Parseval–Goldstein-Type Theorems for Lebedev–Skalskaya Transforms

Abstract

This paper investigates Parseval–Goldstein-type relationships in the framework of Lebedev–Skalskaya transforms. The research also examines the continuity properties of these transforms, along with their adjoint counterparts over weighted Lebesgue spaces. Furthermore, the behavior of Lebedev–Skalskaya transforms and their adjoint transforms in the context of weighted Lebesgue spaces is analyzed. This study aims to provide deeper insights into the functional properties and applications of these transforms in mathematical analysis.

1. Introduction and Preliminaries

Lebedev–Skalskaya transforms serve as prominent integral transforms utilized to address boundary value problems and integral equations within the realms of Science and Engineering [1,2,3,4]. The characteristics and properties of an integral transform are predominantly determined by its associated kernel. In the domain of mathematical analysis, specific functions have been systematically classified based on their distinctive behaviors or properties. These categorized functions, recognized as special functions, encompass entities such as the Bessel function, modified Bessel function (or MacDonald function), Legendre function, and hypergeometric function, among others.

Furthermore, by employing special functions as kernels, a distinct class of integral transforms has emerged, referred to as the index integral transform or simply the index transform. Renowned examples of this category include the Kontorovich–Lebedev transform, the Mehler–Fock transform, and the Whittaker transform. Notably, the impetus for the development of the Kontorovich–Lebedev transform was rooted in the challenges encountered within electrodynamic and diffraction theory. This transform incorporates the modified Bessel function of the second kind, $K_{\alpha +i\tau }\left(x\right)$, with a purely imaginary index as its defining kernel, as delineated in references [5,6]:

$$\begin{array}{ccc}\hfill \left(\mathfrak{T}f\right)\left(\tau \right)& =& \underset{0}{\overset{\infty }{\int }}{K}_{i\tau }\left(x\right)f\left(x\right)dx,\tau >0.\hfill \end{array}$$

Subsequently, the Kontorovich–Lebedev transform found widespread acceptance among researchers globally, undergoing various modifications in its kernel. Notable contributions in this regard include the works of Yakubovich et al. [7,8,9,10], Srivastava et al. [11], Banerjee et al. [12], Glaeske et al. [13], González et al. [14,15], Prasad et al. [16,17], Maan et al. [18], and others.

In 1974, Lebedev–Skalskaya transforms were introduced by N. N. Lebedev and I. P. Skalskaya, utilizing the modified Bessel function of the second kind (MacDonald function) with a complex number index comprising both real and imaginary parts [19]. Subsequently, Rappoport [20], Poruchikov et al. [21], and Yakubovich et al. [6,22] delved into a brief exploration of these integral transforms.

From [23] (p. 82, Entry 21) we have

$$\begin{array}{ccc}\hfill K_{\alpha +i\tau }\left(x\right)& =& {\int }_0^{\infty }{e}^{-x cosh u}cosh\left((\alpha +i\tau )u\right)du\hfill \\ & =& {\displaystyle \frac{1}{2}}{\int }_{-\infty }^{\infty }{e}^{-x cosh u}{e}^{-(\alpha +i\tau )u}du,x>0,\alpha ,\tau \in \mathbb{R}.\hfill \end{array}$$

(1)

The definition of Lebedev–Skalskaya transforms, as provided by Yakubovich in [6] (p. 182), is recalled here:

$$\begin{array}{ccc}\hfill \left(\mathfrak{R}f\right)\left(\tau \right)& =& \underset{0}{\overset{\infty }{\int }}\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)f\left(x\right)dx,\tau >0,\hfill \end{array}$$

(2)

$$\begin{array}{ccc}\hfill \left(\mathfrak{F}g\right)\left(\tau \right)& =& \underset{0}{\overset{\infty }{\int }}\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)g\left(x\right)dx,\tau >0,\hfill \end{array}$$

(3)

where $\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)$ and $\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)$ are the real and imaginary parts of the MacDonald function $K_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)$, respectively.

Now, from (1) one has

$$\begin{array}{ccc}\hfill |K_{\alpha +i\tau }\left(x\right)|& \le & {\displaystyle \frac{1}{2}}{\int }_{-\infty }^{\infty }{e}^{-x cosh u}{e}^{-\alpha u}du\hfill \end{array}$$

(4)

$$\begin{array}{ccc}& =& K_{\alpha }\left(x\right),x>0,\hfill \end{array}$$

(5)

Also, one has

$$\begin{array}{ccc}& & |\mathfrak{R}{K}_{\alpha +i\tau }\left(x\right)| \le |K_{\alpha +i\tau }\left(x\right)| \le K_{\alpha }\left(x\right),x>0,\hfill \end{array}$$

(6)

$$\begin{array}{ccc}& & |\mathfrak{F}{K}_{\alpha +i\tau }\left(x\right)| \le |K_{\alpha +i\tau }\left(x\right)| \le K_{\alpha }\left(x\right),x>0.\hfill \end{array}$$

