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On a class of Hilbert-type inequalities in the whole plane related to exponent function

Abstract

By introducing a kernel involving an exponent function with multiple parameters, we establish a new Hilbert-type inequality and its equivalent Hardy form. We also prove that the constant factors of the obtained inequalities are the best possible. Furthermore, by introducing the Bernoulli number, Euler number, and the partial fraction expansion of cotangent function and cosecant function, we get some special and interesting cases of the newly obtained inequality.

1 Introduction

Suppose that \(f(x)\) and \(\mu (x)\) (>0) are two measurable functions defined on a measurable set Ω, and \(p>1\). Define

$$\begin{aligned} L^{p}_{\mu }(\Omega ):= \biggl\{ f: \Vert f \Vert _{p,\mu }:= \biggl({ \int _{\Omega } \mu (x) \bigl\vert f(x) \bigr\vert ^{p}\,\mathrm{d}m} \biggr)^{\frac{1}{p}}< \infty \biggr\} . \end{aligned}$$
(1.1)

Specially, for \(\mu (x)=1\), we have the abbreviated forms: \(\|f\|_{p,\mu }:=\|f\|_{p}\) and \(L^{p}_{\mu }(\Omega ):= L^{p}(\Omega )\).

Consider two real-valued functions \(f, g\geq 0\) and \(f, g\in L^{p}(\mathbb{R}_{+})\). Suppose that \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1\). Then we have the following two classical Hilbert-type inequalities [1]:

$$\begin{aligned}& \int _{\mathbb{R}_{+}} \int _{\mathbb{R}_{+}} \frac{f(x)g(y)}{x+y}{ \,\mathrm{d} x\,\mathrm{d} y} < \frac{\pi }{\sin \frac{\pi }{p}} \Vert f \Vert _{p} \Vert g \Vert _{q}, \end{aligned}$$
(1.2)
$$\begin{aligned}& \int _{\mathbb{R}_{+}} \int _{\mathbb{R}_{+}} \frac{f(x)g(y)}{\max \{x,y\}}{\,\mathrm{d} x\,\mathrm{d} y} < pq \Vert f \Vert _{p} \Vert g \Vert _{q}, \end{aligned}$$
(1.3)

where the constant factors \(\frac{\pi }{\sin \frac{\pi }{p}}\) and pq in (1.2) and (1.3) are the best possible.

In the past 100 years, especially after the 1990s, by the introduction of several parameters and special functions such as β-function and Γ-function, some classical Hilbert-type integral inequalities like (1.2) and (1.3) as well as their discrete forms were extended to more general forms (see [212]). The inequality below is a typical extension of (1.2) which was established by Yang [13] in 2004:

$$ \int _{0}^{\infty } \int _{0}^{\infty } \frac{f(x)g(y)}{x^{\beta }+y^{\beta }}\,\mathrm{d} x \,\mathrm{d} y< \frac{\pi }{\beta \sin {\frac{\pi }{r}}} \Vert f \Vert _{p,\mu } \Vert g \Vert _{q,\nu }, $$
(1.4)

where \(\rho >0\), \(\mu (x)=x^{p(1-\frac{\beta }{r})-1}\), \(\nu (x)=x^{q(1-\frac{\beta }{s})-1}\), \(\frac{1}{r}+\frac{1}{s}=1\), and the constant factor is the best possible.

In recent years, by constructing new kernel functions and studying their discrete form, half-discrete form, reverse form, multi-dimensional extension and coefficient refinement, researchers have established a large number of new Hilbert-type inequalities (see [1425]). It should be noted that the establishment of these inequalities fully demonstrates the techniques of modern analysis and proves to be critical in the development of modern analysis [26].

In these numerous publications related to the Hilbert inequality, we will present some results with the kernels involving exponent function, and the motivation of this work is precisely from these results. The first result presented below was established by Yang [27] in 2012, that is,

$$ \int _{\mathbb{R}_{+}} \int _{\mathbb{R}_{+}}e^{-\frac{ax}{y}}f(x)g(y){ \,\mathrm{d} x \, \mathrm{d} y}< a^{-\beta }\Gamma (\beta ) \Vert f \Vert _{2, \mu } \Vert g \Vert _{2,\mu }, $$
(1.5)

where \(a>0\), \(\beta >0\), \(\mu (x)=x^{-2\beta +1}\) and \(\nu (y)=y^{2\beta +1}\).

In addition, Liu [28] established an inequality with the kernel involving hyperbolic secant function in 2013, and Yang [29] established an inequality with the kernel involving hyperbolic cosecant function in 2014. The two inequalities can be written as follows:

$$\begin{aligned}& \int _{\mathbb{R}_{+}} \int _{\mathbb{R}_{+}}\operatorname{sech}(xy) f(x)g(y){ \,\mathrm{d} x \, \mathrm{d} y}< 2c_{0} \Vert f \Vert _{2,\mu } \Vert g \Vert _{2, \mu }, \end{aligned}$$
(1.6)
$$\begin{aligned}& \int _{\mathbb{R}_{+}} \int _{\mathbb{R}_{+}}\operatorname{csch} (xy)f(x)g(y) {\,\mathrm{d} x \, \mathrm{d} y}< \frac{\pi ^{2}}{4} \Vert f \Vert _{2, \mu } \Vert g \Vert _{2, \mu } , \end{aligned}$$
(1.7)

where \(\operatorname{sech} (t)=\frac{2}{e^{t}+e^{-t}}\), \(\operatorname{csch} (t)=\frac{2}{e^{t}-e^{-t}}\), \(\mu (x)=x^{-3}\) and \(c_{0}=\sum_{k=0}^{\infty }\frac{(-1)^{k}}{(2k+1)^{2}}=0.91596559^{+}\), which is the Catalan constant.

In this work, the integral interval of inequality (1.6) and (1.7) will be extended to the whole plane, and the following new inequalities will be established:

$$\begin{aligned}& \int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech}(xy)f(x)g(y) { \,\mathrm{d} x \, \mathrm{d} y} < \frac{E_{n}}{4^{n}}\pi ^{2n+1} \Vert f \Vert _{2,\mu } \Vert g \Vert _{2, \mu }, \end{aligned}$$
(1.8)
$$\begin{aligned}& \int _{\mathbb{R}} \int _{\mathbb{R}} \bigl\vert \operatorname{csch}(xy) \bigr\vert f(x)g(y) { \,\mathrm{d} x\,\mathrm{d} y} < \frac{B_{n}}{n} \bigl(2^{2n}-1 \bigr)\pi ^{2n} \Vert f \Vert _{2,\nu } \Vert g \Vert _{2, \nu }, \end{aligned}$$
(1.9)

where \(\mu (x)=| x| ^{-4n-1}\), \(\nu (x)=| x| ^{-4n+1}\), \(E_{n}\) is an Euler number [30, 31] and \(B_{n}\) is a Bernoulli number [30, 31].

Furthermore, we also present some interesting inequalities involving other hyperbolic functions in this paper. More generally, a new kernel function including both the homogeneous case and the non-homogeneous case is constructed, and a Hilbert-type inequality involving this new kernel is established. It will be shown that the newly obtained inequality is a unified extension of (1.8), (1.9) and some other special Hilbert-type inequalities.

