Skip to main content

On Hardy-type integral inequalities in the whole plane related to the extended Hurwitz-zeta function

Abstract

Using weight functions, we establish a few equivalent statements of two kinds of Hardy-type integral inequalities with nonhomogeneous kernel in the whole plane. The constant factors related to the extended Hurwitz-zeta function are proved to be the best possible. In the form of applications, we deduce some special cases involving homogeneous kernel. We additionally consider some particular inequalities and operator expressions.

Introduction

If \(f(x),g(y)\geq0\),

$$0< \int_{0}^{\infty}f^{2}(x)\,dx< \infty \quad\mbox{and}\quad 0< \int_{0}^{\infty }g^{2}(y)\,dy< \infty , $$

we have the following well-known Hilbert integral inequality (see [1]):

$$ \int_{0}^{\infty} \int_{0}^{\infty}\frac{f(x)g(y)}{x+y}\,dx\,dy< \pi \biggl( \int_{0}^{\infty}f^{2}(x)\,dx \int_{0}^{\infty}g^{2}(y)\,dy \biggr) ^{\frac{1}{2}}, $$
(1)

with the best possible constant factor π.

Recently, by the use of weight functions, several extensions of (1) have been established in [2] and [3]. Some Hilbert-type inequalities were also presented in [49]. Furthermore, Hong [10] considered as well an equivalent condition between a Hilbert-type inequality with homogenous kernel and a few parameters. Some additional kinds of Hilbert-type inequalities were also obtained in [1119]. Most of these results are constructed in the quarter plane of the first quadrant.

In 2007, Yang [20] proved the following Hilbert-type integral inequality in the whole plane:

$$\begin{aligned} & \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}\frac{f(x)g(y)}{(1+e^{x+y})^{\lambda}}\,dx\,dy \\ &\quad< B \biggl(\frac{\lambda}{2},\frac{\lambda}{2} \biggr) \biggl( \int _{-\infty}^{\infty }e^{-\lambda x}f^{2}(x) \,dx \int_{-\infty}^{\infty}e^{-\lambda y}g^{2}(y) \,dy \biggr) ^{\frac{1}{2}}, \end{aligned}$$
(2)

with the best possible constant factor \(B (\frac{\lambda}{2},\frac {\lambda}{2} )\) (\(\lambda>0\), where \(B(u,v)\) stands for the beta function) (see [21]). He et al. [2235] also established some Hilbert-type integral inequalities in the whole plane with the best possible constant factors.

In the present paper, using weight functions, we establish a few equivalent statements of two kinds of Hardy-type integral inequalities with nonhomogeneous kernel and multi-parameters in the whole plane. The constant factors related to the extended Hurwitz-zeta function are proved to be the best possible. In the form of applications, we deduce a few equivalent statements of two kinds of Hardy-type integral inequalities with homogeneous kernel in the whole plane. As corollaries, we also consider some particular cases and operator expressions.

An example and two lemmas

Example 1

We set

$$H(xy):=\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }, $$

wherefrom

$$\begin{gathered} H(-xy)=\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy+1 \vert }\quad \bigl(x,y\in\mathbf{R}=(-\infty ,\infty)\bigr), \\ H(u)=\frac{(\min\{ \vert u \vert ,1\})^{1+\alpha} \vert \ln \vert u \vert \vert ^{\beta}}{(\max \{ \vert u \vert ,1\})^{\lambda+\alpha} \vert u-1 \vert },\end{gathered} $$

and

$$H(-u)=\frac{(\min\{ \vert u \vert ,1\})^{1+\alpha} \vert \ln \vert u \vert \vert ^{\beta}}{(\max \{ \vert u \vert ,1\})^{\lambda+\alpha} \vert u+1 \vert }\quad(u\in\mathbf{R}). $$

For \(\beta>0\), \(\sigma>-\alpha-1\), it follows that

$$\begin{aligned} K^{(1)}(\sigma) :=& \int_{-1}^{1}H(u) \vert u \vert ^{\sigma -1}\,du= \int_{0}^{1}(H(-u)+H(u)u^{\sigma-1}\,du \\ =& \int_{0}^{1}\frac{(\min\{u,1\})^{1+\alpha}(-\ln u)^{\beta }u^{\sigma-1}}{(\max\{u,1\})^{\lambda+\alpha}} \biggl( \frac{1}{u+1}+\frac{1}{ \vert u-1 \vert } \biggr) \,du \\ =& \int_{0}^{1}(-\ln u)^{\beta} \biggl( \frac{1}{u+1}+\frac{1}{1-u} \biggr) u^{\sigma+\alpha}\,du \\ =&2 \int_{0}^{1}(-\ln u)^{\beta} \frac{u^{\sigma+\alpha}}{1-u^{2}}\,du=2 \int_{0}^{1}(-\ln u)^{\beta}\sum _{k=0}^{\infty}u^{2k+\sigma +\alpha }\,du. \end{aligned}$$

By the Lebesgue term-by-term integration theorem (cf. [36]), for \(v=-(2k+\sigma+\alpha+1)\ln u\), we obtain

$$\begin{aligned} K^{(1)}(\sigma) =&2\sum_{k=0}^{\infty} \int_{0}^{1}(-\ln u)^{\beta }u^{2k+\sigma+\alpha} \,du \\ =&2\sum_{k=0}^{\infty} \frac{1}{(2k+\sigma+\alpha+1)^{\beta+1}}\int_{0}^{\infty}v^{\beta}e^{-v}\,dv \\ =&\frac{1}{2^{\beta}}\sum_{k=0}^{\infty} \frac{1}{[k+(\sigma+\alpha +1)/2]^{\beta+1}} \int_{0}^{\infty}v^{(\beta+1)-1}e^{-v}\,dv \\ =&\frac{\varGamma(\beta+1)}{2^{\beta}}\zeta \biggl(\beta+1,\frac {\sigma+\alpha +1}{2} \biggr)\in \mathbf{R}_{+}, \end{aligned}$$
(3)

where

$$\zeta(s,a)=\sum_{k=0}^{\infty} \frac{1}{(k+a)^{s}}\quad (\operatorname{Re} s>1;0< a\leq1) $$

stands for the Hurwitz-zeta function. Note that

$$\zeta (s,1)=\zeta(s):=\sum_{k=1}^{\infty} \frac{1}{k^{s}} $$

is the Riemann-zeta function. Moreover,

$$\zeta \biggl(\beta+1,\frac{\sigma+\alpha+1}{2} \biggr) $$

stands for the extended Hurwitz-zeta function (cf. [21]).

In particular, for \(\sigma=-\alpha+1\) (\(>-\alpha-1\)), it follows that

$$K^{(1)}(-\alpha+1)= \int_{-1}^{1}H(u) \vert u \vert ^{-\alpha}\,du=\frac{\varGamma (\beta+1)}{2^{\beta}}\zeta(\beta+1). $$

Similarly, for \(\beta>0\), \(\mu>-\alpha-1\) (\(\sigma+\mu=\lambda\)), we obtain that

$$\begin{aligned}& \begin{aligned}[b] K^{(2)}(\sigma) &:= \int_{\{u^{\prime} \vert u \vert \geq1\}}H(u) \vert u \vert ^{\sigma -1}\,du\\&= \int_{1}^{\infty}\bigl(H(-u)+H(u) \bigr)u^{\sigma-1}\,du \\ &= \int_{-1}^{1}\frac{(\min\{ \vert v \vert ,1\})^{1+\alpha} \vert \ln \vert v \vert \vert ^{\beta }}{(\max \{ \vert v \vert ,1\})^{\lambda+\alpha} \vert v-1 \vert } \vert v \vert ^{\mu-1}\,dv \\ &=\frac{\varGamma(\beta+1)}{2^{\beta}}\zeta\biggl(\beta+1,\frac{\mu +\alpha+1}{2} \biggr)=K^{(1)}(\mu)\in\mathbf{R}_{+}, \end{aligned} \\& K^{(2)}(\lambda+\alpha-1)= \int_{\{u^{\prime} \vert u \vert \geq1\} }H(u) \vert u \vert ^{\lambda +\alpha-2}\,du= \frac{\varGamma(\beta+1)}{2^{\beta}}\zeta(\beta+1). \end{aligned}$$
(4)

Remark 1

For \(\sigma+\mu=\lambda\), it is clear that

$$K^{(1)}(\sigma)< \infty \quad\bigl(\mbox{resp. }K^{(2)}( \sigma)=K^{(1)}(\mu)< \infty\bigr) $$

if and only if \(\sigma>-\alpha-1\) and \(\beta>0\) (resp. \(\mu>-\alpha-1\) and \(\beta >0\)).

In the sequel, we assume that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(\sigma+\mu =\lambda\).

Lemma 1

If\(\sigma_{1}\in\mathbf{R}\), there exists a constant\(M_{1}\)such that, for any nonnegative measurable functions\(f(x)\)and\(g(y)\)inR, the following inequality

$$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{-\frac{1}{ \vert y \vert }}^{\frac {1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x) \,dx \biggr] \,dy \\ &\quad\leq M_{1} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma )-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\sigma _{1})-1}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}} \end{aligned}$$
(5)

holds true, then we have\(\sigma_{1}=\sigma>-\alpha-1\)and\(\beta>0\).

