On the Abscissa of Convergence of Laplace–Stieltjes Integrals in the Euclidean Real Vector Space ${\mathbb{R}}^p$

Abstract

New estimates for the convergence abscissas of the multiple Laplace–Stieltjes integral are obtained. There is described the relationship between the integrand function, the Lebesgue–Stieltjes measure, and the abscissa of convergence of the multiple Laplace–Stieltjes integral. Since the multiple Laplace–Stieltjes integral is a direct generalization of the Laplace integral and multiple Dirichlet series, known results about convergence domains for the multiple Dirichlet series are obtained as corollaries of the presented more general statements for the multiple Laplace–Stieltjes integral.

1. Introduction

Let ${\mathbb{R}}^p$ be the Euclidean real vector space and ${\mathbb{R}}_{+}=(0,+\infty ).$ For $a=(a_1,\dots ,a_p)\in {\mathbb{R}}^p$ and $b=(b_1,\dots ,b_p)\in {\mathbb{R}}^p$, we denote

$$\langle a,b\rangle =\sum _{k=1}^p{a}_k{b}_k,|a|={\left(\sum _{k=1}^p{a}_k^2\right)}^{{\displaystyle \frac{1}{2}}},\parallel a\parallel =\sum _{k=1}^p{a}_k,$$

and we will write $a<b$ and $a\le b$ in the case that $a_k<b_k$ for all $k,1\le k\le p$ or $a_k<b_k$ for all $k,1\le k\le p$, respectively.

We suppose that $\nu$ is a countably additive non-negative measure on ${\mathbb{R}}_{+}^p$ with unbounded $\mathrm{supp}\nu$, and that $f\left(x\right)$ is an arbitrary non-negative $\nu$-measurable function on ${\mathbb{R}}_{+}^p.$ By $\nu \left(E\right)$, we denote the $\nu$-measure of a $\nu$-measurable set $E\subset {\mathbb{R}}^p.$

Let us denote by $\mathrm{supp}\nu$ the support of a measure $\nu$ in ${\mathbb{R}}^p$, which is a closed set $E\equiv \mathrm{supp}\nu$ such that $\nu ({\mathbb{R}}^p\setminus E)=0$ and $\nu \left(\right\{x\in {\mathbb{R}}^p:|x-x_0|<r\})>0$ for every $x_0\in E$ and $r>0$.

By ${\mathcal{I}}^p\left(\nu \right)$, we denote the class of the functions $F:{\mathbb{R}}^p\to [0,+\infty )$ of the form

$$I\left(\sigma \right)={\int }_{{\mathbb{R}}_{+}^p}f\left(x\right)e^{\langle \sigma ,x\rangle }\nu \left(dx\right),\sigma \in {\mathbb{R}}^p.$$

(1)

Let $V_p$ be the class of non-negative, separately non-decreasing functions in every variable $x_j$ on $[0,+\infty )$ for any non-negative fixed values of other variables. Let $dF={\nu }_F\left(dx\right)$ be the Lebesgue–Stieltjes measure, generated by the function $F\in V_p$, and f be ${\nu }_F$-measurable and non-negative on the ${\mathbb{R}}_{+}^p$ function.

Note that many results concerning convergence were obtained in the case $p=1$ at first for the Stieltjes integrals of the form

$$\underset{0}{\overset{+\infty }{\int }}{e}^{st}d\alpha \left(t\right),s=\sigma +it$$

(2)

with a different definition of the function $\alpha \left(t\right).$ In Knopp, 1950 [1], Ugaheri, 1951 [2], and Delange, 1957–1958 [3,4], the authors investigated the convergence of such integrals and obtained formulas for the abscissas of convergence. Bosanguet [5] and Mayer-Kalkschmidt [6] studied summability of the Laplace–Stieltjes integrals. In 1989, Miskelevicius [7,8] obtained an analogue of Riesz’ theorem on the behavior on the boundary of the domain of convergence of the integral (2) and an analogue of Tauber’s theorem which gives necessary and sufficient conditions of the convergence of such an integral at a point of the boundary of its domain of convergence. Similarly, Tauber’s theorem was also deduced in [9]. Das [10] introduced a generalized Laplace–Stieltjes integral. Later, in 1982, Voigt [11] considered the Laplace transform of vector-valued measures and obtained conditions when the abscissa of convergence is the same for the uniform, strong, and weak operator topologies. Vang [12,13], for the analytic function, defined by a Laplace–Stieltjes transform, introduced the $P\left(R\right)$ type $T\left(p\right)$ and the lower $P\left(R\right)$ type $t\left(p\right)$ and discussed the growth of functions of the same $P\left(R\right)$ order under specific conditions. Among recent investigations concerning the Laplace–Stieltjes integral of form (1), it is worth mentioning several papers by Sheremeta, Skaskiv, Posiko, and Zadorozhna [14,15,16,17]. In these articles, one can find results concerning domains of the convergence of the Laplace–Stieltjes integral of form (1). Additionally, the authors of [18] investigated a connection between the growth of the logarithms of the Laplace–Stiltjes integral and the maximum of its integrand. Its partial logical continuation is reference [19], which is devoted to the logarithmic growth of entire functions represented by Laplace–Stieltjes transforms of zero order.

