1 Introduction
Suppose that \(f(x)\) and \(\mu (x)\) (>0) are two measurable functions defined on a measurable set Ω, and \(p>1\). Define Specially, for \(\mu (x)=1\), we have the abbreviated forms: \(\|f\|_{p,\mu }:=\|f\|_{p}\) and \(L^{p}_{\mu }(\Omega ):= L^{p}(\Omega )\).
$$\begin{aligned} L^{p}_{\mu }(\Omega ):= \biggl\{ f: \Vert f \Vert _{p,\mu }:= \biggl({ \int _{\Omega } \mu (x) \bigl\vert f(x) \bigr\vert ^{p}\,\mathrm{d}m} \biggr)^{\frac{1}{p}}< \infty \biggr\} . \end{aligned}$$
(1.1)
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]: where the constant factors \(\frac{\pi }{\sin \frac{\pi }{p}}\) and pq in (1.2) and (1.3) are the best possible.
$$\begin{aligned}& \int _{\mathbb{R}_{+}} \int _{\mathbb{R}_{+}} \frac{f(x)g(y)}{x+y}{ \,\mathrm{d} x\,\mathrm{d} y} < \frac{\pi }{\sin \frac{\pi }{p}} \Vert f \Vert _{p} \Vert g \Vert _{q}, \end{aligned}$$
(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)
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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 [2‐12]). The inequality below is a typical extension of (1.2) which was established by Yang [13] in 2004: 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.
$$ \int _{0}^{\infty } \int _{0}^{\infty } \frac{f(x)g(y)}{x^{\beta }+y^{\beta }}\,\mathrm{d} x \,\mathrm{d} y< \frac{\pi }{\beta \sin {\frac{\pi }{r}}} \Vert f \Vert _{p,\mu } \Vert g \Vert _{q,\nu }, $$
(1.4)
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 [14‐25]). 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, where \(a>0\), \(\beta >0\), \(\mu (x)=x^{-2\beta +1}\) and \(\nu (y)=y^{2\beta +1}\).
$$ \int _{\mathbb{R}_{+}} \int _{\mathbb{R}_{+}}e^{-\frac{ax}{y}}f(x)g(y){ \,\mathrm{d} x \, \mathrm{d} y}< a^{-\beta }\Gamma (\beta ) \Vert f \Vert _{2, \mu } \Vert g \Vert _{2,\mu }, $$
(1.5)
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: 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.
$$\begin{aligned}& \int _{\mathbb{R}_{+}} \int _{\mathbb{R}_{+}}\operatorname{sech}(xy) f(x)g(y){ \,\mathrm{d} x \, \mathrm{d} y}< 2c_{0} \Vert f \Vert _{2,\mu } \Vert g \Vert _{2, \mu }, \end{aligned}$$
(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)
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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: 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].
$$\begin{aligned}& \int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech}(xy)f(x)g(y) { \,\mathrm{d} x \, \mathrm{d} y} < \frac{E_{n}}{4^{n}}\pi ^{2n+1} \Vert f \Vert _{2,\mu } \Vert g \Vert _{2, \mu }, \end{aligned}$$
(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)
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]: and find the \((2n-1)\)th derivative of \(\varphi (x)\), then we obtain Set \(x=r\pi \) in (2.3). Since \(r+s=1\), it follows that Therefore, (2.1) is proved, and similar computation yields (2.2). □
$$ \varphi (x)=\frac{1}{x}+\sum_{k=1}^{\infty } \biggl\{ \frac{1}{x+k\pi }+ \frac{1}{x-k\pi } \biggr\} , $$
$$ \varphi ^{(2n-1)}(x)=-(2n-1)! \Biggl\{ \sum _{k=0}^{\infty }\frac{1}{(x+k\pi )^{2n}}+\sum _{k=1}^{\infty }\frac{1}{(x-k\pi )^{2n}} \Biggr\} . $$
(2.3)
$$\begin{aligned} \varphi ^{(2n-1)}(r\pi )&=-\frac{(2n-1)!}{\pi ^{2n}} \Biggl\{ \sum _{k=0}^{\infty }\frac{1}{(k+r)^{2n}}+\sum _{k=1}^{\infty }\frac{1}{(k-r)^{2n}} \Biggr\} \\ &=-\frac{(2n-1)!}{\pi ^{2n}}\sum_{k=0}^{\infty } \biggl\{ \frac{1}{(k+r)^{2n}}+ \frac{1}{(k+s)^{2n}} \biggr\} . \end{aligned}$$
Furthermore, by considering the following rational fraction expansion of \(\psi (x)=\csc x\) [30]: we can obtain Lemma 2.2.
