1 Introduction

If \(0 < \sum_{m = 1}^{\infty} a_{m}^{2} < \infty\) and \(0 < \sum_{n = 1}^{\infty} b_{n}^{2} < \infty\), then we have the following discrete Hilbert inequality with the best possible constant factor π ([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)

Assuming that \(0 < \int_{0}^{\infty} f^{2}(x)\,dx < \infty\) and \(0 < \int_{0}^{\infty} g^{2}(y) \,dy < \infty\), we still have the following integral analogue of (1) ([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) are playing an important role in analysis and its applications [2–13].

The following half-discrete Hilbert-type inequality was provided in 1934 ([1], Theorem 351): If \(K(x)\) (\(x > 0\)) is a decreasing function, \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(0 < \phi (s) = \int_{0}^{\infty} K(x)x^{s - 1} \,dx < \infty\), \(f(x) \ge 0\), and \(0 < \int_{0}^{\infty} f^{p} (x)\,dx < \infty\), then

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

In recent years, some new extensions of (3) were given by [14–19].

In 2006, using the Euler–Maclaurin summation formula, Krnic et al. [20] gave an extension of (1) with the kernel \(\frac{1}{(m + n)^{\lambda}}\) (\(0 < \lambda \le 14\)). In 2019, following [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] considered some equivalent statements of the extensions of (1) and (2) with a few parameters. Some similar works were provided in [24–26].

In this paper, following [21, 22], by the use of the weight functions and the idea of introduced parameters, we give a new Hilbert-type integral inequality with the kernel \(\frac{1}{(x + y)^{\lambda}}\) (\(\lambda > 0\)) involving the upper limit functions and the beta and gamma functions. We consider the equivalent statements of the best possible constant factor related to a few parameters. As applications, we obtain a corollary in the case of nonhomogeneous kernel and some particular inequalities.

2 Some lemmas

In what follows, we assume that \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(\lambda > 0\), \(\lambda_{1},\lambda_{2} \in (0,\lambda + 1)\), \(f(x)\) and \(g(y)\) are nonnegative measurable functions in \(R_{ +} = (0,\infty )\), \(f(x) = o(e^{x})\), \(g(y) = o(e^{y})\) (\(x,y \to \infty \)), such that for any \(A = (0,a)\) (\(a > 0\)), \(f,g \in L^{1}(A)\), and the upper limit functions are defined by

$$F(x): = \int_{0}^{x} f(t)\,dt\quad (x \ge 0)\quad \mbox{and} \quad G(y): = \int_{0}^{y} g(t)\,dt\quad (y \ge 0), $$

satisfying

$$0 < \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx < \infty \quad \mbox{and}\quad 0 < \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy < \infty. $$

By the definition of the gamma function, for \(\lambda,x,y > 0\), the following expression holds:

$$ \frac{1}{(x + y)^{\lambda}} = \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} t^{\lambda - 1} e^{ - (x + y)t}\,dt. $$
(4)

Lemma 1

For \(t > 0\), we have the following expressions:

$$\begin{aligned}& \int_{0}^{\infty} e^{ - tx} f(x)\,dx = t \int_{0}^{\infty} e^{ - tx} F(x) \,dx, \end{aligned}$$
(5)
$$\begin{aligned}& \int_{0}^{\infty} e^{ - ty} g(y)\,dy = t \int_{0}^{\infty} e^{ - ty} G(y) \,dy. \end{aligned}$$
(6)

Proof

We find

$$\begin{aligned} \int_{0}^{\infty} e^{ - tx} f(x)\,dx &= \int_{0}^{\infty} e^{ - tx}\, dF(x) \\ &= e^{ - tx}F(x)|_{0}^{\infty} - \int_{0}^{\infty} F (x)\,de^{ - tx} \\ &= \lim _{x \to \infty} \frac{F(x)}{e^{tx}} + t \int_{0}^{\infty} e^{ - tx} F(x)\,dx. \end{aligned}$$

If \(F(\infty ) =\mbox{constant}\), then \(\lim_{x \to \infty} \frac{F(x)}{e^{tx}} = 0\), and (5) follows; if \(F(\infty ) = \infty\), in view of \(f(x) = o(e^{x})\) (\(x \to \infty \)), we find

$$\begin{aligned} \int_{0}^{\infty} e^{ - tx} f(x)\,dx &= \lim _{x \to \infty} \frac{F'(x)}{(e^{tx})'_{x}} + t \int_{0}^{\infty} e^{ - tx} F(x)\,dx \\ &= \lim_{x \to \infty} \frac{f(x)}{te^{tx}} + t \int_{0}^{\infty} e^{ - tx} F(x)\,dx \\ &= 0 + t \int_{0}^{\infty} e^{ - tx} F(x)\,dx, \end{aligned}$$

and then (5) follows. In the same way, we have (6).