(7)

For the case where $\alpha ={\displaystyle \frac{1}{2}}$ and having taken into account that [23] (p. 10, Entry 42)

$$\begin{array}{c}\hfill K_{{\displaystyle \frac{1}{2}}}\left(x\right)=\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}},\end{array}$$

(8)

one has

$$\begin{array}{ccc}& & |\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)| \le |K_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)| \le K_{{\displaystyle \frac{1}{2}}}\left(x\right)=\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}},x>0,\hfill \end{array}$$

(9)

$$\begin{array}{ccc}& & |\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)| \le |K_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)| \le K_{{\displaystyle \frac{1}{2}}}\left(x\right)=\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}},x>0.\hfill \end{array}$$

(10)

Observe that $\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)$ and $\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)$ are solutions of the differential equations [2] (Formulas (9) and (10), respectively).

$$\begin{array}{c}\hfill x^2{\displaystyle \frac{d^2{y}_1}{d{x}^2}}+2x{\displaystyle \frac{d{y}_1}{dx}}-x(x-1)y_1+\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)y_1=0,\end{array}$$

(11)

$$\begin{array}{c}\hfill x^2{\displaystyle \frac{d^2{y}_2}{d{x}^2}}+2x{\displaystyle \frac{d{y}_2}{dx}}-x(x+1)y_2+\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)y_2=0,\end{array}$$

(12)

Now, we represent the differential operators as follows:

$$\begin{array}{ccc}\hfill A_x& =& x^2{\displaystyle \frac{d^2}{d{x}^2}}+2x{\displaystyle \frac{d}{dx}}-x(x-1),\hfill \end{array}$$

(13)

$$\begin{array}{ccc}\hfill B_x& =& x^2{\displaystyle \frac{d^2}{d{x}^2}}+2x{\displaystyle \frac{d}{dx}}-x(x+1).\hfill \end{array}$$

(14)

Therefore, from (11)–(14), we can write

$$\begin{array}{ccc}\hfill A_x\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)& =& -\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right),\hfill \end{array}$$

(15)

$$\begin{array}{ccc}\hfill B_x\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)& =& -\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right).\hfill \end{array}$$

(16)

The adjoint to Lebedev–Skalskaya transforms (2) and (3), respectively, can be defined as follows:

$$\begin{array}{ccc}\hfill \left({\mathfrak{R}}^{\prime }f\right)\left(x\right)& =& \underset{0}{\overset{\infty }{\int }}\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)f\left(\tau \right)d\tau ,x>0,\hfill \end{array}$$

(17)

$$\begin{array}{ccc}\hfill \left({\mathfrak{F}}^{\prime }g\right)\left(x\right)& =& \underset{0}{\overset{\infty }{\int }}\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)g\left(\tau \right)d\tau ,x>0.\hfill \end{array}$$

(18)

For $f,g\in L^1\left({\mathbb{R}}_{+}\right)$, the integrals (17) and (18) converge for each $x>0$. In fact, using (9) and (10), respectively, one has $|\left({\mathfrak{R}}^{\prime }f\right)\left(x\right)| \le {\int }_0^{\infty }\left|f\left(\tau \right)\right|d\tau ·\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}<\infty$ and also $|\left({\mathfrak{F}}^{\prime }g\right)\left(x\right)| \le {\int }_0^{\infty }\left|g\left(\tau \right)\right|d\tau ·\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}<\infty$.

Now, observe that for each $\tau >0$ and $f,g\in L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$ one has

$$\begin{array}{c}\hfill \left|\left(\mathfrak{R}f\right)\left(\tau \right)\right| \le {\int }_0^{\infty }\left|f\left(x\right)\right|\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx<\infty ,\end{array}$$

and

$$\begin{array}{c}\hfill \left|\left(\mathfrak{F}g\right)\left(\tau \right)\right| \le {\int }_0^{\infty }\left|g\left(x\right)\right|\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}<\infty .\end{array}$$

The Parseval–Goldstein relations for integral transforms establish a connection between the norm in the original domain and its transformed counterpart [18,24,25,26,27,28]. The present article deals with the study of Parseval–Goldstein-type relations for Lebedev–Skalskaya transforms (2) and (3).

$C_c^k\left({\mathbb{R}}_{+}\right),k\in \mathbb{N}$, denotes as usual the space of compactly supported functions on ${\mathbb{R}}_{+}$, which are k-times differentiable with continuity. This article is structured as follows: In Section 1, we present the definitions and essential results that form the foundation for the entire subsequent discussion. Section 2 focuses on the continuity aspects within Lebesgue spaces for both Lebedev–Skalskaya transforms and their adjoint transforms. Section 3 is dedicated to deriving Parseval–Goldstein-type relations specific to Lebedev–Skalskaya transforms. Section 4 provides concluding remarks.

