1 Introduction

Assuming that $$0 < \sum_{m = 1}^{\infty } a_{m}^{2} < \infty$$ and $$0 < \sum_{n = 1}^{\infty } b_{n}^{2} < \infty$$, we have the following Hilbert’s inequality with the best possible constant factor π (cf. [1], Theorem 315):

$$\sum_{m = 1}^{\infty } \sum _{n = 1}^{\infty } \frac{a_{m}b_{n}}{m + n} < \pi \Biggl(\sum _{m = 1}^{\infty } a_{m}^{2} \sum_{n = 1}^{\infty } b_{n}^{2} \Biggr)^{1/2}.$$
(1)

If $$0 < \int _{0}^{\infty } f^{2}(x)\,dx < \infty$$ and $$0 < \int _{0}^{\infty } g^{2}(y) \,dy < \infty$$, then we still have the following integral analogue of (1), named Hilbert’s integral inequality (cf. [1], Theorem 316):

$$\int _{0}^{\infty } \int _{0}^{\infty } \frac{f(x)g(y)}{x + y}\,dx\,dy < \pi \biggl( \int _{0}^{\infty } f^{2}(x)\,dx \int _{0}^{\infty } g^{2}(y) \,dy \biggr)^{1/2},$$
(2)

where the constant factor π is the best possible. Inequalities (1) and (2) play an important role in analysis and its applications (cf. [213]).

The following half-discrete Hilbert-type inequality was provided: If $$K(x)\ (x > 0)$$ is a decreasing function, $$p > 1,\frac{1}{p} + \frac{1}{q} = 1,0 < \phi (s) = \int _{0}^{\infty } K(x)x^{s - 1} \,dx < \infty$$, $$f(x) \ge 0, 0 < \int _{0}^{\infty } f^{p} (x)\,dx < \infty$$, then (cf. [1], Theorem 351)

$$\sum_{n = 1}^{\infty } n^{p - 2}\biggl( \int _{0}^{\infty } K (nx)f(x)\,dx\biggr)^{p} < \phi ^{p}\biggl(\frac{1}{q}\biggr) \int _{0}^{\infty } f^{p} (x)\,dx.$$
(3)

In recent years, some new extensions of (3) were provided by [1419].

In 2006, by using Euler–Maclaurin summation formula, Krnic et al. [20] gave an extension of (1) with the kernel $$\frac{1}{(m + n)^{\lambda }}\ (0 < \lambda \le 4)$$. In 2019, following the result of [20], Adiyasuren et al. [21] considered an extension of (1) involving the partial sums. In 2016–2017, by applying the weight functions, Hong [22, 23] obtained some equivalent statements of the extensions of (1) and (2) with a few parameters. A few similar works were provided by [2438].

In this paper, following the idea of [21], by using the weight functions, the way of introducing parameters and the technique of real analysis, a new multiple Hilbert-type integral inequality with the kernel $$\frac{1}{(x{}_{1} + \cdots + x_{n})^{\lambda }}\ (\lambda > 0)$$ involving the upper limit functions is given. The constant factor related to the gamma function is proved to be the best possible in a condition. A corollary about the case of the nonhomogeneous kernel and some particular inequalities are obtained.

2 Some lemmas

In what follows, we assume that $$n \in \mathrm{N}\backslash \{ 1\}: = \{ 2,3, \ldots \},p_{i},r_{i} > 1\ (i = 1, \ldots,n),\sum_{i = 1}^{n} \frac{1}{p_{i}} = 1,\lambda > 0$$, $$c_{\lambda }: = (1 - \sum_{j = 1}^{n} \frac{1}{r_{j}} )\lambda, f_{i}(x)$$ ($$i = 1, \ldots,n$$) are nonnegative measurable functions in $$R_{ +} = (0,\infty )$$ such that $$f_{i}(x) = o(e^{tx})\ (t > 0;x \to \infty )$$, and for any $$A = (0,a)\ (a > 0), f_{i} \in L^{1}(A)$$, the upper limit functions are defined by $$F_{i}(x): = \int _{0}^{x} f_{i}(t)\,dt\ (x \ge 0)$$, satisfying

$$0 < \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - c_{\lambda } - 1} F_{i}^{p_{i}}(x_{i})\,dx_{i} < \infty \quad (i = 1, \ldots,n).$$