The generalized convergence classes for the Laplace–Stiltjes integral were defined in [20]. The authors presented conditions under which the integral of the Laplace–Stieltjes type belongs to a specific class.

Other researchers considered various approximation problems concerning Laplace–Stieltjes transforms [21,22,23,24] and the growth of these functions [25,26,27]. A description of the growth of analytic functions defining the Laplace–Stieltjes transform by the generalized orders and types was presented in [28,29]. Moreover, the concepts of pseudostarlikeness and pseudoconvexity were also introduced for Laplace–Stieltjes integrals [30]. Reference [31] contains a coherent and rigorous overview of the Laplace–Stieltjes transform on $H^p$ spaces.

The most significant contribution in theory of the Laplace–Stieltjes integral was made by M. Sheremeta and his co-authors over the last decade. They investigated the following topics concerning the Laplace–Stieltjes integral: the Banach spaces [32,33], the growth of series in systems of functions [34,35], its application to the lacunary power series [36], and lower and upper estimates for the integral [37,38,39].

In reference [40], some new estimates were obtained for the abscissa of the convergence of the integral (1) and, therefore, for the Laplace integrals and Dirichlet series. Moreover, we analyze connections between our results and those previously known. The results from articles [14,15,16,17] follow from the statements in [40].

Below, in Section 2, “Some results about integrals in the case $p=1$”, we present formulated propositions from [16]. They present conditions providing an equality between an abscissa of the convergence of the Laplace–Stieltjes integral and an abscissa of the existence of the maximum of the integrand of the Laplace-Stieltjes integral. Section 3, “New Estimates”, is devoted to the relationship between the abscissa of the convergence and the new characteristic ${\tau }_{*}$. It is defined as the lower limit over the support of the sum of the logarithm of a certain function and the logarithm of the Laplace–Stilles integral given by this function. Section 4, “Some results for multiple integrals”, presents new statements which are counterparts for multiple integrals of corresponding one-dimensional statements.

2. Some Results About Integrals in the Case $\mathbf{p}=\mathbf{1}$

The number ${\sigma }_{\mathrm{c}}$ is called [1,16,40] the abscissa of convergence of integral (1) if it converges for all $\sigma <{\sigma }_{\mathrm{c}}$ and diverges for all $\sigma >{\sigma }_{\mathrm{c}}.$ If the integral (2) converges for all $\sigma \in \mathbb{R},$ then we obtain ${\sigma }_{\mathrm{c}}=+\infty$; if it diverges for all $\sigma \in \mathbb{R},$ then we obtain ${\sigma }_{\mathrm{c}}=-\infty .$

The number ${\sigma }_a$ is called [1,16] the abscissa of absolute convergence of integral (2) if it is absolutely convergent for all $\sigma <{\sigma }_a$ and is not absolutely convergent for all $\sigma >{\sigma }_a.$ If this integral absolutely converges for all $\sigma \in \mathbb{R},$ then we obtain ${\sigma }_a=+\infty$; if it is not absolutely convergent for all $\sigma \in \mathbb{R},$ then we obtain ${\sigma }_a=-\infty .$

Knopp [1] considered integrals of the form

$$I_1\left(\sigma \right)=\underset{0}{\overset{+\infty }{\int }}{e}^{\sigma x}d\chi \left(x\right),$$

(3)

where $\chi$ is the function of bounded variation on every finite interval $[0,A],A>0.$ For such integrals, Knopp [1] obtained the following formula to calculate the abscissa of the convergence of integral (3):

$${\sigma }_{\mathrm{c}}=-\underset{n\to +\infty }{\overline{lim}}{\displaystyle \frac{ln{\chi }_n}{n}},$$

(4)

where ${\chi }_n=sup\{|\chi \left(x\right)-\chi \left(n\right)|:n\le x\le n+1\}.$

Let $S=\mathrm{supp}dF\cap \mathrm{supp}f$, where $\mathrm{supp}f=\{x\in [0,+\infty ):f(x)>0\}$ be a support of the function $f,$$\mathrm{supp}dF$ is a support of the Lebesque–Stieltjes measure $dF$, and $\mu \left(\sigma \right)=\mu (\sigma ,I)=\mathrm{ess}sup\{f\left(x\right)e^{\sigma x}:x\in S\},$$\sigma \in \mathbb{R}$ is a supremum of the integrand of the integral (1) $(p=1)$.

The number ${\sigma }_{\mu }$ is called [16,40] an abscissa of the existence of the maximum of the integrand of integral (1) if $\mu \left(\sigma \right)<+\infty$ for all $\sigma <{\sigma }_{\mu }$ and $\mu \left(\sigma \right)=+\infty$ for all $\sigma >{\sigma }_{\mu }.$ If $\mu \left(\sigma \right)<+\infty$ for all $\sigma \in \mathbb{R},$ then we obtain ${\sigma }_{\mu }=+\infty$; if $\mu \left(\sigma \right)=+\infty$ for all $\sigma \in \mathbb{R},$ then we obtain ${\sigma }_{\mu }=-\infty .$

It is easy to prove [16,40] that

$${\sigma }_{\mu }=\underline{\alpha }:=\underset{x\to +\infty ,x\in S}{\underline{lim}}{\displaystyle \frac{1}{x}}ln{\displaystyle \frac{1}{f\left(x\right)}}.$$

The following relations between ${\sigma }_{\mathrm{c}}$ and ${\sigma }_{\mu }$ were obtained in [16].