2. Lebedev–Skalskaya Transforms and Their Adjoints over Lebesgue Spaces

In this section, we examine the continuity properties of Lebedev–Skalskaya transforms and their adjoint operators over weighted Lebesgue spaces $L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$ and $L^1\left({\mathbb{R}}_{+}\right)$. Additionally, the behavior of these transforms, along with their adjoint counterparts, is analyzed in the context of weighted Lebesgue spaces $L^p({\mathbb{R}}_{+},e^{rx}dx)$ and $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$, $2<p<\infty$. These results are utilized in Section 3 to obtain further findings related to Lebedev–Skalskaya transforms and to establish Parseval–Goldstein-type relations.

2.1. The $\mathfrak{R}$ and $\mathfrak{F}$ Transforms over $L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$

In this subsection, we study the $\mathfrak{R}$ and $\mathfrak{F}$ transforms over $L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$ and establish their continuity properties.

Proposition 1.

The Lebedev–Skalskaya transforms $\mathfrak{R}$ and $\mathfrak{F}$ given by (2) and (3), respectively, are bounded linear operators from $L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$. If $f,g\in L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$ then

$$\begin{array}{c}\hfill {\parallel \mathfrak{R}f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)} \le \sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)}\end{array}$$

and

$$\begin{array}{c}\hfill {\parallel \mathfrak{F}g\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)} \le \sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel g\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)},\end{array}$$

also, $\mathfrak{R}f$ and $\mathfrak{F}g$ are continuous functions on ${\mathbb{R}}_{+}$. Moreover, the Lebedev–Skalskaya transforms $\mathfrak{R}f$ and $\mathfrak{F}g$ are continuous maps from $L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$ to the Banach space of bounded continuous functions on ${\mathbb{R}}_{+}$.

Proof. 

Let ${\tau }_0>0$ be arbitrary. Since the map $\tau \to \mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)$ is continuous for each fixed $x>0$, we have

$$\begin{array}{c}\hfill \mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)\to \mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i{\tau }_0}\left(x\right)as\tau \to {\tau }_0.\end{array}$$

Further, we have that $|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)-\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i{\tau }_0}\left(x\right)|\left|f\left(x\right)\right|$ is dominated by the integrable function $2\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}\left|f\left(x\right)\right|$. Therefore, by using dominated convergence theorem, we obtain

$$\begin{array}{c}\hfill \left|\left(\mathfrak{R}f\right)\left(\tau \right)-\left(\mathfrak{R}f\right)\left({\tau }_0\right)\right| \le {\int }_0^{\infty }\left|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)-\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i{\tau }_0}\left(x\right)\right|\left|f\left(x\right)\right|dx\to 0,as\tau \to {\tau }_0.\end{array}$$

Thus, $\mathfrak{R}f$ is a continuous function on ${\mathbb{R}}_{+}$.

Since for each $\tau >0$

$$\begin{array}{ccc}\hfill \left|\left(\mathfrak{R}f\right)\left(\tau \right)\right|& \le & {\int }_0^{\infty }\left|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)\right|\left|f\left(x\right)\right|dx\hfill \\ & \le & {\int }_0^{\infty }\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}\left|f\left(x\right)\right|dx=\sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)},\hfill \end{array}$$

(19)

one has that $\mathfrak{R}f$ is a bounded function.

The linearity of the integral operator implies that the $\mathfrak{R}$ transform is linear. Also, from (19) we have ${\parallel \mathfrak{R}f\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)} \le \sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)}$ and so $\mathfrak{R}:L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)\to L^{\infty }\left({\mathbb{R}}_{+}\right)$ is a continuous linear map. The scheme of proof is similar for $\mathfrak{F}g$. Thus, we have ${\parallel \mathfrak{F}g\parallel }_{L^{\infty }\left({\mathbb{R}}_{+}\right)} \le \sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel g\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)}$. □

Proposition 2.

The Lebedev–Skalskaya transforms $\mathfrak{R}$ and $\mathfrak{F}$ are bounded linear operators from $L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$ into $L^q({\mathbb{R}}_{+},w\left(x\right)dx)$, $0<q<\infty$, when $w>0$ a.e. on ${\mathbb{R}}_{+}$ and ${\int }_0^{\infty }w\left(x\right)dx<\infty$.

Proof. 

Observe that from (19) for each $\tau >0$

$$\begin{array}{ccc}\hfill \left|\left(\mathfrak{R}f\right)\left(\tau \right)\right|& \le & {\int }_0^{\infty }\left|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)\right|\left|f\left(x\right)\right|dx\hfill \\ & \le & {\int }_0^{\infty }\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}\left|f\left(x\right)\right|dx=\sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)}.\hfill \end{array}$$

Then, for $0<q<\infty$, one has

$$\begin{array}{ccc}\hfill {\left({\int }_0^{\infty }{\left|\left(\mathfrak{R}f\right)\left(x\right)\right|}^{q}w\left(x\right)dx\right)}^{{\displaystyle \frac{1}{q}}}& \le & \sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)}{\left({\int }_0^{\infty }w\left(x\right)dx\right)}^{{\displaystyle \frac{1}{q}}}<\infty .\hfill \end{array}$$

The scheme of proof is similar for $\mathfrak{F}$. □

Remark 1.

