孙文兵, 刘琼
邵阳学院理学与信息科学系, 湖南 邵阳 422000
摘要: 引入两个参数λ₁和λ₂，利用权函数方法和泛函分析技巧，定义一个双参数Hilbert型积分算子，并给出算子的范数.作为应用，获得一些改进的结果，并得到一些具有特殊核的新的Hilbert型积分不等式.
关键词: 双参数的Hilbert型积分算子范数权函数最佳常数因子Hilbert型积分不等式
If p>1, $\frac{1}{p}+\frac{1}{q}$=1, f, g≥0, f∈L^p(0, ∞), g∈L^q(0, ∞), ‖f‖_p={∫₀^∞f^p(x)dx}^1p>0, ‖g‖_q>0, then we have the following famous Hardy-Hilbert integral inequality and its equivalent form^[1]:

$\int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}\frac{f\left( x \right)g\left( y \right)}{x+y}dxdy<\frac{\pi }{sin\left( \frac{\pi }{p} \right)}\|f{{\|}_{p}}\|g{{\|}_{q}},$

(1)

${{\left\{ \int_{0}^{\infty }{{}}{{\left[ \int_{0}^{\infty }{{}}\frac{f\left( x \right)}{x+y}dx \right]}^{p}}dy \right\}}^{\frac{1}{p}}}<\frac{\pi }{sin\left( \frac{\pi }{p} \right)}\|f{{\|}_{p}}\|g{{\|}_{q}},$

(2)

where the constant factor π/sin(π/p) is the best possible. Inequalities (1) and (2) are important in analysis and its applications^[1-2]. Define the Hardy-Hilbert’s integral operator T:L^p(0, ∞)→L^p(0, ∞) as follows. For f∈L^p(0, ∞), corresponding to the only

$T\left( f \right)\left( y \right):=\int\limits_{0}^{\infty }{{}}\frac{f\left( x \right)}{x+y}dx,y\in \left( 0,\infty \right),$

(3)

by (2), we have ‖Tf‖_p<π/sin(π/p)‖f‖_p and ‖T‖≤π/sin(π/p). Since the constant factor in (2) is the best possible, we find that ‖T‖=π/sin(π/p)^[3].
In recent years, the relevant operators were defined by using bilinear function k(x, y)(≥0) to replace $\frac{1}{x+y}$ in (3) by Yang et al. and they obtained two equivalent inequalities similar to (1) and (2)^[4-9]. In this work, we use the nonnegative measurable function h(x₁^λy₂^λ) instead of the kernel $\frac{1}{x+y}$, and a Hilbert-type integral operator with the two-parameter kernels is defined. By using the method of weight function and the techniques of real analysis and functional analysis, the norm of the integral operator is found and two equivalent inequalities related to the norm of the operator are given. As applications, some new inequalities with the particular kernels are obtained.
1 Definitions and lemmasWe need the following special functions^[10]:
Beta-function

$\begin{align} & B\left( u,v \right)=\int_{0}^{\infty }{{}}\frac{{{t}^{u-1}}dt}{{{\left( 1+t \right)}^{u+v}}} \\ & =\int_{0}^{1}{{}}{{\left( 1-t \right)}^{u-1}}{{t}^{v-1}}dt\left( u,v>0 \right) \\ \end{align}$

(4)

Γ-function

$\Gamma \left( z \right)=\int_{0}^{\infty }{{}}{{e}^{-u}}{{u}^{z-1}}du\left( z>0 \right),$

(5)

Riemann’s zeta-function

$\zeta \left( x \right)=\sum\limits_{k=1}^{\infty }{{}}\frac{1}{{{k}^{x}}}\left( x>1 \right),$

(6)

and the extended ζ-function

$\zeta \left( s,a \right)=\sum\limits_{k=0}^{\infty }{{}}\frac{1}{{{\left( k+a \right)}^{s}}},$

(7)

where Re(s)>1, a is not equal to zero or negative integer. Obviously, ζ(s, 1)=ζ(s).
If a is not equal to zero or negative integer, 0＜s≤1, we define the following value of the convergent series as.

$S\left( s,a \right)=\sum\limits_{k=0}^{\infty }{{}}\frac{{{\left( -1 \right)}^{k}}}{{{\left( k+a \right)}^{s}}},$

(8)

Lemma 1.1? If s>0, a is not equal to zero or negative integer, we have the summation formula as

$\begin{array}{l}C\left( {s,a} \right) = \sum\limits_{k = 0}^\infty {} \frac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {k + a} \right)}^s}}} = \\\left\{ \begin{array}{l}S\left( {s,a} \right),0 < s \le 1\\\frac{1}{{{2^s}}}\left[ {\zeta \left( {s,\frac{a}{2}} \right) - \zeta \left( {s,\frac{{a + 1}}{2}} \right),} \right]{\rm{ }}s > 1.\end{array} \right.\end{array}$

