## 1 Introduction

If f(x), g(x) ≥ 0, such that $0<{\int }_{0}^{\infty }{f}^{2}\left(x\right)\mathsf{\text{d}}x<\infty$ and $0<{\int }_{0}^{\infty }{g}^{2}\left(x\right)\mathsf{\text{d}}x<\infty$, then we have (cf. [1]):

$\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{x+y}\mathsf{\text{d}}x\mathsf{\text{d}}y<\pi {\left(\underset{0}{\overset{\infty }{\int }}{f}^{2}\left(x\right)\mathsf{\text{d}}x\underset{0}{\overset{\infty }{\int }}{g}^{2}\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{2}},$
(1)

where the constant factor π is the best possible. Inequality (1) is well known as Hilbert's integral inequality, which is important in Mathematical Analysis and its applications [2].

If p, r > 1, $\frac{1}{p}+\frac{1}{q}=1$, $\frac{1}{r}+\frac{1}{s}=1$ λ > 0, f(x), g(x) ≥ 0, such that $0<{\int }_{0}^{\infty }{x}^{p\left(1+\frac{\lambda }{r}\right)}{f}^{p}\left(x\right)\mathsf{\text{d}}x<\infty$ and, $0<{\int }_{0}^{\infty }{x}^{q\left(1+\frac{\lambda }{s}\right)}{g}^{q}\left(x\right)\mathsf{\text{d}}x<\infty$, then we have [3]:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{\left(min\left\{x,y\right\}\right)}^{\lambda }f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <\phantom{\rule{1em}{0ex}}\frac{rs}{\lambda }{\left(\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1+\frac{\lambda }{r}\right)-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{0}{\overset{\infty }{\int }}{x}^{q\left(1+\frac{\lambda }{s}\right)-1}{g}^{q}\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{q}},\end{array}$
(2)

where the constant factor $\frac{rs}{\lambda }$ is the best possible. By using the way of weight functions, we can get two Hilbert-type integral inequalities with non-homogeneous kernels similar to (1) and (2) as follows [4, 5]:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\frac{f\left(x\right)g\left(y\right)}{\mid 1+xy{\mid }^{\lambda }}\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <\phantom{\rule{1em}{0ex}}B\left(\frac{\lambda }{2},\phantom{\rule{2.77695pt}{0ex}}\frac{\lambda }{2}\right){\left\{\underset{-\infty }{\overset{\infty }{\int }}{x}^{p\left(1-\frac{\lambda }{2}\right)-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\underset{-\infty }{\overset{\infty }{\int }}{x}^{q\left(1-\frac{\lambda }{2}\right)-1}{g}^{q}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}\left(\lambda >0\right),\end{array}$
(3)
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\underset{0}{\overset{\infty }{\int }}\underset{0}{\overset{\infty }{\int }}{\left(min\left\{1,xy\right\}\right)}^{\lambda }f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <\phantom{\rule{1em}{0ex}}\frac{\lambda }{\alpha \left(\lambda -\alpha \right)}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1+\alpha \right)-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{p}}{\left\{\underset{0}{\overset{\infty }{\int }}{x}^{p\left(1+\alpha \right)-1}{g}^{q}\left(x\right)\mathsf{\text{d}}x\right\}}^{\frac{1}{q}}\left(0<\alpha \phantom{\rule{2.77695pt}{0ex}}<\lambda \right)\end{array}.$
(4)

Some inequalities with the non-homogenous kernels have been studied in [68].

In this paper, by using the way of weight functions and the technique of real analysis, a new Hilbert-type integral inequality in the whole plane with the non-homogenous kernel and a best constant factor is built. As applications, the equivalent forms, the reverses and some particular cases are obtained.

## 2 Some lemmas

Lemma 1 If 0 < α1 < α2 < π, define the weight functions ω(y) and $\stackrel{̃}{\omega }\left(x\right)$ as follow:

$\omega \left(y\right):=\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\frac{1}{\mid x\mid }\mathsf{\text{d}}x,\left(y\in \left(-\infty ,\infty \right)\right),$
(5)
$\stackrel{̃}{\omega }\left(x\right):=\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\frac{1}{\mid y\mid }\mathsf{\text{d}}y,\left(x\in \left(-\infty ,\infty \right)\right).$
(6)

Then, we have $\omega \left(y\right)=\stackrel{̃}{\omega }\left(x\right)=k\left(x,y\ne 0\right)$ where

