)}(t)\, dt&= \int _0^{\infty }\gamma (\lambda ,xt){\mathcal {L}}(s^nd\mu (s))(t)\, dt\\&=\Gamma (\lambda )x^{\lambda }\int _0^{\infty }\frac{s^{n-1}}{(s+x)^{\lambda }}\, d\mu (s)\\&=x^{\lambda }\int _0^{\infty }e^{-xt}t^{\lambda -1}{\mathcal {L}}(s^{n-1}d\mu (s))(t)\, dt\\&=x^{\lambda }\int _0^{\infty }e^{-xt}t^{\lambda -1}(-1)^{n-1}\varphi ^{(n-1)}(t)\, dt\\&=x^{\lambda }\int _0^{\infty }e^{-xt}t^{\lambda -1-\alpha }{\mathcal {L}}(\sigma _{n})(t)\, dt\\&=\Gamma (\lambda -\alpha )x^{\lambda }\int _0^{\infty }\frac{d\sigma _{n}(t)}{(x+t)^{\lambda -\alpha }}, \end{aligned}$$

where $\sigma _{n}$ is the positive measure given in (5) and thus satisfies

$$\begin{aligned}(-1)^{n-1}\varphi ^{(n-1)}(t)=t^{-\alpha }{\mathcal {L}}(\sigma _{n})(t). \end{aligned}$$

The proof of the other direction follows by retracing these steps. This concludes the proof of the proposition. $\square$

A $C^{\infty }$-function $f:(0,\infty )\rightarrow (0,\infty )$ is said to be logarithmically completely monotonic if $-(\log f(x))'=-f'(x)/f(x)$ is completely monotonic. A measure or function on the positive half line is called infinitely divisible if its Laplace transform is a logarithmically completely monotonic function. For more details see [3].

Corollary 5.3

Let $\sigma$ and $\sigma _n$ be as in the proposition above. Then the function $t^{\lambda -\alpha -1}{\mathcal {L}}(\sigma _n)(t)$ is infinitely divisible if $1<\lambda -\alpha \le 2$.

Proof

It follows from the relation

$$\begin{aligned} \int _0^{\infty }e^{-xt}t^{\lambda -1-\alpha }\mathcal L(\sigma _{n})(t)\, dt=\Gamma (\lambda -\alpha )\int _0^{\infty }\frac{d\sigma _{n}(t)}{(x+t)^{\lambda -\alpha }} \end{aligned}$$

in the proof above that the Laplace transform of $t^{\lambda -\alpha -1}{\mathcal {L}}(\sigma _n)(t)$ belongs to $\mathcal S_{\lambda -\alpha }$ and thus to ${\mathcal {S}}_2$. Therefore it is logarithmically completely monotonic by a result of Kristiansen. See [7] and [3]. $\square$

Let us end this section with a few additional observations on infinite divisibility.

First of all, if $f\in C_{\alpha }$ for some $\alpha \ge -1$ then f is infinitely divisible. This is well-known for $\alpha \ge 0$ and for $\alpha \in [-1,0)$ it follows from the relation

$$\begin{aligned} {\mathcal {L}}(f)(x)=\int _0^{\infty }e^{-xt}t^{1-\alpha -1}f(t)t^{\alpha }\, dt \end{aligned}$$

and Kristiansen’s theorem.

The next results deal with products of completely monotonic functions of given real orders. The following is a consequence of Lemma 2.6.

Proposition 5.4

If $f\in {\mathcal {C}}_{\alpha }$, $g\in {\mathcal {C}}_{\beta }$ with $\alpha +\beta <1$ and

$$\begin{aligned} f(t)=t^{-\alpha }{\mathcal {L}}(\mu )(t),\quad g(t)=t^{-\beta }{\mathcal {L}}(\nu )(t) \end{aligned}$$

then ${\mathcal {L}}(fg)\in {\mathcal {S}}_{1-\alpha -\beta }$ and

$$\begin{aligned} \int _0^{\infty }e^{-xt}f(t)g(t)\, dt=\Gamma (1-\alpha -\beta )\int _0^{\infty }\frac{d(\mu *\nu )(t)}{(x+t)^{1-\alpha -\beta }}. \end{aligned}$$

Corollary 5.5

If $f\in {\mathcal {C}}_{\alpha }$, $g\in {\mathcal {C}}_{\beta }$ and $-1\le \alpha +\beta <1$, then the function fg is infinitely divisible.

