ABSTRACT

The transition from a known Taylor series $f\left(x\right)={\displaystyle {\sum }_{n=0}^{\infty }}\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n$ of a known function $f\left(x\right)$ to a new function $\hat{f}\left(x\right)$ primarily defined by the infinite power series $\hat{f}\left(x\right)\equiv {\displaystyle {\sum }_{n=0}^{\infty }}{f}^{\left(n\right)}\left(0\right)x^n$ with coefficients $f^{\left(n\right)}\left(0\right)$ from the Taylor series of the function $f\left(x\right)$ can be made by an integral transformation which is a modified Laplace transformation and is called Sumudu transformation. It makes the transition from the Exponential series to the Geometric series and may help to evaluate new infinite power series from known Taylor series. The Sumudu transformation is demonstrated to be a limiting case of Fractional integration. Apart from the basic Sumudu integral transformation we discuss a modification where the coefficients $\frac{f^{\left(n\right)}\left(0\right)}{n!}$ from the Taylor series are not changed to $f^{\left(n\right)}\left(0\right)$ but only to $\frac{f^{\left(n\right)}\left(0\right)}{\left(\frac{n}{2}\right)\mathrm{<mo>\; !\; </mo>}}$ . Beside simple examples our applications are mainly concerned to calculate new Generating functions for Hermite polynomials from the basic ones.

Keywords:

Mellin Transformation, Fractional Integration, Geometric Series and Exponential Series, Error Function, Laguerre Polynomials, Generating Functions of Hermite Polynomials, Bessel Functions, Asymptotic Series, Operator Identities

1. Introduction

Let us suppose that we know the Taylor series of a function $f(x)$

$f\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n,\left(f^{\left(n\right)}\left(0\right)\equiv {\left\{\frac{{\partial }^n}{\partial x^n}f\left(x\right)\right\}}_{x\to 0}\right).$ (1.1)

It is not rare that we want to evaluate a new primarily unknown function $\hat{f}\left(x\right)$ given by its power series of the following kind

$\hat{f}\left(x\right)\equiv {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{f}^{\left(n\right)}\left(0\right)x^n,$ (1.2)

with coefficients $f^{\left(n\right)}\left(0\right)$ taken from the Taylor series of $f\left(x\right)$ in (1.1). This can be made by a linear integral transformation of the function $f\left(x\right)$ which is closely related to a Laplace transformation and which is now also called Sumudu transformation¹. It was introduced by Watugala [1] and further developed by Belgacem and coauthors, e.g., [2] [3] [4] [5] [6] and many others, e.g., [7] . The article [2] contains basic examples of Sumudu transforms of functions. Since Laplace transformations are widely applied in mathematics for the solution of ordinary differential equations which have important applications in physics (e.g., electrical engineering) and in other sciences and since there exist detailed investigations of these transformations and voluminous tables, e.g., [8] [9] [10] [11] [12] it seems to be possible to enlarge the store of power series which can be evaluated in closed form. The present article applies the Sumudu transformation mainly to obtain new Generating functions for Hermite polynomials. In the following we derive and explain this integral transformation and consider in Section 5 a modified similar transformation which may be considered in a loose sense as half the part of the way from the original function to its Sumudu transform. In Section 3 we demonstrate how Sumudu transformation can also be considered as a limiting case of Fractional integration.

Using the well-known sequence of integrals with the real parameter $x\ge 0$

${\displaystyle {\int }_0^{+\infty }}\text{d}y{y}^n\exp \left(-\frac{y}{x}\right)=n!x^{n+1}\mathrm{,}\left(n=0,1,2,\cdots ;x=\left|x\right|\ge 0\right)\mathrm{,}$ (1.3)

we obtain from the Taylor series of $f\left(x\right)$ the following general identity

$\begin{array}{c}\hat{f}\left(x\right)\equiv {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{f}^{\left(n\right)}\left(0\right)x^n=\frac{1}{x}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{f^{\left(n\right)}\left(0\right)}{n!}\left(n!x^{n+1}\right)\\ =\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\left({\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{f^{\left(n\right)}\left(0\right)}{n!}{y}^n\right)\exp \left(-\frac{y}{x}\right),\end{array}$ (1.4)

and, finally, if we substitute herein (1.1)

$\hat{f}\left(x\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}yf\left(y\right)\exp \left(-\frac{y}{x}\right),\left(x=\left|x\right|>0\right).$ (1.5)

¹Sumudu means “smooth, noble” in Sanscrit and in Sinhala, the language of Sinhalese people of Sri Lanka.