(9)

Proof? When 0＜s≤1, clearly, $\sum\nolimits_{k=0}^{\infty }{{}}\frac{{{\left( -1 \right)}^{k}}}{\left( k+a \right)}$=S(s, a), and when s>1 , we have

$\begin{align} & \sum\limits_{k=0}^{\infty }{{}}\frac{{{\left( -1 \right)}^{k}}}{{{\left( k+a \right)}^{s}}}~=\sum\limits_{k=0}^{\infty }{{}}\frac{1}{{{\left( 2k+a \right)}^{s}}}-\sum\limits_{k=0}^{\infty }{{}}\frac{1}{{{\left( 2k+1+a \right)}^{s}}} \\ & =\frac{1}{{{2}^{s}}}\left[ \sum\limits_{k=0}^{\infty }{{}}\frac{1\text{ }}{{{\left( k+\frac{a}{2} \right)}^{s}}}-\text{ }\sum\limits_{k=0}^{\infty }{{}}\frac{1}{{{\left( k+\frac{a+1}{2} \right)}^{s}}} \right] \\ & =\frac{1}{{{2}^{s}}}\left[ \zeta \left( s,\frac{a}{2} \right)-\zeta \left( s,\frac{a+1}{2} \right) \right]. \\ \end{align}$

If λ₁, λ₂>0, writing down α=$\frac{{{\lambda }_{1}}+{{\lambda }_{2}}}{{{\lambda }_{1}}{{\lambda }_{2}}}$, h(u) is a nonnegative measurable function in (0, ∞), we define k _{λ1, λ2}as

${{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}:=\int_{0}^{\infty }{{}}h\left( u \right){{u}^{\alpha -1}}du,$

(10)

assuming that k_{λ1, λ2}(≥0) is a limited number. Setting u=x^λ₁y^λ₂, we have

$\begin{align} & {{\omega }_{{{\lambda }_{1}},{{\lambda }_{2}}}}\left( x \right):={{\lambda }_{2}}{{x}^{\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}+1}}\int_{0}^{\infty }{{}}~h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}}){{y}^{\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}}}dy \\ & ={{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}},\left( x\in \left( 0,\infty \right) \right), \\ \end{align}$

(11)

$\begin{align} & {{\omega }_{{{\lambda }_{1}},{{\lambda }_{2}}}}\left( y \right):={{\lambda }_{1}}{{y}^{^{\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}+1}}}\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}}){{x}^{\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}}}dx \\ & ={{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}},\left( y\in \left( 0,\infty \right) \right), \\ \end{align}$

(12)

Lemma 1.2? Suppose that p>1, $\frac{1}{p}+\frac{1}{q}$=1, λ₁, λ₂>0, h(u) is a nonnegative measurable function in (0, ∞), k_{λ1 , λ2} (see (10)) is a nonnegative and limited number, f, g≥0, satisfying $\int_{0}^{\infty }{{}}{{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}$f^p(x)dx＜∞, $\int_{0}^{\infty }{{}}{{y}^{-\frac{q{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}$g^q(y)dy＜∞, then we have the following equivalent inequalities:

$\begin{align} & {{I}_{{{\lambda }_{1}},{{\lambda }_{2}}}}:=\int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}{{\left\{ \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right\}}^{p}}dy \\ & \le {{\left[ \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}\int_{0}^{\infty }{{}}{{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}{{f}^{p}}\left( x \right)dx, \\ \end{align}$

(13)

$\begin{align} & {{J}_{{{\lambda }_{1}},{{\lambda }_{2}}}}:=\int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)g\left( y \right)dxdy \\ & \le \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}{{\left\{ \int_{0}^{\infty }{{}}{{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}{{f}^{p}}\left( x \right)dx \right\}}^{\frac{1}{p}}}\times \\ & \left\{ \int_{0}^{\infty }{{}}{{y}^{-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1}}{{g}^{q}}\left( y \right)d{{y}^{\frac{1}{q}}} \right\}. \\ \end{align}$

(14)