The lemma is proved. □

Lemma 2

For \(s > 0,\mu,\sigma \in (0,s)\), define the following weight functions:

$$\begin{aligned}& \varpi (\sigma,x): = x^{s - \sigma} \int_{0}^{\infty} \frac{t^{\sigma - 1}}{(x + t)^{s}}\,dt\quad (x \in \mathrm{R}_{ +} ), \end{aligned}$$
(7)
$$\begin{aligned}& \omega (\mu,y): = y^{s - \mu} \int_{0}^{\infty} \frac{t^{\mu - 1}}{(t + y)^{s}} \,dt\quad (y \in \mathrm{R}_{ +} ). \end{aligned}$$
(8)

We have the following expressions:

$$\begin{aligned}& \varpi (\sigma,x) = B(\sigma,s - \sigma )\quad (x \in \mathrm{R}_{ +} ), \end{aligned}$$
(9)
$$\begin{aligned}& \omega (\mu,y) = B(\mu,s - \mu )\quad (y \in \mathrm{R}_{ +} ), \end{aligned}$$
(10)

where \(B(u,v): = \int_{0}^{\infty} \frac{t^{u - 1}}{(1 + t)^{u + v}} \,dt\) (\(u,v > 0\)) is the beta function satisfying

$$B(u,v) = \frac{1}{\varGamma (u + v)}\varGamma (u)\varGamma (v). $$

Proof

Setting \(u = \frac{t}{x}\), we find

$$\varpi (\sigma,x) = x^{s - \sigma} \int_{0}^{\infty} \frac{(ux)^{\sigma - 1}}{(x + ux)^{s}}x\,du = \int_{0}^{\infty} \frac{u^{\sigma - 1}}{(1 + u)^{s}}\,du = B\quad ( \sigma,s - \sigma ), $$

namely, (9) follows. In the same way, we have (10).

The lemma is proved. □

Lemma 3

Suppose that \(s > 0,\mu,\sigma \in (0,s)\). We have the following inequality:

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{s}} \,dx \le B^{\frac{1}{p}}(\sigma,s - \sigma )B^{\frac{1}{q}}(\mu,s - \mu ) \\& \quad {} \times \biggl[ \int_{0}^{\infty} x^{p(1 - \mu ) - (s - \mu - \sigma ) - 1} f^{p}(x)\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} y^{q(1 - \sigma ) - (s - \mu - \sigma ) - 1} g^{q}(y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$
(11)

For \(\lambda > 0\), \(s = \lambda + 2( > 2)\), \(\lambda_{1} = \mu - 1 \in (0,\lambda + 1)\), \(\lambda_{2} = \sigma - 1 \in (0,\lambda + 1)\), by the substitution \(f(x) = F(x)\)and \(g(y) = G(y)\)in (11) we can reduce it to the following:

$$\begin{aligned}& \begin{gathered}[b] \int_{0}^{\infty} \int_{0}^{\infty} \frac{F(x)G(y)}{(x + y)^{\lambda + 2}} \,dx\,dy < B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\ \quad {} \times \biggl[ \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy\biggr]^{\frac{1}{q}}. \end{gathered} \end{aligned}$$
(12)

Proof

By Hölder’s inequality (see [27]) we obtain

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{s}} \,dx\,dy = \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{(x + y)^{s}} \biggl[ \frac{y^{(\sigma - 1)/p}}{x^{(\mu - 1)/q}}f(x)\biggr] \biggl[\frac{x^{(\mu - 1)/q}}{y^{(\sigma - 1)/p}}g(y)\biggr]\,dx\,dy \\& \quad \le \biggl\{ \int_{0}^{\infty} \biggl[ \int_{0}^{\infty} \frac{1}{(x + y)^{s}} \frac{y^{\sigma - 1}\,dy}{x^{(\mu - 1)(p - 1)}}\biggr]f^{p}(x)\,dx\biggr\} ^{\frac{1}{p}} \\& \qquad {}\times\biggl\{ \int_{0}^{\infty} \biggl[ \int_{0}^{\infty} \frac{1}{(x + y)^{s}} \frac{x^{\mu - 1}\,dx}{y^{(\sigma - 1)(q - 1)}}\biggr]g^{q}(y)\,dy\biggr\} ^{\frac{1}{q}} \\& \quad = \biggl[ \int_{0}^{\infty} \varpi (\sigma,x) x^{p(1 - \mu ) - (\lambda - \mu - \sigma ) - 1}f^{p}(x)\,dx\biggr]^{\frac{1}{p}} \\& \qquad {}\times \biggl[ \int_{0}^{\infty} \omega (\mu,y) y^{q(1 - \sigma ) - (\lambda - \mu - \sigma ) - 1}g^{q}(y) \,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$
(13)