Proposition 1.

Let $F\in V$ and $\underset{x\to +\infty }{\overline{lim}}{\displaystyle \frac{lnF\left(x\right)}{x}}\stackrel{def}{=}\overline{\tau }.$ If ${\sigma }_{\mu }<+\infty$ or $\overline{\tau }<+\infty ,$ then ${\sigma }_{\mathrm{c}}\ge {\sigma }_{\mu }-\overline{\tau }.$

Proposition 2.

Let $F\in V$ and there exist $\underset{x\to +\infty }{lim}{\displaystyle \frac{1}{x}}ln{\displaystyle \frac{1}{f\left(x\right)}}={\sigma }_{\mu }.$ If $lnF\left(x\right)=o\left(x\right),$ or $lnF\left(x\right)=o(lnf(x\left)\right)(x\to +\infty ),$ then ${\sigma }_{\mathrm{c}}={\sigma }_{\mu }.$

Remark 1.

Propositions 1 and 2 can be also obtained from Formula (4).

3. New Estimates

In reference [40], new estimates were found between the integrand function, the Lebesgue–Stieltjes measure, and the abscissa of convergence of the Laplace–Stieltjes integral. Now, we proceed to our results. Denote

$${\tau }_{*}=\underset{x\to +\infty ,x\in S}{\underline{lim}}{\displaystyle \frac{1}{x}}\left[lnf\left(x\right)+lnF(x-0)\right].$$

Proposition 3.

If

$$(\forall \epsilon >0):\underset{x_0}{\overset{+\infty }{\int }}{e}^{\epsilon x}{\displaystyle \frac{dF\left(x\right)}{F(x-0)}}=+\infty ,$$

(5)

then ${\sigma }_{\mathrm{c}}\le -{\tau }_{*}.$

For $h>0$, denote

$${\tau }^{*}=\underset{x\to +\infty ,x\in S}{\overline{lim}}{\displaystyle \frac{1}{x}}\left[lnf\left(x\right)+hlnF(x+0)\right].$$

Proposition 4.

If

$$(\forall \epsilon >0):\underset{x_0}{\overset{+\infty }{\int }}{e}^{-\epsilon x}{\displaystyle \frac{dF\left(x\right)}{{\left(F(x+0)\right)}^h}}<+\infty ,$$

(6)

then ${\sigma }_{\mathrm{c}}\ge -{\tau }^{*}.$

Remark 2.

From Proposition 4, we can obtain Proposition 1.

Remark 3.

From Propositions 3 and 4, we obtain Proposition 2 in the case when $lnF\left(x\right)=o\left(x\right)$ as $x\to +\infty .$

Corollary 1.

(i) 

If the function $F\left(x\right)$ is unbounded, then

$$\begin{array}{c}\underset{h\to 1+0}{lim}\underset{x\to +\infty ,x\in S}{\underline{lim}}{\displaystyle \frac{1}{x}}\left[ln{\displaystyle \frac{1}{f\left(x\right)}}-hlnF(x+0)\right]\le {\sigma }_{\mathrm{c}}\\ \le \underset{x\to +\infty ,x\in S}{\overline{lim}}{\displaystyle \frac{1}{x}}\left[ln{\displaystyle \frac{1}{f\left(x\right)}}-lnF(x-0)\right].\end{array}$$

(ii) 

If the function $F\left(x\right)$ is bounded, then

$$\underset{x\to +\infty ,x\in S}{\underline{lim}}{\displaystyle \frac{1}{x}}ln{\displaystyle \frac{1}{f\left(x\right)}}\le {\sigma }_{\mathrm{c}}.$$

Moreover, if

$$(\forall \epsilon >0):{\int }_{x_0}^{+\infty }{e}^{x\epsilon }dF\left(x\right)=+\infty ,$$

(7)

then

$${\sigma }_{\mathrm{c}}\le \underset{x\to +\infty ,x\in S}{\overline{lim}}{\displaystyle \frac{1}{x}}ln{\displaystyle \frac{1}{f\left(x\right)}}.$$

(8)

Note that inequalities in Propositions 2 and 3 are sharp [40].

Since the Laplace–Stieltjes integral is a direct generalization of the Laplace integral and the Dirichlet series, we provide some results for this objects.

Corollary 2.

For the abscissa of convergence ${\sigma }_{\mathrm{c}}$ of the Laplace integral of the form $\underset{0}{\overset{+\infty }{\int }}f\left(x\right)e^{\sigma x}dx$, it holds that

$$\underset{x\to +\infty ,x\in \mathrm{supp}f}{\underline{lim}}{\displaystyle \frac{1}{x}}ln{\displaystyle \frac{1}{f\left(x\right)}}\le {\sigma }_{\mathrm{c}}\le \underset{x\to +\infty ,x\in \mathrm{supp}f}{\overline{lim}}{\displaystyle \frac{1}{x}}ln{\displaystyle \frac{1}{f\left(x\right)}}.$$

Let $\left({\lambda }_n\right)$ be a sequence of non-negative numbers such that $0={\lambda }_0<{\lambda }_n\uparrow +\infty (1\le n\uparrow +\infty )$ and $\left(a_n\right)$ be a sequence of complex numbers. Consider a Dirichlet series of the form

$$\sum _{n=0}^{+\infty }{a}_n{e}^{z{\lambda }_n}.$$

(9)