Examples of weights w for Proposition 2 are as follows:

$$\begin{array}{ccc}& & \left(\mathrm{i}\right)w\left(x\right)={\displaystyle \frac{e^{-x}}{\sqrt{x}}}.\hfill \\ & & \left(\mathrm{ii}\right)w\left(x\right)={(1+x)}^r, for r<-1.\hfill \\ & & \left(\mathrm{iii}\right)w\left(x\right)=e^{rx}, for r<0.\hfill \end{array}$$

2.2. The ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ Transforms over $L^1\left({\mathbb{R}}_{+}\right)$

In this subsection, we study the ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ transforms over $L^1\left({\mathbb{R}}_{+}\right)$ and establish their continuity properties.

Proposition 3.

The ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ transforms given by (17) and (18), respectively, are bounded linear operators from $L^1\left({\mathbb{R}}_{+}\right)$ into $L^q({\mathbb{R}}_{+},w\left(x\right)dx)$, $0<q<\infty$, when $w>0$ a.e. on ${\mathbb{R}}_{+}$ and ${\displaystyle \frac{e^{-x}}{\sqrt{x}}}\in L^q({\mathbb{R}}_{+},w\left(x\right)dx)$.

Proof. 

Observe that for each $x>0$

$$\begin{array}{ccc}\hfill \left|\left({\mathfrak{R}}^{\prime }f\right)\left(x\right)\right|& \le & {\int }_0^{\infty }\left|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)\right|\left|f\left(\tau \right)\right|d\tau \hfill \\ & \le & \sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}{\int }_0^{\infty }\left|f\left(\tau \right)\right|d\tau .\hfill \end{array}$$

Then, for $0<q<\infty$, one has

$$\begin{array}{ccc}\hfill {\left({\int }_0^{\infty }{\left|\left({\mathfrak{R}}^{\prime }f\right)\left(x\right)\right|}^{q}w\left(x\right)dx\right)}^{{\displaystyle \frac{1}{q}}}& \le & \sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}{\left({\int }_0^{\infty }{\left({\displaystyle \frac{e^{-x}}{\sqrt{x}}}\right)}^{q}w\left(x\right)dx\right)}^{{\displaystyle \frac{1}{q}}}<\infty .\hfill \end{array}$$

The scheme of proof is similar for ${\mathfrak{F}}^{\prime }$. □

Remark 2.

Examples of weights w for Proposition 3 are as follows:

$$\begin{array}{ccc}& & \left(\mathrm{i}\right)w\left(x\right)=x^r, for r>-1+{\displaystyle \frac{q}{2}}.\hfill \\ & & \left(\mathrm{ii}\right)w\left(x\right)={(1+x)}^r, for all r and 0<q<2.\hfill \\ & & \left(\mathrm{iii}\right)w\left(x\right)=e^{rx}, for r<q and 0<q<2.\hfill \end{array}$$

2.3. The $\mathfrak{R}$ and $\mathfrak{F}$ Transforms over $L^p({\mathbb{R}}_{+},e^{rx}dx)$ and $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$, $2<p<\infty$

In this subsection, we study the $\mathfrak{R}$ and $\mathfrak{F}$ transforms over $L^p({\mathbb{R}}_{+},e^{rx}dx)$ and $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$, for $2<p<\infty$, and establish their boundedness properties.

Proposition 4.

Assume $2<p<\infty$. Then, for all $0<q<\infty$ we have the following:

(i)

If $-p<r<0$, then the transforms $\mathfrak{R}$ and $\mathfrak{F}$ given by (2) and (3), respectively, are bounded linear operators from $L^p({\mathbb{R}}_{+},e^{rx}dx)$ into $L^q({\mathbb{R}}_{+},e^{rx}dx)$. Also, if $r>-p$ then the $\mathfrak{R}$ and $\mathfrak{F}$ transforms are bounded linear operators from $L^p({\mathbb{R}}_{+},e^{rx}dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$.

(ii)

If $r<-1$, then the transforms $\mathfrak{R}$ and $\mathfrak{F}$ given by (2) and (3), respectively, are bounded linear operators from $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$ into $L^q({\mathbb{R}}_{+},{(1+x)}^{r}dx)$. Also, if $r\in \mathbb{R}$ then the $\mathfrak{R}$ and $\mathfrak{F}$ transforms are bounded linear operators from $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$.

Proof. 