Then by (9) and (10) we have (11).

By simplifications of (11) we have

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{F(x)G(y)}{(x + y)^{\lambda + 2}} \,dx\,dy \le B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\& \quad {}\times \biggl[ \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$
(14)

If (14) keeps the form of equality, then, in view of the proof of (13), there exist constants A and B such that they are not all zero, satisfying for \(s = \lambda + 2\), \(\lambda_{1} = \mu - 1\), \(\lambda_{2} = \sigma - 1\),

$$\begin{aligned}& Ax^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1}x^{(\lambda - \lambda_{1} - \lambda_{2}) + 1}F^{p}(x) \\& \quad = By^{ - q\lambda_{2}}G^{q}(y)\quad \mbox{a.e. in }(0,\infty ) \times (0,\infty ). \end{aligned}$$

Without loss of generality, we assume that \(A \ne 0\). Then for fixed \(y \in (0,\infty )\), we have

$$x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1}F^{p}(x) = \biggl(\frac{B}{A}y^{ - q\lambda_{2}}G^{q}(y) \biggr)x^{ - 1 - (\lambda - \lambda_{1} - \lambda_{2})}\quad \mbox{a.e. in }(0,\infty ), $$

which contradicts the fact that

$$0 < \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx < \infty, $$

since for any \(\lambda - \lambda_{1} - \lambda_{2} \in\mathbf{R}\), \(\int_{0}^{\infty} x^{ - 1 - (\lambda - \lambda_{1} - \lambda_{2})}\,dx = \infty\). Therefore inequality (12) follows.

The lemma is proved. □

3 Main results

Theorem 1

We have the following inequality:

$$\begin{aligned} I : =& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\ &{} \times \biggl[ \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F^{p}(x)\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$
(15)

In particular, for \(\lambda_{1} + \lambda_{2} = \lambda\) (\(\lambda_{1},\lambda_{2} \in (0,\lambda )\)), we reduce it to the following inequality:

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < & \lambda_{1}\lambda_{2}B(\lambda_{1}, \lambda_{2}) \biggl( \int_{0}^{\infty} x^{ - p\lambda_{1} - 1} F^{p}(x)\,dx\biggr)^{\frac{1}{p}} \\ &{}\times\biggl( \int_{0}^{\infty} y^{ - q\lambda_{2} - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}, \end{aligned}$$
(16)

where the constant factor \(\lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})\)is the best possible.

Proof

Using (4), (5), and (6), we find

$$\begin{aligned} I =& \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} \int_{0}^{\infty} f(x)g(y) \biggl( \int_{0}^{\infty} t^{\lambda - 1} e^{ - (x + y)t}\,dt\biggr)\,dx\,dy \\ =& \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} t^{\lambda - 1} \biggl( \int_{0}^{\infty} e^{ - xt}f(x)\,dx\biggr) \biggl( \int_{0}^{\infty} e^{ - yt} g(y)\,dy\biggr) \,dt \\ =& \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} t^{\lambda + 1} \biggl( \int_{0}^{\infty} e^{ - xt}F(x)\,dx\biggr) \biggl( \int_{0}^{\infty} e^{ - yt} G(y)\,dy\biggr) \,dt \\ =& \frac{1}{\varGamma (\lambda )} \int_{0}^{\infty} \int_{0}^{\infty} F(x)G(y) \biggl[ \int_{0}^{\infty} t^{\lambda + 1}e^{ - (x + y)t} \,dt\biggr] \,dx\,dy \\ =& \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )} \int_{0}^{\infty} \int_{0}^{\infty} \frac{F(x)G(y)}{(x + y)^{\lambda + 2}} \,dx\,dy. \end{aligned}$$
(17)

In view of (12), we have (15).

