## 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 [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.

## 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:

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

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

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

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)

## 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:

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$$\|T_{2}^{(2)}\|=K^{(2)}(\sigma)$$.

## 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.