1 Introduction

The Bessel differential operator is defined as follows:

$$\begin{aligned} {\mathscr {B}}_\mu := \frac{d^2}{dx^2} + \frac{2\mu +1}{x} \frac{d}{dx}, \quad \mu \ge -\frac{1}{2}. \end{aligned}$$
(1.1)

This operator exhibits regular singularities at both 0 and \(\infty\) within the interval \((0,\,\infty )\). The operator \(({\mathscr {B}}_\mu , \,D)\) defines an unbounded self-adjoint operator [11]. Remarkably, its spectrum is simple, continuous, spanning the interval \([0,\,\infty )\), and it serves as the foundation for the Fourier-Bessel transform (as discussed in section 2).

A noteworthy example of a non-local pseudo-differential operator in the realm of fractional calculus is the fractional Bessel operator [5, 6, 15, 16, 27, 28], often denoted as \((-{\mathscr {B}}_\mu )^{\alpha /2}\). This operator can be expressed as a Fourier-Bessel pseudo-differential operator with the symbol \(|x|^{\alpha }\), where \(x\in {\mathbb {R}}\). It’s worth noting that fractional Bessel operators have been extensively studied by various authors, yielding significant contributions to the existing literature.

The focal point of this paper lies in exploring the profound impact of the Mellin integral transform within the context of fractional Bessel operators. The primary objective is to elucidate how the Mellin transform provides a means of defining fractional Bessel derivatives. While the Mellin integral transform has been sparingly employed in prior works centered around fractional calculus, its recognition has been significantly enhanced by noteworthy instances of its application. For instance, in [23], the Mellin integral transform was ingeniously harnessed to deliver a further equivalent definition that can be utilized when the fractional Laplacian is applied to radial functions. Additionally, [17] demonstrated the establishment of Erdélyi-Kober type mixed operators as generators for integral transforms of Mellin convolution type.

The impact of the Mellin integral transform was further exemplified in [19] and [1], where it played a pivotal role in deriving Leibniz-type rules for a variety of fractional calculus operators. Moreover, this transformative technique found a natural synergy with special functions intrinsic to fractional calculus. A plethora of references, including [20, 24, 30, 35], and others, validate this interpretational facet.

Lastly, a nod of acknowledgment is extended to the comprehensive compendium [20], which stands as a testament to the exhaustive exploration of fractional calculus theory. Within its pages, the Mellin integral transform emerges as a cornerstone.

The paper is organized as follows:

Section 2 serves as an initial section that provides an overview of fundamental concepts. Topics covered include the Fourier-Bessel transform, the generalized translation, and Mellin transforms, setting the stage for understanding subsequent content.

In Sect. 3, we delve into Bochner subordination approaches. We examine how they apply to the Fractional Bessel operator. Section 4 presents the primary outcomes of our research. This section offers a concise summary of the significant findings we have achieved.

Section 5 offers a comprehensive proof of the main results. Detailed derivation and explanation establish the validity of our findings, aiming to provide readers with a comprehensive understanding of the mathematical underpinnings.

Concluding our paper, Sect. 6 introduces a transmutation operator. Operating between the fractional Bessel operator and the fractional second derivative.

2 Preliminaries

Before revealing our main results, it is essential to establish the groundwork by introducing key notations and collecting pertinent facts about the Bessel operator. This section serves as a primer, elucidating the significance of the Fourier-Bessel transform and the Delsarte translation, which will be pivotal for the subsequent analysis.

2.1 Fourier-Bessel Transform

The normalized Bessel function is defined as follows:

$$\begin{aligned} {\mathscr {J}}_\mu (x):=\Gamma (\mu +1)\,(2/x)^\mu \,J_\mu (x), \quad \mu >-1, \end{aligned}$$
(2.1)

where \(\Gamma (\cdot )\) is the Gamma function [34] and \(J_\mu (\cdot )\) is the Bessel function of the first kind, see [34, (10.16.9)]. Then

$$\begin{aligned} \mathscr {J}_\mu (x)= \sum _{k=0}^\infty \frac{(-\tfrac{1}{4} x^2)^k}{(\mu +1)_k\,k!} =\,_{0}F_{1}\!\left( \genfrac{}{}{0.0pt}{}{-}{\mu +1};-\tfrac{1}{4}x^2\right) \qquad (\mu >-1). \end{aligned}$$

Here \((\mu +1)_k\) denotes the Pochhammer symbol [32]. The normalized Bessel function arises as the sole solution to the eigenvalue problem linked with the Bessel equation. More precisely, the functions defined as \(x\rightarrow {\mathscr {J}}_\mu (\lambda x)\) stand as the exclusive solution to the eigenvalue problem [34, (10.13.5)]

$$\begin{aligned} {\left\{ \begin{array}{ll} &{}{\mathscr {B}}_\mu u(x) =-\lambda ^2u(x), \\ &{}\displaystyle u(0)=1,\quad u'(0)=0. \end{array}\right. } \end{aligned}$$