In the case of \(\lambda_{1} + \lambda_{2} = \lambda\) (\(\lambda_{1},\lambda_{2} \in (0,\lambda )\)), we find

$$\begin{aligned}& \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\& \quad = \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1, \lambda_{1} + 1)B^{\frac{1}{q}}(\lambda_{1} + 1, \lambda_{2} + 1) \\& \quad = \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B(\lambda_{1} + 1, \lambda_{2} + 1) = \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}\frac{\varGamma (\lambda_{1} + 1)\varGamma (\lambda_{2} + 1)}{\varGamma (\lambda + 2)} \\& \quad = \lambda_{1}\lambda_{2}\frac{\varGamma (\lambda_{1})\varGamma (\lambda_{2})}{\varGamma (\lambda )} = \lambda_{1}\lambda_{2}B(\lambda_{1}, \lambda_{2}), \end{aligned}$$

and then (16) follows.

For any \(0 < \varepsilon < \min \{ p\lambda_{1},q\lambda_{2}\}\), we set

$$\tilde{f}(t): = \textstyle\begin{cases} 0,&0 < t \le 1, \\ t^{\lambda_{1} - \frac{\varepsilon}{p} - 1},&t > 1, \end{cases}\displaystyle \qquad \tilde{g}(t): = \textstyle\begin{cases} 0,&0 < t \le 1, \\ t^{\lambda_{2} - \frac{\varepsilon}{q} - 1},&t > 1. \end{cases} $$

We obtain that \(\tilde{f}(x) = o(e^{x})\), \(\tilde{g}(y) = o(e^{y})\) (\(x,y \to \infty \)), and \(\tilde{F}(x) = \tilde{G}(y) \equiv 0\) (\(0 < x,y \le 1\)), where

$$\begin{aligned}& \tilde{F}(x) = \int_{0}^{x} \tilde{f}(t)\,dt = \int_{1}^{x} t^{\lambda_{1} - \frac{\varepsilon}{p} - 1}\,dt = \frac{x^{\lambda_{1} - \frac{\varepsilon}{p}} - 1}{\lambda_{1} - \frac{\varepsilon}{p}} < \frac{x^{\lambda_{1} - \frac{\varepsilon}{p}}}{\lambda_{1} - \frac{\varepsilon}{p}}\quad (x > 1), \\& \tilde{G}(y) = \int_{0}^{y} \tilde{g}(t)\,dt = \int_{1}^{y} t^{\lambda_{2} - \frac{\varepsilon}{q} - 1}\,dt = \frac{y^{\lambda_{2} - \frac{\varepsilon}{q}} - 1}{\lambda_{2} - \frac{\varepsilon}{q}} < \frac{y^{\lambda_{2} - \frac{\varepsilon}{q}}}{\lambda_{2} - \frac{\varepsilon}{q}}\quad (y > 1). \end{aligned}$$

If there exists a positive constant M (\(M \le \lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})\)) such that (16) is valid when replacing \(\lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})\) by M, then, in particular, by substitution of \(f(x) = \tilde{f}(x)\) and \(g(y) = \tilde{g}(y)\) we have

$$\tilde{I}: = \int_{0}^{\infty} \int_{0}^{\infty} \frac{\tilde{f}(x)\tilde{g}(y)}{(x + y)^{\lambda}} \,dx\,dy < M \biggl( \int_{0}^{\infty} x^{ - p\lambda_{1} - 1} \tilde{F}^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - q\lambda_{2} - 1} \tilde{G}^{q}(y)\,dy\biggr)^{\frac{1}{q}}. $$

We find

$$\begin{aligned} \tilde{J} : =& \biggl( \int_{0}^{\infty} x^{ - p\lambda_{1} - 1} \tilde{F}^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - q\lambda_{2} - 1} \tilde{G}^{q}(y)\,dy\biggr)^{\frac{1}{q}} \\ < & \frac{1}{(\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})}\biggl[ \int_{1}^{\infty} x^{ - p\lambda_{1} - 1} \bigl(x^{\lambda_{1} - \frac{\varepsilon}{p}}\bigr)^{p}\,dx\biggr]^{\frac{1}{p}}\biggl[ \int_{1}^{\infty} y^{ - q\lambda_{2} - 1} \bigl(y^{\lambda_{2} - \frac{\varepsilon}{q}}\bigr)^{q}\,dy\biggr]^{\frac{1}{q}} \\ =& \frac{1}{(\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})}\biggl( \int_{1}^{\infty} x^{ - \varepsilon - 1} \,dx \biggr)^{\frac{1}{p}}\biggl( \int_{1}^{\infty} y^{ - \varepsilon - 1} \,dy \biggr)^{\frac{1}{q}} \\ =& \frac{1}{(\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})} \int_{1}^{\infty} x^{ - \varepsilon - 1} \,dx = \frac{1}{\varepsilon (\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})}. \end{aligned}$$

