On a generalization of a Hilbert-type integral inequality in the whole plane with a hypergeometric function

Abstract

By introducing some parameters, we establish a generalization of the Hilbert-type integral inequality in the whole plane with the homogeneous kernel of degree $-2\lambda$ and the best constant factor which involves the hypergeometric function.

MSC:26D15.

1 Introduction

If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$ and $f(x),g(x)\ge 0$ satisfy

$$0<{\int }_0^{\mathrm{\infty }}{f}^p(x)dx<\mathrm{\infty }\text{and}0<{\int }_0^{\mathrm{\infty }}{g}^q(x)dx<\mathrm{\infty },$$

then

$${\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{f(x)g(y)}{x+y}dxdy<\frac{\pi }{sin(\pi /p)}{\{{\int }_0^{\mathrm{\infty }}{f}^p(x)dx\}}^{\frac{1}{p}}{\{{\int }_0^{\mathrm{\infty }}{g}^q(x)dx\}}^{\frac{1}{q}},$$

(1)

where the constant factor $\pi /(sin\pi /p)$ is the best possible. Inequality (1) is called Hardy-Hilbert’s inequality [1] and is important in analysis and applications [2].

In 2001, Yang gave an extension of (1) involving beta function as (see [3]):

$$\begin{array}{r}{\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{f(x)g(y)}{{(x+y)}^{\lambda }}dxdy\\ <B(\frac{p+\lambda -2}{p},\frac{q+\lambda -2}{q}){\{{\int }_0^{\mathrm{\infty }}{x}^{1-\lambda }{f}^p(x)dx\}}^{\frac{1}{p}}{\{{\int }_0^{\mathrm{\infty }}{x}^{1-\lambda }{g}^q(x)dx\}}^{\frac{1}{q}},\end{array}$$

(2)

where the constant factor $B(\frac{p+\lambda -2}{p},\frac{q+\lambda -2}{q})$ ($\lambda >2-min\{p,q\}$) is the best possible.

Recently, some new Hilbert-type inequalities in the whole plane have been obtained [4, 5]. Xin and Yang in [5] established the following:

If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $|\beta |<1$, $0<{\alpha }_1<{\alpha }_2<\pi$, $f,g\ge 0$, satisfy

$$0<{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p\beta -1}{f}^p(x)dx<\mathrm{\infty }\text{and}0<{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy<\mathrm{\infty },$$

then we have

$$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{x^2+2xycos{\alpha }_i+y^2}\right\}f(x)g(y)dxdy\\ <k(\beta ){({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p\beta -1}{f}^p(x)dx)}^{\frac{1}{p}}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy)}^{\frac{1}{q}},\end{array}$$

(3)

and

$$\begin{array}{r}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{p(1-\beta )-1}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{x^2+2xycos{\alpha }_i+y^2}\right\}f(x)dx)}^{p}dy\\ <k^p(\beta ){\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p\beta -1}{f}^p(x)dx,\end{array}$$

(4)

where the constant factors $k(\beta )=\frac{\pi }{sin\beta \pi }(\frac{sin\beta {\alpha }_1}{sin{\alpha }_1}+\frac{sin\beta (\pi -{\alpha }_2)}{sin{\alpha }_2})$ and $k^p(\beta )$ are the best possible. Inequalities (3) and (4) are equivalent.

By introducing some parameters, we establish generalizations of inequalities (3) and (4) with the homogeneous kernel of degree $-2\lambda$ and the best constant factor which involves the hypergeometric function.

2 Preliminary lemmas

In order to prove our assertions, we need the following lemmas.

Recall that the hypergeometric function $F(\alpha ,\beta ;\gamma ;x)$ is defined [6] by

$$F(\alpha ,\beta ;\gamma ;x)=\sum _{r=0}^{\mathrm{\infty }}\frac{{(\alpha )}_r{(\beta )}_r}{{(\gamma )}_r}\frac{x^r}{r!},$$

(5)

where ${(\alpha )}_r$ is the Pochhammer symbol defined by

$${(\alpha )}_r=\alpha (\alpha +1)\cdots (\alpha +r-1)=\frac{\mathrm{\Gamma }(\alpha +r)}{\mathrm{\Gamma }(\alpha )}.$$