(i)

From (9), the condition (2.1) in Proposition 2.1 of [29] becomes

$$\begin{array}{ccc}& & {\int }_0^{\infty }{\left({\int }_0^{\infty }{|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)|}^{p^{\prime }}{e}^{{\displaystyle \frac{-r{p}^{\prime }x}{p}}}dx\right)}^{{\displaystyle \frac{q}{p^{\prime }}}}{e}^{ry}dy\hfill \\ & & \le {\left({\displaystyle \frac{\pi }{2}}\right)}^{{\displaystyle \frac{q}{2}}}{\left({\int }_0^{\infty }{\left({\displaystyle \frac{e^{-x}}{\sqrt{x}}}\right)}^{p^{\prime }}{e}^{{\displaystyle \frac{-r{p}^{\prime }x}{p}}}dx\right)}^{{\displaystyle \frac{q}{p^{\prime }}}}·{\int }_0^{\infty }{e}^{ry}dy,\hfill \end{array}$$

which converge for $-p<r<0$ and $1<p^{\prime }<2,p+p^{\prime }=p{p}^{\prime }$. Thus, the operator $\mathfrak{R}$ is a bounded linear operator from $L^p({\mathbb{R}}_{+},e^{rx}dx)$ into $L^q({\mathbb{R}}_{+},e^{rx}dx)$ when $-p<r<0$ and $2<p<\infty$.

On the other hand, from (9), the condition (2.2) in Proposition 2.1 of [29] becomes

$$\begin{array}{ccc}& & \underset{\tau \in {\mathbb{R}}_{+}}{ess\; sup}\{{\int }_0^{\infty }|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right){|}^{p^{\prime }}{e}^{{\displaystyle \frac{-r{p}^{\prime }x}{p}}}dx\}\hfill \\ & & \le {\left({\displaystyle \frac{\pi }{2}}\right)}^{{\displaystyle \frac{p^{\prime }}{2}}}{\int }_0^{\infty }{\left({\displaystyle \frac{e^{-x}}{\sqrt{x}}}\right)}^{p^{\prime }}{e}^{{\displaystyle \frac{-r{p}^{\prime }x}{p}}}dx,\hfill \end{array}$$

which converges for $r>-p$ and $1<p^{\prime }<2,p+p^{\prime }=p{p}^{\prime }$. Thus, the operator $\mathfrak{R}$ is a bounded linear operator from $L^p({\mathbb{R}}_{+},e^{rx}dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$ when $r>-p$ and $2<p<\infty$.

The scheme of proof is similar for $\mathfrak{F}$.

(ii)

From (9), the condition (2.1) in Proposition 2.1 of [29] becomes

$$\begin{array}{ccc}& & {\int }_0^{\infty }{\left({\int }_0^{\infty }{|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)|}^{p^{\prime }}{(1+x)}^{{\displaystyle \frac{-r{p}^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{q}{p^{\prime }}}}{(1+y)}^{r}dy\hfill \\ & & \le {\left({\displaystyle \frac{\pi }{2}}\right)}^{{\displaystyle \frac{q}{2}}}{\left({\int }_0^{\infty }{\left({\displaystyle \frac{e^{-x}}{\sqrt{x}}}\right)}^{p^{\prime }}{(1+x)}^{{\displaystyle \frac{-r{p}^{\prime }}{p}}}dx\right)}^{{\displaystyle \frac{q}{p^{\prime }}}}·{\int }_0^{\infty }{(1+y)}^{r}dy,\hfill \end{array}$$

which converge for $r<-1$ and $1<p^{\prime }<2,p+p^{\prime }=p{p}^{\prime }$. Thus, the operator $\mathfrak{R}$ is a bounded linear operator from $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$ into $L^q({\mathbb{R}}_{+},{(1+x)}^{r}dx)$ when $r<-1$ and $2<p<\infty$.

On the other hand, from (9), the condition (2.2) in Proposition 2.1 of [29] becomes

$$\begin{array}{ccc}& & \underset{\tau \in {\mathbb{R}}_{+}}{ess\; sup}\{{\int }_0^{\infty }|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right){|}^{p^{\prime }}{(1+x)}^{{\displaystyle \frac{-r{p}^{\prime }}{p}}}dx\}\hfill \\ & & \le {\left({\displaystyle \frac{\pi }{2}}\right)}^{{\displaystyle \frac{p^{\prime }}{2}}}{\int }_0^{\infty }{\left({\displaystyle \frac{e^{-x}}{\sqrt{x}}}\right)}^{p^{\prime }}{(1+x)}^{{\displaystyle \frac{-r{p}^{\prime }}{p}}}dx,\hfill \end{array}$$

which converges for $1<p^{\prime }<2,p+p^{\prime }=p{p}^{\prime }$. Thus, the operator $\mathfrak{R}$ is a bounded linear operator from $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$ into $L^{\infty }\left({\mathbb{R}}_{+}\right)$ when $2<p<\infty$.