The function \({\mathscr {J}}_\mu (\cdot )\) is an entire analytic function with even symmetry. Notably, there are straightforward special cases that hold:

$$\begin{aligned} {\mathscr {J}}_{-1/2}(x)=\cos x,\quad {\mathscr {J}}_{1/2}(x)=\frac{\sin x}{x}. \end{aligned}$$

We introduce the following notation:

  • \(L^p_\mu (0,\infty )\) \((1\le p)\) represents the Lebesgue space associated with the measure

    $$\begin{aligned} \sigma _\mu (dx)=\frac{x^{2\mu +1}}{2^{\mu }\Gamma (\mu +1)}\,dx. \end{aligned}$$
    (2.2)

    The norm \(\Vert f\Vert _{\mu ,p}\) is the conventional norm given by

    $$\begin{aligned}&\Vert f\Vert _{\mu ,p}=\Big (\int _{0}^\infty |f(x)|^p\,\sigma _\mu (dx)\Big )^{1/p}. \end{aligned}$$

  • \(S_*({\mathbb {R}})\) signifies the space of even functions on \({\mathbb {R}}\) that are infinitely differentiable and decrease rapidly, along with all their derivatives.

Recall that a function f defined on \({\mathbb {R}}^n\) is considered radial if there exists an even function \(f_0\) defined on \({\mathbb {R}}\) such that \(f(x) = f_0(\Vert x\Vert )\), where \(\Vert x\Vert\) represents the norm of the vector x. In the context of radial function, there is a reduction formula for the Fourier transform that can be derived:

$$\begin{aligned} {\mathcal {F}}f(\xi )&= \int _{{\mathbb {R}}^n} f(x)e^{-ix\cdot \xi }dx, \quad \xi \in {\mathbb {R}}^n \nonumber \\&= (2\pi )^{n/2}\int _0^\infty f_0(r){\mathscr {J}}_{n/2-1}(|\xi | r) \sigma _{n/2-1}(dr). \end{aligned}$$
(2.3)

For \(\mu \ge -1/2\), the Fourier-Bessel transform \({\mathscr {H}}_\mu f\) of \(f\in L^1_{\mu }(0,\infty )\) is defined as [33]:

$$\begin{aligned} {\mathscr {H}}_\mu f(x):= \int _{0}^\infty f(t) {\mathscr {J}}_\mu (tx) \sigma _\mu (dx), \quad \mu \ge -1/2. \end{aligned}$$
(2.4)

This integral transform provides us with a \(\mu\)-continuation of the Fourier transform beyond integer dimensions. Notably, it can be extended to establish an isometry of \(L^2_\mu (0,\infty )\). For any function f belonging to \(L^1_{\mu }(0,\infty )\cap L^2_{\mu }(0,\infty )\), the following relationships hold:

$$\begin{aligned} \int _{0}^\infty |f(x)|^2\, \sigma _\mu (dx) = \int _{0}^\infty |{\mathscr {H}}_\mu f(t)|^2\, \sigma _\mu (dt). \end{aligned}$$
(2.5)

Furthermore, its inverse is expressed as:

$$\begin{aligned} f(x) = \int _{0}^\infty {\mathscr {H}}_\mu f(t)\, {\mathscr {J}}_\mu (tx)\, \sigma _\mu (dt). \end{aligned}$$
(2.6)

Moving forward, our focus shifts to the exploration of the generalized translation operator linked with the Bessel operator. This operator is symbolized as \(\tau _\mu ^x\) and operates on functions belonging to \(L^1_\mu (0,\,\infty )\) according to the following expression [4, §3.4.1]:

$$\begin{aligned} \tau ^x_\mu f(y)=\left\{ \begin{array}{l l} \int _{0}^{\pi }f(\sqrt{x^2+y^2+ 2xy\cos \theta } )\sin ^{2\mu } \theta \, d\theta ,\quad \text {if}\,\,\mu >-1/2,\\ \\ \\ \frac{1}{2}(f(x+y)+f(x-y)),\quad \text {if}\,\, \mu =-1/2. \end{array} \right. \end{aligned}$$
(2.7)

With the help of this translation operator, one defines the convolution of \(f\in L^1_\mu (0,\infty )\) and \(g\in L^p_\mu (0,\infty )\) for \(p\in [1,\,\infty )\) as the element \(f*_\mu g\) of \(L^p_\mu (0,\,\infty )\) given by

$$\begin{aligned} (f*_\mu g)(x):=\int _0^\infty (\tau ^x_\mu f)(y)\,g(y)\sigma _\mu (dy),\quad \mu \ge -1/2. \end{aligned}$$
(2.8)

The following properties are obvious

  • \({\mathscr {F}}_\mu (\tau ^x_\mu f)(t)={\mathscr {J}}_\mu (xt){\mathscr {F}}_\mu f(t),\)

  • \({\mathscr {F}}_\mu (f*_\mu g)(x)={\mathscr {F}}_\mu f(x) {\mathscr {F}}_\mu g(x).\)

2.2 Mellin Transform

The Mellin transform of a function \(f(x)\) is defined by the integral:

$$\begin{aligned} {\mathscr {M}}\{f(x);s\} = \int _0^\infty x^{s-1} f(x) \, dx, s\in {\mathbb {C}}. \end{aligned}$$