We inserted $n!x^{n+1}$ in (1.4) according to (1.3) and changed the order of summation and integration due to supposed absolute convergence of the sum. From (1.5) it can be seen that the functions $f\left(x\right)$ have to be determined only for $x>0$ . Setting $f\left(x\right)=f\left(x\right)\theta \left(x\right)$ with $\theta \left(x\right)$ the Heaviside step function ( $\theta \left(x\right)=0$ for $x<0$ , $\theta \left(x\right)=1$ for $x>0$ ) in the sense of generalized functions one may extend the integration in (1.5) from $-\infty$ to $+\infty$ that makes some proofs easier, e.g., inversion of the transformation.

By the substitution of the parameter $x\equiv \frac{1}{s}$ we obtain from (1.5) the form

$\frac{1}{s}\hat{f}\left(\frac{1}{s}\right)={\displaystyle {\int }_0^{+\infty }}\text{d}yf\left(y\right){\text{e}}^{-sy}\mathrm{.}$ (1.6)

If we denote the standard form of the Laplace transformation $f\left(x\right)\to F\left(s\right)$ by [8] [9]

$f\left(x\right)\to F\left(s\right)\equiv \mathcal{L}\left\{f\right\}\left(s\right)\equiv {\displaystyle {\int }_0^{+\infty }}\text{d}yf\left(y\right){\text{e}}^{-sy}\mathrm{,}$ (1.7)

the transition $f\left(x\right)\to \hat{f}\left(x\right)$ is found to be a modified Laplace transformation in the following sense

$f\left(x\right)\to \hat{f}\left(x\right)=\frac{1}{x}F\left(\frac{1}{x}\right)=\frac{1}{x}\mathcal{L}\left\{f\right\}\left(\frac{1}{x}\right)\equiv \mathcal{S}\left\{f\right\}\left(x\right)\mathrm{,}\left(x\ge 0\right)\mathrm{,}$ (1.8)

where $\mathcal{S}\{\dots \}$ stands here for Sumudu transformation. To each pair of original function $f\left(x\right)$ and its Laplace transform $F\left(s\right)$ which one may calculate or can find in tables (e.g., [8] [9] [10]) one obtains a pair of Taylor series of $f\left(x\right)$ and $\hat{f}\left(x\right)$ if one of these series is known².

By the substitution $u\equiv \frac{y}{x}$ with x as positive parameter and using partial integration we can bring (1.5) to the form

$\hat{f}\left(x\right)={\displaystyle {\int }_0^{+\infty }}\text{d}uf\left(xu\right){\text{e}}^{-u}={\displaystyle {\int }_0^{+\infty }}\text{d}u{\text{e}}^{-u}\left(x^{u\frac{\partial }{\partial u}}f\left(u\right)\right)\mathrm{,}$ (1.9)

which by partial integration leads to

$\hat{f}\left(x\right)={\displaystyle {\int }_0^{+\infty }}\text{d}uf\left(u\right)\underset{\frac{1}{x}\exp \left(-\frac{u}{x}\right)}{\underset{︸}{\left(x^{-\frac{\partial }{\partial u}u}{\text{e}}^{-u}\right)}}\mathrm{,}$ (1.10)

that means again to the starting form (1.5). We wrote here the multiplication of the argument of a function $f\left(x\right)\to f\left(ax\right)$ by a factor $a$ in operator form as [14] ³

²The last cited tables [10] contain the so-called Laplace-Carson transforms $\bar{F}\left(p\right)=p{\displaystyle {\int }_0^{+\infty }}\text{d}p{\text{e}}^{-pt}f\left(t\right)$ and thus they are related with the standard Laplace transforms $F\left(p\right)$ simply by $F\left(p\right)\to \bar{F}\left(p\right)=pF\left(p\right)=pL\left\{f\right\}\left(p\right)$ .