It is known the series (5) converges for $|x|<1$ and diverges for $|x|>1$. The hypergeometric function satisfies the integral representation

$$F(\alpha ,\beta ;\gamma ;x)=\frac{\mathrm{\Gamma }(\gamma )}{\mathrm{\Gamma }(\beta )\mathrm{\Gamma }(\gamma -\beta )}{\int }_0^1{t}^{\beta -1}{(1-t)}^{\gamma -\beta -1}{(1-xt)}^{-\alpha }dt,\text{if }\gamma >\beta >0.$$

Lemma 2.1 (See [7])

Suppose that $a,c>0$, $b^2<ac$, $0<\alpha <2\lambda$. Then we have

$${\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{{(a{x}^2+2bx+c)}^{\lambda }}dx=a^{-\frac{\alpha }{2}}{c}^{\frac{\alpha }{2}-\lambda }B(\alpha ,2\lambda -\alpha )F(\frac{\alpha }{2},\lambda -\frac{\alpha }{2};\lambda +\frac{1}{2};1-\frac{b^2}{ac}).$$

Lemma 2.2 Let $a,c,\lambda >0$, $b\ge 0$, $1-2\lambda <\beta <1$ and $0<{\alpha }_1<{\alpha }_2<\pi$ be real parameters such that $b^{2}max\{{cos}^2{\alpha }_1,{cos}^2(\pi -{\alpha }_2)\}<ac$. Define the weight functions $\omega (x)$ and $\varpi (y)$ ($x,y\in (-\mathrm{\infty },\mathrm{\infty })$) as follows:

$$\begin{array}{c}\omega (x):={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}\frac{{|x|}^{\beta +2\lambda -1}}{{|y|}^{\beta }}dy,\hfill \\ \varpi (y):={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}\frac{{|y|}^{1-\beta }}{{|x|}^{-\beta -2\lambda +2}}dx.\hfill \end{array}$$

Then we have $\omega (x)=\varpi (y)=C_{\lambda }$ ($x,y\ne 0$), where

$$\begin{array}{rl}{C}_{\lambda }=& a^{\frac{1-\beta }{2}-\lambda }{c}^{\frac{1-\beta }{2}}B(1-\beta ,2\lambda +\beta -1)\\ \times [F(\frac{1-\beta }{2},\lambda -\frac{1-\beta }{2};\lambda +\frac{1}{2};1-\frac{b^2{cos}^2{\alpha }_1}{ac})\\ +F(\frac{1-\beta }{2},\lambda -\frac{1-\beta }{2};\lambda +\frac{1}{2};1-\frac{b^2{cos}^2(\pi -{\alpha }_2)}{ac})].\end{array}$$

Proof For $x\in (-\mathrm{\infty },0)$, setting $u=y/x$, $u=-y/x$ in the following two integrals, respectively, and using Lemma 2.1, we get

$$\begin{array}{rcl}\omega (x)& =& {\int }_{-\mathrm{\infty }}^0\frac{1}{{(a{x}^2+2bxycos{\alpha }_1+c{y}^2)}^{\lambda }}\frac{{(-x)}^{\beta +2\lambda -1}}{{(-y)}^{\beta }}dy\\ +{\int }_0^{\mathrm{\infty }}\frac{1}{{(a{x}^2+2bxycos{\alpha }_2+c{y}^2)}^{\lambda }}\frac{{(-x)}^{\beta +2\lambda -1}}{y^{\beta }}dy\\ =& {\int }_0^{\mathrm{\infty }}\frac{u^{-\beta }}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du+{\int }_0^{\mathrm{\infty }}\frac{u^{-\beta }}{{(c{u}^2+2bucos(\pi -{\alpha }_2)+a)}^{\lambda }}du\\ =& C_{\lambda }.\end{array}$$

For $x\in (0,\mathrm{\infty })$, setting $u=-y/x$, $u=y/x$ in the following two integrals, respectively, and using Lemma 2.1, we get

$$\begin{array}{rcl}\omega (x)& =& {\int }_{-\mathrm{\infty }}^0\frac{1}{{(a{x}^2+2bxycos{\alpha }_2+c{y}^2)}^{\lambda }}\frac{x^{\beta +2\lambda -1}}{{(-y)}^{\beta }}dy\\ +{\int }_0^{\mathrm{\infty }}\frac{1}{{(a{x}^2+2bxycos{\alpha }_1+c{y}^2)}^{\lambda }}\frac{x^{\beta +2\lambda -1}}{y^{\beta }}dy\\ =& {\int }_0^{\mathrm{\infty }}\frac{u^{-\beta }}{{(c{u}^2+2bucos(\pi -{\alpha }_2)+a)}^{\lambda }}du+{\int }_0^{\mathrm{\infty }}\frac{u^{-\beta }}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du\\ =& C_{\lambda }.\end{array}$$