For $\alpha \in [0,1]$, let

$$F_{\alpha }\left(x\right)=\sum _{{\lambda }_n\le x}|a_n{|}^{\alpha },f\left({\lambda }_n\right)={|a_n|}^{1-\alpha }.$$

Then the Laplace–Stieltjes integral transforms into the Dirichlet series; that is,

$$I_0\left(\sigma \right)=\underset{0}{\overset{+\infty }{\int }}f\left(x\right)e^{\sigma x}d{F}_{\alpha }\left(x\right)=\sum _{n=0}^{+\infty }|a_n|e^{\sigma {\lambda }_n}.$$

(10)

At first, note that the abscissa of the absolute convergence of the Dirichlet series (10) is the same as that of the abscissa of convergence of the Laplace–Stieltjes integral $I_0\left(\sigma \right).$ So, we can use the previous results for the integral (1) with $p=1$.

Corollary 3.

Let $\alpha \in [0,1]$ such that ${\displaystyle \sum _{n=0}^{+\infty }}{|a_n|}^{\alpha }=+\infty .$ Then, for the abscissa of absolute convergence ${\sigma }_a$ of the Dirichlet series (9), it holds that

$$\begin{array}{c}\underset{h\to 1+0}{lim}\underset{n\to +\infty }{\underline{lim}}\left[{\displaystyle \frac{1-\alpha }{{\lambda }_n}}ln{\displaystyle \frac{1}{|a_n|}}-{\displaystyle \frac{h}{{\lambda }_n}}ln\sum _{k=0}^n{|a_k|}^{\alpha }\right]\le {\sigma }_a\\ \le \underset{n\to +\infty }{\overline{lim}}\left[{\displaystyle \frac{1-\alpha }{{\lambda }_n}}ln{\displaystyle \frac{1}{|a_n|}}-{\displaystyle \frac{1}{{\lambda }_n}}ln\sum _{k=0}^{n-1}{|a_k|}^{\alpha }\right].\end{array}$$

In particular, ${\sigma }_a=0$ when ${\sum }_{n=0}^{+\infty }|a_n|=+\infty$ and $\beta =\underset{n\to +\infty }{\overline{lim}}{\displaystyle \frac{1}{{\lambda }_n}}ln{\displaystyle \sum _{k=0}^n}|a_k|=0$ (concerning the last property and other similar proposition on abscissas of convergence, one can find these in [40,41].

Denote

$$\underline{\tau }=\underset{n\to +\infty }{\underline{lim}}{\displaystyle \frac{lnn}{{\lambda }_n}}.$$

Then, from Corollary 3, we immediately obtain

Corollary 4.

For the abscissa of absolute convergence of the Dirichlet series (9), it holds that

$$\underset{n\to +\infty }{\underline{lim}}{\displaystyle \frac{1}{{\lambda }_n}}ln{\displaystyle \frac{1}{a_n}}-\tau \le {\sigma }_a\le \underset{n\to +\infty }{\overline{lim}}{\displaystyle \frac{1}{{\lambda }_n}}ln{\displaystyle \frac{1}{a_n}}-\underline{\tau }.$$

4. Some Results for Multiple Integrals

We consider the integral of form (1) with $\nu ={\nu }_F=dF,$$p\ge 2.$

It is easy to see that if $I({\sigma }^{\left(0\right)}<+\infty$ for some ${\sigma }^{\left(0\right)}$, then $I\left(\sigma \right)<+\infty$ for all $\sigma \le {\sigma }^{\left(0\right)}.$

We recall that ${\sigma }_c=({\sigma }_{c1},\dots ,{\sigma }_{cp})$ is the conjugate abscissas of the convergence of integral (1) with $\nu =dF$, if $I\left(\sigma \right)<+\infty$ in the domain $\{({\sigma }_1,\dots ,{\sigma }_p):{\sigma }_1<{\sigma }_{c1},\dots ,{\sigma }_p<{\sigma }_{cp}\}$ and every domain of the form $\{({\sigma }_1,\dots ,{\sigma }_p):{\sigma }_1<{\sigma }_1^0,\dots ,{\sigma }_p<{\sigma }_p^0\},$ ${\sigma }_i^0>{\sigma }_{ci},i\in I$ and ${\sigma }_j^0\ge {\sigma }_{cj},j\in J,$ where $I\bigsqcup J=\{1,\dots ,p\}$ contains the points ${\sigma }^{\left(0\right)}$ such that $I\left({\sigma }^{\left(0\right)}\right)=+\infty$. As it is easy to understand, the domain, the boundary of which consists of conjugate abscissas of convergence, is convex. In the future, we will call this area the convergence region of the integral (1) with $\nu =dF$ and denote it by $G_{\mathrm{conv}}.$ Actually, $G_{\mathrm{conv}}$ is the interior of the set of those points $\sigma ,$ in which $I\left(\sigma \right)<+\infty .$

In the following, we will assume that $I\left(\mathbf{0}\right)<+\infty .$

Similarly to the case where $p=1$, we denote $\mathrm{supp}f=\{x\in {\mathbb{R}}_{+}^p\cup \left\{\mathbf{0}\right\}:f\left(x\right)>0\}$ and $S=\mathrm{supp}dF\cap \mathrm{supp}f.$