Proof

From Proposition 5.4 we see that ${\mathcal {L}}(fg)$ belongs to ${\mathcal {S}}_{1-\alpha -\beta }$, which is contained in ${\mathcal {S}}_2$, and Kristiansen’s result can thus be used. $\square$

The next corollary deals also with the distribution function

$$\begin{aligned} F_{\lambda }(t)=1-\lambda t^{\lambda }e^t\Gamma (-\lambda ,t)\end{aligned}$$

of a so-called randomized Lomax distribution, defined in terms of the complementary incomplete Gamma function,

$$\begin{aligned} \Gamma (\lambda ,x)=\int _x^{\infty }e^{-u}u^{\lambda -1}\, du. \end{aligned}$$

See [12, Example 3.11], where it was shown that $F_{\lambda }\in \mathcal B_{\lambda }$. A standard computation shows that

$$\begin{aligned} F_{\lambda }(x)=\frac{1}{\Gamma (\lambda )}\int _0^{\infty }\gamma (\lambda ,xt)\frac{dt}{(1+t)^2}, \end{aligned}$$

from which it is immediate that $F_{\lambda }$ even belongs to ${\mathcal {T}}_{\lambda ,0}$.

Corollary 5.6

Suppose that $\lambda >0$ and $\lambda -1\le \alpha < \lambda +1$. If $f\in \mathcal B_{\lambda }$ then $t^{-\alpha }f(t)$ is infinitely divisible. In particular, $t^{-\alpha }\gamma (\lambda ,xt)$ (for fixed x) and $t^{-\alpha }F_{\lambda }(t)$ are infinitely divisible.

Proof

Of course $t^{-\alpha }$ belongs to $\mathcal C_{\alpha }$ and f belongs to ${\mathcal {C}}_{-\lambda }$. The result now follows from Corollary 5.5. $\square$

Next we investigate the image of the subclasses $\mathcal T_{\lambda ,\alpha }$ under the Laplace transform.

Proposition 5.7

If $f\in \mathcal T_{\lambda ,\alpha }$ then $x^{\alpha }{\mathcal {L}}(f)(x)-f(0+)x^{\alpha -1}$ belongs to the class ${\mathcal {S}}_{\lambda +1-\alpha }$.

Remark 5.8

If $\alpha \le 1$ then $x^{\alpha }{\mathcal {L}}(f)(x)$ in the proposition above belongs to ${\mathcal {S}}_{\lambda +1-\alpha }$. In particular, ${\mathcal {L}}(f)$ belongs to ${\mathcal {S}}_{\lambda +1}$ for $f\in {\mathcal {T}}_{\lambda ,0}$. Hence, if $\lambda \le 1$ then any function $f\in {\mathcal {T}}_{\lambda ,0}$ is infinitely divisible.

Proof of Proposition 5.7

Let $f\in \mathcal T_{\lambda ,\alpha }$ and write its representation as

$$\begin{aligned} f(t)=at^{\lambda }+b+\int _0^{\infty }\gamma (\lambda ,ts)s^{-\alpha }\mathcal L(\mu )(s)\, ds \end{aligned}$$

with $b=f(0+)$ (see [12, Proposition 2.2]) and some positive measure $\mu$. This gives, using Corollary 2.11 and Fubini’s theorem,

$$\begin{aligned} \mathcal L(f)(x)=\frac{a\Gamma (\lambda +1)}{x^{\lambda +1}}+\frac{b}{x}+\frac{\Gamma (\lambda )}{x}\int _0^{\infty }s^{-\alpha }\mathcal L(\mu )(s)\frac{s^{\lambda }}{(x+s)^{\lambda }}\, ds. \end{aligned}$$

Notice that $\varphi (s)=s^{\lambda }/(x+s)^{\lambda }$ is in $\mathcal B_{\lambda }$ and that

$$\begin{aligned} s^{-\lambda }\varphi (s)=\frac{1}{\Gamma (\lambda )}\int _0^{\infty }e^{-su}e^{-xu}u^{\lambda -1}\, du. \end{aligned}$$