Proof? By the weighted Holder's inequality^[11] and (12), we find

$\begin{align} & {{\left\{ \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right\}}^{p}} \\ & ={{\left\{ \int_{0}^{\infty }{{}}h({{x}^{\lambda 1}}{{y}^{{{\lambda }_{2}}}})\text{ }\left[ \frac{{{y}^{\frac{{{\lambda }_{2}}}{p{{\lambda }_{1}}}}}}{{{x}^{\frac{{{\lambda }_{1}}}{q{{\lambda }_{2}}}}}}f\left( x \right) \right]\text{ }\left[ \frac{{{x}^{\frac{{{\lambda }_{1}}}{q{{\lambda }_{2}}}}}}{{{y}^{\frac{{{\lambda }_{2}}}{p{{\lambda }_{1}}}}}}~ \right]dx \right\}}^{p}} \\ & \le \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})\frac{{{y}^{\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}}}}{~{{x}^{\frac{\left( p-1 \right){{\lambda }_{1}}}{{{\lambda }_{2}}}}}}~{{f}^{p}}\left( x \right)dx\times \\ & {{\left\{ \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})\frac{{{x}^{\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}}}~}{{{y}^{\frac{\left( q-1 \right){{\lambda }_{2}}}{{{\lambda }_{1}}}}}}dx \right\}}^{p-1}} \\ & =\frac{k_{{{\lambda }_{1}},{{\lambda }_{2}}}^{p-1}}{\lambda _{1}^{p-1}}{{y}^{\left( 1-p \right)\left( \frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1 \right)}}\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})\frac{{{y}^{\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}}}}{{{x}^{\frac{\left( p-1 \right){{\lambda }_{1}}}{{{\lambda }_{2}}}}}}~{{f}^{p}}\left( x \right)dx. \\ \end{align}$

By Fubini’s theorem^[12] and (11), we have

$\begin{align} & {{I}_{{{\lambda }_{1}},{{\lambda }_{2}}}} \\ & \le \frac{k_{{{\lambda }_{1}},{{\lambda }_{2}}}^{p-1}}{\lambda _{1}^{p-1}}\int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})\frac{{{y}^{\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}}}}{{{x}^{\frac{\left( p-1 \right){{\lambda }_{1}}}{{{\lambda }_{2}}}}}}{{f}^{p}}\left( x \right)dxdy \\ & =\frac{k_{{{\lambda }_{1}},{{\lambda }_{2}}}^{p-1}}{\lambda _{1}^{p-1}}\int_{0}^{\infty }{{}}\left[ \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})\frac{{{y}^{\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}}}}{{{x}^{\frac{\left( p-1 \right){{\lambda }_{1}}}{{{\lambda }_{2}}}}}}dy \right]{{f}^{p}}\left( x \right)dx \\ & =\frac{k_{{{\lambda }_{1}},{{\lambda }_{2}}}^{p-1}}{\lambda _{1}^{p-1}}\int_{0}^{\infty }{{}}{{\omega }_{{{\lambda }_{1}},{{\lambda }_{2}}}}\left( y \right){{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}{{f}^{p}}\left( x \right)dx \\ \end{align}$

By H?lder’s inequality, we have

$\begin{align} & {{J}_{{{\lambda }_{1}},{{\lambda }_{2}}}}=\int_{0}^{\infty }{{}}\left\{ {{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{p(q-1)}}}\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right\}\times \\ & \left\{ {{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1}{p(q-1)}}}g\left( y \right) \right\}dy \\ & \le I_{{{\lambda }_{1}},{{\lambda }_{2}}}^{\frac{1}{p}}{{\left\{ \int_{0}^{\infty }{{}}{{y}^{-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1}}{{g}^{q}}\left( y \right)dy \right\}}^{\frac{1}{q}}}. \\ \end{align}$

(15)

By (15) and (13), we obtain (14).
On the contrary, if (14) is true, for y>0, setting the function as

$g\left( y \right):={{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}{{\left[ ~\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right]}^{p-1}},$

by (14) we obtain

$\begin{align} & 0\le \int_{0}^{\infty }{{}}{{y}^{-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1}}{{g}^{q}}\left( y \right)dy \\ & =\int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}{{\left\{ \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right\}}^{p}}dy \\ & ={{I}_{{{\lambda }_{1}},{{\lambda }_{2}}}}={{J}_{{{\lambda }_{1}},{{\lambda }_{2}}}} \\ & \le \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}{{\left\{ \int_{0}^{\infty }{{}}{{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}{{f}^{p}}\left( x \right)dx \right\}}^{\frac{1}{p}}}\times \\ & {{\left\{ \int_{0}^{\infty }{{}}{{y}^{-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1}}{{g}^{q}}\left( y \right)dy \right\}}^{\frac{1}{q}}}. \\ \end{align}$

(16)

By (15), we have that I_{λ1, λ2}<∞. If I_{λ1, λ2}=0, (13) is tenable naturally. If 0<I_{λ1, λ2}<∞, by (16) we have

$\begin{align} & {{\left\{ \int_{0}^{\infty }{{}}{{y}^{-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1}}{{g}^{q}}\left( y \right)dy \right\}}^{\frac{1}{p}}}=I_{{{\lambda }_{1}},{{\lambda }_{2}}}^{\frac{1}{p}} \\ & \le \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}{{\left\{ \int_{0}^{\infty }{{}}{{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}{{f}^{p}}\left( x \right)dx \right\}}^{\frac{1}{p}}}, \\ \end{align}$

namely,

$\begin{align} & \int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}{{\left\{ \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right\}}^{p}}dy \\ & \le {{\left[ \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}\int_{0}^{\infty }{{}}{{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}{{f}^{p}}\left( x \right)dx. \\ \end{align}$