$$ \psi (x)=\frac{1}{x}+\sum_{k=1}^{\infty }(-1)^{k} \biggl( \frac{1}{x+k\pi }+\frac{1}{x-k\pi } \biggr), $$
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 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 Applying (2.7) to (2.6), we obtain In addition, letting \(r=\frac{1}{4}\) in (2.1), we can also obtain 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
$$ \varphi ^{(2n-1)} \biggl(\frac{\pi }{2} \biggr)=- \frac{2^{2n+1}}{\pi ^{2n}}(2n-1)!\sum_{k=0}^{\infty} \frac{1}{(2k+1)^{2n}}. $$
(2.6)
$$ \sum_{k=0}^{\infty } \frac{1}{(2k+1)^{2n}}=\sum_{k=1}^{\infty } \frac{1}{k^{2n}}-\sum_{k=1}^{\infty } \frac{1}{(2k)^{2n}}= \frac{2^{2n}-1}{2(2n)!}B_{n}\pi ^{2n}. $$
(2.7)
$$ \varphi ^{(2n-1)} \biggl(\frac{\pi }{2} \biggr)=-\frac{B_{n}}{n}2^{2n-1} \bigl(2^{2n}-1 \bigr). $$
(2.8)
$$ \varphi ^{(2n-1)} \biggl(\frac{\pi }{4} \biggr)=-\frac{B_{n}}{n}4^{2n-1} \bigl(2^{2n}-1 \bigr). $$
(2.9)
$$ E_{n}=\frac{1}{4^{n}}\varphi ^{(2n)} \biggl(\frac{\pi }{4} \biggr)= \psi ^{(2n)} \biggl(\frac{\pi }{2} \biggr). $$
(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 and Then 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 \(\beta \in {\mathbb{N}}^{+}\).
$$ K(x,y):= \frac{ \vert c^{xy^{\delta }}+\eta _{2}d^{xy^{\delta }} \vert }{ \vert a^{xy^{\delta }}+\eta _{1}b^{xy^{\delta }} \vert } $$
(2.11)
$$\begin{aligned} C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) :={}&\sum _{k=0}^{\infty } \biggl( \frac{(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{a}{c})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{a}{d})^{\beta }} \biggr) \\ & {}+\sum_{k=0}^{\infty } \biggl( \frac{(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{d}{b})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{c}{b})^{\beta }} \biggr). \end{aligned}$$
(2.12)
$$ \int _{\mathbb{R}} K(t,1) \vert t \vert ^{\beta -1}\, \mathrm{d} t= \Gamma ( \beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ), $$
(2.13)
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)
$$ \frac{1}{a^{t}+\eta _{1}b^{t}} = \frac{a^{-t}}{1+\eta _{1} {(a^{-1}b)}^{t}}=\sum _{k=0}^{\infty }(-\eta _{1})^{k} \biggl\{ \biggl(\frac{b}{a} \biggr)^{k} \frac{1}{a} \biggr\} ^{t}. $$
(2.15)
$$\begin{aligned} I_{1}&= \int _{\mathbb{R}_{+}} K(t,1) \vert t \vert ^{\beta -1}{ \, \mathrm{d} t} \\ &= \sum_{k=0}^{\infty }(-\eta _{1})^{k} \int _{ \mathbb{R}_{+}} \biggl\{ \biggl(\frac{b}{a} \biggr)^{k}\frac{c}{a} \biggr\} ^{t} t^{\beta -1}{\,\mathrm{d} t}+\sum_{k=0}^{\infty }(-\eta _{1})^{k} \eta _{2} \int _{\mathbb{R}_{+}} \biggl\{ \biggl(\frac{b}{a} \biggr)^{k}\frac{d}{a} \biggr\} ^{t}t^{ \beta -1} {\,\mathrm{d} t} \\ &:=\sum_{k=0}^{\infty } \bigl\{ (- \eta _{1})^{k}I_{11}+(-\eta _{1})^{k} \eta _{2}I_{12} \bigr\} . \end{aligned}$$
(2.16)