³We mention that the operators $K_{-}\equiv \frac{1}{2}\frac{{\partial }^2}{\partial x^2}\mathrm{,}{K}_0\equiv \frac{1}{4}\left(x\frac{\partial }{\partial x}+\frac{\partial }{\partial x}x\right)=\frac{x}{2}\frac{\partial }{\partial x}+\frac{1}{4}\mathrm{,}{K}_{+}\equiv \frac{1}{2}{x}^2\mathrm{,}$ with the commutation relations $\left[K_{-}\mathrm{,}{K}_{+}\right]=2K_0\mathrm{,}\left[K_0\mathrm{,}{K}_{-}\right]=-K_{-}\mathrm{,}\left[K_0\mathrm{,}{K}_{+}\right]=+K_{+}\mathrm{,}$ form a basis of a realization of the Lie algebra $su\left(\mathrm{1,1}\right)$ to the Lie group $SU\left(\mathrm{1,1}\right)$ with useful applications to Hermite polynomials.

$f\left(ax\right)=a^{x\frac{\partial }{\partial x}}f\left(x\right)\mathrm{.}$ (1.11)

On the other side we may consider the function $f\left(xu\right)$ in (1.9) as multiplication of the argument x of the function $f\left(x\right)$ by a factor u and using then again (1.11) we can represent (1.9) in the form

$\begin{array}{c}\hat{f}\left(x\right)={\displaystyle {\int }_0^{+\infty }}\text{d}u{\text{e}}^{-u}{u}^{x\frac{\partial }{\partial x}}f\left(x\right)=\underset{\text{Mellin}\text{tr}\text{.}\text{of}{\text{e}}^{-u}}{\underset{︸}{{\displaystyle {\int }_0^{+\infty }}\text{d}u{\text{e}}^{-u}{u}^{\frac{\partial }{\partial x}x-1}}}f\left(x\right)\\ =\Gamma \left(\frac{\partial }{\partial x}x\right)f\left(x\right)\equiv \left(x\frac{\partial }{\partial x}\right)!f\left(x\right)\mathrm{.}\end{array}$ (1.12)

The operator which transforms $f\left(x\right)$ into $\hat{f}\left(x\right)$ and which has to be applied to $f\left(x\right)$ is here fully separated from $f\left(x\right)$ and represents the Mellin transformation [9] (chap. VI) of the function ${\text{e}}^{-u}$ with argument substituted by

the operator $\frac{\partial }{\partial x}x$ . The Mellin transform $\mathcal{M}\left\{f\right\}\left(s\right)$ of a function $f\left(u\right)$ is defined by

$\mathcal{M}\left\{f\right\}\left(s\right)\equiv {\displaystyle {\int }_0^{+\infty }}\text{d}uf\left(u\right)u^{s-1}\mathrm{,}$ (1.13)

and its special case of the function $f\left(u\right)={\text{e}}^{-u}$ leads to

$\mathcal{M}\left\{{\text{e}}^{-u}\right\}\left(s\right)={\displaystyle {\int }_0^{+\infty }}\text{d}u{\text{e}}^{-u}{u}^{s-1}=\Gamma \left(s\right)\equiv \left(s-1\right)\mathrm{<mo>\; !\; </mo>,}$ (1.14)

and may be considered as one of the possible basic definitions of the Gamma function. Extensive tables of Mellin transforms of functions we find in cited work of Bateman and Erdélyi [9] . Beside more insight into the nature of Sumudu transformations we hope that the representation (1.12) can be applied in future also more directly using the many known formulae for the Gamma function.

We now consider the inversion of the transformation

$f\left(x\right)\to \hat{f}\left(x\right)={\displaystyle {\int }_0^{+\infty }}\text{d}uf\left(xu\right){\text{e}}^{-u}$ . (1.15)

As it is known the inversion of the Laplace transformation (1.7) is (e.g., [8] [9])

$F\left(s\right)\to f\left(x\right)={\mathcal{L}}^{-1}\left\{F\right\}\left(x\right)=\frac{1}{\text{i}2\text{π}}{\displaystyle {\int }_{c-\text{i}\infty }^{c+\text{i}\infty }}\text{d}sF\left(s\right){\text{e}}^{xs}\mathrm{,}$ (1.16)

where real c is widely arbitrary but must be chosen in the region where the integral is convergent. For the Sumudu transformation (1.5) written in the form (1.6) as a Laplace transformation of $f\left(x\right)$ this leads to the inversion

$f\left(x\right)=\frac{1}{\text{i}2\text{π}}{\displaystyle {\int }_{c-\text{i}\infty }^{c+\text{i}\infty }}\text{d}s\frac{1}{s}\hat{f}\left(\frac{1}{s}\right){\text{e}}^{xs}\mathrm{.}$ (1.17)

This transformation allows to evaluate the Taylor series of a function $f\left(x\right)$ from the known function $\hat{f}\left(x\right)$ if the integral (1.17) can be evaluated.