$k:=2ln\left[\left(1+sec\frac{{\alpha }_{1}}{2}\right)\left(1+csc\frac{{\alpha }_{2}}{2}\right)\right].$
(7)

Proof. Setting u = x · |y| in (5), we find

$\omega \left(y\right)=\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{\mid u\mid ,1\right\}}{\sqrt{{u}^{2}+2u\left(y∕\mid y\mid \right)cos{\alpha }_{i}+1}}\right\}\frac{1}{\mid u\mid }\mathsf{\text{d}}u.$
(8)

For y ∈ (0, ∞), we have

$\begin{array}{c}\omega \left(y\right)=\underset{-\infty }{\overset{-1}{\int }}\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{2}+1}}\frac{-1}{u}\text{d}u+\underset{-1}{\overset{0}{\int }}\frac{\text{d}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{2}+1}}\\ \phantom{\rule{0.5em}{0ex}}\phantom{\rule{0.5em}{0ex}}+\underset{0}{\overset{1}{\int }}\frac{\text{d}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\underset{1}{\overset{\infty }{\int }}\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\frac{\text{d}u}{u}.\end{array}$
(9)

Setting

$\begin{array}{c}{\omega }_{1}:\phantom{\rule{2.77695pt}{0ex}}=\underset{-\infty }{\overset{-1}{\int }}\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{2}+1}}\frac{-1}{u}\mathsf{\text{d}}u,\\ {\omega }_{2}:\phantom{\rule{2.77695pt}{0ex}}=\underset{-1}{\overset{0}{\int }}\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{2}+1}}\mathsf{\text{d}}u,\\ {\omega }_{3}:\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\mathsf{\text{d}}u,\\ {\omega }_{4}:\phantom{\rule{2.77695pt}{0ex}}=\phantom{\rule{2.77695pt}{0ex}}\underset{1}{\overset{\infty }{\int }}\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\frac{1}{u}\mathsf{\text{d}}u,\end{array}$

we find

$\begin{array}{c}{\omega }_{1}\stackrel{v=-u}{=}\underset{1}{\overset{\infty }{\int }}\frac{\mathsf{\text{d}}v}{v\sqrt{{v}^{2}+2vcos\left(\pi -{\alpha }_{2}\right)+1}}\\ \stackrel{z=1∕v}{\phantom{\rule{1em}{0ex}}=}\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}z}{\sqrt{{z}^{2}+2zcos\left(\pi -{\alpha }_{2}\right)+1}},\\ {\omega }_{2}\stackrel{v=-u}{=}\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}v}{\sqrt{{v}^{2}+2vcos\left(\pi -{\alpha }_{2}\right)+1}}={\omega }_{1},\\ {\omega }_{4}\stackrel{z=1∕u}{=}\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}z}{\sqrt{{z}^{2}+2zcos{\alpha }_{1}+1}}={\omega }_{3}.\end{array}$

Then, we have

$\begin{array}{cc}\hfill \omega \left(y\right)& =2\left({\omega }_{1}+{\omega }_{3}\right)\hfill \\ =2\left\{\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}u}{\sqrt{{u}^{2}+2ucos\left(\pi -{\alpha }_{2}\right)+1}}+\underset{0}{\overset{1}{\int }}\frac{\mathsf{\text{d}}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\right\}\hfill \\ =\phantom{\rule{2.77695pt}{0ex}}2\left\{\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{{\left[u+cos\left(\pi -{\alpha }_{2}\right)\right]}^{2}+{sin}^{2}\left(\pi -{\alpha }_{2}\right)}}\mathsf{\text{d}}u\hfill \\ +\phantom{\rule{2.77695pt}{0ex}}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{{\left(u+cos{\alpha }_{1}\right)}^{2}+{sin}^{2}{\alpha }_{1}}}\mathsf{\text{d}}u\right\}\hfill \\ =2\left\{ln\left(1+csc\frac{{\alpha }_{2}}{2}\right)+ln\left(1+sec\frac{{\alpha }_{1}}{2}\right)\right\}\hfill \\ =2ln\left[\left(1+sec\frac{{\alpha }_{1}}{2}\right)\left(1+csc\frac{{\alpha }_{2}}{2}\right)\right]=k.\hfill \end{array}$