$$ I_{11}= \frac{1}{(k\ln \frac{a}{b}+\ln \frac{a}{c})^{\beta }} \int _{ \mathbb{R}_{+}}e^{-u}u^{\beta -1} {\,\mathrm{d} u}= \frac{\Gamma (\beta )}{(k\ln \frac{a}{b}+\ln \frac{a}{c})^{\beta }}. $$
(2.17)
$$ I_{12}= \frac{1}{(k\ln \frac{a}{b}+\ln \frac{a}{d})^{\beta }} \int _{ \mathbb{R}_{+}} e^{-u}u^{\beta -1} { \, \mathrm{d} u}= \frac{\Gamma (\beta )}{(k\ln \frac{a}{b}+\ln \frac{a}{d})^{\beta }}. $$
(2.18)
$$ I_{1}=\Gamma (\beta )\sum _{k=0}^{\infty } \biggl( \frac{(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{a}{c})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{a}{d})^{\beta }} \biggr). $$
(2.19)
$$\begin{aligned} I_{2}&= \int _{\mathbb{R}_{-}} K(t,1) \vert t \vert ^{\beta -1}{ \, \mathrm{d} t} = \int _{\mathbb{R}_{+}} K(-t,1) t^{\beta -1}{ \,\mathrm{d} t} \\ &=\Gamma (\beta )\sum_{k=0}^{\infty } \biggl( \frac{(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{d}{b})^{\beta }}+ \frac{\eta _{2}(-\eta _{1})^{k}}{(k\ln \frac{a}{b}+\ln \frac{c}{b})^{\beta }} \biggr). \end{aligned}$$
(2.20)
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 Then
$$\begin{aligned}& f_{n}(x)= \textstyle\begin{cases} 0, &x\in [-1, 1], \\ \vert x \vert ^{{\beta -1}-\frac{2}{np}}, &x\in \mathbb{R}\setminus [-1, 1], \end{cases}\displaystyle \\& g_{n}(y)= \textstyle\begin{cases} \vert y \vert ^{{\delta \beta -1}+\frac{2\delta }{nq}}, &y\in D_{\delta }, \\ 0, &y\in \mathbb{R}\setminus D_{\delta }. \end{cases}\displaystyle \end{aligned}$$
$$ \frac{1}{n} \int _{\mathbb{R}} \int _{\mathbb{R}}K(x, y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y}=C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )+o(1). $$
(2.21)
Proof
Setting \(D_{\delta }^{+}=\{y: y>0, y\in D_{\delta }\}\), \(D_{\delta }^{-}=\{y: y<0, y\in D_{\delta }\}\), we get Setting \(xy^{\delta }=t\), and using Fubini’s theorem, where δ is equal to 1 or −1, we can get Similarly, setting \(xy^{\delta }=-t\), we can also obtain Applying (2.23) and (2.24) to (2.22), we have Letting \(n\to \infty \) in (2.25), and using (2.13), we get The proof of Lemma 2.5 is completed. □
$$\begin{aligned} & \int _{\mathbb{R}} \int _{\mathbb{R}}K(x, y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad = \int _{1}^{\infty }x^{{\beta -1}-\frac{2}{np}} \int _{D_{\delta }^{+}} K(x, y)y^{\delta \beta -1+\frac{2\delta }{nq}}{\,\mathrm{d} y \, \mathrm{d} x} \\ &\qquad {}+ \int _{1}^{\infty }x^{{\beta -1}-\frac{2}{np}} \int _{D_{\delta }^{-}} K(x, y) \vert y \vert ^{\delta \beta -1+\frac{2\delta }{nq}}{\, \mathrm{d} y \,\mathrm{d} x} \\ &\qquad {}+ \int _{-\infty }^{-1} \vert x \vert ^{{\beta -1}-\frac{2}{np}} \int _{D_{ \delta }^{+}} K(x, y)y^{\delta \beta -1+\frac{2\delta }{nq}}{ \,\mathrm{d} y \, \mathrm{d} x} \\ &\qquad {}+ \int _{-\infty }^{-1} \vert x \vert ^{{\beta -1}-\frac{2}{np}} \int _{D_{ \delta }^{-}} K(x, y) \vert y \vert ^{\delta \beta -1+\frac{2\delta }{nq}}{ \, \mathrm{d} y \,\mathrm{d} x} \\ &\quad =J_{1}+J_{2}+J_{3}+J_{4}. \end{aligned}$$