As a first example which gives a deeper insight into the nature of the Sumudu transformation we consider the functions $f\left(x\right)=x^n$ . Using (1.5) together with (1.3) one calculates that these functions are transformed into functions $\hat{f}\left(x\right)$ according to

$f\left(x\right)=x^n\stackrel{\text{Sumudu}}{\to }\hat{f}\left(x\right)=n!x^n\mathrm{,}\left(n=\mathrm{0,1,2,}\cdots \right)\mathrm{,}$ (1.18)

that due to linearity of the transformation can be immediately applied to the Sumudu transformation of Taylor series. This correspondence possesses yet another aspect. For this purpose we mention that $x^n$ is eigenfunction of the

operator $\frac{\partial }{\partial x}x$ to eigenvalue $n+1$ and we conclude

$\left(\frac{\partial }{\partial x}x\right)x^n=\left(n+1\right)x^n\mathrm{,}\iff \Gamma \left(\frac{\partial }{\partial x}x\right)x^n=\Gamma \left(n+1\right)x^n=n!x^n,$ (1.19)

and from this (existence of Taylor series assumed)

$\Gamma \left(\frac{\partial }{\partial x}x\right)f\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{f^{\left(n\right)}\left(0\right)}{n\mathrm{<mo>\; !\; </mo>}}\Gamma \left(\frac{\partial }{\partial x}x\right)x^n={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{f}^{\left(n\right)}\left(0\right)x^n\equiv \hat{f}\left(x\right)\mathrm{.}$ (1.20)

The second part of (1.19) follows immediately from the representation (1.12) of the Sumudu transformation. However, this representation (1.12) alone is not very useful for summing up Taylor series as result of the transformation if we do not possess in addition integral representations of the form (1.5) which allow, at least in principle, to sum up the arising series by calculating an integral.

As generalization one may consider a transformation $F\left(x\right)\to \hat{F}\left(x\right)$ of the kind

$\hat{F}\left(x\right)=G\left(x\frac{\partial }{\partial x}\right)F\left(x\right),$ (1.21)

where $G\left(u\right)$ is an arbitrary given function and $F\left(x\right)$ the function to transform.

Since $x^n$ are eigenfunctions of the operator $x\frac{\partial }{\partial x}$ to eigenvalue n from (1.20) follows for $F\left(x\right)=x^n$

$\left(x\frac{\partial }{\partial x}\right)x^n=n{x}^n,\iff \hat{F}\left(x\right)=G\left(x\frac{\partial }{\partial x}\right)x^n=G\left(n\right)x^n,$ (1.22)

and consequently

$\hat{F}\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{f^{\left(n\right)}\left(0\right)}{n!}G\left(x\frac{\partial }{\partial x}\right)x^n={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{f}^{\left(n\right)}\left(0\right)\frac{G\left(n\right)}{n!}{x}^n.$ (1.23)

For the evaluation of arising new Taylor series after the transformation it is necessary to possess in addition to (1.21) an integral representation (or an equivalent of it) which allows to calculate more directly $\hat{F}\left(x\right)$ from given $F\left(x\right)$ . In particular, this is also true for the form (1.12) of the Sumudu transformation. In this case as mentioned we have it in the basic form (1.5) of the Sumudu transformation.

2. Some General Rules for Sumudu Transformation

In this Section we derive and discuss some general rules for Sumudu transformations according to (1.1) and (1.2) with their explicit form given in (1.5) and denote the correspondence between $f\left(x\right)$ and $\hat{f}\left(x\right)$ by

$f\left(x\right)\stackrel{\text{Sumudu}}{\to }\hat{f}\left(x\right)\mathrm{.}$ (2.1)

Then the correspondence of a function $f\left(a\mathrm{<mo>\; ;\; </mo>}x\right)\equiv f\left(ax\right)$ with stretching parameter a of the argument to its Sumudu transform according to