In view of the Fubini theorem (see [28]), it follows that

$$\begin{aligned} \tilde{I} =& \int_{1}^{\infty} \biggl[ \int_{1}^{\infty} \frac{y^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(x + y)^{\lambda}} \,dy \biggr]x^{\lambda_{1} - \frac{\varepsilon}{p} - 1}\,dx = \int_{1}^{\infty} x^{ - \varepsilon - 1}\biggl[ \int_{1/x}^{\infty} \frac{u^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du\biggr] \,dx \\ =& \int_{1}^{\infty} x^{ - \varepsilon - 1}\biggl[ \int_{1/x}^{1} \frac{u^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \biggr]\,dx + \int_{1}^{\infty} x^{ - \varepsilon - 1}\biggl[ \int_{1}^{\infty} \frac{u^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \biggr] \,dx \\ =& \int_{0}^{1} \biggl( \int_{1/u}^{\infty} x^{ - \varepsilon - 1} \,dx\biggr) \frac{u^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du + \frac{1}{\varepsilon} \int_{1}^{\infty} \frac{u^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \\ =& \frac{1}{\varepsilon} \biggl[ \int_{0}^{1} \frac{u^{\lambda_{2} + \frac{\varepsilon}{p} - 1}}{(1 + u)^{\lambda}} \,du + \int_{1}^{\infty} \frac{u^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du\biggr]. \end{aligned}$$

So we obtain

$$\int_{0}^{1} \frac{u^{\lambda_{2} + \frac{\varepsilon}{p} - 1}}{(1 + u)^{\lambda}} \,du + \int_{1}^{\infty} \frac{u^{\lambda_{2} - \frac{\varepsilon}{q} - 1}}{(1 + u)^{\lambda}} \,du \le \varepsilon \tilde{I}< \varepsilon M\tilde{J} < \frac{M}{(\lambda_{1} - \frac{\varepsilon}{p})(\lambda_{2} - \frac{\varepsilon}{q})}. $$

As \(\varepsilon \to 0^{ +} \) in this inequality, in view of the continuity of the beta function, we find \(B(\lambda_{1},\lambda_{2}) \le \frac{M}{\lambda_{1}\lambda_{2}}\), namely \(\lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2}) \le M\). Hence \(M = \lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})\) is the best possible constant factor of (14).

The theorem is proved. □

Remark 1

We set \(\hat{\lambda}_{1}: = \lambda_{1} + \frac{\lambda - \lambda_{1} - \lambda_{2}}{p}\), \(\hat{\lambda}_{2}: = \lambda_{2} + \frac{\lambda - \lambda_{1} - \lambda_{2}}{q}\). It follows that \(\hat{\lambda}_{1} + \hat{\lambda}_{2} = \lambda\). For \(\lambda - \lambda_{1} - \lambda_{2}\ \in ( - p\lambda_{1},p(\lambda - \lambda_{1}))\), we find

$$\hat{\lambda}_{1} > \lambda_{1} + \frac{ - p\lambda_{1}}{p} = 0,\qquad \hat{\lambda}_{1} < \lambda_{1} + \frac{p(\lambda - \lambda_{1})}{p} = \lambda, $$

namely, \(0 < \hat{\lambda}_{1} < \lambda\), and then \(0 < \hat{\lambda}_{2} < \lambda\). So we reduce (15) as follows:

$$\begin{aligned} I : =& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\ &{}\times \biggl( \int_{0}^{\infty} x^{ - p\hat{\lambda}_{1} - 1} F^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - q\hat{\lambda}_{2} - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}. \end{aligned}$$
(18)

Theorem 2

If \(\lambda - \lambda_{1} - \lambda_{2} \in ( - p\lambda_{1},p(\lambda - \lambda_{1}))\)and the constant factor

$$\frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) $$

in (18) is the best possible, then \(\lambda_{1} + \lambda_{2} = \lambda\)with \(\lambda_{1},\lambda_{2} \in (0,\lambda )\).