By the definition of the gamma function, for $$x_{i} > 0\ (i = 1, \ldots,n)$$, the following expression holds:

$$\frac{1}{(\sum_{i = 1}^{n} x_{i} )^{\lambda }} = \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } t^{\lambda - 1} e^{ - t\sum _{i = 1}^{n} x_{i}} \,dt.$$
(4)

Lemma 1

For $$t > 0$$, we have the following expressions:

$$\int _{0}^{\infty } e^{ - tx} f_{i}(x) \,dx = t \int _{0}^{\infty } e^{ - tx} F_{i}(x) \,dx\quad (i =,1, \ldots,n),$$
(5)

Proof

In view of $$F_{i}(0) = 0$$,we find

\begin{aligned} \int _{0}^{\infty } e^{ - tx} f_{i}(x) \,dx &= \int _{0}^{\infty } e^{ - tx} \,dF_{i}(x) \\ &= e^{ - tx}F_{i}(x)|_{0}^{\infty } - \int _{0}^{\infty } F_{i} (x) \,de^{ - tx}\\ & = \lim_{x \to \infty } \frac{F_{i}(x)}{e^{tx}} + t \int _{0}^{\infty } e^{ - tx} F_{i}(x) \,dx. \end{aligned}

If $$F_{i}(\infty ) = \mathrm{constant}$$, then $$\lim_{x \to \infty } \frac{F_{i}(x)}{e^{tx}} = 0$$ and (5) follows; if $$F_{i}(\infty ) = \infty$$, since $$f_{i}(x) = o(e^{tx})\ (x \to \infty )$$, we find

\begin{aligned} \int _{0}^{\infty } e^{ - tx} f_{i}(x) \,dx &= \lim_{x \to \infty } \frac{F'_{i}(x)}{(e^{tx})'_{x}} + t \int _{0}^{\infty } e^{ - tx} F{}_{i}(x) \,dx \\ &= \lim_{x \to \infty } \frac{f_{i}(x)}{te^{tx}} + t \int _{0}^{\infty } e^{ - tx} F_{i}(x) \,dx = 0 + t \int _{0}^{\infty } e^{ - tx} F_{i}(x) \,dx, \end{aligned}

and then (5) follows, too.

The lemma is proved. □

Lemma 2

For $$x_{i} > 0\ (i = 1, \ldots,n)$$, the following expression holds:

$$A: = \prod_{i = 1}^{n} \Biggl[ x_{i}^{(\frac{\lambda }{r_{i}} - 1)(1 - p_{i})}\prod_{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr]^{\frac{1}{p_{i}}} = 1.$$
(6)

Proof

We have

\begin{aligned} A &= \prod_{i = 1}^{n} \Biggl[ x_{i}^{(\frac{\lambda }{r_{i}} - 1)(1 - p_{i}) + 1 - \frac{\lambda }{r_{i}}}\prod_{j = 1}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr]^{\frac{1}{p_{i}}} = \prod _{i = 1}^{n} \bigl[ x_{i}^{(\frac{\lambda }{r_{i}} - 1)( - p_{i})} \bigr]^{\frac{1}{p_{i}}} \Biggl( \prod_{j = 1}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr)^{\frac{1}{p_{i}}} \\ &= \prod_{i = 1}^{n} x_{i}^{1 - \frac{\lambda }{r_{i}}} \Biggl( \prod_{j = 1}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr)^{\sum _{i = 1}^{n} \frac{1}{p_{i}}} = \prod_{i = 1}^{n} x_{i}^{1 - \frac{\lambda }{r_{i}}} \Biggl( \prod_{j = 1}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr) = 1, \end{aligned}

and then (6) follows.