By the same way, we still can find that $\omega (x)=\varpi (y)=C_{\lambda }$ ($x,y\ne 0$). The lemma is proved. □

Lemma 2.3 Let p and q be conjugate parameters with $p>1$, and let $a,c,\lambda >0$, $b\ge 0$, $1-2\lambda <\beta <1$, $0<{\alpha }_1<{\alpha }_2<\pi$ and $b^{2}max\{{cos}^2{\alpha }_1,{cos}^2(\pi -{\alpha }_2)\}<ac$, and $f(x)$ be a nonnegative measurable function in $(-\mathrm{\infty },\mathrm{\infty })$, then we have

$$\begin{array}{rl}J& :={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{p(1-\beta )-1}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}f(x)dx)}^{p}dy\\ \le C_{\lambda }^p{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx.\end{array}$$

(6)

Proof By Lemma 2.2 and Hölder’s inequality [8], we have

$$\begin{array}{r}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}f(x)dx)}^p\\ =[{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}\\ \times {(\frac{{|x|}^{(-\beta -2\lambda +2)/q}}{{|y|}^{\beta /p}}f(x))\left(\frac{{|y|}^{\beta /p}}{{|x|}^{(-\beta -2\lambda +2)/q}}\right)dx]}^p\\ \le {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}\frac{{|x|}^{(1-p)(\beta +2\lambda -2)}}{{|y|}^{\beta }}{f}^p(x)dx\\ \times {\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}\frac{{|y|}^{(q-1)\beta }}{{|x|}^{(-\beta -2\lambda +2)}}dx\right)}^{p-1}\\ =C_{\lambda }^{p-1}{|y|}^{p(\beta -1)+1}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}\frac{{|x|}^{(1-p)(\beta +2\lambda -2)}}{{|y|}^{\beta }}{f}^p(x)dx.\end{array}$$

(7)

Then by the Fubini theorem, it follows that

$$\begin{array}{rl}J& \le C_{\lambda }^{p-1}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}[{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}\frac{{|x|}^{(1-p)(\beta +2\lambda -2)}}{{|y|}^{\beta }}{f}^p(x)dx]dy\\ =C_{\lambda }^{p-1}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\omega (x){|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx\\ =C_{\lambda }^p{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx.\end{array}$$

The lemma is proved. □

3 Main results

Theorem 3.1 Let p and q be conjugate parameters with $p>1$, and let $a,c,\lambda >0$, $b\ge 0$, $1-2\lambda <\beta <1$, $0<{\alpha }_1<{\alpha }_2<\pi$ and $b^{2}max\{{cos}^2{\alpha }_1,{cos}^2(\pi -{\alpha }_2)\}<ac$, and $f,g\ge 0$, satisfy $0<{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx<\mathrm{\infty }$ and $0<{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy<\mathrm{\infty }$. Then

$$\begin{array}{rl}I& :={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}f(x)g(y)dxdy\\ <C_{\lambda }{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx)}^{1/p}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy)}^{1/q},\end{array}$$

(8)

and

$$\begin{array}{rl}J& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{p(1-\beta )-1}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}f(x)dx)}^{p}dy\\ <C_{\lambda }^p{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx,\end{array}$$

(9)

where the constant factors $C_{\lambda }$ and $C_{\lambda }^p$ are the best possible and $C_{\lambda }$ is defined in Lemma  2.2. Inequalities (8) and (9) are equivalent.