For y ∈ (-∞, 0), we can obtain

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

Lemma 2 If p > 1, $\frac{1}{p}+\frac{1}{q}=1$, 0 < α1 < α2 < π, f (x) is a nonnegative measurable function in (-∞,∞), then we have

$\begin{array}{l}J\phantom{\rule{0.5em}{0ex}}:\phantom{\rule{0.5em}{0ex}}=\underset{-\infty }{\overset{\infty }{\int }}|y{|}^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{\mathrm{min}}\left\{\frac{min\left\{1,|xy|\right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}f\left(x\right)\text{d}x\right]}^{p}\text{d}y\\ \phantom{\rule{0.5em}{0ex}}\le \phantom{\rule{0.5em}{0ex}}{k}^{p}\underset{-\infty }{\overset{\infty }{\int }}|x{|}^{p-1}{f}^{p}\left(x\right)dx.\end{array}$
(10)

Proof. By Lemma 1 and Hölder's inequality [9], we have

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,|xy|\right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}f\left(x\right)\mathsf{\text{d}}x\right)}^{p}\hfill \\ =\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,|xy|\right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\left[\frac{|x{|}^{1∕q}}{|y{|}^{1∕p}}f\left(x\right)\right]\left[\frac{|y{|}^{1∕p}}{|x{|}^{1∕q}}\right]\mathsf{\text{d}}x\right)}^{p}\hfill \\ \le \phantom{\rule{2.77695pt}{0ex}}\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,|xy|\right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\frac{|x{|}^{p-1}}{|y|}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right]\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}×{\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,|xy|\right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\frac{|y{|}^{q-1}}{|x|}\mathsf{\text{d}}x\right]}^{p-1}\hfill \\ =\phantom{\rule{1em}{0ex}}\phantom{\rule{2.77695pt}{0ex}}\omega {\left(y\right)}^{p-1}|y|\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,|xy|\right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\frac{|x{|}^{p-1}}{|y|}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right]\hfill \end{array}$
(11)

Then, by (6), (11) and Fubini theorem [10], it follows

$\begin{array}{cc}\hfill J& \le {k}^{p-1}\underset{-\infty }{\overset{\infty }{\int }}\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\frac{\mid x{\mid }^{p-1}}{\mid y\mid }{f}^{p}\left(x\right)\mathsf{\text{d}}x\right]\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{d}}y\hfill \\ ={k}^{p-1}\underset{-\infty }{\overset{\infty }{\int }}\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\frac{1}{\mid y\mid }\mathsf{\text{d}}y\right]\mid x{\mid }^{p-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\hfill \\ ={k}^{p-1}\underset{-\infty }{\overset{\infty }{\int }}\stackrel{̃}{\omega }\left(x\right)\mid x{\mid }^{p-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\hfill \\ ={k}^{p}\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^{p}\left(x\right)dx\hfill \end{array}$

The lemma is proved.   □

## 3 Main results and applications

Theorem 3 If p > 1, $\frac{1}{p}+\frac{1}{q}=1$, 0 < α1 < α2 < π, f (x), g(x) ≥ 0, satisfying $0<{\int }_{-\infty }^{\infty }\mid x{\mid }^{p-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x<\infty$ and $0<{\int }_{-\infty }^{\infty }\mid y{\mid }^{q-1}{g}^{q}\left(y\right)\mathsf{\text{d}}y<\infty$, then we have

$\begin{array}{c}I\phantom{\rule{1em}{0ex}}:\phantom{\rule{1em}{0ex}}=\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ \phantom{\rule{1em}{0ex}}<\phantom{\rule{1em}{0ex}}k{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^{q}\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}},\end{array}$
(12)
$\begin{array}{cc}\hfill J& =\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}f\left(x\right)\mathsf{\text{d}}x\right]}^{p}\mathsf{\text{d}}y\hfill \\ <{k}^{p}\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^{p}\left(x\right)dx,\hfill \end{array}$
(13)

where the constant factors k and kp are the best possible (k is defined by (7)). Inequality (12) and (13) are equivalent.