Proof

If \(\sigma_{1}>\sigma\), then for \(n\geq\frac{1}{\sigma _{1}-\sigma}\) (\(n\in\mathbf{N}\)) we consider the following functions:

$$f_{n}(x):=\left \{ \textstyle\begin{array}{l@{\quad}l} \vert x \vert ^{\sigma+\frac{1}{pn}-1},& 0< \vert x \vert \leq1, \\ 0, & \vert x \vert >1,\end{array}\displaystyle \right . \qquad g_{n}(y):=\left \{ \textstyle\begin{array}{l@{\quad}l} 0, &0< \vert y \vert < 1, \\ \vert y \vert ^{\sigma_{1}-\frac{1}{qn}-1}, &y\geq1,\end{array}\displaystyle \right . $$

and derive that

$$\begin{aligned} J_{1} :=& \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma )-1}f_{n}^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\sigma _{1})-1}g_{n}^{q}(y)\,dy \biggr] ^{\frac{1}{q}} \\ =& \biggl( 2 \int_{0}^{1}x^{\frac{1}{n}-1}\,dx \biggr) ^{\frac{1}{p}} \biggl( 2 \int_{1}^{\infty}y^{-\frac{1}{n}-1}\,dy \biggr) ^{\frac{1}{q}}=2n. \end{aligned}$$

We obtain

$$\begin{aligned} I_{1} :=& \int_{-\infty}^{\infty}g_{n}(y) \biggl[ \int_{-\frac {1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f_{n}(x) \,dx \biggr] \,dy \\ =& \int_{-\infty}^{-1} \biggl[ \int_{\frac{1}{y}}^{\frac{-1}{y}}\frac {(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert } \vert x \vert ^{\sigma+\frac{1}{pn}-1}\,dx \biggr] (-y)^{\sigma _{1}-\frac{1}{qn}-1}\,dy \\ &{}+ \int_{1}^{\infty} \biggl[ \int_{\frac{-1}{y}}^{\frac{1}{y}}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert } \vert x \vert ^{\sigma+\frac{1}{pn}-1}\,dx \biggr] y^{\sigma _{1}-\frac{1}{qn}-1}\,dy \\ =& \int_{1}^{\infty} \biggl[ \int_{\frac{-1}{y}}^{\frac{1}{y}}\bigl(H(-xy)+H(xy)\bigr) \vert x \vert ^{\sigma+\frac{1}{pn}-1}\,dx \biggr] y^{\sigma _{1}-\frac{1}{qn}-1}\,dy \quad(u=xy) \\ =&2 \int_{1}^{\infty} \biggl[ \int_{0}^{1}\bigl(H(-u)+H(u) \bigr)u^{(\sigma+\frac{1}{pn})-1}\,du \biggr] y^{(\sigma_{1}-\sigma)-\frac {1}{n}-1}\,dy, \end{aligned}$$
(6)

and then by (5) we get

$$ 2K^{(1)} \biggl(\sigma+\frac{1}{pn} \biggr) \int_{1}^{\infty}y^{(\sigma _{1}-\sigma)-\frac{1}{n}-1} \,dy=I_{1}\leq M_{1}J_{1}=2M_{1}n. $$
(7)

Since \((\sigma_{1}-\sigma)-\frac{1}{n}\geq0\), it follows that

$$\int_{1}^{\infty}y^{(\sigma_{1}-\sigma)-\frac{1}{n}-1}\,dy=\infty. $$

By (7), for \(K^{(1)}(\sigma+\frac{1}{pn})>0\), we have \(\infty\leq 2M_{1}n<\infty\), which is a contradiction.

If \(\sigma_{1}<\sigma\), then for \(n\geq\frac{1}{\sigma-\sigma_{1}}\) (\(n\in\mathbf{N}\)) we consider the following functions:

$$\widetilde{f}_{n}(x):=\left \{ \textstyle\begin{array}{l@{\quad}l} 0,&0< \vert x \vert < 1 ,\\ \vert x \vert ^{\sigma-\frac{1}{pn}-1}, &\vert x \vert \geq1,\end{array}\displaystyle \right . \quad\quad \widetilde{g}_{n}(y):=\left \{ \textstyle\begin{array}{l@{\quad}l} \vert y \vert ^{\sigma_{1}+\frac{1}{qn}-1},&0< \vert y \vert \leq1, \\ 0, &\vert y \vert >1,\end{array}\displaystyle \right . $$

and derive that

$$\begin{aligned} \widetilde{J}_{1} :=& \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma )-1}\widetilde{f}_{n}^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty }^{\infty}y^{q(1-\sigma_{1})-1} \widetilde{g}_{n}^{q}(y)\,dy \biggr] ^{\frac{1}{q}} \\ =& \biggl( 2 \int_{1}^{\infty}x^{-\frac{1}{n}-1}\,dx \biggr) ^{\frac {1}{p}} \biggl( 2 \int_{0}^{1}y^{\frac{1}{n}-1}\,dy \biggr) ^{\frac{1}{q}}=2n. \end{aligned}$$

We obtain

$$\begin{aligned} \widetilde{I}_{1} :=& \int_{-\infty}^{\infty}\widetilde{f}_{n}(x) \biggl[ \int_{-\frac{1}{ \vert x \vert }}^{\frac{1}{ \vert x \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha } \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }\widetilde{g} _{n}(y)\,dy \biggr] \,dx \\ =& \int_{-\infty}^{-1} \biggl[ \int_{\frac{1}{x}}^{\frac{-1}{x}}\frac {(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert } \vert y \vert ^{\sigma_{1}+\frac{1}{qn}-1}\,dy \biggr] (-x)^{\sigma-\frac{1}{pn}-1}\,dx \\ &{}+ \int_{1}^{\infty} \biggl[ \int_{-\frac{1}{x}}^{\frac{1}{x}}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert } \vert y \vert ^{\sigma_{1}+\frac{1}{qn}-1}\,dy \biggr] x^{\sigma -\frac{1}{pn}-1}\,dx \\ =& \int_{1}^{\infty} \biggl[ \int_{-\frac{1}{x}}^{\frac{1}{x}}\bigl(H(-xy)+H(xy)\bigr) \vert y \vert ^{\sigma_{1}+\frac{1}{qn}-1}\,dy \biggr] x^{\sigma -\frac{1}{pn}-1}\,dx \\ =&2 \int_{1}^{\infty} \biggl[ \int_{0}^{1}\bigl(H(-u)+H(u) \bigr)u^{\sigma _{1}+\frac{1}{qn}-1}\,du \biggr] x^{(\sigma-\sigma_{1})-\frac{1}{n}-1}\,dx, \end{aligned}$$
(8)

and thus, by Fubini’s theorem (cf. [36]) and (5), it follows that

$$\begin{aligned} &2K^{(1)}\biggl(\sigma_{1}+\frac{1}{qn}\biggr) \int_{1}^{\infty}x^{(\sigma-\sigma _{1})-\frac{1}{n}-1}\,dx \\ &\quad=\widetilde{I}_{1}= \int_{-\infty}^{\infty}\widetilde{g}_{n}(y) \biggl( \int_{\frac{-1}{|y|}}^{\frac{1}{|y|}}H(xy)\widetilde{f}_{n}(x) \,dx \biggr) \,dy\leq M_{1}\widetilde{J}_{1} \\ &\quad=2M_{1}n. \end{aligned}$$
(9)

Since \((\sigma-\sigma_{1})-\frac{1}{n}\geq0\), it follows that

$$\int_{1}^{\infty}x^{(\sigma-\sigma_{1})-\frac{1}{n}-1}\,dx=\infty. $$

By (9), for \(K^{(1)}(\sigma_{1}+\frac{1}{qn})>0\), we get that \(\infty\leq 2M_{1}n<\infty\), which is a contradiction.

Hence, we conclude that \(\sigma_{1}=\sigma\).

For \(\sigma_{1}=\sigma\), we reduce (9) as follows:

$$K^{(1)} \biggl(\sigma_{1}+\frac{1}{qn} \biggr)= \int _{0}^{1}\bigl(H(-u)+H(u) \bigr)u^{\sigma_{1}+\frac{1}{qn}-1}\,du\leq M_{1}. $$

Since \(\{(H(-u)+H(u))u^{\sigma+\frac{1}{qn}-1}\}_{n=1}^{\infty}\) is increasing in \((0,1)\), by Levi’s theorem (cf. [36]), we obtain that

$$\begin{aligned} K^{(1)}(\sigma) =& \int_{0}^{1}\lim_{n\rightarrow\infty } \bigl(H(-u)+H(u)\bigr)u^{\sigma+\frac{1}{qn}-1}\,du \\ =&\lim_{n\rightarrow\infty} \int_{0}^{1}\bigl(H(-u)+H(u) \bigr)u^{\sigma+\frac {1}{qn}-1}\,du\leq M_{1}< \infty. \end{aligned}$$

By Remark 1, it follows that \(\sigma>-\alpha-1\) and \(\beta>0\).

This completes the proof of the lemma. □

Lemma 2

If\(\sigma_{1}\in\mathbf{R}\)and there exists a constant\(M_{2}\)such that, for any nonnegative measurable functions\(f(x)\)and\(g(y)\)inR, the following inequality

$$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{\{x; \vert x \vert \geq\frac{1}{ \vert y \vert }\}} \frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] \,dy \\ &\quad\leq M_{2} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma )-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\sigma _{1})-1}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}} \end{aligned}$$
(10)

holds true, then we have\(\sigma_{1}=\sigma\), \(\mu>-\alpha-1\), and\(\beta >0\).

Proof

If \(\sigma_{1}<\sigma\), then for \(n\geq\frac{1}{\sigma-\sigma _{1}}\) (\(n\in\mathbf{N}\)) we consider the functions \(\widetilde {f}_{n}(x)\) and \(\widetilde{g}_{n}(y)\) as in Lemma 1 and get

$$\widetilde{J}_{1}= \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}\widetilde{f}_{n}^{p}(x) \,dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty }^{\infty} \vert y \vert ^{q(1-\sigma_{1})-1}\widetilde{g}_{n}^{q}(y)\,dy \biggr] ^{\frac{1}{q}}=2n. $$

We obtain

$$\begin{aligned} \widetilde{I}_{2} :=& \int_{-\infty}^{\infty}\widetilde{g}_{n}(y) \biggl[ \int_{\{x; \vert x \vert \geq\frac{1}{ \vert y \vert }\}}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha } \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }\widetilde{f} _{n}(x)\,dx \biggr] \,dy \\ =& \int_{-1}^{0} \biggl[ \int_{\{x; \vert x \vert \geq\frac{-1}{y}\}}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert } \vert x \vert ^{\sigma-\frac{1}{pn}-1}\,dx \biggr] (-y)^{\sigma _{1}+\frac{1}{qn}-1}\,dy \\ &{}+ \int_{0}^{1} \biggl[ \int_{\{x; \vert x \vert \geq\frac{1}{y}\}}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert } \vert x \vert ^{\sigma-\frac{1}{pn}-1}\,dx \biggr] y^{\sigma _{1}+\frac{1}{qn}-1}\,dy \\ =& \int_{0}^{1} \biggl[ \int_{\{x; \vert x \vert \geq\frac{1}{y}\} }\bigl(H(-xy)+H(xy)\bigr) \vert x \vert ^{\sigma-\frac{1}{pn}-1}\,dx \biggr] y^{\sigma_{1}+\frac{1}{qn}-1}\,dy \\ =&2 \int_{0}^{1} \biggl[ \int_{1}^{\infty}\bigl(H(-u)+H(u) \bigr)u^{\sigma-\frac {1}{pn}-1}\,du \biggr] y^{(\sigma_{1}-\sigma)+\frac{1}{n}-1}\,dy, \end{aligned}$$

and then by (10) it follows that

$$ 2K^{(2)} \biggl(\sigma-\frac{1}{pn} \biggr) \int_{0}^{1}y^{(\sigma _{1}-\sigma)+\frac{1}{n}-1}\,dy= \widetilde{I}_{2}\leq M_{2}\widetilde{J}_{1}=2M_{2}n. $$
(11)

Since \((\sigma_{1}-\sigma)+\frac{1}{n}\leq0\), it follows that

$$\int_{0}^{1}y^{(\sigma_{1}-\sigma)+\frac{1}{n}-1}\,dy=\infty. $$

By (11), for \(K^{(2)}(\sigma-\frac{1}{pn})>0\), we have \(\infty\leq 2M_{2}n<\infty\), which is a contradiction.