So (13) and (14) are equivalent. The lemma is proved.
Lemma 1.3? If p>1, $\frac{1}{p}+\frac{1}{q}$=1, λ₁, λ₂>0, writing down α=$\frac{{{\lambda }_{1}}+{{\lambda }_{2}}}{{{\lambda }_{1}}{{\lambda }_{2}}}$, h(u) is a nonnegative measurable function in (0, ∞), satisfying that k_{λ1, λ2} (see (12)) is a nonnegative and finite number. If there is δ>0, such that k_{λ1, λ2}, δ=∫₀^∞h(u)u^a+δ-1 du is still a nonnegative and limited number. For 0<ε<min {pδ, qδ}, and ε is small enough, we define the following real function

$\begin{array}{l}\tilde f\left( x \right): = \left\{ \begin{array}{l}0,x \in \left( {0,1} \right)\\{x^{\frac{{\frac{{p{\lambda _1}}}{{{\lambda _2}}} - {\lambda _1}\varepsilon }}{p}}},x \in \left[ {1,\infty } \right)\end{array} \right.,\\\tilde g\left( y \right): = \left\{ \begin{array}{l}0,y \in \left( {1,\infty } \right)\\{y^{\frac{{\frac{{q{\lambda _2}}}{{{\lambda _1}}} + {\lambda _2}\varepsilon }}{q}}},y \in \left( {0,1} \right]\end{array} \right..\end{array}$

Then we have

$\begin{align} & {{{\tilde{J}}}_{{{\lambda }_{1}},{{\lambda }_{2}}}}\varepsilon ={{\left[ \int_{0}^{\infty }{{}}{{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}{{{\tilde{f}}}^{p}}\left( x \right)dx \right]}^{\frac{1}{p}}}\times \\ & {{\left[ \int_{0}^{\infty }{{}}{{y}^{-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1}}{{{\tilde{g}}}^{q}}\left( y \right)dy \right]}^{\frac{1}{q}}}\varepsilon \\ & =\frac{1}{\lambda _{1}^{\frac{1}{p}}\lambda _{2}^{\frac{1}{q}}}, \\ \end{align}$

(17)

and

$\begin{align} & {{{\tilde{I}}}_{{{\lambda }_{1}},{{\lambda }_{2}}}}\varepsilon =\varepsilon \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})\tilde{f}\left( x \right)\tilde{g}\left( y \right)dxdy \\ & =\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{{{\lambda }_{1}}{{\lambda }_{2}}}+o\left( 1 \right)(\varepsilon \to {{0}^{+}}). \\ \end{align}$

(18)

Proof? We easily get

$\begin{align} & \tilde{J}\varepsilon ={{\left[ \int_{0}^{\infty }{{}}{{x}^{-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1}}{{{\tilde{f}}}^{p}}\left( x \right)dx \right]}^{\frac{1}{p}}}~{{\left[ \int_{0}^{\infty }{{}}{{y}^{-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1}}{{{\tilde{g}}}^{q}}\left( y \right)dy~ \right]}^{\frac{1}{q}}}\varepsilon \\ & ={{\left[ \int_{1}^{\infty }{{}}{{x}^{-1-{{\lambda }_{1}}\varepsilon }}dx \right]}^{\frac{1}{p}}}{{\left[ \int_{0}^{1}{{}}{{y}^{-1+{{\lambda }_{2}}\varepsilon }}dy \right]}^{\frac{1}{q}}}\varepsilon \\ & =\frac{1}{\lambda _{1}^{\frac{1}{p}}}\lambda _{2}^{\frac{1}{q}}. \\ \end{align}$