Proof. If (11) takes the form of equality for a y ∈ (-∞, 0) ∪ (0, ∞), then there exists constants M and N, such that they are not all zero, and

Hence, there exists a constant C, such that

$M\mid x{\mid }^{p}{f}^{p}\left(x\right)=N\mid y{\mid }^{q}=Ca.e.\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}\left(-\infty ,\phantom{\rule{2.77695pt}{0ex}}\infty \right).$

We suppose M ≠ 0 (otherwise N = M = 0). Then, it follows

$\mid x{\mid }^{p-1}{f}^{p}\left(x\right)=\frac{C}{M\mid x\mid }a.e.\phantom{\rule{2.77695pt}{0ex}}in\phantom{\rule{2.77695pt}{0ex}}\left(-\infty ,\phantom{\rule{2.77695pt}{0ex}}\infty \right),$

which contradicts the fact that $0<{\int }_{-\infty }^{\infty }\mid x{\mid }^{p-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x<\infty$. Hence, (11) takes the form of strict sign-inequality; so does (10), and we have (13).

By Hö lder's inequality [9], we have

$\begin{array}{cc}\hfill I& =\underset{-\infty }{\overset{\infty }{\int }}\left[\mid y{\mid }^{\frac{-1}{p}}\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}f\left(x\right)\mathsf{\text{d}}x\right]\left[\mid y{\mid }^{\frac{1}{p}}g\left(y\right)\mathsf{\text{d}}y\right]\hfill \\ \le {J}^{\frac{1}{p}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^{q}\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}}.\hfill \end{array}$
(14)

By (13), we have (12). On the other hand, suppose that (12) is valid. Setting

$g\left(y\right)=\phantom{\rule{2.77695pt}{0ex}}\mid y{\mid }^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}f\left(x\right)\mathsf{\text{d}}x\right]}^{p-1},$

then $J=\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^{q}\left(y\right)dy$. By (10), it follows J < ∞. If J = 0, then (13) is naturally valid. Assuming that 0 < J < ∞, by (12), we obtain

$\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^{q}\left(y\right)\mathsf{\text{d}}y=J=I
(15)
${J}^{\frac{1}{p}}={\left(\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{q-1}{g}^{q}\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{p}}
(16)

Hence, we have (13), which is equivalent to (12).

If the constant factor k in (12) is not the best possible, then there exists a positive constant K with K < k, such that (12) is still valid as we replace k by K, then we have

$I
(17)

For ε > 0, define functions $\stackrel{̃}{f}\left(x\right)$,$\stackrel{̃}{g}\left(y\right)$ as follows:

$\begin{array}{cc}\hfill \stackrel{̃}{f}\left(x\right):\phantom{\rule{2.77695pt}{0ex}}& =\left\{\begin{array}{cc}\hfill {x}^{-\frac{2\epsilon }{p}-1},\hfill & \hfill x\in \left(1,\infty \right),\hfill \\ \hfill 0,\hfill & \hfill x\in \left[-1,1\right],\hfill \\ \hfill {\left(-x\right)}^{-\frac{2\epsilon }{p}-1},\hfill & \hfill x\in \left(-\infty ,-1\right),\hfill \end{array}\right\\hfill \\ \hfill \stackrel{̃}{g}\left(y\right):\phantom{\rule{2.77695pt}{0ex}}& =\left\{\begin{array}{cc}\hfill \begin{array}{c}{y}^{\frac{2\epsilon }{q}-1},\\ 0,\end{array}\hfill & \hfill \begin{array}{c}y\in \left(0,1\right),\\ y\in \left(-\infty ,-1\right]\cup \left[1,\infty \right),\end{array}\hfill \\ \hfill {\left(-y\right)}^{\frac{2\epsilon }{q}-1},\hfill & \hfill y\in \left(-1,0\right).\hfill \end{array}\right\\hfill \end{array}$

Replacing f(x), g(y) by $\stackrel{̃}{f}\left(x\right)$, $\stackrel{̃}{g}\left(y\right)$ in (17), we obtain

$\begin{array}{cc}\hfill Ĩ:& =\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\stackrel{̃}{f}\left(x\right)\stackrel{̃}{g}\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\hfill \\
(18)
$Ĩ=\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}\stackrel{̃}{f}\left(x\right)\stackrel{̃}{g}\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y=\sum _{i=1}^{4}{I}_{i},$
(19)

