Introduction

Methods of algebraic and umbral nature possess, among the other advantages, the undoubtful merit of simplifying practical computational issues. In a numbers of previous papers [1,2,3,4,5,6,7] it has been established that the one of the possible umbral images of a Bessel function is a Gaussian. This statement can be profitably exploited in the present context, provided we establish some reference tools [6, 8].

Definition 1

The function

$$\begin{aligned} \varphi (\nu ):=\varphi _\nu =\frac{1}{\Gamma (\nu +1)}, \quad \forall \nu \in {\mathbb {R}}. \end{aligned}$$
(1)

defines the Umbral “Vacuum” function or simply vacuum.

Definition 2

We introduce the operator \({\hat{c}}\) called “umbral”,

$$\begin{aligned} {\hat{c}}:=e^{\partial _z}, \end{aligned}$$
(2)

as the vacuum shift operator, with z the domain’s variable of the function on which the operator acts.

Theorem 1

The umbral operator \({\hat{c}}^\nu \), \(\forall \nu \in {\mathbb {R}}\), is represented by the action on the vacuum \(\varphi _{0}\) such thatFootnote 1

$$\begin{aligned} {\hat{c}}^{\nu } \varphi _{0} =\frac{1}{\Gamma (\nu +1)}. \end{aligned}$$
(3)

Proof

\(\forall \nu \in {\mathbb {R}}\), by using Definition 2, we get

$$\begin{aligned} {\hat{c}}^\nu \varphi _{0}= {\hat{c}}^\nu \left. \varphi (z)\right| _{z=0}=e^{\nu \partial _z}\left. \varphi (z)\right| _{z=0}=\varphi (z+\nu ) \left. \right| _{z=0}=\dfrac{1}{\Gamma (z+\nu +1)}\left. \right| _{z=0}= \dfrac{1}{\Gamma (\nu +1)}.\nonumber \\ \end{aligned}$$
(4)

\(\square \)

Through these premises, we define the correspondence quoted below.

Proposition 1

\(\forall x,\nu \in {\mathbb {R}}\), the umbral image of a 0-order Bessel function is a Gaussian function

$$\begin{aligned} J_0(x)=e^{-{\hat{c}}\, \left( \frac{x}{2} \right) ^{2} } \varphi _{0} . \end{aligned}$$
(5)

Proof

\(\forall x,\nu \in {\mathbb {R}}\), let

$$\begin{aligned} u(x)=e^{-{\hat{c}}\, \left( \frac{x}{2} \right) ^{2} } \varphi _{0}, \end{aligned}$$
(6)

the relevant series expansion proofs that \(u(x)=J_0(x)\), indeed

$$\begin{aligned} u(x)=e^{-{\hat{c}}\, \left( \frac{x}{2} \right) ^{2} } \varphi _{0} =\sum _{r=0}^{\infty }\frac{(-{\hat{c}})^{r} }{r!} \left( \frac{x}{2} \right) ^{2\, r} \varphi _{0} =\sum _{r=0}^{\infty }\frac{(-1)^{r} }{\left( r!\right) ^{2} } \left( \frac{x}{2} \right) ^{2\, r} =J_{0} (x). \end{aligned}$$
(7)

\(\square \)

It is accordingly evident that we can exploit all the wealth of properties of Gaussian function to deal with those of Bessel, this has already been discussed in previous papers [3] and won’t be repeated here.

In order to summarize the method we are going to exploit, we consider the following Example.