For \(f\in S_*({\mathbb {R}})\), \({\mathscr {M}}\{f(x);s\}\) is analytic for all \(\Re (s)>0,\) see [21, Lemma 1]. The Mellin convolution of two functions \(f\) and \(g\) is given by:

$$\begin{aligned} (f*^{{\mathscr {M}}}g)(x) = \int _{0}^{\infty } f\left( \frac{x}{t}\right) g(t) \, \frac{dt}{t}. \end{aligned}$$

This convolution operation satisfies:

$$\begin{aligned} {\mathscr {M}}\{(f*^{{\mathscr {M}}}g)(x)\,; \; s\}= {\mathscr {M}}\{(f(x)\,; \; s\} {\mathscr {M}}\{g(x)\,; \; s\}. \end{aligned}$$
(2.9)

3 Fractional Bessel Derivative

In this section our focus shifts to the analysis of the fractional Bessel operator, denoted as \((-{\mathscr {B}}_\mu )^{\alpha /2}\). This operator can be conceptualized as a Fourier-Bessel transform pseudo-differential operator with symbol \(|x|^\alpha\). To be more precise, for a function f belonging to the space \(S_*({\mathbb {R}})\), the relationship is established as follows:

$$\begin{aligned} {\mathscr {H}}_\mu \big ((-{\mathscr {B}}_\mu )^{\alpha } f\big )(x) = |x|^{\alpha } {\mathscr {H}}_\mu f(\lambda ), \quad x\in {\mathbb {R}}. \end{aligned}$$
(3.1)

As we delve deeper into our exploration, let’s recall that the Bessel operator \({\mathscr {B}}_\mu\) generates a contractive strongly continuous semigroup \(({\mathscr {E}}_t)_{t\ge 0}\) on \(X=C_0({\mathbb {R}}), \,L^p_{\mu }(0,\,\infty )\) (\(p\ge 1\)) with the domain

$$\begin{aligned} D({\mathscr {B}}_\mu ):=\{f\in X;\quad {\mathscr {B}}_\mu f \in X\}. \end{aligned}$$

For \(t>0\) and \(x\in {\mathbb {R}}\), we have:

$$\begin{aligned} {\mathscr {E}}_t f(x)={\mathscr {E}}(.,\,t)* f(x)=\int _{0}^{\infty }\tau _{\mu }^x{\mathscr {E}}_\mu (y,t)f(y)\sigma _\mu (dy), \end{aligned}$$
(3.2)

where

$$\begin{aligned} {\mathscr {E}}_\mu (x,t)=\frac{e^{-\frac{x^2}{4t}}}{(2t)^{\mu +1}}. \end{aligned}$$

Moreover, it’s worth noting that the inequality

$$\begin{aligned} \Vert {\mathscr {E}}_tf\Vert _{\mu ,p}\le \Vert f\Vert _{\mu ,p} \end{aligned}$$

holds true.

Note that for \(p\in [1,2)\) and \(f\in C_0({\mathbb {R}})\cap L^p_{\mu }(0,\,\infty )\), the function \(u(t,x)={\mathcal {E}}_t f(x)\) serves as an even, infinitely smooth solution to the Cauchy problem:

$$\begin{aligned} {\left\{ \begin{array}{ll} {\mathscr {B}}_\nu u(x,t)=\frac{\partial u(x,t)}{\partial t}\\ u(x,0)=f(x). \end{array}\right. } \end{aligned}$$

Importantly, when an operator generates a strongly continuous semigroup on a Banach space, its fractional power can be defined utilizing Bochner’s subordination.

For \(0<\alpha <2\), the function \(\lambda ^{\alpha /2}\) emerges as a Bernstein function and can be represented through the following integral representation:

$$\begin{aligned} \lambda ^{\alpha /2}=\frac{1}{|\Gamma (-\tfrac{\alpha }{2})|}\int _{0}^{\infty } \big (1-e^{-t\lambda }\big )\frac{dt}{t^{1+\tfrac{\alpha }{2}}}. \end{aligned}$$

This leads us to the conclusion that:

$$\begin{aligned} \big (-{\mathscr {B}}_\mu \big )^{\alpha /2}f(x)=\frac{1}{|\Gamma (\tfrac{\alpha }{2})|}\int _{0}^{\infty } \big (f(x)-e^{t{\mathscr {B}}_\mu }f(x)\big )\frac{dt}{t^{1+\tfrac{\alpha }{2}}}. \end{aligned}$$
(3.3)

By utilizing the fact that

$$\begin{aligned} \int _{0}^{\infty }{\mathscr {E}}_\mu (x,t)\sigma _\mu (dx)=1 \end{aligned}$$

and referencing the integral representation (3.2), we can reexpress (3.3) as follows:

$$\begin{aligned} \big (-{\mathscr {B}}_\mu \big )^{\alpha /2}f(x)&=\frac{1}{|\Gamma (-\tfrac{\alpha }{2})|}\int _{0}^{\infty } \big (f(x)-{\mathscr {E}}_t f(x)\big )\frac{dt}{t^{1+\tfrac{\alpha }{2}}}\\&=\frac{1}{2^{\mu +1}|\Gamma (-\tfrac{\alpha }{2})|}\int _{0}^{\infty }\int _{0}^{\infty } \big (f(x)-\tau _\mu ^{x}f(y)\big )\frac{e^{-\frac{y^2}{4t}}}{t^{\mu +2+\tfrac{\alpha }{2}}} \sigma _\mu (dy)dt. \end{aligned}$$