where,

$\begin{array}{cc}\hfill {I}_{1}:& =\underset{-1}{\overset{0}{\int }}{\left(-y\right)}^{\frac{2\epsilon }{q}-1}\left[\underset{-\infty }{\overset{-1}{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}{\left(-x\right)}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y,\hfill \\ \hfill {I}_{2}:& =\underset{-1}{\overset{0}{\int }}{\left(-y\right)}^{\frac{2\epsilon }{q}-1}\left[\underset{1}{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}{x}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y,\hfill \\ \hfill {I}_{3}:& =\underset{0}{\overset{1}{\int }}{y}^{\frac{2\epsilon }{q}-1}\left[\underset{-\infty }{\overset{-1}{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}{\left(-x\right)}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y,\hfill \\ \hfill {I}_{4}:& =\underset{0}{\overset{1}{\int }}{y}^{\frac{2\epsilon }{q}-1}\left[\underset{1}{\overset{\infty }{\int }}\underset{i\in \left\{1,2\right\}}{min}\left\{\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos{\alpha }_{i}+{\left(xy\right)}^{2}}}\right\}{x}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y.\hfill \end{array}$

By Fubini theorem [10], we obtain

$\begin{array}{cc}\hfill {I}_{1}& ={I}_{4}=\underset{0}{\overset{1}{\int }}{y}^{\frac{2\epsilon }{q}-1}\left[\underset{1}{\overset{\infty }{\int }}\frac{min\left\{1,xy\right\}}{\sqrt{1+2xycos{\alpha }_{1}+{\left(xy\right)}^{2}}}{x}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y\hfill \\ \phantom{\rule{1em}{0ex}}\stackrel{u=xy}{=}\underset{0}{\overset{1}{\int }}{y}^{2\epsilon -1}\left[\underset{y}{\overset{\infty }{\int }}\frac{min\left\{u,1\right\}}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u\right]\mathsf{\text{d}}y\hfill \\ =\underset{0}{\overset{1}{\int }}{y}^{2\epsilon -1}\left[\underset{y}{\overset{1}{\int }}\frac{{u}^{-\frac{2\epsilon }{p}}}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\mathsf{\text{d}}u\right]\mathsf{\text{d}}y\hfill \\ \phantom{\rule{1em}{0ex}}+\underset{0}{\overset{1}{\int }}{y}^{2\epsilon -1}\left[\underset{1}{\overset{\infty }{\int }}\frac{{u}^{-\frac{2\epsilon }{p}-1}}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\mathsf{\text{d}}u\right]\mathsf{\text{d}}y\hfill \\ =\underset{0}{\overset{1}{\int }}\left(\underset{0}{\overset{u}{\int }}{y}^{2\epsilon -1}\mathsf{\text{d}}y\right)\frac{{u}^{-\frac{2\epsilon }{p}}\mathsf{\text{d}}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\frac{1}{2\epsilon }\underset{1}{\overset{\infty }{\int }}\frac{{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\hfill \\ =\frac{1}{2\epsilon }\left[\underset{0}{\overset{1}{\int }}\frac{{u}^{\frac{2\epsilon }{q}}}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\mathsf{\text{d}}u+\underset{1}{\overset{\infty }{\int }}\frac{{u}^{-\frac{2\epsilon }{p}-1}}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\mathsf{\text{d}}u\right],\hfill \\ \hfill {I}_{2}& ={I}_{3}=\underset{0}{\overset{1}{\int }}{y}^{\frac{2\epsilon }{q}-1}\left[\underset{1}{\overset{\infty }{\int }}\frac{min\left\{1,xy\right\}}{\sqrt{1-2xycos{\alpha }_{2}+{\left(xy\right)}^{2}}}{x}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}x\right]\mathsf{\text{d}}y\hfill \\ =\frac{1}{2\epsilon }\left[\underset{0}{\overset{1}{\int }}\frac{{u}^{\frac{2\epsilon }{q}}}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\mathsf{\text{d}}u+\underset{1}{\overset{\infty }{\int }}\frac{{u}^{-\frac{2\epsilon }{p}-1}}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\mathsf{\text{d}}u\right].\hfill \end{array}$

In view of the above results, by using (18) and (19), it follows

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\underset{0}{\overset{1}{\int }}\frac{{u}^{\frac{2\epsilon }{q}}}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\mathsf{\text{d}}u+\underset{1}{\overset{\infty }{\int }}\frac{{u}^{-\frac{2\epsilon }{p}-1}}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\mathsf{\text{d}}u\\ \phantom{\rule{1em}{0ex}}+\underset{0}{\overset{1}{\int }}\frac{{u}^{\frac{2\epsilon }{q}}}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\mathsf{\text{d}}u+\underset{1}{\overset{\infty }{\int }}\frac{{u}^{-\frac{2\epsilon }{p}-1}}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\mathsf{\text{d}}u\\ =\epsilon Ĩ<\epsilon \cdot \phantom{\rule{2.77695pt}{0ex}}\frac{K}{\epsilon }=K.\end{array}$
(20)