Proof

As regards to the assumptions, we find \(0 < \hat{\lambda}_{1},\hat{\lambda}_{2} < \lambda\). By (16) the unified best possible constant factor in (18) must be of the form

$$\hat{\lambda}_{1}\hat{\lambda}_{2}B(\hat{ \lambda}_{1},\hat{\lambda}_{2}) \biggl(= \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B(\hat{\lambda}_{1} + 1,\hat{ \lambda}_{2} + 1)\biggr), $$

namely, it follows that

$$B(\hat{\lambda}_{1} + 1,\hat{\lambda}_{2} + 1)= B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}). $$

By Hölder’s inequality (see [27]) we obtain

$$\begin{aligned} B(\hat{\lambda}_{1} + 1,\hat{\lambda}_{2} + 1) =& \int_{0}^{\infty} \frac{u^{(\hat{\lambda}_{1} + 1) - 1}}{(1 + u)^{\lambda + 2}}\,du \\ =& \int_{0}^{\infty} \frac{1}{(1 + u)^{\lambda + 2}}u^{\frac{\lambda + 1 - \lambda_{2}}{p} + \frac{\lambda_{1} + 1}{q} - 1} \,du = \int_{0}^{\infty} \frac{1}{(1 + u)^{\lambda + 2}} \bigl(u^{\frac{\lambda - \lambda_{2}}{p}}\bigr) \bigl(u^{\frac{\lambda_{1}}{q}}\bigr)\,du \\ \le& \biggl[ \int_{0}^{\infty} \frac{u^{\lambda - \lambda_{2}}}{(1 + u)^{\lambda + 2}}\,du \biggr]^{\frac{1}{p}}\biggl[ \int_{0}^{\infty} \frac{u^{\lambda_{1}}}{(1 + u)^{\lambda + 2}}\,du \biggr]^{\frac{1}{q}} \\ =& B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}). \end{aligned}$$
(19)

We observe that (19) becomes equality if and only if there exist constants A and B such that they are not all zero and

$$Au^{\lambda - \lambda_{2}} = Bu^{\lambda_{1}}\quad \mbox{a.e. in }R_{ +} $$

(see [26]). Without loss of generality, we suppose \(A \ne 0\). It follows that \(u^{\lambda - \lambda_{2} - \lambda_{1}} = \frac{B}{A}\) a.e. in \(R_{ +} \), namely, \(\lambda - \lambda_{1} - \lambda_{2} = 0\), and then \(\lambda_{1} + \lambda_{2} = \lambda\) with \(\lambda_{1},\lambda_{2} \in (0,\lambda )\).

The theorem is proved. □

Theorem 3

The following statements are equivalent:

  1. (i)

    \(B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1})\)is independent ofp, q;

  2. (ii)

    \(B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1})\)is expressible as a single integral;

  3. (iii)

    If \(\lambda - \lambda_{1} - \lambda_{2} \in ( - p\lambda_{1},p(\lambda - \lambda_{1}))\), then \(\lambda_{1} + \lambda_{2} = \lambda\) (\(\lambda_{1},\lambda_{2} \in (0,\lambda )\));

  4. (iv)

    The constant factor

    $$\frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) $$

    in (15) is the best possible.

Proof

(i) ⇒ (ii). Since \(B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1})\) is independent of p, q, we find

$$\begin{aligned}& B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\& \quad = \lim_{p \to \infty} \lim_{q \to 1^{ +}} B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\& \quad = B(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) = \int_{0}^{\infty} \frac{u^{\lambda_{1}}}{(1 + u)^{\lambda + 2}} \,du, \end{aligned}$$

which is a single integral.

(ii) ⇒ (iii). Suppose that \(B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1})\) is expressible as a single integral \(\int_{0}^{\infty} \frac{1}{(1 + u)^{\lambda + 2}}u^{\frac{\lambda + 1 - \lambda_{2}}{p} + \frac{\lambda_{1} + 1}{q} - 1}\,du\). Then (19) keeps the form of equality. By the proof of Theorem 2 we have \(\lambda_{1} + \lambda_{2} = \lambda\) (\(\lambda_{1},\lambda_{2} \in (0,\lambda )\)).

(iii) ⇒ (iv). If \(\lambda_{1} + \lambda_{2} = \lambda\) (\(\lambda_{1},\lambda_{2} \in (0,\lambda )\)), then by Theorem 1 the constant factor

$$\frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \bigl( = \lambda_{1} \lambda_{2}B(\lambda_{1},\lambda_{2})\bigr) $$

in (13) is the best possible.