Proof If (7) takes the form of the equality for a $y\in (-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$, then there exist constants A and B such that they are not all zero, and

$$A\frac{{|x|}^{(1-p)(\beta +2\lambda -2)}}{{|y|}^{\beta }}{f}^p(x)=B\frac{{|y|}^{(q-1)\beta }}{{|x|}^{(-\beta -2\lambda +2)}}\text{a.e. in }(-\mathrm{\infty },\mathrm{\infty })\times (-\mathrm{\infty },\mathrm{\infty }).$$

Hence, there exists a constant K such that

$$A{|x|}^{p(\beta +2\lambda -2)}{f}^p(x)=B{|y|}^{q\beta }=K\text{a.e. in }(-\mathrm{\infty },\mathrm{\infty })\times (-\mathrm{\infty },\mathrm{\infty }).$$

We suppose $A\ne 0$ (otherwise $B=A=0$). Then ${|x|}^{p(\beta +2\lambda -2)-1}{f}^p(x)=K/(A|x|)$ a.e. in $(-\mathrm{\infty },\mathrm{\infty })$, which contradicts the fact that $0<{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx<\mathrm{\infty }$. Hence, (7) takes the form of a strict inequality, so does (6), and we have (9).

By Hölder’s inequality [8], we have

$$\begin{array}{rl}I=& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}({|y|}^{1/q-\beta }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}f(x)dx)\\ \times ({|y|}^{\beta -1/q}g(y))dy\\ \le & J^{1/p}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy)}^{1/q}.\end{array}$$

(10)

By (9), we have (8). On the other hand, suppose that (8) is valid. Set

$$g(y)={|y|}^{p(1-\beta )-1}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}f(x)dx)}^{p-1},$$

then it follows $J={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy$. By (6), we have $J<\mathrm{\infty }$. If $J=0$, then (9) is obviously valid. If $0<J<\mathrm{\infty }$, then by (8), we obtain

$$\begin{array}{rcl}0& <& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy=J=I\\ <& C_{\lambda }{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx)}^{1/p}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy)}^{1/q},\end{array}$$

and

$$\begin{array}{rcl}{J}^{1/p}& =& {({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy)}^{1/p}\\ <& C_{\lambda }{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{f}^p(x)dx)}^{1/p}.\end{array}$$

Hence, we have (9), which is equivalent to (8).

For $\epsilon >0$, define functions $\tilde{f}(x)$, $\tilde{g}(y)$ as follows:

$$\begin{array}{rcl}\tilde{f}(x)& :=& \{\begin{array}{ll}{x}^{(\beta +2\lambda -2)-2\epsilon /p},& x\in (1,\mathrm{\infty }),\\ 0,& x\in [-1,1],\\ {(-x)}^{(\beta +2\lambda -2)-2\epsilon /p},& x\in (-\mathrm{\infty },-1),\end{array}\\ \tilde{g}(y)& :=& \{\begin{array}{ll}{y}^{-\beta -2\epsilon /q},& y\in (1,\mathrm{\infty }),\\ 0,& y\in [-1,1],\\ {(-y)}^{-\beta -2\epsilon /q},& y\in (-\mathrm{\infty },-1).\end{array}\end{array}$$

Then

$$\tilde{L}:={({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta +2\lambda -2)-1}{\tilde{f}}^p(x)dx)}^{1/p}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{\tilde{g}}^q(y)dy)}^{1/q}=\frac{1}{\epsilon },$$

and

$$\begin{array}{rcl}\tilde{I}& :=& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{{(a{x}^2+2bxycos{\alpha }_i+c{y}^2)}^{\lambda }}\right\}\tilde{f}(x)\tilde{g}(y)dxdy\\ =& I_1+I_2+I_3+I_4,\end{array}$$

where

$$\begin{array}{rcl}{I}_1& :=& {\int }_{-\mathrm{\infty }}^{-1}{(-x)}^{(\beta +2\lambda -2)-2\epsilon /p}\left[{\int }_{-\mathrm{\infty }}^{-1}\frac{{(-y)}^{-\beta -2\epsilon /q}}{{(a{x}^2+2bxycos{\alpha }_1+c{y}^2)}^{\lambda }}dy\right]dx,\\ I_2& :=& {\int }_{-\mathrm{\infty }}^{-1}{(-x)}^{(\beta +2\lambda -2)-2\epsilon /p}\left[{\int }_1^{\mathrm{\infty }}\frac{y^{-\beta -2\epsilon /q}}{{(a{x}^2+2bxycos{\alpha }_2+c{y}^2)}^{\lambda }}dy\right]dx,\\ I_3& :=& {\int }_1^{\mathrm{\infty }}{x}^{(\beta +2\lambda -2)-2\epsilon /p}\left[{\int }_{-\mathrm{\infty }}^{-1}\frac{{(-y)}^{-\beta -2\epsilon /q}}{{(a{x}^2+2bxycos{\alpha }_2+c{y}^2)}^{\lambda }}dy\right]dx,\end{array}$$