2 Some lemmas

Lemma 2.1

Let \(r,s>0\), \(r+s=1\), \(\varphi (x)=\cot {x}\). Then

$$\begin{aligned}& \varphi ^{(2n-1)}(r\pi )=-\frac{(2n-1)!}{\pi ^{2n}}\sum _{k=0}^{\infty } \biggl\{ \frac{1}{(k+r)^{2n}}+ \frac{1}{(k+s)^{2n}} \biggr\} ,\quad n\in {\mathbb{N}}^{+}; \end{aligned}$$
(2.1)
$$\begin{aligned}& \varphi ^{(2n)}(r\pi )=\frac{(2n)!}{\pi ^{2n+1}}\sum _{k=0}^{\infty } \biggl\{ \frac{1}{(k+r)^{2n+1}}-\frac{1}{(k+s)^{2n+1}} \biggr\} ,\quad n \in {\mathbb{N}}. \end{aligned}$$
(2.2)

Proof

Consider the rational fraction expansion of \(\varphi (x)=\cot x\) [30]:

$$ \varphi (x)=\frac{1}{x}+\sum_{k=1}^{\infty } \biggl\{ \frac{1}{x+k\pi }+ \frac{1}{x-k\pi } \biggr\} , $$

and find the \((2n-1)\)th derivative of \(\varphi (x)\), then we obtain

$$ \varphi ^{(2n-1)}(x)=-(2n-1)! \Biggl\{ \sum _{k=0}^{\infty }\frac{1}{(x+k\pi )^{2n}}+\sum _{k=1}^{\infty }\frac{1}{(x-k\pi )^{2n}} \Biggr\} . $$
(2.3)

Set \(x=r\pi \) in (2.3). Since \(r+s=1\), it follows that

$$\begin{aligned} \varphi ^{(2n-1)}(r\pi )&=-\frac{(2n-1)!}{\pi ^{2n}} \Biggl\{ \sum _{k=0}^{\infty }\frac{1}{(k+r)^{2n}}+\sum _{k=1}^{\infty }\frac{1}{(k-r)^{2n}} \Biggr\} \\ &=-\frac{(2n-1)!}{\pi ^{2n}}\sum_{k=0}^{\infty } \biggl\{ \frac{1}{(k+r)^{2n}}+ \frac{1}{(k+s)^{2n}} \biggr\} . \end{aligned}$$

Therefore, (2.1) is proved, and similar computation yields (2.2). □

Furthermore, by considering the following rational fraction expansion of \(\psi (x)=\csc x\) [30]:

$$ \psi (x)=\frac{1}{x}+\sum_{k=1}^{\infty }(-1)^{k} \biggl( \frac{1}{x+k\pi }+\frac{1}{x-k\pi } \biggr), $$

we can obtain Lemma 2.2.

Lemma 2.2

Let \(r,s>0\), \(r+s=1\) and \(\psi (x)=\csc {x}\). Then

$$\begin{aligned}& \psi ^{(2n-1)}(r\pi )=-\frac{(2n-1)!}{\pi ^{2n}}\sum _{k=0}^{\infty }(-1)^{k} \biggl\{ \frac{1}{(k+r)^{2n}}-\frac{1}{(k+s)^{2n}} \biggr\} ,\quad n \in { \mathbb{N}}^{+}; \end{aligned}$$
(2.4)
$$\begin{aligned}& \psi ^{(2n)}(r\pi )=\frac{(2n)!}{\pi ^{2n+1}}\sum _{k=0}^{\infty }(-1)^{k} \biggl\{ \frac{1}{(k+r)^{2n+1}}+\frac{1}{(k+s)^{2n+1}} \biggr\} ,\quad n\in {\mathbb{N}}. \end{aligned}$$
(2.5)

Remark 2.3

Let \(r=\frac{1}{2}\) in (2.1). For \(n\in {\mathbb{N}}^{+}\), we have

$$ \varphi ^{(2n-1)} \biggl(\frac{\pi }{2} \biggr)=- \frac{2^{2n+1}}{\pi ^{2n}}(2n-1)!\sum_{k=0}^{\infty} \frac{1}{(2k+1)^{2n}}. $$
(2.6)

By using the equality [30, 31] \(\sum_{k=1}^{\infty }\frac{1}{k^{2n}}=\frac{(2\pi )^{2n}}{2(2n)!}B_{n} \), where \(B_{n}\) is a Bernoulli number, we have

$$ \sum_{k=0}^{\infty } \frac{1}{(2k+1)^{2n}}=\sum_{k=1}^{\infty } \frac{1}{k^{2n}}-\sum_{k=1}^{\infty } \frac{1}{(2k)^{2n}}= \frac{2^{2n}-1}{2(2n)!}B_{n}\pi ^{2n}. $$
(2.7)

Applying (2.7) to (2.6), we obtain

$$ \varphi ^{(2n-1)} \biggl(\frac{\pi }{2} \biggr)=-\frac{B_{n}}{n}2^{2n-1} \bigl(2^{2n}-1 \bigr). $$
(2.8)

In addition, letting \(r=\frac{1}{4}\) in (2.1), we can also obtain

$$ \varphi ^{(2n-1)} \biggl(\frac{\pi }{4} \biggr)=-\frac{B_{n}}{n}4^{2n-1} \bigl(2^{2n}-1 \bigr). $$
(2.9)

Furthermore, let \(r=\frac{1}{4}\) in (2.2) and \(r=\frac{1}{2}\) in (2.5). In view of [30] \(\sum_{k=0}^{\infty }\frac{(-1)^{k}}{(2k+1)^{2n+1}}= \frac{\pi ^{2n+1}}{2^{2n+2}(2n)!}E_{n} \), where \(E_{n}\) is an Euler number, we obtain

$$ E_{n}=\frac{1}{4^{n}}\varphi ^{(2n)} \biggl(\frac{\pi }{4} \biggr)= \psi ^{(2n)} \biggl(\frac{\pi }{2} \biggr). $$
(2.10)

Lemma 2.4

Let \(\eta _{1},\eta _{2}, \delta \in \{1,-1\}\), \(a>c\geq d>b>0 \) and \(c\neq d\) for \(\eta _{2}=-1\). Let β be such that \(\beta \geq 1 \), and \(\beta \neq 1\) for \(\eta _{1}=-1\), \(\eta _{2}=1\). Define

$$ K(x,y):= \frac{ \vert c^{xy^{\delta }}+\eta _{2}d^{xy^{\delta }} \vert }{ \vert a^{xy^{\delta }}+\eta _{1}b^{xy^{\delta }} \vert } $$
(2.11)

and

$$\begin{aligned} C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) :={}&\sum _{k=0}^{\infty } \biggl( \frac{(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{a}{c})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{a}{d})^{\beta }} \biggr) \\ & {}+\sum_{k=0}^{\infty } \biggl( \frac{(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{d}{b})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{c}{b})^{\beta }} \biggr). \end{aligned}$$
(2.12)

Then

$$ \int _{\mathbb{R}} K(t,1) \vert t \vert ^{\beta -1}\, \mathrm{d} t= \Gamma ( \beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ), $$
(2.13)

where \(\Gamma (\beta )=\int _{\mathbb{R}_{+}}x^{\beta -1}e^{-x}\,\mathrm{d} x\) (\(\beta >0\)) is the second type of Euler integral (Γ-function) [30, 31], and \(\Gamma (\beta )=(\beta -1)!\) for \(\beta \in {\mathbb{N}}^{+}\).