Example 1

We prove that the previous prescription allows to reduce the evaluation of the integral

$$\begin{aligned} I_J(a,b)=\int _{-\infty }^\infty J_0\left( 2\sqrt{a x^2 +b x}\right) dx, \quad \forall a,b\in {\mathbb {R}}:a x^2 +b x>0 \end{aligned}$$
(8)

to a straightforward Gaussian integration. We note indeed that

$$\begin{aligned} I_J(a,b)=\int _{-\infty }^\infty e^{-{\hat{c}}\left( a x^2 +b x \right) } dx\; \varphi _0 \end{aligned}$$
(9)

and, according to the well-known integral identity

$$\begin{aligned} I_G(a,b)=\int _{-\infty }^\infty e^{-\left( a x^2 +b x \right) } dx=\sqrt{\dfrac{\pi }{a}}e^{\frac{b^2}{4a}}, \end{aligned}$$
(10)

we can conclude that

$$\begin{aligned} I_J(a,b)=I_G({\hat{c}}a,{\hat{c}}b)\;\varphi _0= \sqrt{\dfrac{\pi }{{\hat{c}}\;a}}\;e^{\frac{{\hat{c}}\;b^2}{4a}}\varphi _0. \end{aligned}$$
(11)

In deriving Eq. (11) we have treated the umbral operator as an ordinary algebraic quantity. It is now necessary to manipulate Eq. (11) to get a meaningful result, we should indeed provide a non-formal expression for the operator function on the r.h.s. of Eq. (11). To this aim we expand in series the exponential on the r.h.s. of (11) and by using the properties of the \({\hat{c}}\)-operator we find

$$\begin{aligned} I_J(a,b)= \sqrt{\dfrac{\pi }{a}}\sum _{r=0}^\infty \dfrac{1}{r!}\left( \dfrac{b}{2\sqrt{a}}\right) ^{2r} {\hat{c}}^{r-\frac{1}{2}}\varphi _0 =\sqrt{\dfrac{\pi }{a}}\sum _{r=0}^\infty \dfrac{1}{r!\Gamma \left( r+\frac{1}{2} \right) }\left( \dfrac{b}{2\sqrt{a}}\right) ^{2r}. \end{aligned}$$
(12)

The series on this last equation can easily be recognized as a Bessel function and therefore we end up with

$$\begin{aligned} I_J(a,b)=\sqrt{\dfrac{\pi }{a}}\sqrt{\dfrac{b}{2\sqrt{a}}} I_{-\frac{1}{2}}\left( \dfrac{b}{\sqrt{a}} \right) \end{aligned}$$
(13)

where \(I_\nu (x)\) is a modified Bessel of first kind [9].

This result, not straightforwardly achievable by standard means, is a naïve consequence of the previously outlined procedure.

The example we have presented is sufficient to clarify the essential features of the methods we are going to use in the following parts of the paper. We will see below that, by stretching the formalism, we can provide further and more interesting results.

Integrals and Non Gaussian Umbral Images

Before entering the specific aspects of this section, we like to stress that the methods we are reporting here are useful and flexible but not unique. We will check the relevant usefulness by deriving well established forms obtained in the past [10, 11]

Example 2

We consider the further example

$$\begin{aligned} I_{J,G}(a,b)=\int _{-\infty }^\infty J_0\left( 2\sqrt{a}x\right) e^{-b x^2}dx \end{aligned}$$
(14)

which, according to the previous formalism, can be written as

$$\begin{aligned} I_{J,G}(a,b)=\int _{-\infty }^\infty e^{-{\hat{c}} ax^2}e^{-bx^2}dx\;\varphi _0=I_G\left( {\hat{c}}a+b,0 \right) \varphi _0 \end{aligned}$$
(15)

and, therefore, we end up withFootnote 2

$$\begin{aligned} I_{J,G}(a,b)=\sqrt{\dfrac{\pi }{b}}\dfrac{1}{\sqrt{1+\dfrac{a}{b}{\hat{c}}}}\varphi _0= \sqrt{\dfrac{\pi }{b}}\dfrac{1}{\Gamma \left( \frac{1}{2}\right) }\int _0^\infty e^{-s}s^{-\frac{1}{2}}e^{-\frac{as}{b}{\hat{c}}}ds\;\varphi _0. \end{aligned}$$
(16)