By Fatou lemma [10] and (20), we find

$\begin{array}{c}\phantom{\rule{1em}{0ex}}k=\omega \left(y\right)=\underset{0}{\overset{\infty }{\int }}\frac{\mathrm{min}\left\{u,1\right\}\mathsf{\text{d}}u}{u\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\underset{0}{\overset{\infty }{\int }}\frac{min\left\{u,1\right\}\mathsf{\text{d}}u}{u\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\\ =\underset{0}{\overset{1}{\int }}\underset{\epsilon \to {0}^{+}}{lim}\frac{{u}^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\underset{1}{\overset{\infty }{\int }}\underset{\epsilon \to {0}^{+}}{lim}\frac{{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\\ \phantom{\rule{1em}{0ex}}+\underset{0}{\overset{1}{\int }}\underset{\epsilon \to {0}^{+}}{lim}\frac{{u}^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}+\underset{1}{\overset{\infty }{\int }}\underset{\epsilon \to {0}^{+}}{lim}\frac{{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\\ \le {\underset{¯}{\mathrm{lim}}}_{\epsilon \to {0}^{+}}\left[\underset{0}{\overset{1}{\int }}\frac{{u}^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\underset{1}{\overset{\infty }{\int }}\frac{{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}\\ \phantom{\rule{1em}{0ex}}+\underset{0}{\overset{1}{\int }}\frac{{u}^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}+\underset{1}{\overset{\infty }{\int }}\frac{{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\right]\le K,\end{array}$
(21)

which contradicts the fact that K < k. Hence, the constant factor k in (12) is the best possible.

If the constant factor in (13) is not the best possible, then by (14), we may get a contradiction that the constant factor in (12) is not the best possible. Thus, the theorem is proved.   □

Theorem 4 As the assumptions of Theorem 3, replacing p > 1 by 0 < p < 1, we have the equivalent reverse of (12) and (13) with the best constant factors.

Proof. The way of proving of Theorem 4 is similar to Theorem 3. By the reverse Hö lder's inequality [9], we have the reverse of (10) and (14). It is easy to obtain the reverse of (13). In view of the reverses of (13) and (14), we obtain the reverse of (12). On the other hand, suppose that the reverse of (12) is valid. Setting the same g(y) as theorem 3, by the reverse of (10), we have J > 0. If J = ∞, then the reverse of (13) is obvious value; if J < ∞, then by the reverse of (12), we obtain the reverses of (15) and (16). Hence, we have the reverse of (13), which is equivalent to the reverse of (12).

If the constant factor k in the reverse of (12) is not the best possible, then there exists a positive constant , such that the reverse of (12) is still valid as we replace k by $\stackrel{̃}{K}$. By the reverse of (20), we have

$\begin{array}{c}\underset{0}{\overset{1}{\int }}\left[\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\frac{1}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\right]\phantom{\rule{2.77695pt}{0ex}}{u}^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u\\ +\underset{1}{\overset{\infty }{\int }}\left[\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\frac{1}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\right]\phantom{\rule{2.77695pt}{0ex}}{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u\\ >\stackrel{̃}{K}.\end{array}$
(22)

For $0<{\epsilon }_{0}<\frac{\mid q\mid }{2}$, we have $\frac{2{\epsilon }_{0}}{q}>-1$. For 0 < εε0, we obtain ${u}^{\frac{2\epsilon }{q}}\le {u}^{\frac{2{\epsilon }_{0}}{q}}\left(u\in \left(0,1\right]\right)$ and

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\underset{0}{\overset{1}{\int }}\left[\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\frac{1}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\right]\phantom{\rule{2.77695pt}{0ex}}u\frac{2{\epsilon }_{0}}{q}\mathsf{\text{d}}u\\ \le \left(\frac{1}{sin{\alpha }_{1}}+\frac{1}{sin{\alpha }_{2}}\right)\underset{0}{\overset{1}{\int }}u\frac{2{\epsilon }_{0}}{q}\mathsf{\text{d}}u=\left(\frac{1}{sin{\alpha }_{1}}+\frac{1}{sin{\alpha }_{2}}\right)\frac{1}{1+\left(2{\epsilon }_{0}\right)∕q}<\infty .\end{array}$