According to Corollary 2.8 we obtain

$$\begin{aligned} \int _0^{\infty }s^{-\alpha }{\mathcal {L}}(\mu )(s)\varphi (s)\, ds = \frac{\Gamma (\lambda +1-\alpha )}{\Gamma (\lambda )}\int _0^{\infty }\int _0^{\infty }\frac{u^{\lambda -1}e^{-xu}}{(u+s)^{\lambda +1-\alpha }}\, du \, d\mu (s), \end{aligned}$$

and thus

$$\begin{aligned} x^{\alpha }{\mathcal {L}}(f)(x)&=\frac{a\Gamma (\lambda +1)}{x^{\lambda +1-\alpha }}+bx^{\alpha -1}\\&\quad +\Gamma (\lambda +1-\alpha )x^{\alpha -1}\int _0^{\infty }\int _0^{\infty }\frac{u^{\lambda -1}e^{-xu}}{(u+s)^{\lambda +1-\alpha }}\, du \, d\mu (s). \end{aligned}$$

In this last integral we make a change of variable $v=(x/s)u$ and get by Fubini’s theorem

$$\begin{aligned} x^{\alpha }{\mathcal {L}}(f)(x)&=\frac{a\Gamma (\lambda +1)}{x^{\lambda +1-\alpha }}+bx^{\alpha -1}\\&\quad +\Gamma (\lambda +1-\alpha )\int _0^{\infty }\int _0^{\infty }s^{\alpha -1}e^{-vs}\, d\mu (s)\frac{v^{\lambda -1}dv}{(x+v)^{\lambda +1-\alpha }}. \end{aligned}$$

It is now clear that $x^{\alpha }{\mathcal {L}}(f)(x)-bx^{\alpha -1}$ belongs to ${\mathcal {S}}_{\lambda +1-\alpha }$. $\square$

We remark that in general not all functions from $\mathcal S_{\lambda +1-\alpha }$ are of the form $x^{\alpha }{\mathcal {L}}(f)(x)$, for some $f\in \mathcal T_{\lambda ,\alpha }$. For example, since $\mathcal T_{\lambda ,1}\subsetneq \mathcal B_{\lambda }$ it follows that $\Phi (\mathcal T_{\lambda ,1})\subsetneq \Phi (\mathcal B_{\lambda })={\mathcal {S}}_{\lambda }$($\Phi$ as in the introduction). Below an example is given when $\lambda =\alpha \in \{1,2,3,\ldots \}$.

Example 5.9

Let $n\ge 1$. To see that $\{x^n{\mathcal {L}}(g)(x)\, |\, g\in {\mathcal {T}}_{n,n}\}$ is not all of ${\mathcal {S}}$ let $\varphi$ be a completely monotonic function and suppose that there is $g\in {\mathcal {T}}_{n,n}$ such that $x^n{\mathcal {L}}(g)(x)=\mathcal L(\varphi )(x)$. This entails

$$\begin{aligned} g(x)=\int _0^x(x-s)^{n-1}\varphi (s)\, ds. \end{aligned}$$

According to Theorem 2.14, g belongs to ${\mathcal {T}}_{n,n}$ if and only if $x^{1-n}g'(x)$ belongs to ${\mathcal {S}}$. We obtain

$$\begin{aligned} x^{1-n}g'(x)=(n-1)\int _0^1(1-t)^{n-2}\varphi (xt)\, dt, \end{aligned}$$

which is not in general a Stieltjes function. As an example, take $\varphi (s)=e^{-s}$. In this case, $x^{1-n}g'(x)$ is the Laplace transform of the non-completely monotonic function $(n-1)(1-t)^{n-2} \mathbbm {1}_{(0,1)}(t)$.

Let us conclude the paper by giving some examples related to the gene-ralized exponential integral, $E_p$, defined for $p>0$, as

$$\begin{aligned} E_p(x)=x^{p-1}\int _x^{\infty }\frac{e^{-t}}{t^p}\, dt. \end{aligned}$$

We write

$$\begin{aligned} E_p(x)=\int _1^{\infty }e^{-xt}\frac{dt}{t^{p}} \end{aligned}$$

and this yields that $E_p$ is a completely monotonic function for any $p>0$. We observe that

$$\begin{aligned} 0\le tE_p(t)\le \int _t^{\infty }e^{-s}\, ds=e^{-t}, \end{aligned}$$

so $E_p(t)$ decays exponentially for t tending to $\infty$ and is bounded by 1/t as t tends to 0.