and

$$I_4:={\int }_1^{\mathrm{\infty }}{x}^{(\beta +2\lambda -2)-2\epsilon /p}\left[{\int }_1^{\mathrm{\infty }}\frac{y^{-\beta -2\epsilon /q}}{{(a{x}^2+2bxycos{\alpha }_1+c{y}^2)}^{\lambda }}dy\right]dx.$$

Taking $u=y/x$, by the Fubini theorem, we obtain

$$\begin{array}{rl}{I}_1=& I_4={\int }_1^{\mathrm{\infty }}{x}^{-1-2\epsilon }{\int }_{1/x}^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}dudx\\ =& {\int }_1^{\mathrm{\infty }}{x}^{-1-2\epsilon }({\int }_{1/x}^1\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du)dx\\ =& {\int }_0^1\left({\int }_{1/u}^{\mathrm{\infty }}{x}^{-1-2\epsilon }dx\right)\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du\\ +\frac{1}{2\epsilon }{\int }_1^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du\\ =& \frac{1}{2\epsilon }({\int }_0^1\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du),\end{array}$$

and

$$I_2=I_3=\frac{1}{2\epsilon }({\int }_0^1\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du).$$

In view of the above results, if the constant factor $C_{\lambda }$ in (8) is not the best possible, then there exists a positive number $\tilde{C}$ with $\tilde{C}<C_{\lambda }$ such that

$$\begin{array}{r}{\int }_0^1\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du\\ +{\int }_0^1\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du\\ =\epsilon \tilde{I}<\epsilon \tilde{C}\cdot \tilde{L}=\tilde{C}.\end{array}$$

(11)

By the Fatou lemma and (11), we have

$$\begin{array}{rcl}{C}_{\lambda }& =& {\int }_0^{\mathrm{\infty }}\frac{u^{-\beta }}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du+{\int }_0^{\mathrm{\infty }}\frac{u^{-\beta }}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du\\ =& {\int }_0^1\underset{\epsilon \to 0^{+}}{lim}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\underset{\epsilon \to 0^{+}}{lim}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du\\ +{\int }_0^1\underset{\epsilon \to 0^{+}}{lim}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\underset{\epsilon \to 0^{+}}{lim}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du\\ \le & \underset{\epsilon \to 0^{+}}{lim}[{\int }_0^1\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2+2bucos{\alpha }_1+a)}^{\lambda }}du\\ +{\int }_0^1\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du+{\int }_1^{\mathrm{\infty }}\frac{u^{-\beta -2\epsilon /q}}{{(c{u}^2-2bucos{\alpha }_2+a)}^{\lambda }}du]\\ \le & \tilde{C},\end{array}$$

which contradicts the fact that $\tilde{C}<C_{\lambda }$. Hence, the constant factor $C_{\lambda }$ in (8) is the best possible.

If the constant factor in (9) is not the best possible, then by (10), we may get a contradiction that the constant factor in (8) is not the best possible. Thus the theorem is proved. □

Remark 1 Setting $\lambda =a=b=c=1$ in Theorem 3.1, we have (3) and (4).

Remark 2 Setting $\lambda =1/2$ in Theorem 3.1, we have the following particular results:

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{\sqrt{a{x}^2+2bxycos{\alpha }_i+c{y}^2}}\right\}f(x)g(y)dxdy\\ <& C_{1/2}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta -1)-1}{f}^p(x)dx)}^{1/p}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|y|}^{q\beta -1}{g}^q(y)dy)}^{1/q},\end{array}$$

and

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}& {|y|}^{p(1-\beta )-1}{({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\underset{i\in \{1,2\}}{min}\left\{\frac{1}{\sqrt{a{x}^2+2bxycos{\alpha }_i+c{y}^2}}\right\}f(x)dx)}^{p}dy\\ <& C_{1/2}^p{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|x|}^{-p(\beta -1)-1}{f}^p(x)dx.\end{array}$$