(2.22)
$$\begin{aligned} J_{1}&=J_{4}= \int _{1}^{\infty }x^{-1-\frac{2}{n}} \biggl( \int _{0}^{x} K(t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} \biggr){ \,\mathrm{d} x} \\ &= \int _{1}^{\infty }x^{-1-\frac{2}{n}} \int _{0}^{1} K(t, 1)t^{ \beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} {\,\mathrm{d} x} \\ &\quad {}+ \int _{1}^{\infty }x^{-1-\frac{2}{n}} \int _{1}^{x} K(t, 1)t^{\beta -1+ \frac{2}{nq}}{ \, \mathrm{d} t} {\,\mathrm{d} x} \\ &= \frac{n}{2} \int _{0}^{1} K(t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} + \int _{1}^{\infty }K(t, 1)t^{\beta -1+ \frac{2}{nq}} \int _{t}^{\infty }x^{-1-\frac{2}{n}}{\,\mathrm{d} x} { \,\mathrm{d} t} \\ &= \frac{n}{2} \int _{0}^{1} K(t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} +\frac{n}{2} \int _{1}^{\infty }K(t, 1)t^{ \beta -1-\frac{2}{np}}{ \, \mathrm{d} t}. \end{aligned}$$
(2.23)
$$\begin{aligned} J_{2}&=J_{3}= \int _{1}^{\infty }x^{-1-\frac{2}{n}} \int _{0}^{x} K(-t, 1)t^{ \beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} {\,\mathrm{d} x} \\ &= \frac{n}{2} \int _{0}^{1} K(-t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} +\frac{n}{2} \int _{1}^{\infty }K(-t, 1)t^{ \beta -1-\frac{2}{np}}{ \, \mathrm{d} t}. \end{aligned}$$
(2.24)
$$\begin{aligned} & \frac{1}{n} \int _{\mathbb{R}} \int _{\mathbb{R}}K(x, y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad = \int _{0}^{1} K(t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} + \int _{1}^{\infty }K(t, 1)t^{\beta -1-\frac{2}{np}}{ \, \mathrm{d} t} \\ &\qquad {}+ \int _{0}^{1} K(-t, 1)t^{\beta -1+\frac{2}{nq}}{ \, \mathrm{d} t} + \int _{1}^{\infty }K(-t, 1)t^{\beta -1-\frac{2}{np}}{ \, \mathrm{d} t}. \end{aligned}$$
(2.25)
$$\begin{aligned} & \frac{1}{n} \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad = \int _{\mathbb{R}^{+}} K(t, 1)t^{\beta -1}{\,\mathrm{d} t} + \int _{\mathbb{R}^{+}}K(-t, 1)t^{\beta -1}{\,\mathrm{d} t}+o(1) \\ &\quad = \int _{\mathbb{R}}K(t, 1)t^{\beta -1}{\,\mathrm{d} t}+o(1)=C_{ \eta _{1},\eta _{2}}(a, b, c, d, \beta )+o(1). \end{aligned}$$
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 where the constant factor \(\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )\) is the best possible.
$$ \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f(x)g(y){\,\mathrm{d} x\,\mathrm{d} y} < \Gamma ( \beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }, $$
(3.1)
Proof
By Hölder’s inequality, we obtain where \(\omega (x)=\int _{\mathbb{R}} K(x, y)| y| ^{\delta \beta -1} \,\mathrm{d} y\), and \(\varpi (y)=\int _{\mathbb{R}} K(x, y)| x| ^{\beta -1}{ \,\mathrm{d} x} \).