By intertwining the last two integrals for \(f\in S_*({\mathbb {R}})\), we obtain:

$$\begin{aligned} \big (-{\mathscr {B}}_\mu \big )^{\alpha /2}f(x)&=\frac{1}{2^{\mu +1}|\Gamma (-\tfrac{\alpha }{2})|} \int _{0}^{\infty } \big (f(x)-\tau _\mu ^{x}f(y)\big )\int _{0}^{\infty }\frac{e^{-\frac{y^2}{4t}}}{t^{\mu +2+\tfrac{\alpha }{2}}} dt\sigma _\mu (dy)\\ {}&=\frac{2^{\alpha +1}\Gamma (\mu +\tfrac{\alpha }{2}+1)}{\Gamma (\mu +1) |\Gamma (-\tfrac{\alpha }{2})|}\int _{0}^{\infty }\big (f(x)-\tau _{\mu }^xf(y)\big ) \frac{dy}{y^{\alpha +1}}. \end{aligned}$$

Theorem 3.4

Let \(\mu \ge -1/2\) and \(0<\alpha < 2.\) For \(f\in S_*({\mathbb {R}}),\) we have

$$\begin{aligned} (-{\mathscr {B}}_\mu )^{\alpha /2}f(x)=\frac{2^{\alpha +1}\Gamma (\mu +\tfrac{\alpha }{2}+1)}{\Gamma (\mu +1) |\Gamma (-\tfrac{\alpha }{2})|}\int _{0}^{\infty }\frac{f(x)-\tau ^x_{\mu }f(y)}{y^{\alpha +1}}\,dy. \end{aligned}$$
(3.5)

Upon applying the Fourier-Bessel transform to (3.5), we can compellingly demonstrate the equivalence between the fractional Bessel operator obtained in (3.1) and the one derived through Bochner subordination.

Consider the special case when \(\mu = -1/2\). In this particular scenario, the Bessel operator simplifies to a second-order derivative, \(\frac{d^2}{dx^2}\). Remarkably, the Fourier-Bessel transform aligns, up to normalized constants, with both the Fourier and Fourier Cosine transforms. These relationships are expressed as follows:

$$\begin{aligned} {\mathscr {H}}_{-1/2}f(x)&= \sqrt{\frac{2}{\pi }}\int _{0}^{\infty }f(t)\cos (xt)\,dt \\&= \frac{1}{\sqrt{2\pi }}\int _{{\mathbb {R}}} f(t)e^{-ixt}\,dt. \end{aligned}$$

From Theorem 3.4, the fractional power of the second derivative, \(-\left( -\frac{d^2}{dx^2}\right) ^{\alpha /2}\), takes on the following form:

$$\begin{aligned} \left( -\frac{d^2}{dx^2}\right) ^{\alpha /2} f(x)&= \frac{2^{\alpha +1}\Gamma \left( \frac{\alpha +1}{2}\right) }{\Gamma \left( \frac{1}{2}\right) |\Gamma \left( -\frac{\alpha }{2}\right) |} \int _{0}^{\infty } \frac{f(x)-\tau _{-1/2}^x f(y)}{y^{\alpha +1}}dy \\&= \frac{2^{\alpha }\Gamma \left( \frac{\alpha +1}{2}\right) }{\sqrt{\pi } |\Gamma \left( -\frac{\alpha }{2}\right) |} \int _{0}^{\infty }\frac{ 2f(x)-f(x+y)-f(x-y)}{y^{\alpha +1}}dy. \end{aligned}$$

By utilizing the following formulas:

$$\begin{aligned} \Gamma (2z)&= \frac{2^{2z-1/2}}{\sqrt{2\pi }} \Gamma (z)\Gamma \left( z+\frac{1}{2}\right) , \\ \Gamma (z)\Gamma (1-z)&= \frac{\pi }{\sin \pi z}, \end{aligned}$$

we can deduce that

$$\begin{aligned} \frac{2^{\alpha }\Gamma \left( \frac{\alpha +1}{2}\right) }{\Gamma \left( \frac{1}{2}\right) |\Gamma \left( -\frac{\alpha }{2}\right) |}&= \frac{\Gamma (\alpha +1)}{\Gamma \left( 1-\frac{\alpha }{2}\right) \Gamma \left( \frac{\alpha }{2}\right) } \\&= \frac{\Gamma (\alpha +1)}{\pi }\sin \left( \frac{\alpha \pi }{2}\right) . \end{aligned}$$

As a result,

$$\begin{aligned} \left( -\frac{d^2}{dx^2}\right) ^{\alpha /2} f(x) =\frac{\Gamma (1+\alpha )}{\pi }\sin \left( \frac{\alpha \pi }{2}\right) \int _{0}^{\infty }\frac{2f(x)-f(x+y)+f(x-y)}{y^{\alpha +1}}dy. \end{aligned}$$

4 Statement of Main Results

The ensuing result functions as an intermediate step, elucidating the Fourier-Bessel transform’s representation through the Mellin transform. This finding extends the established result by N. Ormerod [21, Theorem 1] to encompass a broader parameter range \(\mu\). It is noteworthy that N. Ormerod’s work established a link between the Fourier transform of radial functions in \({\mathbb {R}}^n\) and the Mellin transform.