$f\left(a\mathrm{<mo>\; ;\; </mo>}x\right)\equiv f\left(ax\right)\stackrel{\text{Sumudu}}{\to }\hat{f}\left(a\mathrm{<mo>\; ;\; </mo>}x\right)=\hat{f}\left(ax\right)\mathrm{,}$ (2.2)

is determined by

$\begin{array}{c}\hat{f}\left(a\mathrm{<mo>\; ;\; </mo>}x\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\underset{=f\left(ay\right)}{\underset{︸}{f\left(a\mathrm{<mo>\; ;\; </mo>}y\right)}}\exp \left(-\frac{y}{x}\right)\\ =\frac{1}{ax}{\displaystyle {\int }_0^{+\infty }}\text{d}zf\mathrm{<mo\; stretchy="false">\; (\; </mo>}z\mathrm{<mo\; stretchy="false">\; )\; </mo>}\exp \left(-\frac{z}{ax}\right)=\hat{f}\left(ax\right)\mathrm{,}\end{array}$ (2.3)

where $\hat{f}\left(x\right)$ denotes the Sumudu transform of $f\left(x\right)$ according to (2.1). Thus the multiplication of the argument of a function by a number a leads to a simple relation of its Sumudu transform to the Sumudu transform of the primary function and it is not necessary to consider separately Sumudu transformations of functions $f\left(ax\right)$ . For the case of an argument translation of a function this is a little more complicated (see below).

To the differentiation $f^{\left(1\right)}\left(x\right)\equiv \frac{\partial }{\partial x}f\left(x\right)\equiv f_{\left(-1\right)}\left(x\right)$ of a function corresponds a function ${\hat{f}}_{\left(-1\right)}\left(x\right)$ using partial integration according to

$\begin{array}{c}{\hat{f}}_{\left(-1\right)}\left(x\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\left(\frac{\partial }{\partial y}f\left(y\right)\right)\exp \left(-\frac{y}{x}\right)\\ =\frac{1}{x}{\left\{f\left(y\right)\exp \left(-\frac{y}{x}\right)\right\}}_{y=0}^{y\to +\infty }-\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}yf\left(y\right)\left(\frac{\partial }{\partial y}\exp \left(-\frac{y}{x}\right)\right)\mathrm{,}\end{array}$ (2.4)

with the result of this correspondence

$f_{\left(-1\right)}\left(x\right)\equiv f^{\left(1\right)}\left(x\right)\stackrel{\text{Sumudu}}{\to }{\hat{f}}_{\left(-1\right)}\left(x\right)=\frac{1}{x}\left(\hat{f}\left(x\right)-f\left(0\right)\right),$ (2.5)

if the upper boundary ${\lim }_{y\to +\infty }f\left(y\right)\exp \left(-\frac{y}{x}\right)$ does not give a contribution. In

analogous way one finds after a few calculations for low numbers of n the general formula ( $n=1,2,\cdots$ )

$f_{\left(-n\right)}\left(x\right)\equiv f^{\left(n\right)}\left(x\right)\stackrel{\text{Sumudu}}{\to }{\hat{f}}_{\left(-n\right)}\left(x\right)=\frac{1}{x^n}\left(\hat{f}\left(x\right)-{\displaystyle \underset{k=0}{\overset{n-1}{\sum }}}{x}^k{f}^{\left(k\right)}\left(0\right)\right).$ (2.6)

It is not difficult to prove it by complete induction.

For the displacement of the argument of a function $f\left(x\right)$ by $x_0$ as a parameter

$f\left(x_0\mathrm{<mo>\; ;\; </mo>}x\right)\equiv f\left(x-x_0\right)=\exp \left(-x_0\frac{\partial }{\partial x}\right)f\left(x\right)\mathrm{.}$ (2.7)

we find with intermediate substitution $u=y-x_0$ of the integration variable for the correspondence to the Sumudu transform

$\begin{array}{l}f\left(x_0\mathrm{<mo>\; ;\; </mo>}x\right)\stackrel{\text{Sumudu}}{\to }\hat{f}\left(x_0\mathrm{<mo>\; ;\; </mo>}x\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}yf\left(y-x_0\right)\exp \left(-\frac{y}{x}\right)\\ =\exp \left(-\frac{x_0}{x}\right)\hat{f}\left(x\right)+\frac{1}{x}{\displaystyle {\int }_0^{x_0}}\text{d}yf\left(y-x_0\right)\exp \left(-\frac{y}{x}\right)\mathrm{.}\end{array}$ (2.8)