For $t>0$, let us also denote

$$F_0\left(t\right):={\nu }_F\{x=(x_1,\dots ,x_p)\in {\mathbb{R}}^p:\parallel x\parallel <t\}$$

the one-dimensional distribution function of the Lebesgue–Stieltjes measure ${\nu }_F,$ generated by the function $F:{\mathbb{R}}_{+}^p\cup \left\{\mathbf{0}\right\}\to {\mathbb{R}}_{+}.$

For $x=(x_1,\dots ,x_p)\in {\mathbb{R}}^p$, let us denote

$$E_{-}\left(x\right):=\left\{t\in {\mathbb{R}}^p:t<x\right\},$$

$$F_{-}\left(x\right):=\underset{t\to x,t\in E_{-}\left(x\right)}{\overline{lim}}F\left(t\right).$$

For a set $A\subset {\mathbb{R}}_{+}^p$, we denote

$$D\left(A\right):=\bigcup _{\tau \in \partial A}{E}_{-}(-\tau ),$$

and

$$\underline{A}\stackrel{def}{=}\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{\stackrel{||x||\to +\infty ,}{x\in S}}{\underline{lim}}{\displaystyle \frac{1}{\langle \tau ,x\rangle }}\left(lnf\left(x\right)+ln{F}_{-}\left(x\right)\right)\ge 1\right\}.$$

(11)

Proposition 5.

If

$$(\forall \epsilon \in {\mathbb{R}}_{+}^p):\underset{||x||\ge x_0}{\int }{e}^{\langle \epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{F_{-}\left(x\right)}}=+\infty ,$$

then $G_{\mathrm{conv}}\subset D\left(\underline{A}\right).$

Proof. 

If $\tau \in \partial \underline{A},$ then, for $\epsilon \in {\mathbb{R}}_{+}^p$, we obtain

$$lnf\left(x\right)\ge \langle \tau -\epsilon ,x\rangle -ln{F}_{-}\left(x\right)(||x||\ge x_0,x\in S).$$

So, for $\sigma =-\tau +2\epsilon$,

$$\underset{x:\parallel x\parallel \ge x_0}{\int }f\left(x\right)e^{\langle \sigma ,x\rangle }dF\left(x\right)\ge \underset{x:\parallel x\parallel \ge x_0}{\int }{e}^{\langle \epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{F_{-}\left(x\right)}}=+\infty ,$$

that is, $I(-\tau +2\epsilon )=+\infty$; hence, by the arbitrariness of $\epsilon \in {\mathbb{R}}_{+}^p,$ we obtain $G_{\mathrm{conv}}\subset D\left(\underline{A}\right).$ □

For $x=(x_1,\dots ,x_p)\in {\mathbb{R}}^p$, we denote

$$E_{+}\left(x\right):=\left\{t\in {\mathbb{R}}^p:t>x\}\right\},$$

$$F_{+}\left(x\right):=\underset{t\to x,t\in E_{+}\left(x\right)}{\overline{lim}}F\left(t\right).$$

For $h>0$, let us denote

$${\overline{A}}_h:=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{\stackrel{||x||\to +\infty ,}{x\in S}}{\overline{lim}}{\displaystyle \frac{1}{\langle \tau ,x\rangle }}\left(lnf\left(x\right)+hln{F}_{+}\left(x\right)\right)\le 1\right\}.$$

(12)

Proposition 6.

If

$$(\exists h>0)(\forall \epsilon \in {\mathbb{R}}_{+}^p):\underset{||x||\ge x_0}{\int }{e}^{\langle -\epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{{\left(F_{+}\left(x\right)\right)}^h}}<+\infty ,$$

then $G_{\mathrm{conv}}\supset D\left({\overline{A}}_h\right).$

Proof. 

If $\tau \in \partial {\overline{A}}_h,$ then, for $\epsilon \in {\mathbb{R}}_{+}^p$, we have

$$lnf\left(x\right)\le \langle \tau +\epsilon ,x\rangle -hln{F}_{+}\left(x\right)(||x||\ge x_0,x\in S).$$

Therefore, for $\sigma =-\tau -2\epsilon$,

$$\underset{x:\parallel x\parallel \ge x_0}{\int }f\left(x\right)e^{\langle \sigma ,x\rangle }dF\left(x\right)\le \underset{x:\parallel x\parallel \ge x_0}{\int }{e}^{\langle -\epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{{\left(F_{+}\left(x\right)\right)}^h}}<+\infty ,$$

i.e., $I(-\tau -2\epsilon )<+\infty .$ So, due to the arbitrariness of the choice $\epsilon \in {\mathbb{R}}_{+}^p,$ we obtain $G_{\mathrm{conv}}\supset D\left({\overline{A}}_h\right).$ □

We now denote

$${\underline{A}}^0:=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{\stackrel{||x||\to +\infty ,}{x\in S}}{\underline{lim}}{\displaystyle \frac{1}{\langle \tau ,x\rangle }}\left(lnf\left(x\right)+ln{F}_0(\parallel x\parallel -0)\right)\ge 1\right\},$$

$${\overline{A}}_h^0:=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{\stackrel{||x||\to +\infty ,}{x\in S}}{\overline{lim}}{\displaystyle \frac{1}{\langle \tau ,x\rangle }}\left(lnf\left(x\right)+hln{F}_0(\parallel x\parallel +0)\right)\le 1\right\},h>0.$$

Proposition 7.