Theorem 4.1

Let \(\mu \ge -1/2\). For \(f\in S_*({\mathbb {R}})\), the function \({\mathscr {M}}\{f(x);s\}\) possesses an analytic continuation that is valid for all \(s \ne 0\), and it satisfies the functional equation:

$$\begin{aligned} {\mathscr {M}}\{f(x);\,s\}= \frac{2^{s-\mu -1}\Gamma (\frac{s}{2})}{\Gamma (\mu +1-\frac{s}{2})}{\mathscr {M}}\{{\mathscr {H}}_\mu f(x);\,2\mu +2-s\}. \end{aligned}$$

Theorem 4.2

Let \(\mu \ge -1/2\) and \(\alpha \in (0,\,2).\) For \(f\in S_*({\mathbb {R}}),\) the Mellin transform of the function \((-{\mathscr {B}}_\mu )^{\alpha /2}f\) is given by

$$\begin{aligned} {\mathscr {M}}\{(-{\mathscr {B}}_\mu )^{\alpha /2}f(x);\,s\}=2^{\alpha } \frac{\Gamma (\frac{s}{2})\Gamma (\mu +1-\frac{s-\alpha }{2})}{\Gamma (\mu +1-\frac{s}{2}) \Gamma (\frac{s-\alpha }{2})} {\mathscr {M}}\{f(x);\,s-\alpha \}, \end{aligned}$$

for \(s\ne 0\) and \(0<\Re (s)<2\mu +2.\)

From Theorem 4.1 with \(\mu =-\frac{1}{2}\), we have the following Corollary that holds for the Riesz fractional derivative.

Corollary 4.3

Let \(\alpha \in (0,\,2).\) For \(f\in S_*({\mathbb {R}}),\) the Mellin transform of the function \(\left(-\frac{d^2}{dx^2}\right)^{\alpha /2}f\) is given by

$${\mathcal{M}}\left\{ {\left( { - \frac{{d^{2} }}{{dx^{2} }}} \right)^{{\alpha /2}} f{\mkern 1mu} ;\,s} \right\} = \frac{{\Gamma (s)\cos \left( {\frac{\pi }{2}s} \right)}}{{\Gamma (s - \alpha )\cos \left[ {\frac{\pi }{2}(s - \alpha )} \right]}}{\mathcal{M}}\{ f(x){\mkern 1mu} ;\,s - \alpha \} ,\quad s \in \mathbb{C},$$

for \(s \ne 0\) and \(0< \Re (s)<1\).

Remark 4.4

When \(\mu = \frac{n}{2} - 1\) for \(n \in {\mathbb {N}}\), the Bessel operator \({\mathscr {B}}_\mu\) aligns with the radial component of the Laplace operator \(\Delta\) in Euclidean space \({\mathbb {R}}^n\). The Laplace operator is given by:

$$\begin{aligned} \Delta = \sum _{i=1}^{n} \frac{\partial ^2}{\partial x_i^2} = \frac{n^2}{dr^2} + \frac{n-1}{r} \frac{d}{dr} + \frac{1}{r^2} \Delta _{{\mathbb {S}}^{n-1}}, \end{aligned}$$

where \(\Delta _{{\mathbb {S}}^{n-1}}\) represents the Laplace operator on the unit sphere \({\mathbb {S}}^{n-1}\). In particular, for a radial function \(f(x) = f_0(r)\) where \(r=\Vert x\Vert\), we have:

$$\begin{aligned} \Delta f(x) = \frac{d^2f_0}{dr^2} + \frac{n-1}{r} \frac{df_0}{dr}. \end{aligned}$$

Reacll that for \(0< \alpha < 2\), the fractional Laplacian of order \(\alpha\) \((- \Delta )^{\alpha /2}\) can be defined for functions \(f: {\mathbb {R}}^n \rightarrow {\mathbb {R}}\) as a Fourier multiplier using the following formula:

$$\begin{aligned} {\mathcal {F}}((- \Delta )^{\alpha /2} f)(\xi ) = \Vert \xi \Vert ^{\alpha } {\mathcal {F}}(f)(\xi ). \end{aligned}$$

Here, \({\mathcal {F}}(f)\) represents the Fourier transform of a function \(f: {\mathbb {R}}^n \rightarrow {\mathbb {R}}\), which is given by:

$$\begin{aligned} {\mathcal {F}}(f)(\xi ) = \int _{{\mathbb {R}}^n} f(x) e^{-ix \cdot \xi } \, dx. \end{aligned}$$