The lemma is proved. □

Lemma 3

For $$n \in \mathrm{N}\backslash \{ 1\}$$, defining the following weight functions:

$$\omega _{\lambda }^{(i)}(x_{i}): = x_{i}^{\frac{\lambda }{r_{i}} + c_{\lambda }} \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i} )^{\lambda }} \prod _{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \,dx_{1} \cdots \,dx_{i - 1}\,dx_{i + 1} \cdots \,dx_{n},$$
(7)

we have

$$\omega _{\lambda }^{(i)}(x_{i}) = k_{\lambda }^{(i)}: = \frac{\Gamma (\lambda (1 - \frac{1}{r_{i}}))}{\Gamma (\sum_{j = 1(j \ne i)}^{n} \frac{\lambda }{r_{j}} )} \cdot \frac{\prod_{j = 1}^{n} \Gamma (\frac{\lambda }{r_{j}} )}{\Gamma (\lambda )} \in \mathrm{R}_{ +}\quad (i = 1, \ldots,n).$$
(8)

In particular, for $$\sum_{i = 1}^{n} \frac{1}{r_{i}} = 1$$, we have

$$k_{\lambda }^{(i)} = k_{\lambda }: = \frac{1}{\Gamma (\lambda )}\prod _{j = 1}^{n} \Gamma \biggl(\frac{\lambda }{r_{j}} \biggr)\quad (i = 1, \ldots,n).$$
(9)

Proof

For $$j \ne i$$, setting $$u_{j} = \frac{x_{j}}{x_{i}}$$ in (7), we have

\begin{aligned} \omega _{\lambda }^{(i)}(x_{i}) ={}& \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(u_{1} + \cdots u_{i - 1} + 1 + u_{i + 1} + \cdots + u{}_{n})^{\lambda }} \\ &{}\times \prod _{j = 1(j \ne i)}^{n} u_{j}^{\frac{\lambda }{r_{j}} - 1} \,du_{1} \cdots \,du_{i - 1}\,du_{i + 1} \cdots \,du_{n}. \end{aligned}

Then by Lemma 9.15 and (9.1.19) (cf. [2], p. 341–342), we obtain (8).

The lemma is proved. □

Lemma 4

We have the following inequality:

\begin{aligned} H_{\lambda }&: = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} F_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< \prod_{i = 1}^{n} \biggl[ k_{\lambda }^{(i)} \int _{0}^{\infty } x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr]^{\frac{1}{p_{i}}}. \end{aligned}
(10)

Proof

By (6) and Hölder’s integral inequality (cf. [39]), we obtain

\begin{aligned} H_{\lambda } ={}& \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} \Biggl[ x_{i}^{(\frac{\lambda }{r_{i}} - 1)(1 - p_{i})} \prod_{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \Biggr]^{\frac{1}{p_{i}}}F_{i} (x_{i})\,dx_{1} \cdots \,dx_{n} \\ \le{}& \prod_{i = 1}^{n} \Biggl\{ \int _{0}^{\infty } \Biggl[ \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} x_{i}^{\frac{\lambda }{r_{i}} + c_{\lambda }} \prod_{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} \,dx_{1} \cdots \,dx_{i - 1} \,dx_{i + 1} \cdots \,dx_{n} \Biggr] \\ &{} \times x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \Biggr\} ^{\frac{1}{p_{i}}} \\ = {}&\prod_{i = 1}^{n} \biggl[ \int _{0}^{\infty } \omega _{\lambda }^{(i)}(x_{i})x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr]^{\frac{1}{p_{i}}}. \end{aligned}
(11)

If (11) takes the form of an equality, then there exist constants $$C_{i},C_{k}\ (i \ne k)$$ such that they are not all zero and

\begin{aligned} &C_{i}x_{i}^{\frac{\lambda }{r_{i}} + c_{\lambda }} \prod _{j = 1(j \ne i)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \\ &\quad = C_{k}x_{k}^{\frac{\lambda }{r_{i}} + c_{\lambda }} \prod _{j = 1(j \ne k)}^{n} x_{j}^{\frac{\lambda }{r_{j}} - 1} x_{k}^{p_{k}(1 - \frac{\lambda }{r_{k}}) - c_{\lambda } - 1}F_{k}^{p_{k}}(x_{k})\quad\text{a.e. in } \mathrm{R}_{ +}. \end{aligned}

namely, $$C_{i}x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}})}F_{i}^{p_{i}}(x_{i}) = C_{k}x_{k}^{p_{k}(1 - \frac{\lambda }{r_{k}})}F_{k}^{p_{k}}(x_{k}) = C\text{ a.e. in }\mathrm{R}_{ +}$$. Assuming that $$C_{i} \ne 0$$, we have

$$x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) = \frac{C}{C_{i}}x_{i}^{ - c_{\lambda } - 1}\quad \text{a.e. in }\mathrm{R}_{ +},$$

which contradicts the fact that $$0 < \int _{0}^{\infty } x_{i}^{p_{i}(1 - \frac{\lambda }{r_{i}}) - c_{\lambda } - 1} F_{i}^{p_{i}}(x_{i})\,dx_{i} < \infty$$, in view of $$\int _{0}^{\infty } x_{i}^{ - c_{\lambda } - 1} \,dx_{i} = \infty$$. Then by (8) and (11), we have (10).