$$\begin{aligned} & \int _{\mathbb{R}} \int _{\mathbb{R}} K(x, y)f(x)g(y){\,\mathrm{d} x\,\mathrm{d} y} \\ & \quad = \int _{\mathbb{R}} \int _{\mathbb{R}} \bigl( \bigl( K(x, y) \bigr)^{\frac{1}{p}} \vert y \vert ^{\frac{\delta \beta -1}{p}} \vert x \vert ^{\frac{1-\beta }{q}}f(x) \bigr) \\ &\qquad {}\times \bigl( \bigl( K(x, y) \bigr)^{\frac{1}{q}} \vert x \vert ^{ \frac{\beta -1}{q}} \vert y \vert ^{\frac{1-\delta \beta }{p}}g(y) \bigr){ \, \mathrm{d} x\,\mathrm{d} y} \\ & \quad \leq \biggl( \int _{\mathbb{R}} \int _{\mathbb{R}} K(x, y) \vert y \vert ^{\delta \beta -1} \vert x \vert ^{\frac{p(1-\beta )}{q}}f^{p}(x){ \,\mathrm{d} x\, \mathrm{d} y} \biggr)^{\frac{1}{p}} \\ & \qquad {} \times \biggl( \int _{\mathbb{R}} \int _{\mathbb{R}} K(x, y) \vert x \vert ^{\beta -1} \vert y \vert ^{\frac{q(1-\delta \beta )}{p}}g^{q}(y){ \,\mathrm{d} x\, \mathrm{d} y} \biggr)^{\frac{1}{q}} \\ &\quad = \biggl( \int _{\mathbb{R}} \omega (x) \vert x \vert ^{\frac{p(1-\beta )}{q}}f^{p}(x){ \,\mathrm{d} x} \biggr)^{\frac{1}{p}} \biggl( \int _{\mathbb{R}} \varpi (y) \vert y \vert ^{\frac{q(1-\delta \beta )}{p}}g^{q}(y)dy \biggr)^{\frac{1}{q}}, \end{aligned}$$
(3.2)
Setting \(xy^{\eta }=t\), and using (2.13), we have and Plugging (3.3) and (3.4) back to (3.2), we have 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.e. in \(\mathbb{R}\times \mathbb{R}\), that is, 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.
$$\begin{aligned} \omega (x)&= \vert x \vert ^{-\beta } \int _{\mathbb{R}}K(t,1) \vert t \vert ^{ \beta -1}{\, \mathrm{d} t} \\ &=\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \vert x \vert ^{-\beta } \quad (x\neq 0) \end{aligned}$$
(3.3)
$$\begin{aligned} \varpi (y)&= \vert y \vert ^{-\delta \beta } \int _{\mathbb{R}}K(t,1) \vert t \vert ^{\beta -1}{\, \mathrm{d} t} \\ &=\Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \vert y \vert ^{-\delta \beta }\quad (y\neq 0). \end{aligned}$$
(3.4)
$$ \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f(x)g(y){\,\mathrm{d} x \,\mathrm{d} y} \leq \Gamma (\beta )C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(3.5)
$$ A_{1} K(x, y) \vert y \vert ^{\delta \beta -1} \vert x \vert ^{ \frac{p(1-\beta )}{q}}f^{p}(x)=A_{2} K(x, y) \vert x \vert ^{\beta -1} \vert y \vert ^{\frac{q(1-\delta \beta )}{p}}g^{q}(y), $$
$$ A_{1} \vert x \vert ^{p(1-\beta )}f^{p}(x)=A_{2} \vert y \vert ^{q(1-\delta \beta )}g^{q}(y), $$
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, Replacing f and g in (3.6) with \(f_{n}\) and \(g_{n}\) defined in Lemma 2.5, respectively, we obtain 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. □
$$ \int _{\mathbb{R}} \int _{\mathbb{R}}K(x, y)f(x)g(y){\,\mathrm{d} x\,\mathrm{d} y} < k \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(3.6)
$$\begin{aligned} & \frac{1}{n} \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f_{n}(x)g_{n}(y){ \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < \frac{k}{n} \biggl( \int _{1}^{\infty }x^{-\frac{2}{n}-1}{ \,\mathrm{d} x}+ \int _{-\infty }^{-1} \vert x \vert ^{- \frac{2}{n}-1}{ \,\mathrm{d} x} \biggr)^{\frac{1}{p}} \\ &\qquad {}\times \biggl( \int _{D_{\delta }} \vert x \vert ^{\frac{2\delta }{n}-1}{ \,\mathrm{d} x} \biggr)^{\frac{1}{q}}=k. \end{aligned}$$
(3.7)
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 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).