The second sum term is some kind of incomplete Sumudu transformation of the function $f\left(x-x_0\right)$ and it vanishes for $x_0=0$ . By the substitution $y-x_0=z$ this can be transformed to

$f\left(x_0\mathrm{<mo>\; ;\; </mo>}x\right)\stackrel{\text{Sumudu}}{\to }\hat{f}\left(x_0\mathrm{<mo>\; ;\; </mo>}x\right)=\exp \left(-\frac{x_0}{x}\right){\displaystyle {\int }_{-x_0}^{\infty }}\text{d}zf\left(z\right)\exp \left(-\frac{z}{x}\right)\mathrm{.}$ (2.9)

We mention that for even functions $f_{+}\left(x\right)=+f_{+}\left(-x\right)$ with existing Taylor series we find the correspondence

$\begin{array}{c}{\hat{f}}_{+}\left(x\right)={\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}{f}_{+}^{\left(2m\right)}\left(0\right)x^{2m}\\ =\frac{1}{2x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\left(f\left(y\right)+f\left(-y\right)\right)\exp \left(-\frac{y}{x}\right)\\ =\frac{1}{2\left|x\right|}{\displaystyle {\int }_{-\infty }^{+\infty }}\text{d}yf\left(y\right)\exp \left(-\frac{y}{\left|x\right|}\right)=+{\hat{f}}_{+}\left(-x\right)\mathrm{,}\end{array}$ (2.10)

and for odd functions $f_{-}\left(x\right)=-f_{-}\left(-x\right)$ the correspondence

$\begin{array}{c}{\hat{f}}_{-}\left(x\right)={\displaystyle \underset{m=0}{\overset{\infty }{\sum }}}{f}_{-}^{\left(2m+1\right)}\left(0\right)x^{2m+1}\\ =\frac{1}{2x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\left(f\left(y\right)-f\left(-y\right)\right)\exp \left(-\frac{y}{x}\right)\\ =\frac{1}{2\left|x\right|}{\displaystyle {\int }_{-\infty }^{+\infty }}\text{d}y\text{sign}\left(y\right)f\left(y\right)\exp \left(-\frac{y}{\left|x\right|}\right)=-{\hat{f}}_{-}\left(-x\right)\mathrm{.}\end{array}$ (2.11)

Sumudu transformation preserves the lateral symmetry of a function.

3. Fractional Integration and Sumudu Transformation as a Certain Limiting Case

We now consider fractional integration and differentiation (e.g., Bateman and Erdélyi [15], Vladimirov [16], Oldham and Spanier [17], Samko, Kilbas, Marichev [18]) which is often useful in connection with Sumudu transformation. Furthermore, Sumudu transformation can be represented as a certain limiting case of fractional integration as we will show in this Section. First, we shortly introduce fractional integration.

It is known (in an elementary form, e.g., Kranzer [19], p. 205) that the n-fold integration of a function $f\left(x\right)$ can be represented by a convolution integral as follows

$\begin{array}{c}{f}_{\left(n\right)}\left(x\right)\equiv {\displaystyle {\int }_0^x}\text{d}{x}_{n-1}{\displaystyle {\int }_0^{x_{n-1}}}\text{d}{x}_{n-2}\dots {\displaystyle {\int }_0^{x_2}}\text{d}{x}_1{\displaystyle {\int }_0^{x_1}}\text{d}{x}_{0}f\left(x_0\right)\\ =\frac{1}{\left(n-1\right)!}{\displaystyle {\int }_0^x}\text{d}y{\left(x-y\right)}^{n-1}f\left(y\right),\left(n=1,2,\cdots \right)\end{array}$ . (3.1)

This can be proved by complete induction first changing the order of integration in the arising double integral and then calculating the inner integral

$f_{\left(n+1\right)}\left(x\right)={\displaystyle {\int }_0^x}\text{d}y{f}_{\left(n\right)}\left(y\right)={\displaystyle {\int }_0^x}\text{d}y\frac{1}{\left(n-1\right)!}{\displaystyle {\int }_0^y}\text{d}z{\left(y-z\right)}^{n-1}f(z)$

$\begin{array}{c}={\displaystyle {\int }_0^x}\text{d}zf\left(z\right){\displaystyle {\int }_z^x}\text{d}y\frac{{\left(y-z\right)}^{n-1}}{\left(n-1\right)!}={\displaystyle {\int }_0^z}\text{d}zf\left(z\right){\left\{\frac{{\left(y-z\right)}^n}{n!}\right\}}_{y=z}^{y=x}\\ =\frac{1}{n!}{\displaystyle {\int }_0^x}\text{d}z{\left(x-z\right)}^{n}f\left(z\right),\end{array}$ (3.2)

that together with the correct $f_{\left(1\right)}\left(x\right)$ as initial member proves the proposition.