If \(\sigma_{1}>\sigma\), then for \(n\geq\frac{1}{\sigma_{1}-\sigma}\) (\(n\in\mathbf{N}\)) we consider the functions \(f_{n}(x)\) and \(g_{n}(y)\) as in Lemma 1 and derive that

$$J_{1}= \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f_{n}^{p}(x)\, dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\sigma _{1})-1}g_{n}^{q}(y)\,dy \biggr] ^{\frac{1}{q}}=2n. $$

We obtain

$$\begin{aligned} I_{2} :=& \int_{-\infty}^{\infty}f_{n}(x) \biggl[ \int_{\{y; \vert y \vert \geq\frac {1}{ \vert x \vert }\}}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }g_{n}(y)\,dy \biggr] \,dx \\ =& \int_{-1}^{0} \biggl[ \int_{\{y; \vert y \vert \geq\frac{-1}{x}\}}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert } \vert y \vert ^{\sigma_{1}-\frac{1}{qn}-1}\,dy \biggr] (-x)^{\sigma+\frac{1}{pn}-1}\,dx \\ &{}+ \int_{0}^{1} \biggl[ \int_{\{y; \vert y \vert \geq\frac{1}{x}\}}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert } \vert y \vert ^{\sigma_{1}-\frac{1}{qn}-1}\,dy \biggr] x^{\sigma +\frac{1}{pn}-1}\,dx \\ =& \int_{0}^{1} \biggl[ \int_{\{y; \vert y \vert \geq\frac{1}{x}\} }\bigl(H(-xy)+H(xy)\bigr) \vert y \vert ^{\sigma_{1}-\frac{1}{qn}-1}\,dy \biggr] x^{\sigma+\frac{1}{pn}-1}\,dx \\ =&2 \int_{1}^{\infty}\bigl(H(-u)+H(u) \bigr)u^{(\sigma_{1}-\frac{1}{qn})-1}\,du \int_{0}^{1}x^{(\sigma-\sigma_{1})+\frac{1}{n}-1}\,dx, \end{aligned}$$

and then, by Fubini’s theorem (cf. [36]) and (8), it follows that

$$\begin{aligned} &2K_{2} \biggl(\sigma_{1}-\frac{1}{qn} \biggr) \int_{0}^{1}x^{(\sigma -\sigma_{1})+\frac{1}{n}-1} \,dx=I_{2} \\ &\quad= \int_{0}^{\infty}g_{n}(y) \biggl( \int_{\{x;|x|\geq\frac{1}{|y|}\}}H(xy)f_{n}(x)\,dx \biggr) \,dy\leq M_{2}J_{1}=2M_{2}n. \end{aligned}$$
(12)

Since \((\sigma-\sigma_{1})+\frac{1}{n}\leq0\), we get that

$$\int_{0}^{1}x^{(\sigma-\sigma_{1})+\frac{1}{n}-1}\,dx=\infty. $$

By (12), for \(K^{(2)}(\sigma_{1}-\frac{1}{qn})>0\), we deduce that \(\infty\leq 2M_{2}n<\infty\), which is a contradiction.

Hence, we conclude the fact that \(\sigma_{1}=\sigma\).

For \(\sigma_{1}=\sigma\), we reduce (12) as follows:

$$ K^{(2)} \biggl(\sigma-\frac{1}{qn} \biggr)= \int_{1}^{\infty }\bigl(H(-u)+H(u) \bigr)u^{\sigma-\frac{1}{qn}-1}\,du\leq M_{2}. $$
(13)

Since \(\{(H(-u)+H(u))u^{\sigma-\frac{1}{qn}-1}\}_{n=1}^{\infty}\) is increasing in \([1,\infty)\), applying again Levi’s theorem (cf. [36]), we have that

$$\begin{aligned} K^{(2)}(\sigma) =& \int_{1}^{\infty}\lim_{n\rightarrow\infty } \bigl(H(-u)+H(u)\bigr)u^{\sigma-\frac{1}{qn}-1}\,du \\ =&\lim_{n\rightarrow\infty} \int_{1}^{\infty}\bigl(H(-u)+H(u) \bigr)u^{\sigma -\frac{1}{qn}-1}\,du\leq M_{2}< \infty. \end{aligned}$$

By Remark 1, we get that \(\mu>-\alpha-1\) and \(\beta>0\).

This completes the proof of the lemma. □

Main results and some corollaries

Theorem 1

If\(\sigma_{1}\in\mathbf{R,}\)then the following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{1}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty , $$

    we have the following Hardy-type integral inequality of the first kind with the nonhomogeneous kernel:

    $$\begin{aligned} J :=& \biggl\{ \int_{-\infty}^{\infty} \vert y \vert ^{p\sigma_{1}-1} \biggl[ \int_{ \frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\, dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ < &M_{1} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (14)
  2. (ii)

    There exists a constant\(M_{1}\)such that, for any\(f(x),g(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty\quad \textit{and}\quad 0< \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\sigma_{1})-1}g^{q}(y)\,dy< \infty , $$

    we have the following inequality:

    $$\begin{aligned} I :=& \int_{-\infty}^{\infty}g(y) \biggl[ \int_{\frac{-1}{ \vert y \vert }}^{\frac {1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] \,dy \\ < &M_{1} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\sigma _{1})-1}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}. \end{aligned}$$
    (15)
  3. (iii)

    \(\sigma_{1}=\sigma>-\alpha-1\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{1}=K^{(1)}(\sigma )\) (\(\in \mathbf{R}_{+}\)) in (14) and (15) (for\(\sigma _{1}=\sigma\)) is the best possible.

Proof

(i) (ii). By Hölder’s inequality (cf. [37]), we have

$$\begin{aligned} I =& \int_{-\infty}^{\infty} \biggl( \vert y \vert ^{\sigma_{1}-\frac{1}{p}} \int_{ \frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}H(xy)f(x)\,dx \biggr) \bigl( \vert y \vert ^{\frac {1}{p}-\sigma_{1}}g(y) \bigr) \,dy \\ \leq&J \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\sigma _{1})-1}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}. \end{aligned}$$
(16)

Then by (14) we deduce (15).

(ii) (iii). By Lemma 1, we have \(\sigma_{1}=\sigma >-\alpha-1\) and \(\beta>0\).

(iii) (i). We obtain the following weight function:

For \(y\neq0\),

$$\begin{aligned} \omega_{1}(\sigma,y) :=& \vert y \vert ^{\sigma} \int_{\frac{-1}{ \vert y \vert }}^{\frac {1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert } \vert x \vert ^{\sigma-1}\,dx \\ =& \vert y \vert ^{\sigma} \int_{\frac{-1}{ \vert y \vert }}^{0}H(xy) (-x)^{\sigma-1}\, dx+ \vert y \vert ^{\sigma } \int_{0}^{\frac{1}{ \vert y \vert }}H(xy)x^{\sigma-1}\,dx \\ =& \vert y \vert ^{\sigma} \int_{0}^{\frac{1}{ \vert y \vert }}H(-xy)x^{\sigma-1}\, dx+ \vert y \vert ^{\sigma } \int_{0}^{\frac{1}{ \vert y \vert }}H(xy)x^{\sigma-1}\,dx \\ =& \vert y \vert ^{\sigma} \int_{0}^{\frac{1}{ \vert y \vert }}(H\bigl(-x \vert y \vert \bigr)+H\bigl(x \vert y \vert \bigr)x^{\sigma-1}\,dx \\ =& \int_{0}^{1}\bigl(H(-u)+H(u) \bigr)u^{\sigma-1}\,du \\ =&K^{(1)}(\sigma). \end{aligned}$$
(17)

By the weighted Hölder inequality and (17), we obtain

$$\begin{aligned} & \biggl[ \int_{\frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] ^{p} \\ &\quad= \biggl\{ \int_{\frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}H(xy) \biggl[ \frac{ \vert y \vert ^{(\sigma-1)/p}}{ \vert x \vert ^{(\sigma-1)/q}}f(x) \biggr] \biggl[ \frac{ \vert x \vert ^{(\sigma-1)/q}}{ \vert y \vert ^{(\sigma-1)/p}} \biggr] \,dx \biggr\} ^{p} \\ &\quad\leq \int_{\frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}H(xy)\frac{ \vert y \vert ^{\sigma -1}f^{p}(x)}{ \vert x \vert ^{(\sigma-1)p/q}}\,dx \biggl[ \int_{\frac {-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}h(xy)\frac{ \vert x \vert ^{\sigma-1}}{ \vert y \vert ^{(\sigma-1)q/p}}\,dx \biggr] ^{p-1} \\ &\quad= \int_{\frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}H(xy)\frac{ \vert y \vert ^{\sigma-1}}{ \vert x \vert ^{(\sigma-1)p/q}}f^{p}(x) \,dx\cdot \bigl[ \omega_{1}(\sigma ,y) \vert y \vert ^{q(1-\sigma)-1} \bigr] ^{p-1} \\ &\quad=\bigl(K^{(1)}(\sigma)\bigr)^{p-1} \vert y \vert ^{-p\sigma+1} \int_{\frac{-1}{ \vert y \vert }}^{\frac {1}{ \vert y \vert }}H(xy)\frac{ \vert y \vert ^{\sigma-1}}{ \vert x \vert ^{(\sigma-1)p/q}}f^{p}(x) \,dx. \end{aligned}$$
(18)

If (18) takes the form of equality for some \(y\in\mathbf {R}\setminus \{0\}\), then (cf. [37]) there exist constants A and B such that they are not both zero and

$$A\frac{ \vert y \vert ^{\sigma-1}}{ \vert x \vert ^{(\sigma-1)p/q}}f^{p}(x)=B\frac { \vert x \vert ^{\sigma-1}}{ \vert y \vert ^{(\sigma-1)q/p}} \quad \mbox{a.e. in }\mathbf{R}. $$

Let us assume that \(A\neq0\) (otherwise \(B=A=0\)). It follows that

$$\vert x \vert ^{p(1-\sigma)-1}f^{p}(x)= \vert y \vert ^{q(1-\sigma)}\frac{B}{A \vert x \vert }\quad\mbox{a.e. in }\mathbf{R}, $$

which contradicts the fact that

$$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma )-1}f^{p}(x)\,dx< \infty. $$

Hence, (18) takes the form of strict inequality.