Setting u=x^λ₁y^λ₂, by Fubini’s theorem, we have

$\begin{array}{l}\tilde I\varepsilon = \varepsilon \int_0^\infty {} \int_0^\infty {} h({x^{{\lambda _1}}}{y^{{\lambda _2}}})\tilde f\left( x \right)\tilde g\left( y \right)dxdy\\ = \varepsilon \int_1^\infty {} {x^{\frac{{\frac{{p{\lambda _1}}}{{{\lambda _2}}} - {\lambda _1}\varepsilon }}{p}}}dx{\rm{ }}\left[ {\int_0^1 {} h({x^{{\lambda _1}}}{y^{{\lambda _2}}}){y^{\frac{{\frac{{q{\lambda _2}}}{{{\lambda _1}}} + {\lambda _2}\varepsilon }}{q}}}dy} \right]\\ = \frac{\varepsilon }{{{\lambda _2}}}\int_1^\infty {} {x^{ - 1 - {\lambda _1}\varepsilon }}dx{\rm{ }}\left[ {\int_1^{{x^{{\lambda _1}}}} {} h\left( u \right){u^{\alpha + \frac{\varepsilon }{q} - 1}}du} \right]\\ = \frac{\varepsilon }{{{\lambda _2}}}\int_1^\infty {} {x^{ - 1 - {\lambda _1}\varepsilon }}dx\left[ \begin{array}{l}\int_0^1 {} {\rm{ }}h\left( u \right){u^{\alpha + \frac{\varepsilon }{q} - 1}}du + \\\int_1^{{x^{{\lambda _1}}}} {} h\left( u \right){u^{\alpha + \frac{\varepsilon }{q} - 1}}du\end{array} \right]\\ = \frac{1}{{{\lambda _1}{\lambda _2}}}\int_0^1 {} h\left( u \right){u^{\alpha + \frac{\varepsilon }{q} - 1}}du + \\\frac{\varepsilon }{{{\lambda _2}}}\int_1^\infty {} {x^{ - 1 - {\lambda _1}\varepsilon }}dx{\rm{ }}\left[ {\int_1^{{x^{{\lambda _1}}}} {} h\left( u \right){u^{\alpha + \frac{\varepsilon }{q} - 1}}du} \right]\\ = \frac{1}{{{\lambda _1}{\lambda _2}}}\int_0^1 {} h\left( u \right){u^{\alpha + \frac{\varepsilon }{q} - 1}}du + \\\frac{\varepsilon }{{{\lambda _2}}}\int_1^\infty {} \left( {\int_u^\infty {} {x^{ - 1 - {\lambda _1}\varepsilon }}dx} \right)h\left( u \right){u^{\alpha + \frac{\varepsilon }{q} - 1}}du\\ = \frac{1}{{{\lambda _1}{\lambda _2}}}\int_0^1 {} h\left( u \right){u^{\alpha + \frac{\varepsilon }{q} - 1}}du + \\\frac{1}{{{\lambda _1}{\lambda _2}}}\int_1^\infty {} h\left( u \right){u^{\alpha - \frac{\varepsilon }{p} - 1}}du.\end{array}$

(19)

Since h(u)u^{α+$\frac{\varepsilon }{q}$-1}≤h(u)u^α-1, u∈(0, 1), h(u)u^{α-$\frac{\varepsilon }{q}$-1}≤h(u)u^α+δ-1, u∈(1, ∞), k_{λ1, λ2} and k_{λ1, λ2, δ} are limited numbers, by Lebesgue's dominated convergence theorem^[12], when ε→0⁺, we obtain

$\int_{0}^{1}{{}}h\left( u \right){{u}^{\alpha +\frac{\varepsilon }{q}-1}}du=\int_{0}^{1}{{}}h\left( u \right){{u}^{\alpha -1}}du+{{o}_{1}}\left( 1 \right)$

(20)

$\int_{1}^{\infty }{{}}h\left( u \right){{u}^{\alpha -\frac{\varepsilon }{q}-1}}du=\int_{1}^{\infty }{{}}h\left( u \right){{u}^{\alpha -1}}du+{{o}_{2}}\left( 1 \right).$

(21)

Putting (20) and (21) into (19), we get (18).
2 Main results and applicationsIf θ(x)(>0) is a measurable function, ρ≥1, the function space is set as

$\begin{align} & L_{\theta }^{\rho }\left( 0,\infty \right):= \\ & \left\{ h\ge 0;\|h{{\|}_{\rho ,\theta }}:={{\left\{ \int_{0}^{\infty }{{}}\theta \left( x \right){{h}^{\rho }}\left( x \right)dx \right\}}^{\frac{1}{p}}}<\infty \right\}. \\ \end{align}$

If p>1, $\frac{1}{p}+\frac{1}{q}$=1, λ₁, λ₂>0, writing down α=$\frac{{{\lambda }_{1}}+{{\lambda }_{2}}}{{{\lambda }_{1}}{{\lambda }_{2}}}$, h(u) is a nonnegative measurable function in (0, ∞), satisfying that k_{λ1, λ2}(see (10)) is a nonnegative and finite number, setting φ(x):=x^{$-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1$}, ψ(y):=y^{$-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1$}, x, y∈(0, ∞), and ψ^1-p(y)=y^{$\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}$}, we define an operator

$\begin{align} & T:L_{\varphi }^{p}\left( 0,\infty \right)\to L_{{{\psi }^{1-p}}}^{p}\left( 0,\infty \right),\text{ }for\text{ }f\in L_{\varphi }^{p}\left( 0,\infty \right) \\ & \left( Tf \right)\left( y \right):=\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx,\left( y\in \left( 0,\infty \right) \right). \\ \end{align}$

(22)

In view of (13), it follows Tf∈Lψ1-pp. For g∈Lψq(0, ∞), we define the formal inner of Tf and g

$\left( Tf,g \right):=\int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)g\left( y \right)dxdy.$

(23)