Proof

$$\begin{aligned} & \int _{\mathbb{R}} K(t,1) \vert t \vert ^{\beta -1}\, \mathrm{d} t \\ &\quad = \int _{\mathbb{R}_{+}} K(t,1) \vert t \vert ^{\beta -1}\, \mathrm{d} t+ \int _{\mathbb{R}_{-}} K(t,1) \vert t \vert ^{\beta -1} \, \mathrm{d} t:=I_{1}+I_{2}. \end{aligned}$$
(2.14)

Observing that \(a>b>0 \) and \(\eta _{1}\in \{1,-1\}\), we obtain

$$ \frac{1}{a^{t}+\eta _{1}b^{t}} = \frac{a^{-t}}{1+\eta _{1} {(a^{-1}b)}^{t}}=\sum _{k=0}^{\infty }(-\eta _{1})^{k} \biggl\{ \biggl(\frac{b}{a} \biggr)^{k} \frac{1}{a} \biggr\} ^{t}. $$
(2.15)

Hence

$$\begin{aligned} I_{1}&= \int _{\mathbb{R}_{+}} K(t,1) \vert t \vert ^{\beta -1}{ \, \mathrm{d} t} \\ &= \sum_{k=0}^{\infty }(-\eta _{1})^{k} \int _{ \mathbb{R}_{+}} \biggl\{ \biggl(\frac{b}{a} \biggr)^{k}\frac{c}{a} \biggr\} ^{t} t^{\beta -1}{\,\mathrm{d} t}+\sum_{k=0}^{\infty }(-\eta _{1})^{k} \eta _{2} \int _{\mathbb{R}_{+}} \biggl\{ \biggl(\frac{b}{a} \biggr)^{k}\frac{d}{a} \biggr\} ^{t}t^{ \beta -1} {\,\mathrm{d} t} \\ &:=\sum_{k=0}^{\infty } \bigl\{ (- \eta _{1})^{k}I_{11}+(-\eta _{1})^{k} \eta _{2}I_{12} \bigr\} . \end{aligned}$$
(2.16)

Setting \(t=\frac{u}{k\ln \frac{a}{b}+\ln \frac{a}{c}}\), we have

$$ I_{11}= \frac{1}{(k\ln \frac{a}{b}+\ln \frac{a}{c})^{\beta }} \int _{ \mathbb{R}_{+}}e^{-u}u^{\beta -1} {\,\mathrm{d} u}= \frac{\Gamma (\beta )}{(k\ln \frac{a}{b}+\ln \frac{a}{c})^{\beta }}. $$
(2.17)

Similarly, we can obtain

$$ I_{12}= \frac{1}{(k\ln \frac{a}{b}+\ln \frac{a}{d})^{\beta }} \int _{ \mathbb{R}_{+}} e^{-u}u^{\beta -1} { \, \mathrm{d} u}= \frac{\Gamma (\beta )}{(k\ln \frac{a}{b}+\ln \frac{a}{d})^{\beta }}. $$
(2.18)

Applying (2.17) and (2.18) to (2.16), we obtain

$$ I_{1}=\Gamma (\beta )\sum _{k=0}^{\infty } \biggl( \frac{(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{a}{c})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{a}{d})^{\beta }} \biggr). $$
(2.19)

Similarly, we can obtain

$$\begin{aligned} I_{2}&= \int _{\mathbb{R}_{-}} K(t,1) \vert t \vert ^{\beta -1}{ \, \mathrm{d} t} = \int _{\mathbb{R}_{+}} K(-t,1) t^{\beta -1}{ \,\mathrm{d} t} \\ &=\Gamma (\beta )\sum_{k=0}^{\infty } \biggl( \frac{(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{d}{b})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{c}{b})^{\beta }} \biggr). \end{aligned}$$
(2.20)

Plugging (2.19) and (2.20) into (2.14), and using (2.12), we have (2.13). □

Lemma 2.5

Let \(\eta _{1},\eta _{2}, \delta \in \{1,-1\}\), \(a>c\geq d>b>0\) and \(c\neq d\) for \(\eta _{2}=-1\). Let β be such that \(\beta \geq 1 \), and \(\beta \neq 1\) for \(\eta _{1}=-1\), \(\eta _{2}=1\). \(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \) is defined by Lemma 2.4. Suppose \(D_{\delta }=\{y: | y| ^{\delta }<1\}\), and, for arbitrary natural number n which is large enough, set

$$\begin{aligned}& f_{n}(x)= \textstyle\begin{cases} 0, &x\in [-1, 1], \\ \vert x \vert ^{{\beta -1}-\frac{2}{np}}, &x\in \mathbb{R}\setminus [-1, 1], \end{cases}\displaystyle \\& g_{n}(y)= \textstyle\begin{cases} \vert y \vert ^{{\delta \beta -1}+\frac{2\delta }{nq}}, &y\in D_{\delta }, \\ 0, &y\in \mathbb{R}\setminus D_{\delta }. \end{cases}\displaystyle \end{aligned}$$

Then

$$ \frac{1}{n} \int _{\mathbb{R}} \int _{\mathbb{R}}K(x, y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y}=C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )+o(1). $$
(2.21)

Proof

Setting \(D_{\delta }^{+}=\{y: y>0, y\in D_{\delta }\}\), \(D_{\delta }^{-}=\{y: y<0, y\in D_{\delta }\}\), we get

$$\begin{aligned} & \int _{\mathbb{R}} \int _{\mathbb{R}}K(x, y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad = \int _{1}^{\infty }x^{{\beta -1}-\frac{2}{np}} \int _{D_{\delta }^{+}} K(x, y)y^{\delta \beta -1+\frac{2\delta }{nq}}{\,\mathrm{d} y \, \mathrm{d} x} \\ &\qquad {}+ \int _{1}^{\infty }x^{{\beta -1}-\frac{2}{np}} \int _{D_{\delta }^{-}} K(x, y) \vert y \vert ^{\delta \beta -1+\frac{2\delta }{nq}}{\, \mathrm{d} y \,\mathrm{d} x} \\ &\qquad {}+ \int _{-\infty }^{-1} \vert x \vert ^{{\beta -1}-\frac{2}{np}} \int _{D_{ \delta }^{+}} K(x, y)y^{\delta \beta -1+\frac{2\delta }{nq}}{ \,\mathrm{d} y \, \mathrm{d} x} \\ &\qquad {}+ \int _{-\infty }^{-1} \vert x \vert ^{{\beta -1}-\frac{2}{np}} \int _{D_{ \delta }^{-}} K(x, y) \vert y \vert ^{\delta \beta -1+\frac{2\delta }{nq}}{ \, \mathrm{d} y \,\mathrm{d} x} \\ &\quad =J_{1}+J_{2}+J_{3}+J_{4}. \end{aligned}$$
(2.22)

