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.

1 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 [4–9]. 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 [11–19]. 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. [22–35] 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.

2 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. □

3 Main results and some corollaries

Theorem 1

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

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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

• (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)

• (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)

• (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)

4 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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (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:

• (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)

• (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)

• (iii) $\mu>-\alpha-1$and$\beta>0$.

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

5 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.