$$ \int _{\mathbb{R}} \vert y \vert ^{p\delta \beta -1} \biggl( \int _{\mathbb{R}}K(x,y)f(x){ \,\mathrm{d} x} \biggr)^{p} \,\mathrm{d} y< \bigl(C_{\eta _{1}, \eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \bigr)^{p}, $$
(3.8)
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 Therefore 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 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. □
$$\begin{aligned} 0&< \bigl( \Vert g \Vert _{q,\nu } \bigr)^{pq}= \biggl( \int _{\mathbb{R}} \vert y \vert ^{q(1-\delta \beta )-1}g^{q}(y) \,\mathrm{d} y \biggr)^{p} \\ &= \biggl( \int _{\mathbb{R}} \vert y \vert ^{p\delta \beta -1} \biggl( \int _{\mathbb{R}}K(x,y)f(x){\,\mathrm{d} x} \biggr)^{p} \,\mathrm{d} y \biggr)^{p} \\ &= \biggl( \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f(x)g(y){\,\mathrm{d} x\,\mathrm{d} y} \biggr)^{p} \leq \bigl(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q,\nu } \bigr)^{p}. \end{aligned}$$
(3.9)
$$\begin{aligned} 0&< \int _{\mathbb{R}} \vert y \vert ^{p\delta \beta -1} \biggl( \int _{\mathbb{R}}K (x,y)f(x){\,\mathrm{d} x} \biggr)^{p} \,\mathrm{d} y \\ &= \bigl( \Vert g \Vert _{q,\nu } \bigr)^{q} \leq \bigl(C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta ) \Vert f \Vert _{p,\mu } \bigr)^{p}. \end{aligned}$$
(3.10)
$$\begin{aligned} & \int _{\mathbb{R}} \int _{\mathbb{R}}K(x,y)f(x)g(y){\,\mathrm{d} x \,\mathrm{d} y} \\ &\quad = \int _{\mathbb{R}} \biggl( \vert y \vert ^{-(1-\delta \beta -\frac{1}{q})} \int _{\mathbb{R}} K(x,y)f(x){\,\mathrm{d} x} \biggr) \bigl( \vert y \vert ^{1-\delta \beta -\frac{1}{q}}g(y) \bigr)\,\mathrm{d} y \\ &\quad \leq \biggl( \int _{\mathbb{R}} \vert y \vert ^{p\delta \beta -1} \biggl( \int _{\mathbb{R}} K(x,y)f(x){\,\mathrm{d} x} \biggr)^{p} \,\mathrm{d} y \biggr)^{\frac{1}{p}} \Vert g \Vert _{q,\nu }. \end{aligned}$$
(3.11)
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 Therefore, the following corollary holds.
$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{-2}{(2n-1)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\varphi ^{(2n-1)} \biggl( \frac{\pi \ln a}{\ln \frac{a}{b}} \biggr). $$
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 Let \(p=q=2 \), \(\delta =1\) in (4.2), then we have (1.9).
$$ \int _{\mathbb{R}} \int _{\mathbb{R}} \bigl\vert \operatorname{csch} \bigl(xy^{\delta } \bigr) \bigr\vert f(x)g(y) {\,\mathrm{d} x\,\mathrm{d} y}< \frac{B_{n}}{n} \bigl(2^{2n}-1 \bigr) \pi ^{2n} \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(4.2)
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 Therefore, we obtain the following corollary.
$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{2}{(2n)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n+1}\psi ^{(2n)} \biggl( \frac{\pi \ln a}{\ln \frac{a}{b}} \biggr). $$
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 Let \(p=q=2\) and \(\delta =1\) in (4.4), then we get (1.8).
$$ \int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech} \bigl(xy^{\delta } \bigr)f(x)g(y) {\,\mathrm{d} x\,\mathrm{d} y}< \frac{E_{n}}{4^{n}}\pi ^{2n+1} \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
(4.4)
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 Hence, we can obtain another corollary as follows.