In a more concrete sense, the fractional Laplacian can be expressed as a singular integral operator defined by:

$$\begin{aligned} (- \Delta )^{\alpha /2} f(x) = c_{n,\alpha } \int _{{\mathbb {R}}^n} \frac{f(x) - f(y)}{\Vert x-y\Vert ^{n+2\alpha }} \, dy \end{aligned}$$

Here, \(c_{n,\alpha }\) is given by:

$$\begin{aligned} c_{n,\alpha } = \frac{4^\alpha \Gamma (n/2 + \alpha )}{\pi ^{n/2} |\Gamma (-\alpha )|}. \end{aligned}$$

These two definitions, as well as several other definitions, are proven to be equivalent [14].

The following theorem is attributed to Pagnini and Runfola [23]. However, in this section, we present an alternative proof that offers a fresh perspective on the theorem’s validity.

Theorem 4.5

For a radial function \(f=f_0(r)\) in \(S({\mathbb {R}}^n)\), where \(S({\mathbb {R}}^n)\) is the Schwartz space, the fractional Laplace operator \((-\Delta )^{\alpha /2}f(x)\) with \(\alpha \in (0,2)\) is also a radial function. Furthermore,

$$\begin{aligned} {\mathscr {M}}\{(-\Delta )^{\alpha /2} f(x);\,s\}= 2^\alpha \frac{\Gamma \left( \frac{s}{2}\right) \Gamma \left( \frac{n-(s-\alpha )}{2}\right) }{\Gamma \left( \frac{n-s}{2}\right) \Gamma \left( \frac{s-\alpha }{2}\right) } {\mathscr {M}}\{ f_0(r);\,s-\alpha \} , \quad s \in {\mathbb {C}} , \end{aligned}$$

for \(s \ne 0\) and \(0< \Re (s)<n\).

5 Proof of Main Results

5.1 Proof of Theorem 4.1

The subsequent lemma serves as an analogue to Tate’s Lemma 2.4.2 [8, p.314], and its relevance is also highlighted by Ormerod [21, Lemma 2].

Lemma 5.1

Let \(f,\,g\in S_*({\mathbb {R}}).\) For \(0<\Re (s)<2\mu +2,\) the following equation holds:

$$\begin{aligned} {\mathscr {M}}\{f(x);\,s\}{\mathscr {M}}\{({\mathscr {H}}_\mu g)(x);\,2\mu +2-s\} = {\mathscr {M}}\{({\mathscr {H}}_\mu f)(x);\,2\mu +2-s\}{\mathscr {M}}\{g(x);\,s\} \end{aligned}$$
(5.2)

Proof

We start by recalling the following formula [34, Ch.12 & 13]:

$$\begin{aligned} \int _{0}^{\infty }J_\nu (at)t^{\nu +1}e^{-p^2t^2}dt=\frac{a^\nu }{(2p^2)^{\nu +1}}e^{-a^2/4p^2},\quad \Re (\nu >-1). \end{aligned}$$
(5.3)

Let’s consider the function

$$\begin{aligned} g(x)=e^{-x^2/2}. \end{aligned}$$

Applying the formula provided above (5.3), we obtain:

$$\begin{aligned} ({\mathscr {H}}_\mu g)(x)=g(x),\quad x\in {\mathbb {R}}. \end{aligned}$$
(5.4)

The Schwartz space \(S_*({\mathbb {R}})\) is invariant under the Fourier-Bessel transform. Therefore, by [21, Lemma 1], both the right and left sides of (5.2) are analytic functions in the region:

$$\begin{aligned} 0<\Re (s)<2\mu +2. \end{aligned}$$

For \(f, g \in S_*({\mathbb {R}})\), we have:

$$\begin{aligned} {\mathscr {M}}\{f(x);\,s\}{\mathscr {M}}\{({\mathscr {H}}_\mu g)(x);\,2\mu +2-s\} =\int _{0}^{\infty }\int _{0}^{\infty }f(t)t^{s-1}\,({\mathscr {H}}_\mu g)(x)x^{2\mu +1-s}\,dt \, dx. \end{aligned}$$
(5.5)

Under the transformation:

$$\begin{aligned} t\rightarrow t,\qquad x\rightarrow tx, \end{aligned}$$

the right-hand side of (5.5) becomes

$$\begin{aligned} \int _{0}^{\infty }\int _{0}^{\infty }f(t)t^{2\mu +1}\,({\mathscr {H}}_\mu g)(tx)x^{2\mu +1-s}\,dtdx. \end{aligned}$$

By utilizing (2.4), we can deduce that

$$\begin{aligned} \int _{0}^{\infty }\int _{0}^{\infty }f(t)&t^{2\mu +1}\,{\mathscr {H}}_\mu g(tx)x^{2\mu +1-s}\,dtdx \\ {}&=\frac{1}{2^{\mu }\Gamma (\mu +1)}\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }f(t) g(u){\mathscr {J}}_\mu (tux)(tu)^{2\mu +1}x^{2\mu +1-s}\,dudtdx. \end{aligned}$$

It’s evident that this expression is symmetric in f and g, thereby establishing the validity of our lemma. It’s clear that:

$$\begin{aligned} {\mathscr {M}}\{g(x);\,s\}={\mathscr {M}}\{({\mathscr {H}}_\mu g)(x);\,s\}=2^{\frac{s}{2}-1}\Gamma (\tfrac{s}{2}). \end{aligned}$$

\(\square\)

Now, we can prove Theorem 4.1.