Setting \(xy^{\delta }=t\), and using Fubini’s theorem, where δ is equal to 1 or −1, we can get

$$\begin{aligned} J_{1}&=J_{4}= \int _{1}^{\infty }x^{-1-\frac{2}{n}} \biggl( \int _{0}^{x} K(t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} \biggr){ \,\mathrm{d} x} \\ &= \int _{1}^{\infty }x^{-1-\frac{2}{n}} \int _{0}^{1} K(t, 1)t^{ \beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} {\,\mathrm{d} x} \\ &\quad {}+ \int _{1}^{\infty }x^{-1-\frac{2}{n}} \int _{1}^{x} K(t, 1)t^{\beta -1+ \frac{2}{nq}}{ \, \mathrm{d} t} {\,\mathrm{d} x} \\ &= \frac{n}{2} \int _{0}^{1} K(t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} + \int _{1}^{\infty }K(t, 1)t^{\beta -1+ \frac{2}{nq}} \int _{t}^{\infty }x^{-1-\frac{2}{n}}{\,\mathrm{d} x} { \,\mathrm{d} t} \\ &= \frac{n}{2} \int _{0}^{1} K(t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} +\frac{n}{2} \int _{1}^{\infty }K(t, 1)t^{ \beta -1-\frac{2}{np}}{ \, \mathrm{d} t}. \end{aligned}$$
(2.23)

Similarly, setting \(xy^{\delta }=-t\), we can also obtain

$$\begin{aligned} J_{2}&=J_{3}= \int _{1}^{\infty }x^{-1-\frac{2}{n}} \int _{0}^{x} K(-t, 1)t^{ \beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} {\,\mathrm{d} x} \\ &= \frac{n}{2} \int _{0}^{1} K(-t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} +\frac{n}{2} \int _{1}^{\infty }K(-t, 1)t^{ \beta -1-\frac{2}{np}}{ \, \mathrm{d} t}. \end{aligned}$$
(2.24)

Applying (2.23) and (2.24) to (2.22), we have

$$\begin{aligned} & \frac{1}{n} \int _{\mathbb{R}} \int _{\mathbb{R}}K(x, y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad = \int _{0}^{1} K(t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} + \int _{1}^{\infty }K(t, 1)t^{\beta -1-\frac{2}{np}}{ \, \mathrm{d} t} \\ &\qquad {}+ \int _{0}^{1} K(-t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} + \int _{1}^{\infty }K(-t, 1)t^{\beta -1-\frac{2}{np}}{ \, \mathrm{d} t}. \end{aligned}$$
(2.25)

Letting \(n\to \infty \) in (2.25), and using (2.13), we get

$$\begin{aligned} & \frac{1}{n} \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad = \int _{\mathbb{R}^{+}} K(t, 1)t^{\beta -1}{\,\mathrm{d} t} + \int _{\mathbb{R}^{+}}K(-t, 1)t^{\beta -1}{\,\mathrm{d} t}+o(1) \\ &\quad = \int _{\mathbb{R}}K(t, 1)t^{\beta -1}{\,\mathrm{d} t}+o(1)=C_{ \eta _{1},\eta _{2}}(a, b, c, d, \beta )+o(1). \end{aligned}$$

The proof of Lemma 2.5 is completed. □

3 Main results

Theorem 3.1

Let \(\eta _{1},\eta _{2}, \delta \in \{1,-1\}\), \(a>c\geq d >b>0\) and \(c\neq d\) for \(\eta _{2}=-1\). Let β be such that \(\beta \geq 1 \), and \(\beta \neq 1\) for \(\eta _{1}=-1\), \(\eta _{2}=1\). Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\mu (x)=| x| ^{p(1-\beta )-1}\) and \(\nu (y)=| y| ^{q(1-\delta \beta )-1}\). Let \(f(x)\), \(g(y)\geq 0\) with \(f(x)\in L_{\mu }^{p}(\mathbb{R})\) and \(g(y)\in L_{\nu }^{q}(\mathbb{R})\). \(K(x,y)\) and \(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )\) are defined via Lemma 2.4. Then

$$ \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f(x)g(y){\,\mathrm{d} x\,\mathrm{d} y} < \Gamma ( \beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }, $$
(3.1)

where the constant factor \(\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )\) is the best possible.

Proof

By Hölder’s inequality, we obtain

$$\begin{aligned} & \int _{\mathbb{R}} \int _{\mathbb{R}} K(x, y)f(x)g(y){\,\mathrm{d} x\,\mathrm{d} y} \\ & \quad = \int _{\mathbb{R}} \int _{\mathbb{R}} \bigl( \bigl( K(x, y) \bigr)^{\frac{1}{p}} \vert y \vert ^{\frac{\delta \beta -1}{p}} \vert x \vert ^{\frac{1-\beta }{q}}f(x) \bigr) \\ &\qquad {}\times \bigl( \bigl( K(x, y) \bigr)^{\frac{1}{q}} \vert x \vert ^{ \frac{\beta -1}{q}} \vert y \vert ^{\frac{1-\delta \beta }{p}}g(y) \bigr){ \, \mathrm{d} x\,\mathrm{d} y} \\ & \quad \leq \biggl( \int _{\mathbb{R}} \int _{\mathbb{R}} K(x, y) \vert y \vert ^{\delta \beta -1} \vert x \vert ^{\frac{p(1-\beta )}{q}}f^{p}(x){ \,\mathrm{d} x\, \mathrm{d} y} \biggr)^{\frac{1}{p}} \\ & \qquad {} \times \biggl( \int _{\mathbb{R}} \int _{\mathbb{R}} K(x, y) \vert x \vert ^{\beta -1} \vert y \vert ^{\frac{q(1-\delta \beta )}{p}}g^{q}(y){ \,\mathrm{d} x\, \mathrm{d} y} \biggr)^{\frac{1}{q}} \\ &\quad = \biggl( \int _{\mathbb{R}} \omega (x) \vert x \vert ^{\frac{p(1-\beta )}{q}}f^{p}(x){ \,\mathrm{d} x} \biggr)^{\frac{1}{p}} \biggl( \int _{\mathbb{R}} \varpi (y) \vert y \vert ^{\frac{q(1-\delta \beta )}{p}}g^{q}(y)dy \biggr)^{\frac{1}{q}}, \end{aligned}$$
(3.2)

where \(\omega (x)=\int _{\mathbb{R}} K(x, y)| y| ^{\delta \beta -1} \,\mathrm{d} y\), and \(\varpi (y)=\int _{\mathbb{R}} K(x, y)| x| ^{\beta -1}{ \,\mathrm{d} x} \).