Example 5.10

We have, using Proposition 2.10,

$$\begin{aligned} \int _0^{\infty }\gamma (\lambda ,xt)E_p(t)\, dt=\Gamma (\lambda )x^{\lambda }\int _1^{\infty }\frac{1}{(x+s)^{\lambda }}\frac{ds}{s^{p+1}}.\end{aligned}$$

The change of variable $t=x/(x+s)$ transforms the right-hand side into

$$\begin{aligned} \frac{\Gamma (\lambda )}{x^p}B(\lambda +p,-p; x/(x+1)), \end{aligned}$$

(6)

which is thus an example of a function in ${\mathcal {T}}_{\lambda ,0}$, for any $p>0$. Notice that it is infinitely divisible for $\lambda \le 1$.

Example 5.11

According to [5, 8.19.4] we have

$$\begin{aligned} x^{1-p}E_p(x)=\frac{1}{\Gamma (p)}\int _0^{\infty }\frac{e^{-x(t+1)}t^{p-1}}{1+t}\, dt, \end{aligned}$$

and it follows that $E_p\in {\mathcal {C}}_{1-p}$ with $E_p(x)=x^{p-1}{\mathcal {L}}(h_p)(x)$, where

$$\begin{aligned} h_p(t)=\mathbbm {1}_{(1,\infty )}(t)\frac{(t-1)^{p-1}}{\Gamma (p)t}. \end{aligned}$$

This gives, applying Proposition 2.10 followed by the change of variable $u=1-1/t$

$$\begin{aligned} \int _0^{\infty }\gamma (\lambda ,xt)E_p(t)\, dt&= \Gamma (\lambda +p)\int _0^{\infty }B\left( \lambda ,p; \frac{x}{x+t}\right) \frac{h_p(t)}{t^{p}}\, dt\\&=\frac{\Gamma (\lambda +p)}{\Gamma (p)}\int _0^{1}B\left( \lambda ,p; \frac{x(1-u)}{1+x(1-u)}\right) u^{p-1}\, du. \end{aligned}$$

Moreover, using the relation (6) we obtain the identity

$$\begin{aligned} \frac{\Gamma (\lambda +p)}{\Gamma (p)}\int _0^{1}B\left( \lambda ,p; \frac{x(1-u)}{1+x(1-u)}\right) u^{p-1}\, du= \frac{\Gamma (\lambda )}{x^p}B(\lambda +p,-p; x/(x+1)) \end{aligned}$$

and this function belongs to ${\mathcal {T}}_{\lambda ,1-p}$.

Example 5.12

The function

$$\begin{aligned} C_p(x)=\int _0^{\infty }\gamma (\lambda ,xt)t^{1-p}E_p(t)\, dt \end{aligned}$$

is defined when $p<\lambda +1$. Furthermore, $C_p(x)$ belongs to ${\mathcal {T}}_{\lambda ,0}$ and it has the representation

$$\begin{aligned} C_p(x)= & {} \Gamma (\lambda )x^{\lambda }\int _0^{\infty }\frac{1}{(x+s)^{\lambda }}\frac{h_p(s)}{s}\, ds\\= & {} \frac{\Gamma (\lambda )x^{\lambda }}{\Gamma (p)}\int _1^{\infty }\frac{(s-1)^{p-1}}{(x+s)^{\lambda }s^2}\, ds. \end{aligned}$$

The change of variable $t=(s-1)/s$, Euler’s integral representation of ${}_2F_1$ and Pfaff’s transformation yield

$$\begin{aligned} C_p(x)&= \frac{\Gamma (\lambda )}{\Gamma (p)}x^{\lambda }\int _0^1t^{\lambda -p+1}(1-t)^{p-1}(1+xt)^{-\lambda }\, dt\\&=\frac{\Gamma (\lambda -p+2)}{\lambda (\lambda +1)}x^{\lambda } {}_2F_1\left( \lambda ,\lambda -p+2; \lambda +2 ;-x\right) \\&=\frac{\Gamma (\lambda -p+2)}{\lambda (\lambda +1)}\left( \frac{x}{x+1}\right) ^{\lambda }{}_2F_1\left( \lambda , p; \lambda +2 ;\frac{x}{x+1}\right) . \end{aligned}$$