(i) 

If

$$(\forall \epsilon \in {\mathbb{R}}_{+}^p):\underset{||x||\ge x_0}{\int }{e}^{\langle \epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{F_0(\parallel x\parallel -0)}}=+\infty ,$$

then $G_{\mathrm{conv}}\subset D\left({\underline{A}}^0\right).$

(ii) 

If

$$(\exists h>0)(\forall \epsilon \in {\mathbb{R}}_{+}^p):\underset{||x||\ge x_0}{\int }{e}^{\langle -\epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{(F_0(\parallel x\parallel +{0\left)\right)}^h}}<+\infty ,$$

then $G_{\mathrm{conv}}\supset D\left({\overline{A}}_h^0\right).$

Proof. 

$\left(i\right)$ The same as in the proof of Proposition 5, we have, if $\tau \in \partial {\underline{A}}^0,$ then for $\epsilon \in {\mathbb{R}}_{+}^p$,

$$lnf\left(x\right)\ge \langle \tau -\epsilon ,x\rangle -ln{F}_0(\parallel x\parallel -0\left)\right(||x||\ge x_0,x\in S),$$

Hence, for $\sigma =-\tau +2\epsilon$,

$$\underset{x:\parallel x\parallel \ge x_0}{\int }f\left(x\right)e^{\langle \sigma ,x\rangle }dF\left(x\right)\ge \underset{x:\parallel x\parallel \ge x_0}{\int }{e}^{\langle \epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{F_0(\parallel x\parallel -0)}}=+\infty ,$$

that is, $I(-\tau +2\epsilon )=+\infty .$ Therefore, due to the arbitrariness of the choice $\epsilon \in {\mathbb{R}}_{+}^p,$ we obtain $G_{\mathrm{conv}}\subset D\left({\underline{A}}^0\right).$

The proof of $\left(ii\right)$ is similar to the proof of Proposition 6. □

Remark 4.

In the case ${\nu }_F\left({\mathbb{R}}_{+}^p\right)=+\infty ,$ that is, $F_0\left(t\right)\nearrow +\infty$$(t\to +\infty ),$ the conditions of Proposition 7 are fulfilled even without the multipliers $e^{\langle \epsilon ,x\rangle }$ and $e^{-\langle \epsilon ,x\rangle }$, respectively. Namely,

$${\int }_{||x||\ge ||x_0||}{\displaystyle \frac{dF\left(x\right)}{F_0(\parallel x\parallel -0)}}=+\infty ,{\int }_{||x||\ge ||x_0||}{\displaystyle \frac{dF\left(x\right)}{(F_0(\parallel x\parallel +{0\left)\right)}^h}}<+\infty (h>1).$$

Therefore, in this case,

$$D\left({\overline{A}}_h^0\right)\subset G_{\mathrm{conv}}\subset D\left({\underline{A}}^0\right)(h>1),$$

where ${\overline{A}}_h^0$ and ${\underline{A}}^0$ are defined in Proposition 7.

Indeed, we consider $\alpha =(h+1)/\left(2\right(h-1\left)\right)$ and $t_k=min\{t\ge x_0:F_0\left(t\right)\ge k^{\alpha }\},$$k\in \mathbb{N}.$ Thus, $A_k:=\{t\in {\mathbb{R}}_{+}:k^{\alpha }\le F_0\left(t\right)<{(k+1)}^{\alpha }\}\subset$$[t_k,t_{k+1}).$ Since $F\left(x\right)\le F_0(t_{k+1}-0)<{(k+1)}^{\alpha }$ for $\parallel x\parallel \in A_k,$ we get

$$\underset{x_1}{\overset{+\infty }{\int }}{\displaystyle \frac{dF\left(x\right)}{(F_0(\parallel x\parallel +{0\left)\right)}^h}}=\sum _{k=1}^{+\infty }\underset{\parallel x\parallel \in A_k}{\int }{\displaystyle \frac{dF\left(x\right)}{(F_0(\parallel x\parallel +{0\left)\right)}^h}}$$

$$\le \sum _{k=1}^{+\infty }{\displaystyle \frac{{\nu }_F\left(\right\{x:\parallel x\parallel \in A_k\left\}\right)}{k^{\alpha h}}}\le \sum _{k=1}^{+\infty }{\displaystyle \frac{F_0(t_{k+1}-0)}{k^{\alpha h}}}$$

$$\le 2^{\alpha }\sum _{k=1}^{+\infty }{\displaystyle \frac{1}{k^{(h+1)/2}}}<+\infty .$$

On the other hand, we input $B_k:=\{t\in {\mathbb{R}}_{+}:k\le F_0\left(t\right)\le (k+1)\}.$ Then, we obtain

$$\underset{x_1}{\overset{+\infty }{\int }}{\displaystyle \frac{dF\left(x\right)}{F_0(\parallel x\parallel -0)}}=\sum _{k=1}^{+\infty }\underset{\parallel x\parallel \in B_k}{\int }{\displaystyle \frac{dF\left(x\right)}{F_0(\parallel x\parallel -0)}}$$

$$\ge \sum _{k=1}^{+\infty }{\displaystyle \frac{{\nu }_F\left(\right\{x:\parallel x\parallel \in B_{2k}\left\}\right)}{2k+1}}\ge \sum _{k=1}^{+\infty }{\displaystyle \frac{F_0(t_{2k+1}+0)-F_0(t_{2k}-0)}{2k+1}}$$

$$\ge \sum _{k=1}^{+\infty }{\displaystyle \frac{(2k+1)-2k}{2k+1}}=+\infty .$$

Remark 5.