The lemma is proved. □

Remark 1

Replacing λ (resp. $$\frac{\lambda }{r_{i}}$$) by $$\lambda + n$$ (resp. $$\frac{\lambda }{r_{i}} + 1$$) in (10), we have

\begin{aligned} H_{\lambda + n}& = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda + n}} \prod _{i = 1}^{n} F_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< \prod_{i = 1}^{n} \biggl( \tilde{k}_{\lambda + n}^{(i)} \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr)^{\frac{1}{p_{i}}}, \end{aligned}
(12)

where we denote

$$\tilde{k}_{\lambda + n}^{(i)}: = \frac{\Gamma (\lambda (1 - \frac{1}{r_{i}}) + n - 1)}{\Gamma (\sum_{j = 1(j \ne i)}^{n} (\frac{\lambda }{r_{j}} + 1))} \cdot \frac{\prod_{j = 1}^{n} \Gamma (\frac{\lambda }{r_{j}} + 1 )}{\Gamma (\lambda + n)} \in \mathrm{R}_{ +} \quad(i = 1, \ldots,n).$$

3 Main results and a corollary

Theorem 1

We have the following inequality:

\begin{aligned} I&: = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}\prod_{i = 1}^{n} \biggl( \tilde{k}_{\lambda + n}^{(i)} \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - c_{\lambda } - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr)^{\frac{1}{p_{i}}}. \end{aligned}
(13)

In particular, for $$\sum_{i = 1}^{n} \frac{1}{r_{i}} = 1$$, we have

$$0 < \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - 1} F_{i}^{p_{i}}(x_{i})\,dx_{i} < \infty\quad (i = 1, \ldots,n),$$

and the following inequality:

\begin{aligned} I &= \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma \biggl(\frac{\lambda }{r_{i}}\biggr) \biggl( \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr)^{\frac{1}{p_{i}}}. \end{aligned}
(14)

Proof

By (4) and (5), we have

\begin{aligned} I& = \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } \cdots \int _{0}^{\infty } \prod_{i = 1}^{n} f_{i} (x_{i}) \int _{0}^{\infty } t^{\lambda - 1} e^{ - t(x_{1} + \cdots + x_{n})}\,dt \,dx_{1} \cdots \,dx_{n} \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } t^{\lambda - 1} \prod _{i = 1}^{n} \int _{0}^{\infty } e^{ - tx_{i}} f_{i} (x_{i})\,dx_{i}\,dt \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } t^{\lambda + n - 1} \prod _{i = 1}^{n} \int _{0}^{\infty } e^{ - tx_{i}} F_{i} (x_{i})\,dx_{i}\,dt \\ &= \frac{1}{\Gamma (\lambda )} \int _{0}^{\infty } \cdots \int _{0}^{\infty } \prod_{i = 1}^{n} F_{i} (x_{i}) \int _{0}^{\infty } t^{\lambda + n - 1} e^{ - t(x_{1} + \cdots + x_{n})}\,dt \,dx_{1} \cdots \,dx_{n} \\ &= \frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}H_{\lambda + n}. \end{aligned}

Then by (12), we have (13).

The theorem is proved. □

Theorem 2

The constant factor $$\frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}})$$ in (14) is the best possible.