Hence the equivalent inequalities (13) and (14) may be rewritten in the following abstract forms

$\|Tf\|_{p,{{\psi }^{1-p}}}^{p}\le \text{ }{{\left[ \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}\|f\|_{p,\varphi }^{p},$

(24)

$\left( Tf,g \right)\le \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}\|f{{\|}_{p,\varphi }}\|g{{\|}_{q,\psi }}.$

(25)

T is obviously bounded^[13], and ‖T‖≤$\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$. We call that T is two-parameters Hilbert-type integral operator.
Theorem 2.1? If p>1, $\frac{1}{p}+\frac{1}{q}$=1, λ₁, λ₂>0, h(u) is a nonnegative measurable function in (0, ∞), k_{λ1, λ2}(see (10)) is a nonnegative and limited number , φ(x)=x^{$-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1$}, ψ(y)=y^{$-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1$}, f∈L_φ^p(0, ∞), g∈L_ψ^q(0, ∞), ‖f‖_{p, φ}, ‖g‖_{q, ψ}>0, then we have ‖T‖=$\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$, and the constant factors in (24) and (25) are the best possible.
Proof? Assuming that the constant factor $\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$ in (25) is not the best possible, then there exists a positive k<$\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$, such that inequality (25) is still valid if we replace $\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$ by k. Specially, putting f(x) and g(y) in Lemma 1.3 instead of f(x) and g(y), then by (17) and (18) we have

$\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{{{\lambda }_{1}}{{\lambda }_{2}}}+o\left( 1 \right)<\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}.$

Letting ε→0⁺, we get k≥$\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$, which contradicts the fact that k<$\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$. So the constant factor $\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$ in (25) is the best possible one. If the constant factor ${{\left[ \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}$ in (24) is not the best possible, then by (15) we get a contradictory conclusion that the constant factor $\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$ in (25) is not the best possible. So, ‖T‖=$\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$.

$\begin{align} & \|Tf\|_{p,{{\psi }^{1-p}}}^{p}=\int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}{{\left\{ \int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right\}}^{p}}dy \\ & <{{\left[ \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}\|f\|_{p,\varphi }^{p}, \\ \end{align}$

(26)

$\begin{align} & \left( Tf,g \right)=\int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}h({{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)g\left( y \right)dxdy \\ & <\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}\|f{{\|}_{p,\varphi }}\|g{{\|}_{q,\psi }}, \\ \end{align}$

(27)

where the constant factors ${{\left[ \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}$ and $\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$ are the best possible ones.
Proof? If inequality (25) keeps the form of an equality, by Lemma1.2 there exist two constants A and B such that they are not all zeroes^[11], and they satisfy

$A\frac{{{y}^{\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}}}}{{{x}^{\frac{p{{\lambda }_{1}}}{q{{\lambda }_{2}}}}}}{{f}^{p}}\left( x \right)=B\frac{{{x}^{\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}}}}{{{y}^{\frac{q{{\lambda }_{2}}}{p{{\lambda }_{1}}}}}}{{g}^{q}}\left( y \right)a.e.in\left( 0,\infty \right)\times \left( 0,\infty \right).$

It follows that Ax^{$-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}$}f^p(x)=By^{$-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}$}g^q(y)a.e. in (0, ∞)×(0, ∞). Assuming that A≠0, there exists y>0, such that x^{$-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1$}f^p(x)=[By^{$-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}$}g^q(y)]$\frac{1}{Ax}$ a.e. in (0, ∞), which contradicts the fact that 0<‖f‖_{p, φ}<∞. Then inequality (25) keeps the strict form. So (27) is true. Since inequalities (24) and (25) are equivalent, inequality (24) keeps the strict form too. So (26) is true.By Theorem 2.1, the constant factors ${{\left[ \frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}$ and $\frac{{{k}_{{{\lambda }_{1}},{{\lambda }_{2}}}}}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}$ are the best possible ones.
Assuming that p>1, $\frac{1}{p}+\frac{1}{q}$=1, λ₁, λ₂>0, α=$\frac{{{\lambda }_{1}}+{{\lambda }_{2}}}{{{\lambda }_{1}}{{\lambda }_{2}}}$, φ(x)=x^{$-\frac{p{{\lambda }_{1}}}{{{\lambda }_{2}}}-1$}, ψ(y)=y^{$-\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}-1$}, f∈L_φ^p(0, ∞), g∈L_ψ^q(0, ∞), ‖f‖_{p, φ}, ‖g‖_{q, ψ}>0, we obtain some useful inequalities by selecting some special kernels in (24) and (25).
1) If h(u)=$\frac{1}{{{\left( 1+u \right)}^{\beta }}}$, β>α, such that k_{λ1, λ2}=$\int_{0}^{\infty }{{}}\frac{1}{{{\left( 1+u \right)}^{\beta }}}$u^α-1du=B(β-α, α), by Theorem 2.2 we have the following equivalent inequalities with the best constant factors