The scheme of proof is similar for $\mathfrak{F}$.

□

2.4. The ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ Transforms over $L^p({\mathbb{R}}_{+},e^{rx}dx)$ and $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$, $2<p<\infty$

In this subsection, we study the ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ transforms over $L^p({\mathbb{R}}_{+},e^{rx}dx)$ and $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$, for $2<p<\infty$, and establish their boundedness properties.

Proposition 5.

Assume $2<p<\infty$ and $p+p^{\prime }=p{p}^{\prime }$. Then, we have the following

(i) 

For all $0<r<p^{\prime }$, the transforms ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ given by (17) and (18), respectively, are bounded linear operators from $L^p({\mathbb{R}}_{+},e^{rx}dx)$ into $L^{p^{\prime }}({\mathbb{R}}_{+},e^{rx}dx)$.

(ii) 

For all $r>p-1$, the transforms ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ given by (17) and (18), respectively, are bounded linear operators from $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$ into $L^{p^{\prime }}({\mathbb{R}}_{+},{(1+x)}^{r}dx)$.

Proof. 

(i)

For $r>0$, one has

$$\begin{array}{c}\hfill {\int }_0^{\infty }{e}^{{\displaystyle \frac{-r{p}^{\prime }y}{p}}}dy={\displaystyle \frac{p}{r{p}^{\prime }}},\end{array}$$

and ${\displaystyle \frac{e^{-x}}{\sqrt{x}}}\in L^{p^{\prime }}({\mathbb{R}}_{+},e^{rx}dx)$ for $r<p^{\prime }$ and $1<p^{\prime }<2$. Then, from Proposition 2.1 in [30] the results hold.

(ii)

For $r>p-1$, one has

$$\begin{array}{c}\hfill {\int }_0^{\infty }{(1+y)}^{{\displaystyle \frac{-r{p}^{\prime }}{p}}}dy={\displaystyle \frac{p}{r{p}^{\prime }}},\end{array}$$

and ${\displaystyle \frac{e^{-x}}{\sqrt{x}}}\in L^{p^{\prime }}({\mathbb{R}}_{+},{(1+x)}^{r}dx)$ for $1<p^{\prime }<2$. Then, from Proposition 2.1 in [30] the results hold.

□

3. Parseval–Goldstein-Type Theorems

Parseval–Goldstein relations for integral transforms establish a link between the norms of the original domain and their transformed counterparts. These relations play a crucial role in the theory of integral transforms, as they highlight the energy-conserving properties and consistency between different domains. Such analysis is essential for understanding the fundamental characteristics and applications of integral transforms in mathematical analysis (see [18,24,25,26,27,28]). In this section, we derive Parseval–Goldstein-type relations specifically for the Lebedev–Skalskaya transforms.

As a consequence of Proposition 2.2 in [30], one has

Theorem 1.

Assume that $2<p<\infty$ and $p+p^{\prime }=p{p}^{\prime }$. Then, the following Parseval–Goldstein-type relations hold:

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left(\mathfrak{R}f\right)\left(x\right)g\left(x\right)dx={\int }_0^{\infty }f\left(x\right)\left({\mathfrak{R}}^{\prime }g\right)\left(x\right)dx,\end{array}$$

(20)

and

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left(\mathfrak{F}f\right)\left(x\right)g\left(x\right)dx={\int }_0^{\infty }f\left(x\right)\left({\mathfrak{F}}^{\prime }g\right)\left(x\right)dx,\end{array}$$

(21)

(i) 

for $f,g\in L^p({\mathbb{R}}_{+},e^{rx}dx)$ with $0<r<p^{\prime }$, or, alternatively,

(ii) 

for $f,g\in L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$ with $r>p-1$.

Remark 3.

From this result, the transforms ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ become the adjoints of Lebedev–Skalskaya transforms $\mathfrak{R}$ and $\mathfrak{F}$, respectively, over $L^p({\mathbb{R}}_{+},e^{rx}dx)$, $2<p<\infty$ and $0<r<p^{\prime }$, and $L^p({\mathbb{R}}_{+},{(1+x)}^{r}dx)$, $2<p<\infty$ and $r>p-1$.

Theorem 2.

If $f\in L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$ and $g\in L^1\left({\mathbb{R}}_{+}\right)$, then the following Parseval–Goldstein-type relations hold:

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left(\mathfrak{R}f\right)\left(x\right)g\left(x\right)dx={\int }_0^{\infty }f\left(x\right)\left({\mathfrak{R}}^{\prime }g\right)\left(x\right)dx,\end{array}$$

(22)

and

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left(\mathfrak{F}f\right)\left(x\right)g\left(x\right)dx={\int }_0^{\infty }f\left(x\right)\left({\mathfrak{F}}^{\prime }g\right)\left(x\right)dx.\end{array}$$

(23)

Proof. 