Setting \(xy^{\eta }=t\), and using (2.13), we have

$$\begin{aligned} \omega (x)&= \vert x \vert ^{-\beta } \int _{\mathbb{R}}K(t,1) \vert t \vert ^{ \beta -1}{\, \mathrm{d} t} \\ &=\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \vert x \vert ^{-\beta } \quad (x\neq 0) \end{aligned}$$
(3.3)

and

$$\begin{aligned} \varpi (y)&= \vert y \vert ^{-\delta \beta } \int _{\mathbb{R}}K(t,1) \vert t \vert ^{\beta -1}{\, \mathrm{d} t} \\ &=\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \vert y \vert ^{-\delta \beta }\quad (y\neq 0). \end{aligned}$$
(3.4)

Plugging (3.3) and (3.4) back to (3.2), we have

$$ \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f(x)g(y){\,\mathrm{d} x \,\mathrm{d} y} \leq \Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(3.5)

If (3.5) takes the form of an equality, then there must be two constants \(A_{1}\) and \(A_{2}\) that are not both equal to zero, such that

$$ A_{1} K(x, y) \vert y \vert ^{\delta \beta -1} \vert x \vert ^{ \frac{p(1-\beta )}{q}}f^{p}(x)=A_{2} K(x, y) \vert x \vert ^{\beta -1} \vert y \vert ^{\frac{q(1-\delta \beta )}{p}}g^{q}(y), $$

a.e. in \(\mathbb{R}\times \mathbb{R}\), that is,

$$ A_{1} \vert x \vert ^{p(1-\beta )}f^{p}(x)=A_{2} \vert y \vert ^{q(1-\delta \beta )}g^{q}(y), $$

a.e. in \(\mathbb{R}\times \mathbb{R}\). Therefore, there exists a constant A such that \(A_{1}| x| ^{p(1-\beta )}f^{p}(x)=A\), a.e. in \(\mathbb{R}\), and \(A_{2}| y| ^{q(1-\delta \beta )}g^{q}(y)=A\), a.e. in \(\mathbb{R}\). Without loss of generality, we assume that \(A_{1}\neq 0\), then we can obtain \(x^{p(1-\beta )-1}f^{p}(x)=\frac{A}{A_{1}x}\) a.e. in \(\mathbb{R}\), which contradicts the fact \(f(x)\in L_{\mu }^{p}(\mathbb{R})\). Hence, (3.5) keeps the form of a strict inequality, and (3.1) is obtained.

What we need to prove next is that the constant factor \(\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )\) in inequality (3.1) is the best possible. Suppose that there exists a positive constant \(k<\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )\), such that (3.1) still holds if \(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )\) is replaced by k. That is,

$$ \int _{\mathbb{R}} \int _{\mathbb{R}}K(x, y)f(x)g(y){\,\mathrm{d} x\,\mathrm{d} y} < k \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(3.6)

Replacing f and g in (3.6) with \(f_{n}\) and \(g_{n}\) defined in Lemma 2.5, respectively, we obtain

$$\begin{aligned} & \frac{1}{n} \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < \frac{k}{n} \biggl( \int _{1}^{\infty }x^{-\frac{2}{n}-1}{ \,\mathrm{d} x}+ \int _{-\infty }^{-1} \vert x \vert ^{- \frac{2}{n}-1}{ \,\mathrm{d} x} \biggr)^{\frac{1}{p}} \\ &\qquad {}\times \biggl( \int _{D_{\delta }} \vert x \vert ^{\frac{2\delta }{n}-1}{ \,\mathrm{d} x} \biggr)^{\frac{1}{q}}=k. \end{aligned}$$
(3.7)

Combining (3.7) and (2.21), we have \(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )+o(1)< k \). Let \(n\rightarrow \infty \), then we have \(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )\leq k\). This contradicts the hypothesis obviously. Therefore, the constant factor in (3.1) is the best possible. □

Theorem 3.2

Let \(\eta _{1},\eta _{2}, \delta \in \{1,-1\}\), \(a>c\geq d> b>0\) and \(c\neq d\) for \(\eta _{2}=-1\). Let β be such that \(\beta \geq 1 \), and \(\beta \neq 1\) for \(\eta _{1}=-1\), \(\eta _{2}=1\). Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\mu (x)=| x| ^{p(1-\beta )-1}\) and \(\nu (y)=| y | ^{q(1-\delta \beta )-1}\). Let \(f(x)\geq 0\) with \(f(x)\in L_{\mu }^{p}(0, \infty )\). \(K(x,y)\) and \(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )\) be defined via Lemma 2.4. Then

$$ \int _{\mathbb{R}} \vert y \vert ^{p\delta \beta -1} \biggl( \int _{\mathbb{R}}K(x,y)f(x){ \,\mathrm{d} x} \biggr)^{p} \,\mathrm{d} y< \bigl(C_{\eta _{1}, \eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \bigr)^{p}, $$
(3.8)

where the constant factor \((\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) )^{p}\) is the best possible, and (3.8) is equivalent to (3.1).

Proof

Consider \(g(y):=| y| ^{p\delta \beta -1} (\int _{\mathbb{R}}K(x, y)f(x){ \,\mathrm{d} x} )^{p-1}\). By Theorem 3.1, we can get

$$\begin{aligned} 0&< \bigl( \Vert g \Vert _{q,\nu } \bigr)^{pq}= \biggl( \int _{\mathbb{R}} \vert y \vert ^{q(1-\delta \beta )-1}g^{q}(y) \,\mathrm{d} y \biggr)^{p} \\ &= \biggl( \int _{\mathbb{R}} \vert y \vert ^{p\delta \beta -1} \biggl( \int _{\mathbb{R}}K(x,y)f(x){\,\mathrm{d} x} \biggr)^{p} \,\mathrm{d} y \biggr)^{p} \\ &= \biggl( \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f(x)g(y){\,\mathrm{d} x\,\mathrm{d} y} \biggr)^{p} \leq \bigl(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q,\nu } \bigr)^{p}. \end{aligned}$$
(3.9)

Therefore

$$\begin{aligned} 0&< \int _{\mathbb{R}} \vert y \vert ^{p\delta \beta -1} \biggl( \int _{\mathbb{R}}K (x,y)f(x){\,\mathrm{d} x} \biggr)^{p} \,\mathrm{d} y \\ &= \bigl( \Vert g \Vert _{q,\nu } \bigr)^{q} \leq \bigl(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \bigr)^{p}. \end{aligned}$$
(3.10)

Since \(f(x)\in L_{\mu }^{p}(\mathbb{R})\), by using (3.10), we obtain \(g(y)\in L_{\nu }^{q}(\mathbb{R})\). By using Theorem 3.1 again, we can deduce that both (3.9) and (3.1) take the form of a strict inequality, and (3.8) is proved. On the other hand, if (3.8) is valid, by Hölder’s inequality, we obtain

$$\begin{aligned} & \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f(x)g(y){\,\mathrm{d} x \,\mathrm{d} y} \\ &\quad = \int _{\mathbb{R}} \biggl( \vert y \vert ^{-(1-\delta \beta -\frac{1}{q})} \int _{\mathbb{R}} K(x,y)f(x){\,\mathrm{d} x} \biggr) \bigl( \vert y \vert ^{1-\delta \beta -\frac{1}{q}}g(y) \bigr)\,\mathrm{d} y \\ &\quad \leq \biggl( \int _{\mathbb{R}} \vert y \vert ^{p\delta \beta -1} \biggl( \int _{\mathbb{R}} K(x,y)f(x){\,\mathrm{d} x} \biggr)^{p} \,\mathrm{d} y \biggr)^{\frac{1}{p}} \Vert g \Vert _{q,\nu }. \end{aligned}$$
(3.11)

Combining (3.8) and (3.11), we can get (3.1). Therefore, (3.1) is equivalent to (3.8). From the equivalence of (3.1) and (3.8), we can easily show that the constant factor \((\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) )^{p}\) in (3.8) is the best possible. Theorem 3.2 is proved. □

4 Applications

Setting \(\eta _{1}=-1\), \(\eta _{2}=1\), \(c=d=1\) and \(\beta =2n\) (\(n\in {\mathbb{N}}^{+}\)) in Theorem 3.1, and using (2.1), we can obtain

$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{-2}{(2n-1)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\varphi ^{(2n-1)} \biggl( \frac{\pi \ln a}{\ln \frac{a}{b}} \biggr). $$

Therefore, the following corollary holds.