Therefore, after noting that [6]

$$\begin{aligned} \begin{aligned}&\int _0^\infty e^{-s}s^{-\frac{1}{2}}e^{-xs{\hat{c}}}ds\; \varphi _0=\sum _{r=0}^\infty \dfrac{\Gamma \left( r+\frac{1}{2}\right) (-x)^r }{r!^2}=c^{(1)}_{0,-\frac{1}{2}}(x),\\&\quad c_{\mu ,\alpha }^{(\nu )}(x)=\sum _{r=0}^\infty \dfrac{\Gamma (\nu r+\alpha +1)}{r!\Gamma (r+\mu +1)}(-x)^r, \quad \forall \alpha ,\mu ,\nu \in {\mathbb {R}}^+, \forall x\in {\mathbb {R}} \end{aligned} \end{aligned}$$
(17)

we find

$$\begin{aligned} I_{J,G}(a,b)=\dfrac{1}{\sqrt{b}}c^{(1)}_{0,-\frac{1}{2}}\left( \dfrac{a}{b} \right) . \end{aligned}$$
(18)

We will comment on the properties of the auxiliary function \(c_{\mu ,\alpha }^{(\nu )}(x)\) [6] in the concluding section.

The use of a Gaussian function as umbral image is useful but not unique. The umbral method has allowed to reduce a higher transcendental function, like a Bessel function, to an ordinary trascendental, like a Gaussian. By following the same “down-grading” criterion we can use rational functions as umbral image of exponentials. We note indeed that a Lorentzian function is the umbral image of the Gaussian function according to the identities

$$\begin{aligned} e^{-ax^2}=\sum _{r=0}^\infty \dfrac{(-a)^r}{r!}x^{2r}=\sum _{r=0}^\infty (-a{\hat{c}})^r x^{2r}\varphi _0=\dfrac{1}{1+a{\hat{c}}x^2}\varphi _0. \end{aligned}$$
(19)

We will exploit the above umbral restyling by deriving an integral involving Gaussian and Lorentzian functions.

Example 3

Let

$$\begin{aligned} I_L(a,b)=\int _{-\infty }^\infty \dfrac{e^{-\frac{ax^2}{1+bx^2}}}{1+bx^2}dx, \qquad \quad \forall a,b\in {\mathbb {R}}^+, \end{aligned}$$
(20)

the use of the outlined formalism, eventually yieldsFootnote 3

$$\begin{aligned} \begin{aligned} I_L(a,b)&=\int _{-\infty }^\infty \dfrac{1}{(1+bx^2)\left( 1+{\hat{c}}\dfrac{ax^2}{1+bx^2} \right) }dx\;\varphi _0= \int _{-\infty }^\infty \dfrac{1}{1+(b+a{\hat{c}})x^2}dx\;\varphi _0\\&=I_L(0, b+a{\hat{c}})\varphi _0= \dfrac{\pi }{\sqrt{b+a{\hat{c}}}}\varphi _0=\dfrac{\sqrt{\pi }}{\sqrt{b}} c^{(1)}_{0,-\frac{1}{2}}\left( \dfrac{a}{b} \right) . \end{aligned} \end{aligned}$$
(21)

The use of the same formalism yields \(\forall b>0\)

$$\begin{aligned} \begin{aligned} I_E(a,b)&=\int _{-\infty }^\infty \dfrac{e^{-\frac{ax}{1+bx^2}}}{1+bx^2}dx= \int _{-\infty }^\infty \dfrac{1}{1+a{\hat{c}}x+bx^2}dx\;\varphi _0\\&= \sqrt{\dfrac{\pi }{b}}J_{0,-\frac{1}{2}}^{(2)}\left( \dfrac{a}{\sqrt{b}} \right) , \\ J_{\mu ,\nu }^{(\alpha )}(x)&=\sum _{r=0}^\infty \dfrac{\Gamma \left( r+\nu +1 \right) }{r!\Gamma \left( \alpha r+\mu +1 \right) }\left( \dfrac{x}{2} \right) ^{2r+\mu } , \qquad \quad \forall \alpha ,\mu ,\nu \in {\mathbb {R}}^+, \forall x\in {\mathbb {R}}. \end{aligned} \end{aligned}$$
(22)

We can now apply the method to more complicated functions.