Proof

By the Lemma 5.1, for \(0<\Re (s)<2\mu +2\), we have:

$$\begin{aligned} {\mathscr {M}}\{f(x);\,s\}= \frac{{\mathscr {M}}\{g(x);\,s\}}{{\mathscr {M}}\{({\mathscr {H}}_\mu g)(x);\,2\mu +2-s\}}{\mathscr {M}}\{({\mathscr {H}}_\mu f)(x);\,2\mu +2-s\}. \end{aligned}$$

Substituting the expressions we derived earlier, we get:

$$\begin{aligned} {\mathscr {M}}\{f(x);\,s\}= \frac{2^{s-\mu -1}\Gamma (\tfrac{s}{2})}{\Gamma (\mu +1-\tfrac{s}{2})}{\mathscr {M}}\{({\mathscr {H}}_\mu f)(x);\,2\mu +2-s\}. \end{aligned}$$

This completes the proof. \(\square\)

5.2 Proof of Theorem 4.2

We shall now establish Theorem 4.2.

Proof

We begin with Theorem 4.1, leading to the following expression:

$${\mathcal{M}}\left\{ {\left( { - {\mathcal{B}}_{\mu } } \right)^{{\alpha /2}} f(x);\,s} \right\} = \frac{{2^{{s - \mu - 1}} \Gamma (\frac{s}{2})}}{{\Gamma (\mu + 1 - \frac{s}{2})}}{\mathcal{M}}\left\{ {{\mathcal{H}}_{\mu } \left( {\left( { - {\mathcal{B}}_{\mu } } \right)^{{\alpha /2}} f} \right)(x);\,2\mu + 2 - s} \right\}.$$
(5.6)

Since we have the relation \({\mathscr {H}}_\mu \Big ((-{\mathscr {B}}_\mu )^{\alpha /2}f\Big )(x)=x^{\alpha }({\mathscr {H}}_\mu f)(x),\) we can simplify the expression further:

$$\begin{aligned} {\mathscr {M}}\{{\mathscr {H}}_\mu \Big ((-{\mathscr {B}}_\mu )^{\alpha /2}f\Big )(x);\,2\mu +2-s\}&= {\mathscr {M}}\{x^\alpha ({\mathscr {H}}_\mu f)(x);\,2\mu +2-s\}\\ {}&= {\mathscr {M}}\{({\mathscr {H}}_\mu f)(x);\,2\mu +2-(s-\alpha )\}. \end{aligned}$$

Now, by utilising Theorem 4.1 again, we arrive at:

$$\begin{aligned} {\mathscr {M}}\{{\mathscr {H}}_\mu \Big ((-{\mathscr {B}}_\mu )^{\alpha /2}f\Big )(x);\,2\mu +2-s\}= \frac{\Gamma (\mu +1-\frac{s-\alpha }{2})}{2^{s-\alpha -\mu -1}\Gamma (\frac{s-\alpha }{2})} {\mathscr {M}}\{f(x);\,s-\alpha \}. \end{aligned}$$
(5.7)

Combining equations (5.6) and (5.7), we obtain the final result:

$$\begin{aligned} {\mathscr {M}}\{(-{\mathscr {B}}_\mu )^{\alpha /2}f(x);\,s\}= \frac{2^{\alpha }\Gamma (\frac{s}{2})\Gamma (\frac{\alpha -s}{2})}{\Gamma (\mu +1-\frac{s}{2})\Gamma (\frac{s-\alpha }{2})} {\mathscr {M}}\{f(x);\,s-\alpha \}. \end{aligned}$$

This concludes the proof. \(\square\)

6 Application

In this section, we present a transmutation operator that operates between the fractional Bessel operator and the fractional second derivative, highlighting their interrelationship in the realm of fractional calculus.

Let \(f\) be an even continuous function on \({\mathbb {R}}\), the Riemann-Liouville integral transform is defined by [33]

$$\begin{aligned} {\mathscr {R}}_\mu f(x)&:= \frac{2\Gamma (\mu +1)}{\sqrt{\pi }\Gamma (\mu +1/2)}\int _{0}^{1}f(xt)\,(1-t^2)_{+}^{\mu -1/2}dt, \quad \text {Re}(\mu ) > -1/2 \\&= \frac{2\Gamma (\mu +1)}{\sqrt{\pi }\Gamma (\mu +1/2)x^{2\mu }}\int _{0}^{x}f(t)\,(x^2-t^2)_{+}^{\mu -1/2}dt. \end{aligned}$$

It has been demonstrated in [33, Theorem 2.1.1] that the Riemann-Liouville integral transform serves as a topological isomorphism from \(S_*({\mathbb {R}})\) onto itself. Moreover, this transform adheres to the transmutation relation as follows:

$$\begin{aligned} {\mathscr {B}}_\mu {\mathscr {R}}_\mu f = {\mathscr {R}}_\mu D^2 f, \quad D = \frac{d}{dx}. \end{aligned}$$
(6.1)

In the following theorem, we extend the previously established transmutation relation to the fractional setting.