If $0<{\nu }_F\left({\mathbb{R}}_{+}^p\right)<+\infty ,$ then condition $\left(ii\right)$ from Proposition 7 is fulfilled, and condition $\left(i\right)$ from Proposition 7 holds if and only if

$$(\forall \epsilon \in {\mathbb{R}}_{+}^p):\underset{||x||\ge x_0}{\int }{e}^{\langle \epsilon ,x\rangle }dF\left(x\right)=+\infty .$$

The integral (1) with $\nu =dF$ transforms in the multiple Laplace integral if we input $dF(x_1,\dots ,x_p)=d{x}_1\times \dots \times d{x}_p.$ Since, in the case, we have

$$F(x_1,\dots ,x_p)=\prod _{j=1}^p{x}_j,F_0\left(t\right)=c_p{t}^p(t>0),$$

then, from Propositions 5, 6, or 7, we immediately obtain this consequence.

Corollary 5.

For multiple Laplace integral

$$\underset{{\mathbb{R}}_{+}^p}{\int }f\left(x\right)e^{\langle \sigma ,x\rangle }d{x}_1\cdots d{x}_p$$

(13)

the following inclusion, $D\left(\overline{A}\right)\subset G_{\mathrm{conv}}\subset D\left(\underline{A}\right)$, is true, where

$$\underline{A}=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{||x||\to +\infty ,x\in \mathrm{supp}f}{\underline{lim}}{\displaystyle \frac{1}{\langle \tau ,x\rangle }}lnf\left(x\right)\ge 1\right\},$$

$$\overline{A}=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{||x||\to +\infty ,x\in \mathrm{supp}f}{\overline{lim}}{\displaystyle \frac{1}{\langle \tau ,x\rangle }}lnf\left(x\right)\le 1\right\}.$$

Proof. 

For any $\tau \in {\mathbb{R}}_{+}^p$, one has

$$\underset{||x||\to +\infty }{lim}{\displaystyle \frac{1}{\langle \tau ,x\rangle }}ln{F}_0(\parallel x\parallel )=\underset{||x||\to +\infty }{lim}{\displaystyle \frac{p}{\langle \tau ,x\rangle }}ln\parallel x\parallel =0.$$

Hence, in view of item $\left(i\right)$ of Proposition 7, we obtain the desired inclusions. □

Let $\left({\lambda }_k^{\left(i\right)}\right)$ be sequences of non-negative numbers such that $0={\lambda }_0^{\left(i\right)}\le {\lambda }_k^{\left(i\right)}\uparrow +\infty (1\le k\uparrow +\infty ),(i\in \{1,2,\dots ,p\}),$ $\left(a_n\right)$ is a sequence of complex numbers,

$$n=(n_1,n_2,\dots ,n_p),n_i\in {\mathbb{Z}}_{+}=\mathbb{N}\cup \left\{0\right\}(i\in \{1,\dots ,p\}),$$

$$||n||=n_1+\dots +n_p,\langle {\lambda }_n,t\rangle ={\lambda }_{n_1}^{\left(1\right)}{t}_1+\dots +{\lambda }_{n_p}^{\left(p\right)}{t}_p.$$

For $\alpha \in [0,1],x\in {\mathbb{R}}_{+}^p$, we input

$$F\left(x\right)=F^{\alpha }\left(x\right)=\sum _{{\lambda }_n\in E_{-}\left(x\right)}|a_n{|}^{\alpha },f\left(x\right)=f^{\alpha }\left(x\right)={|a_n|}^{1-\alpha }.$$

Then, its Lebesgue–Stieltjes measure for every bounded set $A\subset {\mathbb{R}}_{+}^p$ is evaluated in the following formula:

$${\nu }_{F^{\alpha }}\left(A\right)=\sum _{{\lambda }_n\in A}{|a_n|}^{\alpha }{\delta }_{{\lambda }_n}\left(A\right),$$

where ${\delta }_{\lambda }$ is the unit Dirac measure, which is concentrated at a single point, $\lambda ,$ i.e., ${\delta }_{\lambda }\left(A\right)=1,$ if $\lambda \in A$, and ${\delta }_{\lambda }\left(A\right)=0$ otherwise. The distribution function of the Lebesgue–Stieltjes measure admits the following representation:

$$F_0\left(t\right)=\sum _{\parallel {\lambda }_k\parallel <t}{|a_k|}^{\alpha }(t>0).$$

Then, $I\left(\sigma \right)={\displaystyle \sum _{||n||=0}^{+\infty }}|a_n|e^{\langle \sigma ,{\lambda }_n\rangle }$, and from Proposition 7, in view of Remark 4, we deduce the following corollary.

Corollary 6.