(iv) ⇒ (i). In this case, by Theorem 2 we have \(\lambda_{1} + \lambda_{2} = \lambda\), and

$$B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) = B(\lambda_{1} + 1, \lambda_{2} + 1) $$

is independent of p, q.

Hence statements (i), (ii), (iii), and (iv) are equivalent.

The theorem is proved. □

Remark 2

If \(\mu + \sigma = s\) (\(\mu,\sigma \in (0,s)\)), then inequality (11) reduces to

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{s}} \,dx\,dy \le& B(\mu,\sigma )\biggl[ \int_{0}^{\infty} x^{p(1 - \mu ) - 1} f^{p}(x)\,dx\biggr]^{\frac{1}{p}} \\ &{}\times\biggl[ \int_{0}^{\infty} y^{q(1 - \sigma ) - 1}g^{q} (y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$
(20)

We confirm that the constant factor \(B(\mu,\sigma )\) in (20) is the best possible. Otherwise, we would reach a contradiction by (17) that the constant factor in (16) is not the best possible.

Replacing x by \(\frac{1}{x}\) and then \(x^{s - 2}f(\frac{1}{x})\) by \(f(x)\) in (20), we have the following Hardy–Hilbert’s integral inequality with a nonhomogeneous kernel and the best possible constant factor \(B(s - \sigma,\sigma )\):

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{s}} \,dx\,dy \le& B(s - \sigma,\sigma )\biggl[ \int_{0}^{\infty} x^{p(1 - \sigma ) - 1} f^{p}(x)\,dx\biggr]^{\frac{1}{p}} \\ &{}\times\biggl[ \int_{0}^{\infty} y^{q(1 - \sigma ) - 1}g^{q} (y)\,dy\biggr]^{\frac{1}{q}}. \end{aligned}$$
(21)

4 A corollary and some particular cases

Replacing x by \(\frac{1}{x}\) in (15) and setting \(\hat{f}(x) = x^{\lambda - 2}f(\frac{1}{x})\), we define

$$F_{\lambda} (x): = \int_{\frac{1}{x}}^{\infty} t^{ - \lambda} \hat{f}(t)\,dt \biggl( = \int_{\frac{1}{x}}^{\infty} f\biggl(\frac{1}{u} \biggr)\frac{1}{u^{2}}\,du = \int_{0}^{x} f(t)\,dt \biggr). $$

Then replacing \(\hat{f}(x)\) by \(f(x)\), we have \(F_{\lambda} (x) = \int_{\frac{1}{x}}^{\infty} t^{ - \lambda} f(t)\,dt\) and the following Hilbert-type integral inequality with nonhomogeneous kernel:

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < & \frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) \\ &{}\times \biggl[ \int_{0}^{\infty} x^{ - p\lambda_{1} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} F_{\lambda}^{p}(x)\,dx\biggr]^{\frac{1}{p}} \\ &{}\times \biggl[ \int_{0}^{\infty} y^{ - q\lambda_{2} - (\lambda - \lambda_{1} - \lambda_{2}) - 1} G^{q}(y)\,dy\biggr]^{\frac{1}{q}}, \end{aligned}$$
(22)

which is equivalent to (15).

In view of Theorem 3, we have the following:

Corollary 1

Assuming that \(\lambda - \lambda_{1} - \lambda_{2} \in ( - p\lambda_{1},p(\lambda - \lambda_{1}))\), the constant factor

$$\frac{\varGamma (\lambda + 2)}{\varGamma (\lambda )}B^{\frac{1}{p}}(\lambda_{2} + 1,\lambda + 1 - \lambda_{2})B^{\frac{1}{q}}(\lambda_{1} + 1,\lambda + 1 - \lambda_{1}) $$

in (22) is the best possible if and only if \(\lambda_{1} + \lambda_{2} = \lambda\) (\(\lambda_{1},\lambda_{2} \in (0,\lambda )\)).