Proof

For any $$0 < \varepsilon < \lambda \min_{1 \le i \le n}\{ \frac{p_{i}}{r_{i}}\}$$, we set

$$\tilde{f}_{i}(x_{i}): = \textstyle\begin{cases} 0,&0 < x_{i} \le 1, \\ x_{i}^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}} - 1},&x_{i} > 1, \end{cases}\displaystyle \quad(i = 1, \ldots,n).$$

We obtain that $$\tilde{f}_{i}(x_{i}) = o(e^{tx_{i}})\ (t > 0;x_{i} \to \infty )$$, and $$\tilde{F}_{i}(x_{i}) \equiv 0\ (0 < x_{i} \le 1)$$,

$$\tilde{F}_{i}(x_{i}) = \int _{0}^{x_{i}} \tilde{f}_{i}(t)\,dt = \int _{1}^{x_{i}} t^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}} - 1}\,dt = \frac{x_{i}^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}}} - 1}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}}} < \frac{x_{i}^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}}}}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}}}\quad (x_{i} > 1;i = 1, \ldots,n).$$

If there exists a positive constant $$M(M \le \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}}))$$ such that (14) is valid when replacing $$\frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}})$$ by M, then in particular, by substitution of $$f_{i}(x_{i}) = \tilde{f}_{i}(x_{i})\text{ and }F_{i}(x_{i}) = \tilde{F}_{i}(x_{i})$$, we have

\begin{aligned} \tilde{I}&: = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} \tilde{f}_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &< M\prod_{i = 1}^{n} \biggl( \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - 1} \tilde{F}_{i}^{p_{i}}(x_{i})\,dx_{i} \biggr)^{\frac{1}{p_{i}}} \\ &= M\prod_{i = 1}^{n} \frac{1}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p{}_{i}}} \biggl( \int _{1}^{\infty } x_{i}^{ - \varepsilon - 1} \,dx_{i} \biggr)^{\frac{1}{p_{i}}} = \frac{M}{\varepsilon } \prod _{i = 1}^{n} \frac{1}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p{}_{i}}}.. \end{aligned}

In view of Lemma 9.1.4 (9.1.5) in [2], we find

$$I_{\varepsilon }: = \varepsilon \tilde{I} = \varepsilon \int _{1}^{\infty } \cdots \int _{1}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} x_{i}^{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p_{i}} - 1} \,dx_{1} \cdots \,dx_{n} = k{}_{\lambda } + o(1)\quad \bigl( \varepsilon \to 0^{ +} \bigr).$$

Hence, we have

$$\frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \Gamma \biggl(\frac{\lambda }{r_{i}} \biggr) + o(1) = k{}_{\lambda } + o(1) = \varepsilon \tilde{I} < M\prod_{i = 1}^{n} \frac{1}{\frac{\lambda }{r_{i}} - \frac{\varepsilon }{p{}_{i}}}.$$

For $$\varepsilon \to 0^{ +}$$, we find

$$\frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma \biggl(\frac{\lambda }{r_{i}} \biggr) \le M,$$

which yields that the constant factor $$M = \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}})$$ in (14) is the best possible.

The theorem is proved. □

Setting $$x = \frac{1}{x_{1}},f(x) = x^{\lambda - 2}f_{1}(\frac{1}{x})$$ in I of (14), we have

$$I = \int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{f(x)}{(1 + \sum_{i = 2}^{n} xx_{i})^{\lambda }} \prod _{i = 2}^{n} f_{i} (x_{i})\,dx \,dx_{2} \cdots \,dx_{n}.$$

For $$f_{1}(t) = t^{\lambda - 2}f(\frac{1}{t})$$, we find

$$F_{1}(x_{1}) = \int _{0}^{x_{1}} f_{1} (t)\,dt = \int _{0}^{x_{1}} t^{\lambda - 2}f \biggl( \frac{1}{t}\biggr)\,dt.$$

Then, replacing back x (resp. $$f(x)$$) by $$x_{1}$$ (resp. $$f_{1}(x_{1})$$), we have

Corollary 1

If $$\tilde{F}_{1}(x_{1}) = \int _{0}^{x_{1}} t^{\lambda - 2}f_{1} (\frac{1}{t})\,dt$$,

$$\tilde{F}_{i}(x_{i}): = \int _{0}^{x_{i}} f_{i}(t)\,dt\quad (i = 2, \ldots,n),$$

then we have the following inequality with the nonhomogeneous kernel:

\begin{aligned} &\int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(1 + \sum_{i = 2}^{n} x_{1}x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &\quad< \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma \biggl(\frac{\lambda }{r_{i}}\biggr) \biggl( \int _{0}^{\infty } x_{i}^{ - p_{i}\frac{\lambda }{r_{i}} - 1} \tilde{F}_{i}^{p_{i}}(x_{i})\,dx_{i} \biggr)^{\frac{1}{p_{i}}}, \end{aligned}
(15)

where the constant factor $$\frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{r_{i}}\Gamma (\frac{\lambda }{r_{i}})$$ in (15) is the best possible.