Then, by Lebesgue control convergence theorem [10], we have for ε → 0+ that

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\underset{0}{\overset{1}{\int }}\left[\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\frac{1}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\right]\phantom{\rule{2.77695pt}{0ex}}{u}^{\frac{2\epsilon }{q}}\mathsf{\text{d}}u\\ =\underset{0}{\overset{1}{\int }}\left[\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\frac{1}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\right]\phantom{\rule{2.77695pt}{0ex}}\mathsf{\text{d}}u+o\left(1\right).\end{array}$
(23)

By Levi's theorem [10], we find for ε → 0+ that

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\underset{1}{\overset{\infty }{\int }}\left[\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\frac{1}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\right]\phantom{\rule{2.77695pt}{0ex}}{u}^{-\frac{2\epsilon }{p}-1}\mathsf{\text{d}}u\\ =\underset{1}{\overset{\infty }{\int }}\left[\frac{1}{\sqrt{{u}^{2}+2ucos{\alpha }_{1}+1}}+\frac{1}{\sqrt{{u}^{2}-2ucos{\alpha }_{2}+1}}\right]\phantom{\rule{2.77695pt}{0ex}}{u}^{-1}\mathsf{\text{d}}u+\stackrel{˜}{o}\left(1\right).\end{array}$
(24)

By (22), (23) and (24), for ε → 0+ in (22), we have $k\ge \stackrel{̃}{K}$, which contradicts the fact that $k<\stackrel{̃}{K}$. Hence, the constant factor k in the reverse of (12) is the best possible.

If the constant factor in reverse of (13) is not the best possible, then by the reverse of (14), we may get a contradiction that the constant factor in the reverse of (12) is not the best possible. Thus, the theorem is proved.   □

Remark 1 For α1 = α2 = α ∈ (0, π) in (12) and (13), we have the following equivalent inequalities:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos\alpha +{\left(xy\right)}^{2}}}f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <{k}_{0}{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^{p}\left(x\right)\mathsf{\text{d}}x\right)}^{\frac{1}{p}}{\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{q-1}{g}^{q}\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{q}},\end{array}$
(25)
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+2xycos\alpha +{\left(xy\right)}^{2}}}f\left(x\right)\mathsf{\text{d}}x\right]}^{p}\mathsf{\text{d}}y\\ <{k}_{0}^{p}\underset{-\infty }{\overset{\infty }{\int }}\mid x{\mid }^{p-1}{f}^{p}\left(x\right)dx,\end{array}$
(26)

where the constant factors ${k}_{0}:=2ln\left[\left(1+sec\frac{\alpha }{2}\right)\left(1+csc\frac{\alpha }{2}\right)\right]$ and ${k}_{0}^{p}$ are the best possible.

Remark 2 For ${\alpha }_{1}={\alpha }_{2}=\frac{\pi }{3}$, p = q = 2 in (12) and (13), we have the following equivalent inequalities:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+xy+{\left(xy\right)}^{2}}}f\left(x\right)g\left(y\right)\mathsf{\text{d}}x\mathsf{\text{d}}y\\ <2ln\left(3+2\sqrt{3}\right){\left(\underset{-\infty }{\overset{\infty }{\int }}\mid x\mid {f}^{2}\left(x\right)\mathsf{\text{d}}x\underset{-\infty }{\overset{\infty }{\int }}\mid y\mid {g}^{2}\left(y\right)\mathsf{\text{d}}y\right)}^{\frac{1}{2}},\end{array}$
(27)
$\begin{array}{c}\phantom{\rule{1em}{0ex}}\underset{-\infty }{\overset{\infty }{\int }}\mid y{\mid }^{-1}{\left[\underset{-\infty }{\overset{\infty }{\int }}\frac{min\left\{1,\mid xy\mid \right\}}{\sqrt{1+xy+{\left(xy\right)}^{2}}}f\left(x\right)\mathsf{\text{d}}x\right]}^{2}\mathsf{\text{d}}y\\ <{2}^{2}{\left[ln\left(3+2\sqrt{3}\right)\right]}^{2}\underset{-\infty }{\overset{\infty }{\int }}\mid x\mid {f}^{2}\left(x\right)dx.\end{array}$
(28)