In the case of \(\lambda_{1} + \lambda_{2} = \lambda\), (22) reduces to the following Hilbert-type integral inequality with nonhomogeneous kernel and the best possible constant factor \(\lambda_{1}\lambda_{2}B(\lambda_{1},\lambda_{2})\):

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < & \lambda_{1}\lambda_{2}B(\lambda_{1}, \lambda_{2}) \\ &{}\times \biggl\{ \int_{0}^{\infty} x^{ - p\lambda_{1} - 1} F_{\lambda}^{p}(x)\,dx\biggr\} ^{\frac{1}{p}}\biggl\{ \int_{0}^{\infty} y^{ - q\lambda_{2} - 1} G^{q}(y)\,dy\biggr\} ^{\frac{1}{q}}, \end{aligned}$$
(23)

which is equivalent to (16).

Remark 3

In (16) and (23), for \(\lambda_{1} = \frac{\lambda}{ q}\), \(\lambda_{2} = \frac{\lambda}{p}\), we have the following equivalent inequalities:

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < \frac{\lambda^{2}}{pq}B\biggl(\frac{\lambda}{p},\frac{\lambda}{q}\biggr) \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy< } {}\times \biggl( \int_{0}^{\infty} x^{(1 - p)\lambda - 1} F^{p}(x)\,dx\biggr)^{\frac{1}{p}} \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < } {}\times \biggl( \int_{0}^{\infty} y^{(1 - q)\lambda - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}, \end{aligned}$$
(24)
$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < \frac{\lambda^{2}}{pq}B\biggl(\frac{\lambda}{p},\frac{\lambda}{q}\biggr) \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < } {}\times \biggl( \int_{0}^{\infty} x^{(1 - p)\lambda - 1} F_{\lambda}^{p}(x)\,dx\biggr)^{\frac{1}{p}} \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < } {}\times \biggl( \int_{0}^{\infty} y^{(1 - q)\lambda - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}, \end{aligned}$$
(25)

and for \(\lambda_{1} = \frac{\lambda}{p}\), \(\lambda_{2} = \frac{\lambda}{q}\), we have the following equivalent inequalities:

$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < \frac{\lambda^{2}}{pq}B\biggl(\frac{\lambda}{p},\frac{\lambda}{q}\biggr) \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < }{}\times \biggl( \int_{0}^{\infty} x^{ - \lambda - 1} F^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - \lambda - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}, \end{aligned}$$
(26)
$$\begin{aligned}& \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < \frac{\lambda^{2}}{pq}B\biggl(\frac{\lambda}{p},\frac{\lambda}{q}\biggr) \\& \hphantom{\int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < } {}\times \biggl( \int_{0}^{\infty} x^{ - \lambda - 1} F_{\lambda}^{p}(x)\,dx\biggr)^{\frac{1}{p}}\biggl( \int_{0}^{\infty} y^{ - \lambda - 1} G^{q}(y)\,dy\biggr)^{\frac{1}{q}}. \end{aligned}$$
(27)

In particular, for \(p = q = 2\), both inequalities (24) and (26) reduce to

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(x + y)^{\lambda}} \,dx\,dy < & \frac{\lambda^{2}}{4}B\biggl(\frac{\lambda}{2},\frac{\lambda}{2}\biggr) \\ &{}\times \biggl( \int_{0}^{\infty} x^{ - \lambda - 1} F^{2}(x)\,dx \int_{0}^{\infty} y^{ - \lambda - 1} G^{2}(y)\,dy\biggr)^{\frac{1}{2}}, \end{aligned}$$
(28)

and both (25) and (27) reduce to the following equivalent form of (25):

$$\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x)g(y)}{(1 + xy)^{\lambda}} \,dx\,dy < & \frac{\lambda^{2}}{4}B\biggl(\frac{\lambda}{2},\frac{\lambda}{2}\biggr) \\ &{}\times \biggl( \int_{0}^{\infty} x^{ - \lambda - 1} F_{\lambda}^{2}(x)\,dx \int_{0}^{\infty} y^{ - \lambda - 1} G^{2}(y)\,dy\biggr)^{\frac{1}{2}}. \end{aligned}$$
(29)

The constant factors in the inequalities of Remark 3 are the best possible.

5 Conclusions

In this paper, following [21, 22], using the weight functions and the idea of introduced parameters, we give a new Hilbert-type integral inequality with the kernel \(\frac{1}{(x + y)^{\lambda}}\) (\(\lambda > 0\)) involving the upper limit functions and the beta and gamma functions (Theorem 1). The preliminaries and the equivalent statements of the best possible constant factor related to a few parameters are considered in Theorems 2 and 3. As applications, we obtain a corollary in the case of nonhomogeneous kernel and some particular inequalities (Corollary 1 and Remark 3). The lemmas and theorems provide an extensive account of inequalities of this type.