Corollary 4.1

Let \(\delta \in \{1,-1\}\), \(a>1> b>0\). Let \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1\). Suppose that \(\varphi (x)=\cot x\), \(\mu (x)=| x| ^{p(1-2n)-1}\) and \(\nu (y)=| y | ^{q(1-2\delta n)-1}\), where \(n\in {\mathbb{N}}^{+}\). Let \(f(x)\), \(g(y)\geq 0\) with \(f(x)\in L_{\mu }^{p}(\mathbb{R})\) and \(g(y)\in L_{\nu }^{q}(\mathbb{R})\). Then

$$ \int _{\mathbb{R}} \int _{\mathbb{R}} \frac{f(x)g(y)}{ \vert a^{xy^{\delta }}-b^{xy^{\delta }} \vert } { \,\mathrm{d} x\,\mathrm{d} y} < - \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\varphi ^{(2n-1)} \biggl( \frac{\pi \ln a}{\ln \frac{a}{b}} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(4.1)

Particularly, let \(a=b^{-1}=e\) in (4.1), by virtue of (2.8), then (4.1) is transformed to

$$ \int _{\mathbb{R}} \int _{\mathbb{R}} \bigl\vert \operatorname{csch} \bigl(xy^{\delta } \bigr) \bigr\vert f(x)g(y) {\,\mathrm{d} x\,\mathrm{d} y}< \frac{B_{n}}{n} \bigl(2^{2n}-1 \bigr) \pi ^{2n} \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(4.2)

Let \(p=q=2 \), \(\delta =1\) in (4.2), then we have (1.9).

Setting \(\eta _{1}=1\), \(\eta _{2}=1\), \(c=d=1\) and \(\beta =2n+1\) (\(n\in {\mathbb{N}}\)) in Theorem 3.1, and using (2.5), we have

$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{2}{(2n)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n+1}\psi ^{(2n)} \biggl( \frac{\pi \ln a}{\ln \frac{a}{b}} \biggr). $$

Therefore, we obtain the following corollary.

Corollary 4.2

Let \(\delta \in \{1,-1\}\) and \(a>1> b>0 \). Let \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1\). Suppose that \(\psi (x)=\csc x\), \(\mu (x)=| x| ^{-2np-1}\) and \(\nu (y)=| y | ^{-2\delta nq-1}\), where \(n\in {\mathbb{N}}\). Let \(f(x)\), \(g(y)\geq 0\) with \(f(x)\in L_{\mu }^{p}(\mathbb{R})\) and \(g(y)\in L_{\nu }^{q}(\mathbb{R})\). Then

$$ \int _{\mathbb{R}} \int _{\mathbb{R}} \frac{f(x)g(y)}{ a^{xy^{\delta }}+b^{xy^{\delta }}} {\,\mathrm{d} x\,\mathrm{d} y} < \biggl(\frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n+1} \psi ^{(2n)} \biggl(\frac{\pi \ln a}{\ln \frac{a}{b}} \biggr) \Vert f \Vert _{p, \mu } \Vert g \Vert _{q, \nu }. $$
(4.3)

Particularly, let \(a=b^{-1}=e\) in (4.3), by virtue of (2.10), then (4.3) is transformed to

$$ \int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech} \bigl(xy^{\delta } \bigr)f(x)g(y) {\,\mathrm{d} x\,\mathrm{d} y}< \frac{E_{n}}{4^{n}}\pi ^{2n+1} \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(4.4)

Let \(p=q=2\) and \(\delta =1\) in (4.4), then we get (1.8).

Setting \(\eta _{1}=-1\), \(\eta _{2}=1\), \(ab=cd\) and \(\beta =2n\) (\(n\in {\mathbb{N}}^{+}\)) in Theorem 3.1, and using (2.1), we obtain

$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{-2}{(2n-1)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\varphi ^{(2n-1)} \biggl( \frac{\pi \ln (a/c)}{\ln (a/b)} \biggr). $$

Hence, we can obtain another corollary as follows.

Corollary 4.3

Let \(\delta \in \{1,-1\}\), \(a>c\geq d >b>0\) and \(ab=cd\). Let \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1\). Suppose that \(\varphi (x)=\cot x\), \(\mu (x)=| x| ^{p(1-2n)-1}\) and \(\nu (y)=| y | ^{q(1-2\delta n)-1}\), where \(n\in {\mathbb{N}}^{+}\). Let \(f(x)\), \(g(y)\geq 0\) with \(f(x)\in L_{\mu }^{p}(\mathbb{R})\) and \(g(y)\in L_{\nu }^{q}(\mathbb{R})\). Then

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}} \frac{ c^{xy^{\delta }}+d^{xy^{\delta }}}{ \vert a^{xy^{\delta }}-b^{xy^{\delta }} \vert } f(x)g(y) {\,\mathrm{d} x \, \mathrm{d} y} \\ &\quad < -2 \biggl(\frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\varphi ^{(2n-1)} \biggl(\frac{\pi \ln (a/c)}{\ln (a/b)} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.5)

Let \(a=e^{\lambda _{1}}\), \(b=e^{-\lambda _{1}}\), \(c=e^{\lambda _{2}}\) and \(d=e^{-\lambda _{2}}\) in (4.5), where \(\lambda _{1}>\lambda _{2}>0\), then we have

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}} \bigl\vert \operatorname{csch} \bigl(\lambda _{1}xy^{ \delta } \bigr) \bigr\vert \operatorname{cosh} \bigl(\lambda _{2}xy^{\delta } \bigr)f(x)g(y) { \,\mathrm{d} x \,\mathrm{d} y} \\ &\quad < -2 \biggl(\frac{\pi }{2\lambda _{1}} \biggr)^{2n}\varphi ^{(2n-1)} \biggl(\frac{(\lambda _{1}-\lambda _{2})\pi }{2\lambda _{1}} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.6)

Letting \(\lambda _{1}=2\) and \(\lambda _{2}=1\) in (4.6), in view of (2.9), we can also obtain (4.2).

Letting \(\lambda _{1}=4\) and \(\lambda _{2}=1\) in (4.6), we have

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech} \bigl(2xy^{\delta } \bigr) \bigl\vert \operatorname{csch}\bigl(xy^{\delta } \bigr) \bigr\vert f(x)g(y) {\,\mathrm{d} x \,\mathrm{d} y} \\ &\quad < \frac{-\pi ^{2n}}{8^{2n-1}}\varphi ^{(2n-1)} \biggl( \frac{3\pi }{8} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.7)

Setting \(\eta _{1}=-1\), \(\eta _{2}=-1\), \(ab=cd\) and \(\beta =2n+1\) (\(n\in {\mathbb{N}}\)) in Theorem 3.1, and using (2.2), we have

$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{2}{(2n)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n+1}\varphi ^{(2n)} \biggl( \frac{\pi \ln (a/c)}{\ln (a/b)} \biggr). $$

Therefore, the following corollary holds obviously.