Example 4

We consider indeed the case

$$\begin{aligned} I_{L,J}(a,b)=\int _{-\infty }^\infty \dfrac{J_0\left( \dfrac{2\sqrt{a}x}{\sqrt{1+bx^2}}\right) }{1+bx^2}dx, \qquad \quad \forall a,b\in {\mathbb {R}}^+, \end{aligned}$$
(23)

which, according to our formalism, can be transformed into a \(G-L\) integral and eventually into a L-integral. Accordingly we get

$$\begin{aligned} I_{L,J}(a,b)=I_L(a{\hat{c}},b)\varphi _0=\sqrt{\dfrac{\pi }{b}}c^{(1)}_{0,-\frac{1}{2}} \left( \dfrac{a\;{\hat{c}}}{b} \right) \varphi _0=\sqrt{\dfrac{\pi }{b}}\sum _{r=0}^\infty \dfrac{\Gamma \left( r+\frac{1}{2}\right) }{r!^3}\left( -\dfrac{a}{b} \right) ^r \end{aligned}$$
(24)

where the series on the r.h.s. is a Humbert type Bessel function, to be commented in the concluding section. Furthermore, if we introduce a new umbral operator \({}_2{\hat{c}}\) such that

$$\begin{aligned} {}_2{\hat{c}}^r\psi _0=\dfrac{1}{\left( \Gamma (r+1) \right) ^2}, \end{aligned}$$
(25)

we end up with

$$\begin{aligned} I_{L,J}(a,b)=\sqrt{\dfrac{\pi }{b}}\sum _{r=0}^\infty \dfrac{\Gamma \left( r+\frac{1}{2}\right) }{r!}\left( -\dfrac{a\;{}_2{\hat{c}}}{b} \right) ^r \psi _0=\dfrac{\pi }{\sqrt{b+a\;{}_2{\hat{c}}}}\psi _0=I_L\left( 0,b+a\;{}_2{\hat{c}} \right) \psi _0.\nonumber \\ \end{aligned}$$
(26)

The technicalities, we have displayed in this section, yield an idea of the usefulness of the methods we have foreseen. Further comments and examples, corroborating the previous results, will be presented in the concluding section.

Final Comments

The results of the previous sections, albeit supported by the check of the relevant correctness, including comparison with known cases and numerical analysis, are hampered by the absence of mathematical rigor. In Ref. [6] a great deal of attention has been devoted to a more appropriate formulation. In the following we will extend the method by including the recourse to families of Special Polynomials.

Example 5

We can provide further elements of discussion by setting

$$\begin{aligned} J_0\left( 2\sqrt{\alpha \;x +\beta \;x^2}\right) =\dfrac{1}{1+\;{}_2{\hat{c}}\left( \alpha x+\beta x^2 \right) }\psi _0, \qquad \quad \forall \alpha , \beta \in {\mathbb {R}}^+\;. \end{aligned}$$
(27)

We now remind that the two variable Cebyshev polynomials of second kind [9],

$$\begin{aligned} U_{n} (x,y)=(-1)^{n} \sum _{r=0}^{\lfloor \frac{n}{2} \rfloor }\frac{(n-r)!\, x^{n-2r} (-y)^{r} }{(n-2\, r)!\, r!},\quad \forall x,y\in {\mathbb {R}}, \forall n\in {\mathbb {N}}, \end{aligned}$$
(28)

are specified by the generating function

$$\begin{aligned} \sum _{n=0}^{\infty }t^{n} U_{n} (x,y)=\frac{1}{1+x\, t+y\, t^{2} }, \quad Re(1+x\, t+y\, t^{2})>0 . \end{aligned}$$
(29)