For \(\sigma_{1}=\sigma>-\alpha-1\) and \(\beta>0\), we have \(K^{(1)}(\sigma )\in\mathbf{R}_{+}\). In view of Fubini’s theorem (cf. [36]), we obtain

$$\begin{aligned} J < &\bigl(K^{(1)}(\sigma)\bigr)^{\frac{1}{q}} \biggl\{ \int_{-\infty}^{\infty } \biggl[ \int_{\frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}H(xy)\frac{ \vert y \vert ^{\sigma-1}}{ \vert x \vert ^{(\sigma-1)p/q}}f^{p}(x) \,dx \biggr] \,dy \biggr\} ^{\frac{1}{p}} \\ =&\bigl(K^{(1)}(\sigma)\bigr)^{\frac{1}{q}} \biggl\{ \int_{-\infty}^{\infty } \biggl[ \int_{\frac{-1}{ \vert x \vert }}^{\frac{1}{ \vert x \vert }}H(xy)\frac{ \vert y \vert ^{\sigma-1}}{ \vert x \vert ^{(\sigma-1)(p-1)}}\,dy \biggr] f^{p}(x)\,dx \biggr\} ^{\frac{1}{p}} \\ =&\bigl(K^{(1)}(\sigma)\bigr)^{\frac{1}{q}} \biggl[ \int_{-\infty}^{\infty }\omega _{1}(\sigma,x) \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}} \\ =&K^{(1)}(\sigma) \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma )-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$

Setting \(M_{1}\geq K^{(1)}(\sigma)\), we have

$$J< K^{(1)}(\sigma) \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma )-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}\leq M_{1} \biggl[ \int_{-\infty }^{\infty } \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}, $$

namely, (14) follows.

Therefore, statements (i), (ii), and (iii) are equivalent.

When statement (iii) is satisfied, if there exists a constant \(M_{1}\leq K^{(1)}(\sigma)\) such that (15) is valid, then by the proof of Lemma 1, we have \(K^{(1)}(\sigma)\leq M_{1}\). It follows that the constant factor \(M_{1}=K^{(1)}(\sigma)\) in (15) is the best possible. The constant factor \(M_{1}=K^{(1)}(\sigma)\) in (14) is still the best possible. Otherwise, by (16) (for \(\sigma _{1}=\sigma\)), we would conclude that the constant factor \(M_{1}=K^{(1)}(\sigma)\) in (15) was not the best possible.

This completes the proof of the theorem. □

In particular, for \(\sigma=\sigma_{1}=\frac{1}{p}>-\alpha-1\) in Theorem 1, the following corollary holds true.

Corollary 1

The following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{1}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p-2}f^{p}(x)\,dx< \infty, $$

    the following inequality is satisfied:

    $$\begin{aligned} \begin{aligned}[b] & \biggl\{ \int_{-\infty}^{\infty} \biggl[ \int_{\frac{-1}{ \vert y \vert }}^{\frac {1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{1} \biggl( \int_{-\infty}^{\infty} \vert x \vert ^{p-2}f^{p}(x)\,dx \biggr) ^{\frac{1}{p}}.\end{aligned} \end{aligned}$$
    (19)
  2. (ii)

    There exists a constant\(M_{1}\)such that, for any\(f(x), g(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p-2}f^{p}(x)\,dx< \infty\quad \textit{and}\quad 0< \int _{-\infty}^{\infty}g^{q}(y)\,dy< \infty, $$

    we have the following inequality:

    $$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{\frac{-1}{ \vert y \vert }}^{\frac {1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x) \,dx \biggr] \,dy \\ &\quad< M_{1} \biggl( \int_{-\infty}^{\infty} \vert x \vert ^{p-2}f^{p}(x)\,dx \biggr) ^{\frac{1}{p}} \biggl( \int_{-\infty}^{\infty}g^{q}(y)\,dy \biggr) ^{\frac{1}{q}}. \end{aligned}$$
    (20)
  3. (iii)

    \(\alpha>-\frac{1}{p}-1\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{1}=K^{(1)}(\frac {1}{p})\) (\(\in\mathbf{R}_{+}\)) in (19) and (20) is the best possible.

Setting \(y=\frac{1}{Y}\), \(G(Y)=g(\frac{1}{Y})\frac{1}{Y^{2}}\) in Theorem 1, and then replacing Y by y, we obtain the following corollary.

Corollary 2

If\(\sigma_{1}\in\mathbf{R,}\)then the following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{1}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty , $$

    we have the following inequality:

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty} \vert y \vert ^{-p\sigma_{1}-1} \biggl[ \int_{- \vert y \vert }^{ \vert y \vert }\frac{(\min\{ \vert x/y \vert ,1\})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta }}{(\max\{ \vert x/y \vert ,1\})^{\lambda+\alpha} \vert x/y-1 \vert }f(x)\,dx \biggr] ^{p}\, dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{1} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (21)
  2. (ii)

    There exists a constant\(M_{1}\)such that, for any\(f(x),G(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty\quad \textit{and}\quad 0< \int_{-\infty}^{\infty}y^{q(1+\sigma_{1})-1}G^{q}(y) \, dy< \infty, $$

    we have the following inequality:

    $$\begin{aligned} \begin{aligned}[b] & \int_{-\infty}^{\infty}G(y) \biggl[ \int_{- \vert y \vert }^{ \vert y \vert }\frac{(\min \{ \vert x/y \vert ,1\})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max\{ \vert x/y \vert ,1\} )^{\lambda +\alpha} \vert x/y-1 \vert }f(x)\,dx \biggr] \,dy \\ &\quad< M_{1} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1+\sigma _{1})-1}G^{q}(y)\,dy \biggr] ^{\frac{1}{q}}.\end{aligned} \end{aligned}$$
    (22)
  3. (iii)

    \(\sigma_{1}=\sigma>-\alpha-1\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{1}=K^{(1)}(\sigma )\) (\(\in \mathbf{R}_{+}\)) in (21) and (22) (for\(\sigma _{1}=\sigma\)) is the best possible.

For \(g(y)=y^{\lambda}G(y)\) and \(\mu_{1}=\lambda-\sigma_{1}\) in Corollary 2, we deduce the following corollary.

Corollary 3

If\(\mu_{1}\in\mathbf{R}\), then the following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{1}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty , $$

    we have the following Hardy-type integral inequality of the first kind with homogeneous kernel:

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty}y^{p\mu_{1}-1} \biggl[ \int _{- \vert y \vert }^{ \vert y \vert }\frac{(\min\{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max \{ \vert x \vert , \vert y \vert \})^{\lambda+\alpha} \vert x-y \vert }f(x) \,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{1} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (23)
  2. (ii)

    There exists a constant\(M_{1}\)such that, for any\(f(x),g(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty\quad \textit{and}\quad 0< \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\mu_{1})-1}g^{q}(y)\, dy< \infty, $$

    we have the following inequality:

    $$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{- \vert y \vert }^{ \vert y \vert }\frac{(\min \{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max\{ \vert x \vert , \vert y \vert \} )^{\lambda +\alpha} \vert x-y \vert }f(x)\,dx \biggr] \,dy \\ &\quad< M_{1} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\mu _{1})-1}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}. \end{aligned}$$
    (24)
  3. (iii)

    \(\mu_{1}=\mu<\lambda+\alpha+1\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{1}=K^{(1)}(\sigma )\) (\(\in \mathbf{R}_{+}\)) in (23) and (24) (for\(\mu_{1}=\mu\)) is the best possible.

In particular, for \(\lambda=1\), \(\sigma=\frac{1}{q}\), \(\mu=\frac{1}{p}\) in Corollary 3, we get the following corollary.

Corollary 4

The following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{1}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty}f^{p}(x)\,dx< \infty, $$

    the following inequality holds true:

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty} \biggl[ \int_{- \vert y \vert }^{ \vert y \vert } \biggl( \frac {\min \{ \vert x \vert , \vert y \vert \}}{\max\{ \vert x \vert , \vert y \vert \}} \biggr) ^{1+\alpha}\frac{ \vert \ln \vert x/y \vert \vert ^{\beta}}{ \vert x-y \vert }f(x)\,dx \biggr] ^{p} \,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{1} \biggl( \int_{-\infty}^{\infty}f^{p}(x)\,dx \biggr) ^{\frac{1}{p}}. \end{aligned}$$
    (25)
  2. (ii)

    There exists a constant\(M_{1}\)such that, for any\(f(x),g(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty}f^{p}(x)\,dx< \infty \quad \textit{and}\quad 0< \int_{-\infty }^{\infty}g^{q}(y)\,dy< \infty, $$

    we have the following inequality:

    $$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{- \vert y \vert }^{ \vert y \vert } \biggl( \frac{\min \{ \vert x \vert , \vert y \vert \}}{\max\{ \vert x \vert , \vert y \vert \}} \biggr) ^{1+\alpha}\frac{ \vert \ln \vert x/y \vert \vert ^{\beta}}{ \vert x-y \vert }f(x)\,dx \biggr] \,dy \\ &\quad< M_{1} \biggl( \int_{-\infty}^{\infty}f^{p}(x)\,dx \biggr) ^{\frac {1}{p}} \biggl( \int_{-\infty}^{\infty}g^{q}(y)\,dy \biggr) ^{\frac{1}{q}}. \end{aligned}$$
    (26)
  3. (iii)

    \(\alpha>-\frac{1}{q}-1\)and\(\beta>0\).