Corollary 4.4

Let \(\delta \in \{1,-1\}\), \(a>c\geq d >b>0\) and \(ab=cd\). Let \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\). Suppose that \(\varphi (x)=\cot x\), \(\mu (x)=| x| ^{-2np-1}\) and \(\nu (y)=| y | ^{-2\delta nq-1}\), where \(n\in {\mathbb{N}}\). Let \(f(x)\), \(g(y)\geq 0\) with \(f(x)\in L_{\mu }^{p}(\mathbb{R})\) and \(g(y)\in L_{\nu }^{q}(\mathbb{R})\). Then

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}} \frac{ c^{xy^{\delta }}-d^{xy^{\delta }}}{ a^{xy^{\delta }}-b^{xy^{\delta }}} f(x)g(y) {\,\mathrm{d} x \, \mathrm{d} y} \\ &\quad < 2 \biggl(\frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n+1}\varphi ^{(2n)} \biggl(\frac{\pi \ln (a/c)}{\ln (a/b)} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.8)

Let \(a=e^{\lambda _{1}}\), \(b=e^{-\lambda _{1}}\), \(c=e^{\lambda _{2}}\) and \(d=e^{-\lambda _{2}}\) in (4.8), where \(\lambda _{1}>\lambda _{2}>0\), then we have

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{csch} \bigl(\lambda _{1}xy^{ \delta } \bigr)\operatorname{sinh} \bigl(\lambda _{2}xy^{\delta } \bigr) f(x)g(y) { \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < 2 \biggl(\frac{\pi }{2\lambda _{1}} \biggr)^{2n+1}\varphi ^{(2n)} \biggl(\frac{(\lambda _{1}-\lambda _{2})\pi }{2\lambda _{1}} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.9)

Let \(\lambda _{1}=2\) and \(\lambda _{2}=1\) in (4.9), then it follows from (2.10) that we also get (4.4).

Let \(\lambda _{1}=4\) and \(\lambda _{2}=1\) in (4.9), then we get

$$ \int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech} \bigl(xy^{\delta } \bigr) \operatorname{sech} \bigl(2xy^{\delta } \bigr)f(x)g(y) {\,\mathrm{d} x \,\mathrm{d} y}< \frac{\pi ^{2n+1}}{8^{2n}}\varphi ^{(2n)} \biggl( \frac{3\pi }{8} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(4.10)

Setting \(\eta _{1}=1\), \(\eta _{2}=-1\), \(ab=cd\) and \(\beta =2n\) (\(n\in {\mathbb{N}}^{+}\)) in Theorem 3.1, and using (2.4), we have

$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{-2}{(2n-1)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\psi ^{(2n-1)} \biggl( \frac{\pi \ln (a/c)}{\ln (a/b)} \biggr). $$

Therefore, we obtain the following corollary.

Corollary 4.5

Let \(\delta \in \{1,-1\}\), \(a>c\geq d >b>0\) and \(ab=cd\). Let \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1\). Suppose that \(\psi (x)=\csc x\), \(\mu (x)=| x| ^{p(1-2n)-1}\) and \(\nu (y)=| y | ^{q(1-2\delta n)-1}\), where \(n\in {\mathbb{N}}^{+}\). Let \(f(x)\), \(g(y)\geq 0\) with \(f(x)\in L_{\mu }^{p}(\mathbb{R})\) and \(g(y)\in L_{\nu }^{q}(\mathbb{R})\). Then

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}} \frac{ \vert c^{xy^{\delta }}-d^{xy^{\delta }} \vert }{ a^{xy^{\delta }}+b^{xy^{\delta }}} f(x)g(y) {\,\mathrm{d} x \, \mathrm{d} y} \\ &\quad < -2 \biggl(\frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\psi ^{(2n-1)} \biggl(\frac{\pi \ln (a/c)}{\ln (a/b)} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.11)

Let \(a=e^{\lambda _{1}}\), \(b=e^{-\lambda _{1}}\), \(c=e^{\lambda _{2}}\) and \(d=e^{-\lambda _{2}}\) in (4.11), where \(\lambda _{1}>\lambda _{2}>0\), then we have

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech}\bigl(\lambda _{1}xy^{ \delta } \bigr) \bigl\vert \operatorname{sinh} \bigl(\lambda _{2}xy^{\delta } \bigr) \bigr\vert f(x)g(y) { \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < -2 \biggl(\frac{\pi }{2\lambda _{1}} \biggr)^{2n}\psi ^{(2n-1)} \biggl( \frac{(\lambda _{1}-\lambda _{2})\pi }{2\lambda _{1}} \biggr) \Vert f \Vert _{p, \mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.12)

Let \(\lambda _{1}=2\) and \(\lambda _{2}=1\) in (4.12), then we can have

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech} \bigl(xy^{\delta } \bigr) \bigl\vert \operatorname{tanh} \bigl(2xy^{\delta } \bigr) \bigr\vert f(x)g(y) { \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < \frac{-\pi ^{2n}}{4^{2n-1}}\psi ^{(2n-1)} \biggl( \frac{\pi }{4} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.13)

At last, setting \(\eta _{1}=1\), \(\eta _{2}=1\), \(ab=cd\) and \(\beta =2n+1\) (\(n\in {\mathbb{N}}\)) in Theorem 3.1, by virtue of (2.5), then the following corollary holds.

Corollary 4.6

Let \(\delta \in \{1,-1\}\), \(a>c\geq d >b>0\) and \(ab=cd\). Let \(p>1\) and \(\frac{1}{p}+\frac{1}{q}=1\). Suppose that \(\psi (x)=\csc x\), \(\mu (x)=| x| ^{-2np-1}\) and \(\nu (y)=| y | ^{-2\delta nq-1}\), where \(n\in {\mathbb{N}}\). Let \(f(x)\), \(g(y)\geq 0\) with \(f(x)\in L_{\mu }^{p}(\mathbb{R})\) and \(g(y)\in L_{\nu }^{q}(\mathbb{R})\). Then

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}} \frac{ c^{xy^{\delta }}+d^{xy^{\delta }}}{ a^{xy^{\delta }}+b^{xy^{\delta }}} f(x)g(y) {\,\mathrm{d} x \, \mathrm{d} y} \\ &\quad < 2 \biggl(\frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n+1}\psi ^{(2n)} \biggl(\frac{\pi \ln (a/c)}{\ln (a/b)} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.14)

Let \(a=e^{\lambda _{1}}\), \(b=e^{-\lambda _{1}}\), \(c=e^{\lambda _{2}}\) and \(d=e^{-\lambda _{2}}\) in (4.14), where \(\lambda _{1}>\lambda _{2}>0\), then we have

$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech} \bigl(\lambda _{1}xy^{ \delta } \bigr)\operatorname{cosh} \bigl(\lambda _{2}xy^{\delta } \bigr) f(x)g(y) { \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < 2 \biggl(\frac{\pi }{2\lambda _{1}} \biggr)^{2n+1}\psi ^{(2n)} \biggl( \frac{(\lambda _{1}-\lambda _{2})\pi }{2\lambda _{1}} \biggr) \Vert f \Vert _{p, \mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.15)