$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{-2}{(2n-1)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\varphi ^{(2n-1)} \biggl( \frac{\pi \ln (a/c)}{\ln (a/b)} \biggr). $$
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 Letting \(\lambda _{1}=2\) and \(\lambda _{2}=1\) in (4.6), in view of (2.9), we can also obtain (4.2).
$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}} \bigl\vert \operatorname{csch} \bigl(\lambda _{1}xy^{ \delta } \bigr) \bigr\vert \operatorname{cosh} \bigl(\lambda _{2}xy^{\delta } \bigr)f(x)g(y) { \,\mathrm{d} x \,\mathrm{d} y} \\ &\quad < -2 \biggl(\frac{\pi }{2\lambda _{1}} \biggr)^{2n}\varphi ^{(2n-1)} \biggl(\frac{(\lambda _{1}-\lambda _{2})\pi }{2\lambda _{1}} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.6)
Letting \(\lambda _{1}=4\) 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 Therefore, the following corollary holds obviously.
$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{2}{(2n)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n+1}\varphi ^{(2n)} \biggl( \frac{\pi \ln (a/c)}{\ln (a/b)} \biggr). $$
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 Let \(\lambda _{1}=2\) and \(\lambda _{2}=1\) in (4.9), then it follows from (2.10) that we also get (4.4).
$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{csch} \bigl(\lambda _{1}xy^{ \delta } \bigr)\operatorname{sinh} \bigl(\lambda _{2}xy^{\delta } \bigr) f(x)g(y) { \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < 2 \biggl(\frac{\pi }{2\lambda _{1}} \biggr)^{2n+1}\varphi ^{(2n)} \biggl(\frac{(\lambda _{1}-\lambda _{2})\pi }{2\lambda _{1}} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.9)
Let \(\lambda _{1}=4\) 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 Therefore, we obtain the following corollary.
$$ C_{\eta _{1},\eta _{2}}(a, b, c, d, \beta )=\frac{-2}{(2n-1)!} \biggl( \frac{\pi }{\ln \frac{a}{b}} \biggr)^{2n}\psi ^{(2n-1)} \biggl( \frac{\pi \ln (a/c)}{\ln (a/b)} \biggr). $$
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 Let \(\lambda _{1}=2\) and \(\lambda _{2}=1\) in (4.12), then we can have
$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech}\bigl(\lambda _{1}xy^{ \delta } \bigr) \bigl\vert \operatorname{sinh} \bigl(\lambda _{2}xy^{\delta } \bigr) \bigr\vert f(x)g(y) { \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < -2 \biggl(\frac{\pi }{2\lambda _{1}} \biggr)^{2n}\psi ^{(2n-1)} \biggl( \frac{(\lambda _{1}-\lambda _{2})\pi }{2\lambda _{1}} \biggr) \Vert f \Vert _{p, \mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(4.12)
$$\begin{aligned} &\int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{sech} \bigl(xy^{\delta } \bigr) \bigl\vert \operatorname{tanh} \bigl(2xy^{\delta } \bigr) \bigr\vert f(x)g(y) { \,\mathrm{d} x\,\mathrm{d} y} \\ &\quad < \frac{-\pi ^{2n}}{4^{2n-1}}\psi ^{(2n-1)} \biggl( \frac{\pi }{4} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. \end{aligned}$$
(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 Letting \(\lambda _{1}=2\), \(\lambda _{2}=1\) in (4.15), 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)
$$ \int _{\mathbb{R}} \int _{\mathbb{R}}\operatorname{csch} \bigl(xy^{\delta } \bigr) \operatorname{tanh} \bigl(2xy^{\delta } \bigr) f(x)g(y) {\, \mathrm{d} x \,\mathrm{d} y}< \frac{\pi ^{2n+1}}{4^{2n}}\psi ^{(2n)} \biggl( \frac{\pi }{4} \biggr) \Vert f \Vert _{p,\mu } \Vert g \Vert _{q, \nu }. $$
Acknowledgements
The author is indebted to the anonymous referees for their valuable suggestions and comments that helped improve the paper significantly.
Competing interests
The author declares that they have no competing interests.
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