If statement (iii) holds true, then the constant factor\(M_{1}=K^{(1)}(\frac{1}{q})\) (\(\in\mathbf{R}_{+}\)) in (25) and (26) is the best possible.

Remark 2

  1. (i)

    For \(\sigma_{1}=\sigma=-\alpha+1\) in (14), we have the following inequality with the best possible constant factor \(\frac{\varGamma(\beta+1)}{2^{\beta}}\zeta(\beta+1)\) (\(\beta>0\)):

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty} \vert y \vert ^{p(1-\alpha)-1} \biggl[ \int _{\frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] ^{p}\, dy \biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{\varGamma(\beta+1)}{2^{\beta}}\zeta(\beta+1) \biggl[ \int _{-\infty }^{\infty} \vert x \vert ^{p\alpha-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (27)
  2. (ii)

    For \(\mu_{1}=\mu=\lambda+\alpha-1\) in (23), we have the following inequality with the best possible constant factor \(\frac {\varGamma (\beta+1)}{2^{\beta}}\zeta(\beta+1)\) (\(\beta>0\)):

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty}y^{p(\lambda+\alpha-1)-1} \biggl[ \int_{- \vert y \vert }^{ \vert y \vert }\frac{(\min\{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta }}{(\max\{ \vert x \vert , \vert y \vert \})^{\lambda+\alpha} \vert x-y \vert }f(x)\,dx \biggr] ^{p}\, dy \biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{\varGamma(\beta+1)}{2^{\beta}}\zeta(\beta+1) \biggl[ \int _{-\infty }^{\infty} \vert x \vert ^{p\alpha-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (28)
  3. (iii)

    For \(\alpha=-1\) in (25), we have the following inequality with the best possible constant factor \(\frac{\varGamma(\beta +1)}{2^{\beta}}\zeta(\beta+1,\frac{1}{2q})\) (\(\beta>0\)):

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty} \biggl[ \int_{- \vert y \vert }^{ \vert y \vert }\frac{ \vert \ln \vert x/y \vert \vert ^{\beta}}{ \vert x-y \vert }f(x)\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{\varGamma(\beta+1)}{2^{\beta}}\zeta\biggl(\beta+1,\frac {1}{2q}\biggr) \biggl( \int_{-\infty}^{\infty}f^{p}(x)\,dx \biggr) ^{\frac{1}{p}}. \end{aligned}$$
    (29)

Similarly, in view of Lemma 2, we obtain the following weight function:

For \(y\neq0\),

$$\begin{aligned} \omega_{2}(\sigma,y) :=& \vert y \vert ^{\sigma} \int_{\{x; \vert x \vert \geq\frac {1}{ \vert y \vert }\}}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert } \vert x \vert ^{\sigma-1}\,dx \\ =& \int_{1}^{\infty}\bigl(H(-u)+H(u) \bigr)u^{\sigma-1}\,du=K^{(2)}(\sigma), \end{aligned}$$

and then similarly, we derive the following results.

Theorem 2

If\(\sigma_{1}\in\mathbf{R,}\)then the following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{2}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty , $$

    the following Hardy-type integral inequality of the second kind with nonhomogeneous kernel is satisfied:

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty}y^{p\sigma_{1}-1} \biggl[ \int_{\{x; \vert x \vert \geq\frac{1}{ \vert y \vert }\}}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha } \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\, dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{2} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (30)
  2. (ii)

    There exists a constant\(M_{2}\)such that, for any\(f(x),g(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty\quad \textit{and}\quad 0< \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\sigma_{1})-1}g^{q}(y)\, dy< \infty , $$

    we have the following inequality:

    $$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{\{x; \vert x \vert \geq\frac{1}{ \vert y \vert }\}} \frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] \,dy \\ &\quad< M_{2} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty}y^{q(1-\sigma _{1})-1}g^{q}(y) \,dy \biggr] ^{\frac{1}{q}}. \end{aligned}$$
    (31)
  3. (iii)

    \(\sigma_{1}=\sigma<\lambda+\alpha+1\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{2}=K^{(2)}(\sigma )\) (\(\in \mathbf{R}_{+}\)) in (30) and (31) (for\(\sigma _{1}=\sigma\)) is the best possible.

In particular, for \(\sigma=\sigma_{1}=\frac{1}{p}\) in Theorem 2, we obtain the following corollary.

Corollary 5

The following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{2}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p-2}f^{p}(x)\,dx< \infty, $$

    we have the following inequality:

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty} \biggl[ \int_{\{x; \vert x \vert \geq\frac {1}{ \vert y \vert }\}}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{2} \biggl( \int_{-\infty}^{\infty} \vert x \vert ^{p-2}f^{p}(x)\,dx \biggr) ^{\frac{1}{p}}. \end{aligned}$$
    (32)
  2. (ii)

    There exists a constant\(M_{2}\)such that, for any\(f(x),g(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p-2}f^{p}(x)\,dx< \infty\quad \textit{and}\quad 0< \int _{-\infty}^{\infty}g^{q}(y)\,dy< \infty, $$

    we have the following inequality:

    $$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{\{x; \vert x \vert \geq\frac{1}{ \vert y \vert }\}} \frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] \,dy \\ &\quad< M_{2} \biggl( \int_{-\infty}^{\infty} \vert x \vert ^{p-2}f^{p}(x)\,dx \biggr) ^{\frac{1}{p}} \biggl( \int_{-\infty}^{\infty}g^{q}(y)\,dy \biggr) ^{\frac{1}{q}}. \end{aligned}$$
    (33)
  3. (iii)

    \(\alpha>-\lambda-\frac{1}{q}\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{2}=K^{(2)}(\frac {1}{p})\) (\(\in\mathbf{R}_{+}\)) in (32) and (33) is the best possible.

Setting \(y=\frac{1}{Y}\), \(G(Y)=g(\frac{1}{Y})\frac{1}{Y^{2}}\) in Theorem 2, and then replacing Y by y, we deduce the following corollary.

Corollary 6

If\(\sigma_{1}\in\mathbf{R,}\)then the following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{2}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty , $$

    we have the following inequality:

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty}y^{-p\sigma_{1}-1} \biggl[ \int_{\{x; \vert x \vert \geq \vert y \vert \}}\frac{(\min\{ \vert x/y \vert ,1\})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max\{ \vert x/y \vert ,1\})^{\lambda+\alpha} \vert x/y-1 \vert }f(x)\, dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{2} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (34)
  2. (ii)

    There exists a constant\(M_{2}\)such that, for any\(f(x),G(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty\quad \textit{and}\quad 0< \int_{-\infty}^{\infty} \vert y \vert ^{q(1+\sigma_{1})-1}G^{q}(y)\, dy< \infty , $$

    we have the following inequality:

    $$\begin{aligned} & \int_{-\infty}^{\infty}G(y) \biggl[ \int_{\{x; \vert x \vert \geq \vert y \vert \}}\frac {(\min \{ \vert x/y \vert ,1\})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max\{ \vert x/y \vert ,1\} )^{\lambda +\alpha} \vert x/y-1 \vert }f(x)\,dx \biggr] \,dy \\ &\quad< M_{2} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1+\sigma _{1})-1}G^{q}(y)\,dy \biggr] ^{\frac{1}{q}}. \end{aligned}$$
    (35)
  3. (iii)

    \(\sigma_{1}=\sigma<\lambda+\alpha+1\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{2}=K^{(2)}(\sigma )\) (\(\in \mathbf{R}_{+}\)) in (34) and (35) (for\(\sigma _{1}=\sigma\)) is the best possible.

For \(g(y)=y^{\lambda}G(y)\) and \(\mu_{1}=\lambda-\sigma_{1}\) in Corollary 6, we deduce the following corollary.

Corollary 7

If\(\mu_{1}\in\mathbf{R,}\)then the following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{2}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty , $$

    we have the following inequality:

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty}y^{p\mu_{1}-1} \biggl[ \int_{\{ x; \vert x \vert \geq \vert y \vert \}}\frac{(\min\{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max \{ \vert x \vert , \vert y \vert \})^{\lambda+\alpha} \vert x-y \vert }f(x)\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{2} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (36)
  2. (ii)

    There exists a constant\(M_{2}\)such that, for any\(f(x),g(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\,dx< \infty\quad \textit{and}\quad 0< \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\mu_{1})-1}g^{q}(y)\, dy< \infty, $$

    we have the following inequality:

    $$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{\{x; \vert x \vert \geq \vert y \vert \}}\frac {(\min \{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max\{ \vert x \vert , \vert y \vert \} )^{\lambda +\alpha} \vert x-y \vert }f(x)\,dx \biggr] \,dy \\ &\quad< M_{2} \biggl[ \int_{-\infty}^{\infty} \vert x \vert ^{p(1-\sigma)-1}f^{p}(x)\, dx \biggr] ^{\frac{1}{p}} \biggl[ \int_{-\infty}^{\infty} \vert y \vert ^{q(1-\mu _{1})-1}g^{q}(y)\,dy \biggr] ^{\frac{1}{q}}. \end{aligned}$$
    (37)
  3. (iii)

    \(\mu_{1}=\mu>-\alpha-1\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{2}=K^{(2)}(\sigma )\) (\(\in \mathbf{R}_{+}\)) in (36) and (37) (for\(\mu_{1}=\mu\)) is the best possible.

In particular, for \(\lambda=1\), \(\sigma=\frac{1}{q}\), \(\mu=\frac{1}{p}\) in Corollary 7, we obtain the following corollary.