In fact, for each $\tau >0$

$$\begin{array}{c}\hfill \left|\left(\mathfrak{R}f\right)\left(\tau \right)\right| \le \sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)}.\end{array}$$

Therefore,

$$\begin{array}{c}\hfill {\int }_0^{\infty }\left|\left(\mathfrak{R}f\right)\left(\tau \right)\right|\left|g\left(\tau \right)\right|d\tau \le \sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)}{\parallel g\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}.\end{array}$$

Also, for each $x>0$

$$\begin{array}{c}\hfill |\left({\mathfrak{R}}^{\prime }g\right)\left(x\right)| \le {\int }_0^{\infty }|\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)||g\left(\tau \right)|d\tau \le \sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}{\parallel g\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}.\end{array}$$

Then

$$\begin{array}{ccc}\hfill {\int }_0^{\infty }\left|f\left(x\right)\right||\left({\mathfrak{R}}^{\prime }g\right)\left(x\right)|dx& & \le {\int }_0^{\infty }\left|f\left(x\right)\right|\sqrt{{\displaystyle \frac{\pi }{2}}}{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx{\parallel g\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}\hfill \\ & & =\sqrt{{\displaystyle \frac{\pi }{2}}}{\parallel f\parallel }_{L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)}{\parallel g\parallel }_{L^1\left({\mathbb{R}}_{+}\right)}.\hfill \end{array}$$

Thus, by using Fubini’s theorem one obtains the relation (22).

The scheme of proof is similar for relation (23). □

Remark 4.

From this result, the transforms ${\mathfrak{R}}^{\prime }$ and ${\mathfrak{F}}^{\prime }$ become the adjoints of the Lebedev–Skalskaya transforms $\mathfrak{R}$ and $\mathfrak{F}$ over $L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$, respectively.

From (15) one has, for $k\in \mathbb{N}$,

$$\begin{array}{c}\hfill A_x^k(\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right))={(-1)}^k{\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)}^k\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right),\end{array}$$

(24)

and so, for $f\in C_c^{2k}\left({\mathbb{R}}_{+}\right)$, $k\in \mathbb{N}$,

$$\begin{array}{c}\hfill \left(\mathfrak{R}\left(A_x^{k}f\right)\right)\left(\tau \right)={(-1)}^k{\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)}^k\left(\mathfrak{R}f\right)\left(\tau \right),\tau >0.\end{array}$$

(25)

Theorem 3.

If $f\in C_c^{2k}\left({\mathbb{R}}_{+}\right)$, $k\in \mathbb{N}$, and $g\in L^1\left({\mathbb{R}}_{+}\right)$, then the following Parseval–Goldstein-type relation holds:

$$\begin{array}{c}\hfill {(-1)}^k{\int }_0^{\infty }\left(\mathfrak{R}f\right)\left(x\right)g\left(x\right){\left({\displaystyle \frac{1}{4}}+x^2\right)}^{k}dx={\int }_0^{\infty }\left(A_x^{k}f\right)\left(x\right)\left({\mathfrak{R}}^{\prime }g\right)\left(x\right)dx.\end{array}$$

(26)

Proof. 

For $f\in C_c^2\left({\mathbb{R}}_{+}\right)$, f and $A_{x}f\in L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$. Also, for $\tau >0$,

$$\begin{array}{ccc}\hfill \left(\mathfrak{R}\left(A_{x}f\right)\right)\left(\tau \right)& =& {\int }_0^{\infty }\left(A_{x}f\right)\left(x\right)\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)dx\hfill \\ & =& {\int }_0^{\infty }f\left(x\right)(A_x(\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)))\left(x\right)dx\hfill \\ & =& -\left({\displaystyle \frac{1}{4}}+{\tau }^2\right){\int }_0^{\infty }f\left(x\right)\mathfrak{R}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)dx\hfill \\ & =& -\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)\left(\mathfrak{R}f\right)\left(\tau \right).\hfill \end{array}$$

Then, for $f\in C_c^2\left({\mathbb{R}}_{+}\right)$ and $g\in L^1\left({\mathbb{R}}_{+}\right)$ and using Theorem 2 above, one has

$$\begin{array}{ccc}& & {\int }_0^{\infty }\left(\mathfrak{R}\left(A_{x}f\right)\right)\left(\tau \right)g\left(\tau \right)d\tau ={\int }_0^{\infty }\left(A_{x}f\right)\left(x\right)\left({\mathfrak{R}}^{\prime }g\right)\left(x\right)dx.\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill -{\int }_0^{\infty }\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)\left(\mathfrak{R}f\right)\left(\tau \right)g\left(\tau \right)d\tau ={\int }_0^{\infty }\left(A_{x}f\right)\left(x\right)\left({\mathfrak{R}}^{\prime }g\right)\left(x\right)dx.\end{array}$$