If $\alpha \in [0,1]$ such that ${\displaystyle \sum _{\parallel n\parallel =0}^{+\infty }}{|a_n|}^{\alpha }=+\infty ,$ then for domain of the convergence of the Dirichlet series ${\sum }_{\parallel n\parallel =0}^{+\infty }{a}_n{e}^{\langle z,{\lambda }_n\rangle },$ we have

$$(\forall h>1):D\left({\overline{A}}_h^0\right)\subset G_a\subset D\left({\underline{A}}^0\right),$$

where

$${\displaystyle \underline{A^0}=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{||n||\to +\infty }{\underline{lim}}\frac{1}{\langle \tau ,{\lambda }_n\rangle }\left[(\alpha -1)ln\frac{1}{|a_n|}+ln\sum _{\parallel {\lambda }_k\parallel <\parallel {\lambda }_n\parallel }{|a_k|}^{\alpha }\right]\right\},}$$

$$\dots$$

$$\overline{A_h^0}=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{||n||\to +\infty }{\overline{lim}}{\displaystyle \frac{1}{\langle \tau ,{\lambda }_n\rangle }}\left[(\alpha -1)ln{\displaystyle \frac{1}{|a_n|}}+hln\sum _{\parallel {\lambda }_k\parallel \le \parallel {\lambda }_n\parallel }{|a_k|}^{\alpha }\right]\right\}.$$

Let $\left(a_n\right)$ be a sequence of the complex numbers

$$F\left(x\right)=\sum _{k=1}^p\sum _{{\lambda }_{n_k}^{\left(k\right)}\le x_k}1,f\left(x\right)=\left\{\begin{array}{c}{a}_n,x={\lambda }_n,\hfill \\ 0,x\ne {\lambda }_n.\hfill \end{array}\right.$$

Then, $F\left({\lambda }_n\right)={\displaystyle \sum _{k=1}^p}{n}_k=||n||$ and $I\left(\sigma \right)={\displaystyle \sum _{||n||=0}^{+\infty }}|a_n|e^{\langle \sigma ,{\lambda }_n\rangle }$ are a multiple Dirichlet series with non-negative exponents and coeficients. For a series such as those from Propositions 5 and 6, we obtain the following corollary.

Corollary 7.

$D\left(\overline{A}\right)\subset G_{\mathrm{conv}}\subset D\left(\underline{A}\right),$ where

$$\underline{A}=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{||n||\to +\infty }{\underline{lim}}{\displaystyle \frac{1}{\langle \tau ,{\lambda }_n\rangle }}\left[ln|a_n|+ln||n||\right]\ge 1\right\},$$

$$\overline{A}=\left\{\tau \in {\mathbb{R}}_{+}^p:\underset{||n||\to +\infty }{\overline{lim}}{\displaystyle \frac{1}{\langle \tau ,{\lambda }_n\rangle }}\left[ln|a_n|+hln||n||\right]\le 1\right\}(h>p\ge 2).$$

Proof. 

Since the number of an integer, non-negative number of solutions of the equation $n_1+n_2+\dots +n_p=k$ is equal to $C_{p+k-1}^k={\displaystyle \frac{(p+k-1)!}{k!·(p-1)!}}\le {\left(2k\right)}^{p-1},k\ge 1,$

$$\underset{||x||\ge x_0}{\int }{e}^{\langle -\epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{{\left(F_{+}\left(x\right)\right)}^h}}=\sum _{||n||=n_0}^{+\infty }{e}^{\langle -\epsilon ,{\lambda }_n\rangle }{\displaystyle \frac{1}{{||n||}^h}}$$

$$\le \sum _{k=1}^{+\infty }\sum _{||n||=k}{e}^{\langle -\epsilon ,{\lambda }_n\rangle }{\displaystyle \frac{1}{k^h}}\le 2^{p-1}\sum _{k=1}^{+\infty }{e}^{\langle -\epsilon ,{\lambda }_n\rangle }{\displaystyle \frac{1}{k^{h-p+1}}}<+\infty ,$$

because $h>p.$ Moreover,

$$\underset{||x||\ge x_0}{\int }{e}^{\langle \epsilon ,x\rangle }{\displaystyle \frac{dF\left(x\right)}{F_{-}\left(x\right)}}=\sum _{||n||=n_0}^{+\infty }{e}^{\langle \epsilon ,x\rangle }{\displaystyle \frac{1}{||n||}}=+\infty .$$

So, the conditions of the Propositions 5 and 6 are valid, from where we obtain the desired inclusions. □

Remark 6.

Regarding the domains of convergence of the multiple Dirichlet series, see articles [42,43,44,45,46,47].

5. Discussion

The results presented in this paper are a natural and logical generalization in a sense of the results which were obtained earlier in a one-dimensional case. They also quite fully reflect the convergence conditions of the considered integrals in this article. However, in the case of a one-dimensional Dirichlet series with an arbitrary sequence of pairwise different complex exponents, presented in [48], there are results containing a complete description of the convergence domains. The proofs of the obtained statements are based on a complete description of the domains of existence of the maximal term of the Dirichlet series. In the case of the integrals, the analog of the maximum term of the Dirichlet series is the essential supremum of the integrand. In essence, our considerations in this article are also based on this approach. Therefore, it is natural to assume that the study of the supremum of the integrand can lead to new results in the case of integrals of the Laplace–Stieltjes type with complex-valued or even vector-valued measure $dF$ (see results for a simpler case of the Laplace transform in [11]).