Corollary 8

The following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{2}\)such that, for any\(f(x)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty}f^{p}(x)\,dx< \infty, $$

    we have the following inequality:

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty} \biggl[ \int_{\{x; \vert x \vert \geq \vert y \vert \}} \biggl( \frac{\min\{ \vert x \vert , \vert y \vert \}}{\max\{ \vert x \vert , \vert y \vert \}} \biggr) ^{1+\alpha}\frac{ \vert \ln \vert x/y \vert \vert ^{\beta}}{ \vert x-y \vert }f(x)\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< M_{2} \biggl( \int_{-\infty}^{\infty}f^{p}(x)\,dx \biggr) ^{\frac{1}{p}}. \end{aligned}$$
    (38)
  2. (ii)

    There exists a constant\(M_{2}\)such that, for any\(f(x),g(y)\geq0\)satisfying

    $$0< \int_{-\infty}^{\infty}f^{p}(x)\,dx< \infty \quad \textit{and}\quad 0< \int_{-\infty }^{\infty}g^{q}(y)\,dy< \infty, $$

    we have the following inequality:

    $$\begin{aligned} & \int_{-\infty}^{\infty}g(y) \biggl[ \int_{\{x; \vert x \vert \geq \vert y \vert \}} \biggl( \frac{\min\{ \vert x \vert , \vert y \vert \}}{\max\{ \vert x \vert , \vert y \vert \}} \biggr) ^{1+\alpha}\frac{ \vert \ln \vert x/y \vert \vert ^{\beta}}{ \vert x-y \vert }f(x)\,dx \biggr] \,dy \\ &\quad< M_{2} \biggl( \int_{-\infty}^{\infty}f^{p}(x)\,dx \biggr) ^{\frac {1}{p}} \biggl( \int_{-\infty}^{\infty}g^{q}(y)\,dy \biggr) ^{\frac{1}{q}}. \end{aligned}$$
    (39)
  3. (iii)

    \(\alpha>-\frac{1}{p}-1\)and\(\beta>0\).

If statement (iii) holds true, then the constant\(M_{2}=K^{(2)}(\frac {1}{q})\) (\(\in\mathbf{R}_{+}\)) in (38) and (39) is the best possible.

Remark 3

  1. (i)

    For \(\sigma_{1}=\sigma=\lambda+\alpha-1\) in (30), we have the following inequality with the best possible constant factor \(\frac{\varGamma(\beta+1)}{2^{\beta}}\zeta(\beta+1)\) (\(\beta>0\)):

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty}y^{p(\lambda+\alpha-1)-1} \biggl[ \int_{\{x; \vert x \vert \geq\frac{1}{ \vert y \vert }\}}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha } \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\, dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{\varGamma(\beta+1)}{2^{\beta}}\zeta(\beta+1) \biggl[ \int _{-\infty }^{\infty} \vert x \vert ^{p(2-\lambda-\alpha)-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (40)
  2. (ii)

    For \(\mu_{1}=\mu=-\alpha+1\) in (36), we have the following inequality with the best possible constant factor \(\frac{\varGamma(\beta +1)}{2^{\beta}}\zeta(\beta+1)\) (\(\beta>0\)):

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty}y^{p(1-\alpha)-1} \biggl[ \int_{\{x; \vert x \vert \geq \vert y \vert \}}\frac{(\min\{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max\{ \vert x \vert , \vert y \vert \})^{\lambda+\alpha} \vert x-y \vert }f(x)\, dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{\varGamma(\beta+1)}{2^{\beta}}\zeta(\beta+1) \biggl[ \int _{-\infty }^{\infty} \vert x \vert ^{p(2-\lambda-\alpha)-1}f^{p}(x)\,dx \biggr] ^{\frac{1}{p}}. \end{aligned}$$
    (41)
  3. (iii)

    For \(\alpha=-1\) in (38), we have the following inequality with the best possible constant factor \(\frac{\varGamma(\beta +1)}{2^{\beta}}\zeta(\beta+1,\frac{1}{2p})\) (\(\beta>0\)):

    $$\begin{aligned} & \biggl\{ \int_{-\infty}^{\infty} \biggl[ \int_{\{x; \vert x \vert \geq \vert y \vert \}}\frac { \vert \ln \vert x/y \vert \vert ^{\beta}}{ \vert x-y \vert }f(x)\,dx \biggr] ^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\quad< \frac{\varGamma(\beta+1)}{2^{\beta}}\zeta\biggl(\beta+1,\frac {1}{2p}\biggr) \biggl( \int_{-\infty}^{\infty}f^{p}(x)\,dx \biggr) ^{\frac{1}{p}}. \end{aligned}$$
    (42)

Operator expressions

We set the following functions:

$$\varphi(x):= \vert x \vert ^{p(1-\sigma)-1},\qquad \psi (y):= \vert y \vert ^{q(1-\sigma)-1}, \qquad\phi(y):= \vert y \vert ^{q(1-\mu)-1}, $$

wherefrom

$$\psi ^{1-p}(y)= \vert y \vert ^{p\sigma-1},\qquad \phi^{1-p}(y)= \vert y \vert ^{p\mu-1}\quad (x,y\in\mathbf {R}). $$

We also define the following real normed linear spaces:

$$L_{p,\varphi}(\mathbf{R}):= \biggl\{ f: \Vert f \Vert _{p,\varphi}:= \biggl( \int_{-\infty}^{\infty}\varphi(x) \bigl\vert f(x) \bigr\vert ^{p}\,dx \biggr) ^{\frac{1}{p}}< \infty \biggr\} , $$

wherefrom

$$\begin{aligned}& L_{q,\psi}(\mathbf{R}) = \biggl\{ g: \Vert g \Vert _{q,\psi}:= \biggl( \int _{-\infty }^{\infty}\psi(y) \bigl\vert g(y) \bigr\vert ^{q}\,dy \biggr) ^{\frac{1}{q}}< \infty \biggr\} , \\ & L_{q,\phi}(\mathbf{R}) = \biggl\{ g: \Vert g \Vert _{q,\phi}:= \biggl( \int _{-\infty }^{\infty}\phi(y) \bigl\vert g(y) \bigr\vert ^{q}\,dy \biggr) ^{\frac{1}{q}}< \infty \biggr\} , \\ & L_{p,\psi^{1-p}}(\mathbf{R}) = \biggl\{ h: \Vert h \Vert _{p,\psi^{1-p}}= \biggl( \int_{-\infty}^{\infty}\psi^{1-p}(y) \bigl\vert h(y) \bigr\vert ^{p}\,dy \biggr) ^{\frac {1}{p}}< \infty \biggr\} , \\ & L_{q,\phi^{1-p}}(\mathbf{R}) = \biggl\{ h: \Vert h \Vert _{p,\phi^{1-p}}= \biggl( \int_{-\infty}^{\infty}\phi^{1-p}(y) \bigl\vert h(y) \bigr\vert ^{p}\,dy \biggr) ^{\frac {1}{p}}< \infty \biggr\} . \end{aligned}$$

(a) In view of Theorem 1, for \(\sigma_{1}=\sigma\) and \(f\in L_{p,\varphi }(\mathbf{R})\), setting

$$h_{1}(y):= \int_{\frac{-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert }f(x)\,dx \quad(y\in\mathbf{R}), $$

by (14), we obtain that

$$ \Vert h_{1} \Vert _{p,\psi^{1-p}}= \biggl[ \int_{-\infty}^{\infty}\psi ^{1-p}(y)h_{1}^{p}(y) \,dy \biggr] ^{\frac{1}{p}}< M_{1} \Vert f \Vert _{p,\varphi }< \infty . $$
(43)

Definition 1

We define a Hardy-type integral operator of the first kind with nonhomogeneous kernel

$$T_{1}^{(1)} : L_{p,\varphi}(\mathbf{R}) \rightarrow L_{p,\psi^{1-p}}(\mathbf{R}) $$

as follows:

For any \(f\in L_{p,\varphi}(\mathbf{R})\), there exists a unique representation

$$T_{1}^{(1)}f=h_{1}\in L_{p,\psi^{1-p}}( \mathbf{R}) $$

satisfying \(T_{1}^{(1)}f(y)=h_{1}(y)\) for any \(y\in\mathbf{R}\).

In view of (43), it follows that

$$\bigl\Vert T_{1}^{(1)}f \bigr\Vert _{p,\psi ^{1-p}}= \Vert h_{1} \Vert _{p,\psi^{1-p}}\leq M_{1} \Vert f \Vert _{p,\varphi}, $$

and thus the operator \(T_{1}^{(1)}\) is bounded satisfying

$$\bigl\Vert T_{1}^{(1)} \bigr\Vert =\sup _{f(\neq\theta)\in L_{p,\varphi}(\mathbf {R})}\frac{ \Vert T_{1}^{(1)}f \Vert _{p,\psi^{1-p}}}{ \Vert f \Vert _{p,\varphi}}\leq M_{1}. $$

If we define the formal inner product of \(T_{1}^{(1)}f\) and g as follows:

$$\bigl(T_{1}^{(1)}f,g\bigr):= \int_{-\infty}^{\infty} \biggl[ \int_{\frac {-1}{ \vert y \vert }}^{\frac{1}{ \vert y \vert }}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta }}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] g(y)\,dy, $$

we can then rewrite Theorem 1 (for \(\sigma_{1}=\sigma\)) as follows.

Theorem 3

The following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{1}\)such that, for any\(f(x)\geq 0\), \(f\in L_{p,\varphi}(\mathbf{R})\), \(\|f\|_{p,\varphi}>0\), we have the following inequality:

    $$ \bigl\Vert T_{1}^{(1)}f \bigr\Vert _{p,\psi^{1-p}}< M_{1} \Vert f \Vert _{p,\varphi}. $$
    (44)
  2. (ii)

    There exists a constant\(M_{1}\)such that, for any\(f(x),g(y)\geq 0\), \(f\in L_{p,\varphi}(\mathbf{R})\), \(g\in L_{q,\psi }(\mathbf{R})\), \(\|f\|_{p,\varphi},\|g\|_{q,\psi}>0\), we have the following inequality:

    $$ \bigl(T_{1}^{(1)}f,g\bigr)< M_{1} \Vert f \Vert _{p,\varphi} \Vert g \Vert _{q,\psi}. $$
    (45)
  3. (iii)

    \(\sigma>-\alpha-1\)and\(\beta>0\).

If statement (iii) holds true, then it holds that\(\|T_{1}^{(1)}\|=K^{(1)}(\sigma)\).

(b) In view of Corollary 3, for \(\mu_{1}=\mu\) and for \(f\in L_{p,\varphi}(\mathbf{R})\), setting

$$h_{2}(y):= \int_{- \vert y \vert }^{ \vert y \vert }\frac{(\min\{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max\{ \vert x \vert , \vert y \vert \})^{\lambda+\alpha} \vert x-y \vert }f(x) \,dx\quad(y\in\mathbf{R}), $$

by (23), we have

$$ \Vert h_{2} \Vert _{p,\phi^{1-p}}= \biggl[ \int_{-\infty}^{\infty}\phi ^{1-p}(y)h_{2}^{p}(y) \,dy \biggr] ^{\frac{1}{p}}< M_{1} \Vert f \Vert _{p,\varphi }< \infty . $$
(46)

Definition 2

We define a Hardy-type integral operator of the first kind with homogeneous kernel

$$T_{1}^{(2)} : L_{p,\varphi}(\mathbf{R}) \rightarrow L_{p,\phi^{1-p}}(\mathbf{R}) $$

as follows:

For any \(f\in L_{p,\varphi}(\mathbf{R})\), there exists a unique representation

$$T_{1}^{(2)}f=h_{2}\in L_{p,\phi^{1-p}}(\mathbf{R}) $$

satisfying \(T_{1}^{(2)}f(y)=h_{2}(y)\) for any \(y\in\mathbf{R}\).