Letting \(\lambda _{1}=2\), \(\lambda _{2}=1\) in (4.15), we have

$$ \int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{csch} \bigl(xy^{\delta } \bigr) \operatorname{tanh} \bigl(2xy^{\delta } \bigr) f(x)g(y) {\, \mathrm{d} x \,\mathrm{d} y}< \frac{\pi ^{2n+1}}{4^{2n}}\psi ^{(2n)} \biggl( \frac{\pi }{4} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$

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References

  1. Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities. Cambridge University Press, London (1952)

    MATH  Google Scholar 

  2. Rassias, M.Th., Yang, B.C.: On a Hilbert-type integral inequality in the whole plane related to the extended Riemann zeta function. Complex Anal. Oper. Theory 13(4), 1765–1782 (2019)

    Article  MathSciNet  Google Scholar 

  3. Rassias, M.Th., Yang, B.C.: On a Hilbert-type integral inequality related to the extended Hurwitz zeta function in the whole plane. Acta Appl. Math. 160(1), 67–80 (2019)

    Article  MathSciNet  Google Scholar 

  4. Rassias, M.Th., Yang, B.C.: A Hilbert-type integral inequality in the whole plane related to the hypergeometric function and the beta function. J. Math. Anal. Appl. 428, 1286–1308 (2015)

    Article  MathSciNet  Google Scholar 

  5. Yang, B.C.: A note on Hilbert’s integral inequalities. Chin. Q. J. Math. 13, 83–85 (1998)

    MathSciNet  MATH  Google Scholar 

  6. Yang, B.C.: On Hilbert’s integral inequality. J. Math. Anal. Appl. 220, 778–785 (1998)

    Article  MathSciNet  Google Scholar 

  7. Krnić, M., Pečarić, J.: General Hilbert’s and Hardy’s inequalities. Math. Inequal. Appl. 8, 29–51 (2005)

    MathSciNet  MATH  Google Scholar 

  8. Krnić, M., Pečarić, J., Perić, I., et al.: Advances in Hilbert-Type Inequalities. Element Press, Zagreb (2012)

    MATH  Google Scholar 

  9. Krnić, M., Pečarić, J., Vuković, P.: Discrete Hilbert-type inequalities with general homogeneous kernels. Rend. Circ. Mat. Palermo 60, 161–171 (2011)

    Article  MathSciNet  Google Scholar 

  10. Krnić, M., Pečarić, J.: Extension of Hilbert’s inequality. J. Math. Anal. Appl. 324, 150–160 (2006)

    Article  MathSciNet  Google Scholar 

  11. Kuang, J.C., Debnath, L.: On new generalizations of Hilbert’s inequality and their applications. J. Math. Anal. Appl. 245, 248–265 (2000)

    Article  MathSciNet  Google Scholar 

  12. You, M.H.: On a new discrete Hilbert-type inequality and its application. Math. Inequal. Appl. 18, 1575–1587 (2015)

    MathSciNet  MATH  Google Scholar 

  13. Yang, B.C.: On an extension of Hilbert’s integral inequality with some parameters. Aust. J. Anal. Appl. 1(1), 1–8 (2004)

    Article  MathSciNet  Google Scholar 

  14. Rassias, M.Th., Yang, B.C.: On an equivalent property of a reverse Hilbert-type integral inequality related to the extended Hurwitz-zeta function. J. Math. Inequal. 13(2), 315–334 (2019)

    Article  MathSciNet  Google Scholar 

  15. Rassias, M.Th., Yang, B.C., Raigorodskii, A.: Two kinds of the reverse Hardy-type integral inequalities with the equivalent forms related to the extended Riemann zeta function. Appl. Anal. Discrete Math. 12, 273–296 (2018)

    Article  MathSciNet  Google Scholar 

  16. Rassias, M.Th., Yang, B.C., Raigorodskii, A.: On a half-discrete Hilbert-type inequality in the whole plane with the kernel of hyperbolic secant function related to the Hurwitz zeta function. In: Trigonometric Sums and Their Applications, pp. 229–259. Springer, Berlin (2020)

    Chapter  Google Scholar 

  17. Rassias, M.Th., Yang, B.C.: On half-discrete Hilbert’s inequality. Appl. Math. Comput. 220, 75–93 (2013)

    MathSciNet  MATH  Google Scholar 

  18. Yang, B.C., Wu, S.H., Wang, A.Z.: On a reverse half-discrete Hardy–Hilbert’s inequality with parameters. Mathematics (2019). https://doi.org/10.3390/math7111054

    Article  Google Scholar 

  19. Yang, B.C.: An extended multi-dimensional Hardy-Hilbert-type inequality with a general homogeneous kernel. Int. J. Nonlinear Anal. Appl. 9, 131–143 (2018)

    MATH  Google Scholar 

  20. Yang, B.C., Debnath, L.: Half-Discrete Hilbert-Type Inequalities. World Scientific, Singapore (2014)

    Book  Google Scholar 

  21. Yang, B.C.: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing (2009)

    Google Scholar 

  22. Krnić, M., Pečarić, J., Vuković, P.: A unified treatment of half-discrete Hilbert-type inequalities with a homogeneous kernel. Mediterr. J. Math. 10, 1697–1716 (2013)

    Article  MathSciNet  Google Scholar 

  23. Gao, X., Gao, M.Z.: A new Hilbert-type integral inequality with parameters. J. Math. Res. Expo. 31, 467–473 (2011)

    MathSciNet  Google Scholar 

  24. Hong, Y., He, B., Yang, B.C.: Necessary and sufficient conditions for the validity of Hilbert-type inequalities with a class of quasi-homogeneous kernels ans its applications in operator theory. J. Math. Inequal. 12, 777–788 (2018)

    Article  MathSciNet  Google Scholar 

  25. He, B., Yang, B.C., Chen, Q.: A new multiple half-discrete Hilbert-type inequality with parameters and a best possible constant factor. Mediterr. J. Math. (2014). https://doi.org/10.1007/s00009-014-0468-0

    Article  MATH  Google Scholar 

  26. Mintrinovic, D.S., Pečarić, J., Fink, A.M.: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic, Boston (1991)

    Book  Google Scholar 

  27. Yang, B.C.: A new Hilbert-type integral inequality with the homogeneous kernel of degree 0. J. Zhejiang Univ. Sci. Ed. 39, 390–392 (2012)

    MathSciNet  Google Scholar 

  28. Liu, Q., Long, S.C.: A Hilbert-type integral inequality with the kernel of hyperbolic secant function. J. Zhejiang Univ. Sci. Ed. 40, 255–259 (2013)

    MathSciNet  Google Scholar 

  29. Yang, B.C., Chen, Q.: A Hilbert-type integral inequality related to Riemann zeta function. J. Jilin Univ. Sci. Ed. 52, 869–872 (2014)

    MathSciNet  MATH  Google Scholar 

  30. Wang, Z.X., Guo, D.R.: Introduction to Special Functions. Higher Education Press, Beijing (2012)

    Google Scholar 

  31. Richard, C.F.J.: Introduction to Calculus and Analysis. Springer, New York (1989)

    Google Scholar 

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You, M. On a class of Hilbert-type inequalities in the whole plane related to exponent function. J Inequal Appl 2021, 33 (2021). https://doi.org/10.1186/s13660-021-02563-5

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