According to Eq. (27) we can conclude that

$$\begin{aligned} J_0\left( 2\sqrt{\alpha \;x +\beta \;x^2}\right) =\sum _{n=0}^{\infty }x^n\;U_n\left( \alpha \; {\hat{b}},\beta \;{\hat{b}} \right) \psi _0. \end{aligned}$$
(30)

and therefore, by noting that

$$\begin{aligned} U_n\left( \alpha \; {\hat{b}},\beta \;{\hat{b}} \right) \psi _0= & {} (-1)^n\sum _{r=0}^{\lfloor \frac{n}{2}\rfloor }\dfrac{(-1)^r (n-r)! \alpha ^{n-2r}\beta ^r}{(n-2r)!r!}{\hat{b}}^{n-r}\psi _0\nonumber \\= & {} (-1)^n\sum _{r=0}^{\lfloor \frac{n}{2}\rfloor }\dfrac{(-1)^r \alpha ^{n-2r}\beta ^r}{(n-2r)!(n-r)!r!}, \end{aligned}$$
(31)

we find

$$\begin{aligned} \begin{aligned}&J_0\left( 2\sqrt{\alpha \;x +\beta \;x^2}\right) =\sum _{n=0}^{\infty }x^n\; {}_2U_n\left( \alpha ,\beta \right) , \quad \forall x,\alpha ,\beta \in {\mathbb {R}}\\&{}_2U_n\left( \alpha ,\beta \right) \;{:}{=}\; (-1)^n\sum _{r=0}^{\lfloor \frac{n}{2}\rfloor }\dfrac{(-1)^r \alpha ^{n-2r}\beta ^r}{(n-2r)!(n-r)!r!}. \end{aligned} \end{aligned}$$
(32)

Observation 1

It is also useful to note that

$$\begin{aligned} \partial _{x}^m \;J_0\left( 2\sqrt{\alpha \;x +\beta \;x^2}\right) =\sum _{n=0}^{\infty }\dfrac{(n+m)!}{n!}x^n\;{}_2U_{n+m}\left( \alpha ,\beta \right) . \end{aligned}$$
(33)

We have shown that the method we have proposed allows a great deal of flexibility, in the sense that, hardly achievable results with ordinary means, are easily derived within the present computational framework. This is perhaps an indication that the study of these methods should be pursued.

Remark 1

Before concluding the paper we go back to the auxiliary functions we have defined as, e.g., \(c_{0,-\frac{1}{2}}(x)\) in Eq. (17). To better clarify the relevant properties we remind that

$$\begin{aligned} e^{-{\hat{c}}x}\varphi _0=C_0(x)=\sum _{r=0}^\infty \dfrac{(-x)^r}{r!^2}, \quad \forall x\in {\mathbb {R}}, \end{aligned}$$
(34)

is the Tricomi-Bessel function of zero order [12]. The relevant properties can be easily drawn from Eq. (17) itself. By kee** the derivative with respect to x, we find

$$\begin{aligned} \partial _x^{\;m} c^{(1)}_{0,-\frac{1}{2}}(x)=(-1)^m\int _0^\infty e^{-s}s^{-\frac{1}{2}+m}\;{\hat{c}}^m e^{-xs{\hat{c}}}ds\;\varphi _0. \end{aligned}$$
(35)

which suggests the introduction of the function \(C_{\mu ,\nu }(x)\) defined as

$$\begin{aligned} \begin{aligned}&C_{\mu ,\nu }(x)=\int _0^\infty e^{-s}s^{\nu }{\hat{c}}^\mu e^{-xs{\hat{c}}}ds\; \varphi _0 = \int _0^\infty e^{-s}s^{\nu } C_{\mu }(xs)ds,\\&C_{\mu }(x)= \sum _{r=0}^\infty \dfrac{(-x)^r}{r!\Gamma \left( r+\mu +1 \right) }. \end{aligned} \end{aligned}$$
(36)