In view of (46), it follows that

$$\bigl\Vert T_{1}^{(2)}f \bigr\Vert _{p,\phi ^{1-p}}= \Vert h_{2} \Vert _{p,\phi^{1-p}}\leq M_{1} \Vert f \Vert _{p,\varphi}, $$

and thus the operator \(T_{1}^{(2)}\) is bounded satisfying

$$\bigl\Vert T_{1}^{(2)} \bigr\Vert =\sup _{f(\neq\theta)\in L_{p,\varphi}(\mathbf {R})}\frac{ \Vert T_{1}^{(2)}f \Vert _{p,\phi^{1-p}}}{ \Vert f \Vert _{p,\varphi}}\leq M_{1}. $$

If we define the formal inner product of \(T_{1}^{(2)}f\) and g in the following manner:

$$\bigl(T_{1}^{(2)}f,g\bigr):= \int_{-\infty}^{\infty} \biggl[ \int_{- \vert y \vert }^{ \vert y \vert }\frac {(\min\{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max \{ \vert x \vert , \vert y \vert \})^{\lambda+\alpha} \vert x-y \vert }f(x)\,dx \biggr] g(y)\,dy, $$

then we can rewrite Corollary 3 (for \(\mu_{1}=\mu\)) as follows.

Corollary 9

The following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{1}\)such that, for any\(f(x)\geq 0\), \(f\in L_{p,\varphi}(\mathbf{R})\), \(\|f\|_{p,\varphi}>0\), we have the following inequality:

    $$ \bigl\Vert T_{1}^{(2)}f \bigr\Vert _{p,\phi^{1-p}}< M_{1} \Vert f \Vert _{p,\varphi}. $$
    (47)
  2. (ii)

    There exists a constant\(M_{1}\)such that, for any\(f(x),g(y)\geq 0\), \(f\in L_{p,\varphi}(\mathbf{R})\), \(g\in L_{q,\phi }(\mathbf{R})\), \(\|f\|_{p,\varphi},\|g\|_{q,\phi}>0\), we have the following inequality:

    $$ \bigl(T_{1}^{(2)}f,g\bigr)< M_{1} \Vert f \Vert _{p,\varphi} \Vert g \Vert _{q,\phi}. $$
    (48)
  3. (iii)

    \(\mu<\lambda+\alpha+1\)and\(\beta>0\).

If statement (iii) holds true, then we have\(\|T_{1}^{(2)}\|=K^{(1)}(\sigma)\).

(c) In view of Theorem 2, for \(\sigma_{1}=\sigma\) and for \(f\in L_{p,\varphi }(\mathbf{R})\), setting

$$H_{1}(y):= \int_{\{x; \vert x \vert \geq\frac{1}{ \vert y \vert }\}}\frac{(\min \{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max\{ \vert xy \vert ,1\})^{\lambda +\alpha} \vert xy-1 \vert }f(x)\,dx\quad(y\in\mathbf{R}), $$

by (30) we obtain that

$$ \Vert H_{1} \Vert _{p,\psi^{1-p}}= \biggl[ \int_{-\infty}^{\infty}\psi ^{1-p}(y)H_{1}^{p}(y) \,dy \biggr] ^{\frac{1}{p}}< M_{2} \Vert f \Vert _{p,\varphi }< \infty . $$
(49)

Definition 3

We define a Hardy-type integral operator of the second kind with nonhomogeneous kernel

$$T_{2}^{(1)} : L_{p,\varphi}(\mathbf{R}) \rightarrow L_{p,\psi^{1-p}}(\mathbf{R}) $$

as follows:

For any \(f\in L_{p,\varphi}(\mathbf{R})\), there exists a unique representation

$$T_{2}^{(1)}f=H_{1}\in L_{p,\psi^{1-p}}(\mathbf{R}) $$

satisfying \(T_{2}^{(1)}f(y)=H_{1}(y)\) for any \(y\in\mathbf{R}\).

In view of (49), it follows that

$$\bigl\Vert T_{2}^{(1)}f \bigr\Vert _{p,\psi ^{1-p}}= \Vert H_{1} \Vert _{p,\psi^{1-p}}\leq M_{2} \Vert f \Vert _{p,\varphi}, $$

and then the operator \(T_{2}^{(1)}\) is bounded satisfying

$$\bigl\Vert T_{2}^{(1)} \bigr\Vert =\sup _{f(\neq\theta)\in L_{p,\varphi}(\mathbf {R})}\frac{ \Vert T_{2}^{(1)}f \Vert _{p,\psi^{1-p}}}{ \Vert f \Vert _{p,\varphi}}\leq M_{2}. $$

If we define the formal inner product of \(T_{2}^{(1)}f\) and g in the following manner:

$$\bigl(T_{2}^{(1)}f,g\bigr):= \int_{-\infty}^{\infty} \biggl[ \int_{\{x; \vert x \vert \geq \frac{1}{ \vert y \vert }\}}\frac{(\min\{ \vert xy \vert ,1\})^{1+\alpha} \vert \ln \vert xy \vert \vert ^{\beta}}{(\max \{ \vert xy \vert ,1\})^{\lambda+\alpha} \vert xy-1 \vert }f(x)\,dx \biggr] g(y)\,dy, $$

then we can rewrite Theorem 2 (for \(\sigma_{1}=\sigma\)) as follows.

Theorem 4

The following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{2}\)such that, for any\(f(x)\geq 0\), \(f\in L_{p,\varphi}(\mathbf{R})\), \(\|f\|_{p,\varphi}>0\), we have the following inequality:

    $$ \bigl\Vert T_{2}^{(1)}f \bigr\Vert _{p,\psi^{1-p}}< M_{2} \Vert f \Vert _{p,\varphi}. $$
    (50)
  2. (ii)

    There exists a constant\(M_{2}\)such that, for any\(f(x),g(y)\geq 0\), \(f\in L_{p,\varphi}(\mathbf{R})\), \(g\in L_{q,\psi }(\mathbf{R})\), \(\|f\|_{p,\varphi},\|g\|_{q,\psi}>0\), we have the following inequality:

    $$ \bigl(T_{2}^{(1)}f,g\bigr)< M_{2} \Vert f \Vert _{p,\varphi} \Vert g \Vert _{q,\psi}. $$
    (51)
  3. (iii)

    \(\sigma<\lambda+\alpha+1\)and\(\beta>0\).

If statement (iii) holds true, then we have\(\|T_{2}^{(1)}\|=K^{(2)}(\sigma)\).

(d) In view of Corollary 7 (\(\mu_{1}=\mu\)), for \(f\in L_{p,\varphi}(\mathbf{R})\), setting

$$H_{2}(y):= \int_{\{x; \vert x \vert \geq \vert y \vert \}}\frac{(\min\{ \vert x \vert , \vert y \vert \})^{1+\alpha } \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max\{ \vert x \vert , \vert y \vert \})^{\lambda+\alpha} \vert x-y \vert }f(x)\,dx\quad(y\in \mathbf{R}), $$

by (36) we obtain that

$$ \Vert H_{2} \Vert _{p,\phi^{1-p}}= \biggl[ \int_{-\infty}^{\infty}\phi ^{1-p}(y)H_{2}^{p}(y) \,dy \biggr] ^{\frac{1}{p}}< M_{2} \Vert f \Vert _{p,\varphi }< \infty . $$
(52)

Definition 4

We define a Hardy-type integral operator of the second kind with homogeneous kernel

$$T_{2}^{(2)} : L_{p,\varphi}(\mathbf{R}) \rightarrow L_{p,\phi^{1-p}}(\mathbf{R}) $$

as follows:

For any \(f\in L_{p,\varphi}(\mathbf{R})\), there exists a unique representation

$$T_{2}^{(2)}f=H_{2}\in L_{p,\phi^{1-p}}(\mathbf{R}) $$

satisfying \(T_{2}^{(2)}f(y)=H_{2}(y)\) for any \(y\in\mathbf{R}\).

In view of (52), it follows that

$$\bigl\Vert T_{2}^{(2)}f \bigr\Vert _{p,\phi ^{1-p}}= \Vert H_{2} \Vert _{p,\phi^{1-p}}\leq M_{2} \Vert f \Vert _{p,\varphi}, $$

and thus the operator \(T_{2}^{(2)}\) is bounded satisfying

$$\bigl\Vert T_{2}^{(2)} \bigr\Vert =\sup _{f(\neq\theta)\in L_{p,\varphi}(\mathbf {R})}\frac{ \Vert T_{2}^{(2)}f \Vert _{p,\phi^{1-p}}}{ \Vert f \Vert _{p,\varphi}}\leq M_{2}. $$

If we define the formal inner product of \(T_{1}^{(2)}f\) and g as follows:

$$\bigl(T_{2}^{(2)}f,g\bigr):= \int_{-\infty}^{\infty} \biggl[ \int_{\{x; \vert x \vert \geq \vert y \vert \}}\frac{(\min\{ \vert x \vert , \vert y \vert \})^{1+\alpha} \vert \ln \vert x/y \vert \vert ^{\beta}}{(\max \{ \vert x \vert , \vert y \vert \})^{\lambda+\alpha} \vert x-y \vert }f(x)\,dx \biggr] g(y)\,dy, $$

then we can rewrite Corollary 7 (for \(\mu_{1}=\mu\)) as follows.