Also, in general, for $f\in C_c^{2k}\left({\mathbb{R}}_{+}\right)$, $k\in \mathbb{N}$, and $g\in L^1\left({\mathbb{R}}_{+}\right)$ one obtains

$$\begin{array}{c}\hfill {(-1)}^k{\int }_0^{\infty }\left(\mathfrak{R}f\right)\left(\tau \right)g\left(\tau \right)\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)d\tau ={\int }_0^{\infty }\left(A_x^{k}f\right)\left(x\right)\left({\mathfrak{R}}^{\prime }g\right)\left(x\right)dx.\end{array}$$

□

From (16) one has, for $k\in \mathbb{N}$,

$$\begin{array}{c}\hfill B_x^k(\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right))={(-1)}^k{\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)}^k\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right),\end{array}$$

(27)

and so, for $f\in C_c^{2k}\left({\mathbb{R}}_{+}\right)$, $k\in \mathbb{N}$,

$$\begin{array}{c}\hfill \left(\mathfrak{F}\left(B_x^{k}f\right)\right)\left(\tau \right)={(-1)}^k{\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)}^k\left(\mathfrak{F}f\right)\left(\tau \right),\tau >0.\end{array}$$

(28)

Theorem 4.

If $f\in C_c^{2k}\left({\mathbb{R}}_{+}\right)$, $k\in \mathbb{N}$, and $g\in L^1\left({\mathbb{R}}_{+}\right)$, then the following Parseval–Goldstein-type relation holds

$$\begin{array}{c}\hfill {(-1)}^k{\int }_0^{\infty }\left(\mathfrak{F}f\right)\left(x\right)g\left(x\right){\left({\displaystyle \frac{1}{4}}+x^2\right)}^{k}dx={\int }_0^{\infty }\left(B_x^{k}f\right)\left(x\right)\left({\mathfrak{F}}^{\prime }g\right)\left(x\right)dx.\end{array}$$

(29)

Proof. 

For $f\in C_c^2\left({\mathbb{R}}_{+}\right)$, f and $B_{x}f\in L^1({\mathbb{R}}_{+},{\displaystyle \frac{e^{-x}}{\sqrt{x}}}dx)$. Also, for $\tau >0$,

$$\begin{array}{ccc}\hfill \left(\mathfrak{F}\left(B_{x}f\right)\right)\left(\tau \right)& =& {\int }_0^{\infty }\left(B_{x}f\right)\left(x\right)\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)dx\hfill \\ & =& {\int }_0^{\infty }f\left(x\right)(B_x(\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)))\left(x\right)dx\hfill \\ & =& -\left({\displaystyle \frac{1}{4}}+{\tau }^2\right){\int }_0^{\infty }f\left(x\right)\mathfrak{F}{K}_{{\displaystyle \frac{1}{2}}+i\tau }\left(x\right)dx\hfill \\ & =& -\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)\left(\mathfrak{F}f\right)\left(\tau \right).\hfill \end{array}$$

Then, for $f\in C_c^2\left({\mathbb{R}}_{+}\right)$ and $g\in L^1\left({\mathbb{R}}_{+}\right)$ and using Theorem 2 above, one has

$$\begin{array}{ccc}& & {\int }_0^{\infty }\left(\mathfrak{F}\left(B_{x}f\right)\right)\left(\tau \right)g\left(\tau \right)d\tau ={\int }_0^{\infty }\left(B_{x}f\right)\left(x\right)\left({\mathfrak{F}}^{\prime }g\right)\left(x\right)dx.\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill -{\int }_0^{\infty }\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)\left(\mathfrak{F}f\right)\left(\tau \right)g\left(\tau \right)d\tau ={\int }_0^{\infty }\left(B_{x}f\right)\left(x\right)\left({\mathfrak{F}}^{\prime }g\right)\left(x\right)dx.\end{array}$$

Also, in general, for $f\in C_c^{2k}\left({\mathbb{R}}_{+}\right)$, $k\in \mathbb{N}$, and $g\in L^1\left({\mathbb{R}}_{+}\right)$ one obtains

$$\begin{array}{c}\hfill {(-1)}^k{\int }_0^{\infty }\left(\mathfrak{F}f\right)\left(\tau \right)g\left(\tau \right)\left({\displaystyle \frac{1}{4}}+{\tau }^2\right)d\tau ={\int }_0^{\infty }\left(B_x^{k}f\right)\left(x\right)\left({\mathfrak{F}}^{\prime }g\right)\left(x\right)dx.\end{array}$$

□

4. Conclusions

The current research article thoroughly explores continuity properties within Lebesgue spaces for the Lebedev–Skalskaya transforms and their adjoints. With a focus on Parseval–Goldstein relations, the investigation unveils energy-preserving characteristics and inter-domain consistency. This comprehensive analysis enhances our comprehension of the fundamental properties and applications of the integral transforms in mathematical analysis. The outcomes put forth in this article pave the way for the exploration of various other integral transforms.