Remark 2

(i) For $$n = 2$$, (14) reduces to (cf. [40])

\begin{aligned} &\int _{0}^{\infty } \int _{0}^{\infty } \frac{f_{1}(x_{1})f_{2}(x_{2})}{(x_{1} + x_{2})^{\lambda }} \,dx_{1} \,dx_{2} \\ &\quad< \frac{\lambda ^{2}}{r_{1}r_{2}}B\biggl(\frac{\lambda }{r_{1}},\frac{\lambda }{r_{2}}\biggr) \biggl( \int _{0}^{\infty } x_{1}^{ - p_{1}\frac{\lambda }{r_{1}} - 1}F_{1}^{p_{1}}(x_{1}) \,dx_{1} \biggr)^{\frac{1}{p_{1}}} \biggl( \int _{0}^{\infty } x_{2}^{ - p_{2}\frac{\lambda }{r_{2}} - 1}F_{2}^{p_{2}}(x_{2}) \,dx_{2} \biggr)^{\frac{1}{p_{2}}}, \end{aligned}
(16)

and (15) reduces to the following new inequality:

\begin{aligned} &\int _{0}^{\infty } \int _{0}^{\infty } \frac{f_{1}(x_{1})f_{2}(x_{2})}{(1 + x_{1}x_{2})^{\lambda }} \,dx_{1} \,dx_{2} \\ &\quad < \frac{\lambda ^{2}}{r_{1}r_{2}}B\biggl(\frac{\lambda }{r_{1}},\frac{\lambda }{r_{2}}\biggr) \biggl( \int _{0}^{\infty } x_{1}^{ - p_{1}\frac{\lambda }{r_{1}} - 1} \tilde{F}_{1}^{p_{1}}(x_{1})\,dx_{1} \biggr)^{\frac{1}{p_{1}}} \biggl( \int _{0}^{\infty } x_{2}^{ - p_{2}\frac{\lambda }{r_{2}} - 1} \tilde{F}_{2}^{p_{2}}(x_{2})\,dx_{2} \biggr)^{\frac{1}{p_{2}}}. \end{aligned}
(17)

(ii) For $$r_{i} = p_{i}\ (i = 1, \ldots,n)$$, (14) reduces to

\begin{aligned} &\int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(\sum_{i = 1}^{n} x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &\quad < \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{p_{i}}\Gamma \biggl(\frac{\lambda }{p_{i}}\biggr) \biggl( \int _{0}^{\infty } x_{i}^{ - \lambda - 1}F_{i}^{p_{i}}(x_{i}) \,dx_{i} \biggr)^{\frac{1}{p_{i}}}, \end{aligned}
(18)

and (15) reduces to

\begin{aligned} &\int _{0}^{\infty } \cdots \int _{0}^{\infty } \frac{1}{(1 + \sum_{i = 2}^{n} x_{1}x_{i})^{\lambda }} \prod _{i = 1}^{n} f_{i} (x_{i}) \,dx_{1} \cdots \,dx_{n} \\ &\quad < \frac{1}{\Gamma (\lambda )}\prod_{i = 1}^{n} \frac{\lambda }{p_{i}}\Gamma \biggl(\frac{\lambda }{p_{i}}\biggr) \biggl( \int _{0}^{\infty } x_{i}^{ - \lambda - 1} \tilde{F}_{i}^{p_{i}}(x_{i})\,dx_{i} \biggr)^{\frac{1}{p_{i}}}. \end{aligned}
(19)

The constant factors in the above inequalities are the best possible.

4 Conclusions

In this paper, following the idea of [21], by the use of the weight functions, the way of introducing parameters and the technique of real analysis, a new multiple Hilbert-type integral inequality with the kernel $$\frac{1}{(x_{1} + \cdots + x_{n})^{\lambda }}\ (\lambda > 0)$$ involving the upper limit functions is given in Theorem 1. In a condition, the best possible constant factor related to the gamma function and a few parameters is proved in Theorem 2. A corollary about the case of nonhomogeneous kernel and some particular inequalities are obtained in Corollary 1 and Remark 2. The lemmas and theorems provide an extensive account of this type of inequalities.