Corollary 10

The following statements (i), (ii), and (iii) are equivalent:

  1. (i)

    There exists a constant\(M_{2}\)such that, for any\(f(x)\geq 0\), \(f\in L_{p,\varphi}(\mathbf{R})\), \(\|f\|_{p,\varphi}>0\), we have the following inequality:

    $$ \bigl\Vert T_{2}^{(2)}f \bigr\Vert _{p,\phi^{1-p}}< M_{2} \Vert f \Vert _{p,\varphi}. $$
    (53)
  2. (ii)

    There exists a constant\(M_{2}\)such that, for any\(f(x),g(y)\geq 0\), \(f\in L_{p,\varphi}(\mathbf{R})\), \(g\in L_{q,\phi }(\mathbf{R})\), \(\|f\|_{p,\varphi},\|g\|_{q,\phi}>0\), we have the following inequality:

    $$ \bigl(T_{2}^{(2)}f,g\bigr)< M_{2} \Vert f \Vert _{p,\varphi} \Vert g \Vert _{q,\phi}. $$
    (54)
  3. (iii)

    \(\mu>-\alpha-1\)and\(\beta>0\).

If statement (iii) holds true, then we have\(\|T_{2}^{(2)}\|=K^{(2)}(\sigma)\).

Conclusions

In the present paper, using weight functions we obtain in Theorems 1, 2 a few equivalent statements of two kinds of Hardy-type integral inequalities with nonhomogeneous kernel and multi-parameters in the whole plane. The constant factors related to the extended Hurwitz-zeta function are proved to be the best possible. In the form of applications, a few equivalent statements of two kinds of Hardy-type integral inequalities with the homogeneous kernel in the whole plane are also deduced in Corollaries 3, 7. We also consider some particular cases in Corollaries 1, 4, 5, 8 and in Remarks 2, 3. We additionally consider operator expressions in Theorems 3, 4 and Corollaries 9, 10. The lemmas and theorems within the present work provide an extensive account of this type of inequalities.

References

  1. 1.

    Hardy, G.H., Littlewood, J.E., Pólya, G.: Inequalities. Cambridge University Press, Cambridge (1934)

    Google Scholar 

  2. 2.

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

    Google Scholar 

  3. 3.

    Yang, B.C.: Hilbert-Type Integral Inequalities. Bentham Science Publishers Ltd., Sharjah (2009)

    Google Scholar 

  4. 4.

    Yang, B.C.: On the norm of an integral operator and applications. J. Math. Anal. Appl. 321, 182–192 (2006)

    MathSciNet  MATH  Article  Google Scholar 

  5. 5.

    Xu, J.S.: Hardy–Hilbert’s inequalities with two parameters. Adv. Math. 36(2), 63–76 (2007)

    MathSciNet  Google Scholar 

  6. 6.

    Yang, B.C.: On the norm of a Hilbert’s type linear operator and applications. J. Math. Anal. Appl. 325, 529–541 (2007)

    MathSciNet  MATH  Article  Google Scholar 

  7. 7.

    Xin, D.M.: A Hilbert-type integral inequality with the homogeneous kernel of zero degree. Math. Theory Appl. 30(2), 70–74 (2010)

    MathSciNet  Google Scholar 

  8. 8.

    Yang, B.C.: A Hilbert-type integral inequality with the homogenous kernel of degree 0. J. Shandong Univ. Nat. Sci. 45(2), 103–106 (2010)

    MathSciNet  Google Scholar 

  9. 9.

    Debnath, L., Yang, B.C.: Recent developments of Hilbert-type discrete and integral inequalities with applications. Int. J. Math. Math. Sci. 2012, Article ID 871845 (2012)

    MathSciNet  MATH  Article  Google Scholar 

  10. 10.

    Hong, Y.: On the structure character of Hilbert’s type integral inequality with homogeneous kernel and applications. J. Jilin Univ. Sci. Ed. 55(2), 189–194 (2017)

    Google Scholar 

  11. 11.

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

    MathSciNet  MATH  Google Scholar 

  12. 12.

    Yang, B.C., Krnic, M.: A half-discrete Hilbert-type inequality with a general homogeneous kernel of degree 0. J. Math. Inequal. 6(3), 401–417 (2012)

    MathSciNet  MATH  Google Scholar 

  13. 13.

    Rassias, M.Th., Yang, B.C.: A multidimensional half-discrete Hilbert-type inequality and the Riemann zeta function. Appl. Math. Comput. 225, 263–277 (2013)

    MathSciNet  MATH  Google Scholar 

  14. 14.

    Rassias, M.Th., Yang, B.C.: On a multidimensional half-discrete Hilbert-type inequality related to the hyperbolic cotangent function. Appl. Math. Comput. 242, 800–813 (2013)

    MathSciNet  MATH  Google Scholar 

  15. 15.

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

    MathSciNet  Article  Google Scholar 

  16. 16.

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

    MathSciNet  MATH  Article  Google Scholar 

  17. 17.

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

    MathSciNet  MATH  Article  Google Scholar 

  18. 18.

    Rassias, M.Th., Yang, B.C.: A multidimensional Hilbert-type integral inequality related to the Riemann zeta function. In: Daras, N.J. (ed.) Applications of Mathematics and Informatics in Science and Engineering, pp. 417–433. Springer, New York (2014)

    Google Scholar 

  19. 19.

    Chen, Q., Yang, B.C.: A survey on the study of Hilbert-type inequalities. J. Inequal. Appl. 2015, Article ID 302 (2015)

    MathSciNet  Article  Google Scholar 

  20. 20.

    Yang, B.C.: A new Hilbert-type integral inequality. Soochow J. Math. 33(4), 849–859 (2007)

    MathSciNet  MATH  Google Scholar 

  21. 21.

    Wang, Z.Q., Guo, D.R.: Introduction to Special Functions. Science Press, Beijing (1979)

    Google Scholar 

  22. 22.

    He, B., Yang, B.C.: On a Hilbert-type integral inequality with the homogeneous kernel of 0-degree and the hypergeometric function. Math. Pract. Theory 40(18), 105–211 (2010)

    MathSciNet  Google Scholar 

  23. 23.

    Yang, B.C.: A new Hilbert-type integral inequality with some parameters. J. Jilin Univ. Sci. Ed. 46(6), 1085–1090 (2008)

    MathSciNet  Google Scholar 

  24. 24.

    Yang, B.C.: A Hilbert-type integral inequality with a non-homogeneous kernel. J. Xiamen Univ. Natur. Sci. 48(2), 165–169 (2008)

    Google Scholar 

  25. 25.

    Zeng, Z., Xie, Z.T.: On a new Hilbert-type integral inequality with the homogeneous kernel of degree 0 and the integral in whole plane. J. Inequal. Appl. 2010, Article ID 256796 (2010)

    MATH  Article  Google Scholar 

  26. 26.

    Wang, A.Z., Yang, B.C.: A new Hilbert-type integral inequality in whole plane with the non-homogeneous kernel. J. Inequal. Appl. 2011, Article ID 123 (2011)

    MathSciNet  MATH  Article  Google Scholar 

  27. 27.

    Xin, D.M., Yang, B.C.: A Hilbert-type integral inequality in whole plane with the homogeneous kernel of degree −2. J. Inequal. Appl. 2011, Article ID 401428 (2011)

    MathSciNet  MATH  Article  Google Scholar 

  28. 28.

    He, B., Yang, B.C.: On an inequality concerning a non-homogeneous kernel and the hypergeometric function. Tamsui Oxf. J. Inf. Math. Sci. 27(1), 75–88 (2011)

    MathSciNet  MATH  Google Scholar 

  29. 29.

    Yang, B.: A reverse Hilbert-type integral inequality with a non-homogeneous kernel. J. Jilin Univ. Sci. Ed. 49(3), 437–441 (2011)

    Google Scholar 

  30. 30.

    Xie, Z.T., Zeng, Z., Sun, Y.F.: A new Hilbert-type inequality with the homogeneous kernel of degree −2. Adv. Appl. Math. Sci. 12(7), 391–401 (2013)

    MathSciNet  MATH  Google Scholar 

  31. 31.

    Huang, Q.L., Wu, S.H., Yang, B.C.: Parameterized Hilbert-type integral inequalities in the whole plane. Sci. World J. 2014, Article ID 169061 (2014)

    Google Scholar 

  32. 32.

    Zhen, Z., Raja Rama Gandhi, K., Xie, Z.T.: A new Hilbert-type inequality with the homogeneous kernel of degree −2 and with the integral. Bull. Math. Sci. Appl. 3(1), 11–20 (2014)

    Google Scholar 

  33. 33.

    Rassias, M.Th., Yang, B.C.: A Hilbert-type integral inequality in the whole plane related to the hyper geometric function and the beta function. J. Math. Anal. Appl. 428(2), 1286–1308 (2015)

    MathSciNet  MATH  Article  Google Scholar 

  34. 34.

    Huang, X.Y., Cao, J.F., He, B., Yang, B.C.: Hilbert-type and Hardy-type integral inequalities with operator expressions and the best constants in the whole plane. J. Inequal. Appl. 2015, Article ID 129 (2015)

    MathSciNet  MATH  Article  Google Scholar 

  35. 35.

    Gu, Z.H., Yang, B.C.: A Hilbert-type integral inequality in the whole plane with a non-homogeneous kernel and a few parameters. J. Inequal. Appl. 2015, 314 (2015)

    MathSciNet  MATH  Article  Google Scholar 

  36. 36.

    Kuang, J.C.: Introduction to Real Analysis. Hunan Educiton Press, Changsha (1996)

    Google Scholar 

  37. 37.

    Kuang, J.C.: Applied Inequalities. Shangdong Science and Technology Press, Jinan (2004)

    Google Scholar 

Download references

Acknowledgements

We are thankful to the mathematicians who have read the manuscript of the paper for their constructive comments that helped improve its presentation.

Availability of data and materials

Not applicable.

Funding

B. Yang’s work is supported by the National Natural Science Foundation (No. 61370186, No. 61640222) and Appropriative Researching Fund for Professors and Doctors, Guangdong University of Education (No. 2015ARF25). We are grateful for this help.

Author information

Affiliations

Authors

Contributions

The authors have contributed equally to the preparation of the present paper. All authors read and approved the final manuscript.

Corresponding author

Correspondence to Michael Th. Rassias.

Ethics declarations

Competing interests

The authors of the present paper do not have any competing interests.

Rights and permissions

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Rassias, M.T., Yang, B. & Raigorodskii, A. On Hardy-type integral inequalities in the whole plane related to the extended Hurwitz-zeta function. J Inequal Appl 2020, 94 (2020). https://doi.org/10.1186/s13660-020-02365-1

Download citation

MSC

  • 26D15
  • 65B10

Keywords

  • Hardy-type integral inequality
  • Weight function
  • Equivalent form
  • Operator
  • Hurwitz-zeta function