$\begin{align} & \int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}\int_{0}^{\infty }{{}}{{\left[ \frac{f\left( x \right)dx}{{{(1+{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})}^{\beta }}} \right]}^{p}}dy \\ & <\left[ \frac{B\left( \beta -\alpha ,\alpha \right)}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]\left\| f \right\|_{p,\varphi }^{p}, \\ \end{align}$

(28)

$\int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}\frac{f\left( x \right)g(y)dxdy}{{{\left( 1+{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}} \right)}^{\beta }}}\frac{B\left( \beta -\alpha ,\alpha \right)}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}\left\| f \right\|_{p,\varphi }^{{}}\left\| g \right\|_{q,\varphi }^{{}}.$

(29)

2) If h(u)=$\frac{1}{{{(max~\left\{ 1,u \right\})}^{\beta }}}$, β>α, such that k_{λ1, λ2}=$\int_{0}^{\infty }{{}}\frac{1}{{{(max~\left\{ 1,u \right\})}^{\beta }}}$u^α-1du=$\frac{1}{\alpha \left( \beta -\alpha \right)}$, by Theorem 2.2 we have the following equivalent inequalities with the best constant factors as

$\begin{align} & \int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}~\left[ \int_{0}^{\infty }{{}}\frac{f\left( x \right)dx}{{{(max~\{1,{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}}\})}^{\beta }}} \right]{{~}^{p}}dy \\ & <{{\left[ \frac{1}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}a\left( \beta -a \right)} \right]}^{p}}\|f{{\|}^{p}}_{p,\varphi }, \\ \end{align}$

(30)

$\begin{align} & \int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}\frac{f\left( x \right)g\left( y \right)dxdy}{{{(max~\{1,{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}}\})}^{\beta }}} \\ & <\frac{1}{\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}a\left( \beta -a \right)}\|f{{\|}_{p,\varphi }}\|g{{\|}_{q,\psi }}. \\ \end{align}$

(31)

3) If h(u)=sechu, by Lemma 1.1 we get that k_{λ1, λ2}=∫₀^∞(sechu)u^α-1du=2$\int_{0}^{\infty }{{}}\frac{{{e}^{-u}}}{1+{{e}^{-2u}}}$u^α-1du=2∑_k=0^∞(-1)^k∫₀^∞e^-(2k+1)uu^α-1du=2Γ(α)∑_k=0^∞(-1)^k$\frac{{{\left( -1 \right)}^{k}}}{{{\left( 2k+1 \right)}^{\alpha }}}=\frac{1}{{{2}^{\alpha -1}}}\Gamma \left( \alpha \right)C\left( \alpha ,\frac{1}{2} \right)$, and by Theorem 2.2 we have the following equivalent inequalities with the best constant factor

$\begin{align} & \int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}~{{\left[ \int_{0}^{\infty }{{}}(sech{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right]}^{p}}dy \\ & <{{\left[ \frac{\Gamma \left( \alpha \right)C\left( \alpha ,\frac{1}{2} \right)}{{{2}^{\alpha -1}}\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}\|f\|_{p,\varphi }^{p}, \\ \end{align}$

(32)

$\begin{align} & \int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}(sech{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)g\left( y \right)dxdy \\ & <{{\left[ \frac{\Gamma \left( \alpha \right)C\left( \alpha ,\frac{1}{2} \right)}{{{2}^{\alpha -1}}\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}\|f{{\|}_{p,\varphi }}\|g{{\|}_{q,\psi }}. \\ \end{align}$

(33)

4) If h(u)=cschu, α>1(λ₁+λ₂>λ₁λ₂), by (7) we get that k_{λ1, λ2}=∫₀^∞(cschu)u^α-1du=2 $\int_{0}^{\infty }{{}}\frac{{{e}^{-u}}}{1+{{e}^{-2u}}}$u^α-1du=2 $\sum\limits_{k=0}^{\infty }{{}}$∫₀^∞e^-(2k+1)uu^α-1du=2Γ(α)$\sum\limits_{k=0}^{\infty }{{}}\frac{1}{{{\left( 2k+1 \right)}^{\alpha }}}=\frac{\Gamma \left( \alpha \right)\zeta \left( \alpha ,\frac{1}{2} \right)}{{{2}^{\alpha -1}}}$, and by Theorem 2.2 we have the following equivalent inequalities with the best constant factors

$\begin{align} & \int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}{{\left[ \int_{0}^{\infty }{{}}(csch{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)dx \right]}^{p}}dy \\ & <{{\left[ \frac{\Gamma \left( \alpha \right)\zeta \left( \alpha ,\frac{1}{2} \right)}{{{2}^{\alpha -1}}\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right]}^{p}}\|f\|_{p,\varphi }^{p}, \\ \end{align}$