In other words Eq. (36) states that \(C_{\mu ,\nu }(x)\) can be viewed as Borel (\(B-\)) transform [13, 14] of the \(\mu \)-order Tricomi function \(C_{\mu }(x)\) [15]. Then, we can get

$$\begin{aligned} \partial _x^{\;m} c^{(1)}_{0,-\frac{1}{2}}(x)=(-1)^m\int _0^\infty e^{-s} s^{m-\frac{1}{2}}C_m(xs)ds=(-1)^mC_{m,m-\frac{1}{2}}(x) \end{aligned}$$
(37)

Analogous considerations apply to the functions \(J_{\mu ,\nu }^{(2)}(x)\) which can be viewed as the B-transform of the Bessel-Wright functions [16].

The B-transform has played an important role for the a rigorous formulation of the umbral methods exploited here. Further comments and in particular the link with the Humbert Bessel will be discussed elsewhere.

In this paper we have dealt with definite integrals between \(-\infty \) and \(\infty \), the umbral method can be exploited to derive definite or indefinite integrals as e.g.,

$$\begin{aligned} I_C(a)=\int _0^1 C_0(ax)dx=\int _0^1 e^{-a{\hat{c}}x}dx \varphi _0 \qquad \forall a\in {\mathbb {R}}:a\ge 0 \end{aligned}$$
(38)

which eventually yields

$$\begin{aligned} I_C(a)=\dfrac{1}{{\hat{c}}a}\left( 1-e^{-{\hat{c}}a} \right) \varphi _0=C_1(a). \end{aligned}$$
(39)

The presented umbral procedure can be further handled and merged to other techniques of operational nature to obtain further generalizations. It is further worth noting that integration methods based on the umbral procedure does not apply to special function only but to special polynomials as well.

Within such a context the two variable Hermite polynomials

$$\begin{aligned} H_n(x,y)=n!\sum _{r=0}^{\lfloor \frac{n}{2}\rfloor } \dfrac{x^{n-2r}y^r}{(n-2r)!r!} \end{aligned}$$
(40)

can be written as an ordinary Newton binomial [2, 6]

$$\begin{aligned} H_n(x,y)=\left( x+ {}_y{\hat{h}}\right) ^n \theta _0, \end{aligned}$$
(41)

where \({}_y{\hat{h}}\) is an umbral operator acting on the vacuum \(\theta _0\) in the following way

$$\begin{aligned} \begin{aligned}&{}_y {\hat{h}}^r\;\theta _0 :=\theta _r, \quad \forall r\in {\mathbb {R}},\\&\theta _r =\dfrac{y^{\frac{r}{2}}r!}{\Gamma \left( \frac{r}{2}+1\right) }\left| \cos \left( r\dfrac{\pi }{2}\right) \right| = \left\{ \begin{array}{ll} 0 , &{} r=2s+1 ,\\ y^s \dfrac{(2s)!}{s!}\;, &{} r=2s, \end{array}\right. \;\;\forall s\in {\mathbb {Z}}. \end{aligned} \end{aligned}$$
(42)

The definite integral

$$\begin{aligned} I_{m,n}(x)=\int _0^x \xi ^m H_n(\xi ,y)d\xi \end{aligned}$$
(43)

can be then reduced, after a straightforward steps (for further comments see Ref. [11]), to

$$\begin{aligned} I_{m,n}(x)=\int _0^x \xi ^m \left( \xi + {}_y{\hat{h}}\right) ^n d\xi \theta _0=m!\sum _{s=0}^{\min (n,m+1)}(-1)^s \dfrac{x^{m+s+1}}{(m+s+1)!}\dfrac{n!}{(n-s)!}H_{n-s}(x,y).\nonumber \\ \end{aligned}$$
(44)

This result is a straightforward consequence of the (umbral) binomial nature of the Hermite polynomials but can also be obtained with standard means.

In a forthcoming investigation we will see how the merging between umbral methods and negative derivative integration procedures [11] may lead to further interesting developments.