Theorem 6.2

Let \(\mu > -1/2\) and \(0< \alpha < 2\). It holds that

$$\begin{aligned} (-{\mathscr {B}}_\mu )^{\alpha /2} {\mathscr {R}}_\mu f = {\mathscr {R}}_\mu \left( -\frac{d^2}{dx^2}\right) ^{\alpha /2} f, \quad f \in S_*({\mathbb {R}}). \end{aligned}$$

Proof

Observe that the Riemann-Liouville transform can be represented in terms of Mellin convolutions:

$$\begin{aligned} {\mathscr {R}}_\mu f(x) = (k*^{{\mathscr {M}}}f)(x), \end{aligned}$$
(6.3)

where

$$\begin{aligned} k(x) = \frac{2\Gamma (\mu +1)}{\sqrt{\pi }\Gamma (\mu +1/2)}x^{-2\mu }(x^2-1)_{+}^{\mu -1/2}. \end{aligned}$$

Using the well-known Mellin transform formula [20]:

$$\begin{aligned} {\mathscr {M}}\{k(x);s\} = \frac{\Gamma (\mu +1)\Gamma (\tfrac{1-s}{2})}{\sqrt{\pi }\Gamma (\mu +1-\tfrac{s}{2})}, \end{aligned}$$

we can deduce that

$$\begin{aligned} {\mathscr {M}}\{{\mathscr {R}}_\mu f(x)\,;s\} = \frac{\Gamma (\mu +1)\Gamma (\tfrac{1-s}{2})}{\sqrt{\pi }\Gamma (\mu +1-\tfrac{s}{2})}{\mathscr {M}}\{f(x)\,;s\}. \end{aligned}$$

By using Theorem 4.2 and Corollary 4.3, we obtain

$$\begin{aligned}&{\mathscr {M}}\{(-{\mathscr {B}}_\mu )^{\alpha /2} {\mathscr {R}}_\mu f(x)\,;s\} \nonumber \\&\quad = \frac{2^{\alpha }\Gamma (\frac{s}{2})\Gamma (\mu +1-\frac{s-\alpha }{2})}{\Gamma (\mu +1-\frac{s}{2})\Gamma (\frac{s-\alpha }{2})}{\mathscr {M}}\{{\mathscr {R}}_\mu f(x);\,s-\alpha \}\nonumber \\&\quad = \frac{2^{\alpha }\Gamma (\mu +1)}{\sqrt{\pi }}\,\,\frac{\Gamma (\frac{s}{2})\Gamma (\frac{1-(s-\alpha )}{2})}{\Gamma (\mu +1-\frac{s}{2})\Gamma (\tfrac{s-\alpha }{2})}{\mathscr {M}}\{ f(x);\,s-\alpha \}. \end{aligned}$$
(6.4)

Similarly,

$$\begin{aligned}&{\mathscr {M}}\{{\mathscr {R}}_\mu \left( -\frac{d^2}{dx^2}\right) ^{\alpha /2} f(x);\,s\} \nonumber \\&\quad = \frac{\Gamma (\mu +1)\Gamma (\tfrac{1-s}{2})}{\sqrt{\pi }\Gamma (\mu +1-\tfrac{s}{2})}{\mathscr {M}}\{\left( -\frac{d^2}{dx^2}\right) ^{\alpha /2} f(x);\,s\}\nonumber \\&\quad = \frac{2^{\alpha }\Gamma (\mu +1)}{\sqrt{\pi }}\,\,\frac{\Gamma (\frac{s}{2})\Gamma (\frac{1-(s-\alpha )}{2})}{\Gamma (\mu +1-\frac{s}{2})\Gamma (\tfrac{s-\alpha }{2})}{\mathscr {M}}\{ f(x);\,s-\alpha \}. \end{aligned}$$
(6.5)

Comparing equations (6.4) and (6.5), and utilizing the injectivity of the Mellin transform, we conclude that

$$\begin{aligned} (-{\mathscr {B}}_\mu )^{\alpha /2} {\mathscr {R}}_\mu f = {\mathscr {R}}_\mu \left( -\frac{d^2}{dx^2}\right) ^{\alpha /2} f. \end{aligned}$$

This completes the proof. \(\square\)

6.1 Concluding Remark

In conclusion, this paper thoroughly explores the fractional Bessel operator. Its significance is determined by its role in Fourier-Bessel pseudo-differential operations, highlighting its importance in mathematical analysis. The paper effectively establishes a connection between this operator and the Mellin integral transform, demonstrating its adaptability in various scenarios. These include defining fractional Bessel derivatives and deriving a transmutation operator that bridges the gap between fractional Bessel and fractional second-order derivatives.