(34)

$\begin{align} & \int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}(csch{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}})f\left( x \right)g\left( y \right)dxdy \\ & <\frac{\Gamma \left( \alpha \right)\zeta \left( \alpha ,\frac{1}{2} \right)}{{{2}^{\alpha -1}}\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}\|f{{\|}_{p,\varphi }}\|g{{\|}_{q,\psi }}. \\ \end{align}$

(35)

5) If h(u)=ln(1+e^-u), by Lemma 1.1 we get that k_{λ1, λ2}=∫₀^∞ ln(1+e^-u)u^α-1du=∑_k=0^∞$\frac{{{\left( -1 \right)}^{k}}}{k+1}$∫₀^∞e^-(k+1)uu^a-1du=Γ(α)∑_k=0^∞$\frac{{{\left( -1 \right)}^{k}}}{{{\left( k+1 \right)}^{\alpha +1}}}=\frac{\Gamma \left( \alpha \right)}{{{2}^{\alpha +1}}}~\left[ \zeta \left( \alpha +1,\frac{1}{2} \right)-\zeta \left( \alpha +1,\frac{3}{2} \right) \right]$, and by Theorem 2.2 we have the following equivalent inequalities with the best constant factors

$\begin{align} & \int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}{{\left\{ \int_{0}^{\infty }{{}}\left[ ln(1+{{e}^{-{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}}}}) \right]f\left( x \right)dx \right\}}^{p}}dy \\ & <{{\frac{\Gamma \left( \alpha \right)\left[ \zeta \left( \alpha +1,\frac{1}{2} \right)-\text{ }\zeta \left( \alpha +1,\frac{3}{2} \right) \right]}{{{2}^{\alpha +1}}\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}}^{p}}\|f{{\|}^{p}}_{p,\varphi } \\ \end{align}$

(36)

$\begin{align} & \int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}\left[ ln(1+{{e}^{-{{x}^{{{\lambda }_{1}}}}{{y}^{{{\lambda }_{2}}}}}}) \right]f\left( x \right)g\left( y \right)dxdy \\ & <{{\frac{\Gamma \left( \alpha \right)\left[ \zeta \left( \alpha +1,\frac{1}{2} \right)-\text{ }\zeta \left( \alpha +1,\frac{3}{2} \right) \right]}{{{2}^{\alpha +1}}\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}}}^{p}}\|f{{\|}_{p,\varphi }}\|g{{\|}_{q,\psi }}. \\ \end{align}$

(37)

6) If h(u)=arctan(e^-u), by Lemma 1.1 we get that k_{λ1, λ2}=∫₀ ^∞arctan(e^-u)u^α-1du=∑_k=0^∞$\frac{{{\left( -1 \right)}^{k}}}{{{\left( 2k+1 \right)}^{\alpha +1}}}$∫₀^∞e^-uu^α-1du=$\frac{\Gamma \left( \alpha \right)}{{{4}^{\alpha +1}}}\left[ \zeta \left( \alpha +1,\frac{1}{4} \right)-\text{ }\zeta \left( \alpha +1,\frac{3}{4} \right) \right]$, 34, and by Theorem 2.2 we have the following equivalent inequalities with the best constant factors as

$\begin{align} & \int_{0}^{\infty }{{}}{{y}^{\frac{\frac{q{{\lambda }_{2}}}{{{\lambda }_{1}}}+1}{q-1}}}{{\left\{ \int_{0}^{\infty }{{}}[arctan({{e}^{-{{x}^{{{\lambda }_{1}}}}}}{{y}^{{{\lambda }_{2}}}})]f\left( x \right)dx \right\}}^{p}}dy \\ & <{{\left\{ \frac{\Gamma \left( \alpha \right)\left[ \zeta \left( \alpha +1,\frac{1}{4} \right)-\text{ }\zeta \left( \alpha +1,\frac{3}{4} \right)\text{ } \right]}{{{4}^{\alpha +1}}\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \right\}}^{p}}\|f\|_{p,\varphi }^{p} \\ \end{align}$

(38)

$\begin{align} & \int_{0}^{\infty }{{}}\int_{0}^{\infty }{{}}[arctan({{e}^{-{{x}^{{{\lambda }_{1}}}}}}{{y}^{{{\lambda }_{2}}}})]f\left( x \right)g\left( y \right)dxdy \\ & <\frac{\Gamma \left( \alpha \right)\left[ \zeta \left( \alpha +1,\frac{1}{4} \right)-\text{ }\zeta \left( \alpha +1,\frac{3}{4} \right)\text{ } \right]}{{{4}^{\alpha +1}}\lambda _{1}^{\frac{1}{q}}\lambda _{2}^{\frac{1}{p}}} \\ & \|\text{ }f{{\|}_{p,\varphi }}\|g{{\|}_{q,\psi }}. \\ \end{align}$

(39)

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