The n-fold integral (3.1) can be generalized in the rank of a definition to a ν-fold fractional integral with arbitrary real parameter $\nu$ by the extension $n\to \nu$ in (3.1) (positive real solution is to insert for the multi-valued function $z^{\nu -1}$ )

$f_{\left(\nu \right)}\left(x\right)\equiv \frac{1}{\left(\nu -1\right)\mathrm{<mo>\; !\; </mo>}}{\displaystyle {\int }_0^x}\text{d}y{\left(x-y\right)}^{\nu -1}f\left(y\right)=\frac{1}{\left(\nu -1\right)\mathrm{<mo>\; !\; </mo>}}{\displaystyle {\int }_0^x}\text{d}z{z}^{\nu -1}f\left(x-z\right)\mathrm{.}$ (3.3)

This form is also called the Riemann-Liouville integral [15] . The case $\nu =0$ as limiting case provides

$\begin{array}{c}{f}_{\left(0\right)}\left(x\right)\equiv \underset{\epsilon \to +0}{\lim }{f}_{\left(\epsilon \right)}\left(x\right)=\underset{\epsilon \to 0}{\lim }\frac{1}{\left(\epsilon -1\right)!}{\displaystyle {\int }_0^x}\text{d}z{z}^{\epsilon -1}f\left(x-z\right)\\ =\underset{\epsilon \to 0}{\lim }\frac{\epsilon }{\epsilon !}{\displaystyle {\int }_0^x}\text{d}z{z}^{\epsilon -1}\left(f\left(x\right)-\frac{z}{1!}{f}^{\left(1\right)}\left(x\right)+\cdots \right)\\ =f\left(x\right)\underset{\epsilon \to 0}{\lim }\frac{x^{\epsilon }-0^{\epsilon }}{\epsilon !}-\frac{f^{\left(1\right)}\left(x\right)}{1!}\underset{\epsilon \to 0}{\lim }\epsilon \frac{z^{\epsilon +1}-0^{\epsilon +1}}{\left(\epsilon +1\right)!}+\cdots =f\left(x\right).\end{array}$ (3.4)

The convolution (3.3) of $f\left(x\right)$ with powers of x is one of the possible basic definitions of fractional integration of order $\nu >0$ of a function $f\left(x\right)$ , e.g., [15] [16] [17] [18] (the last two [17] and [18] with a short historical introduction). In [15] (chap. XIII) and in [18] (pp. 140-142) one finds tables of fractional integrals. Fractional derivatives ( $\nu <0$ ) can be obtained from appropriate fractional integrals ( $\nu \ge 0$ ) by corresponding integer differentiations. Interesting is the view from the theory of the convolution algebra of generalized functions where one can introduce the functions ${\theta }_{\left(\nu \right)}\left(x\right)$ which generalize the step function [16] (p. 142; he uses $f_{\nu }$ instead of our ${\delta }_{\nu }$ ) by

${\delta }_{\nu }\left(x\right)=\{\begin{array}{ll}\theta \left(x\right)\frac{x^{\nu -1}}{\left(\nu -1\right)!},\hfill & \left(\nu >0\right)\hfill \\ {\delta }_{\nu +1}^{\left(1\right)},\hfill & \left(\nu \le 0\right)\hfill \end{array}\Rightarrow {\delta }_{\mu }\left(x\right)\ast {\delta }_{\nu }\left(x\right)={\delta }_{\mu +\nu }\left(x\right),$

${\delta }_0\left(x\right)=\delta \left(x\right),\delta \left(x\right)=\theta \left(x\right),{\delta }_{-n}\left(x\right)={\delta }^{\left(n\right)}\left(x\right),$ (3.5)

and where “ $\ast$ ” stands for the convolution $f\left(x\right)\ast g\left(x\right)\equiv {\displaystyle {\int }_{-\infty }^{+\infty }}\text{d}yf\left(x-y\right)g\left(y\right)$ . Fractional integration of functions $f_0\left(x\right)=f\left(x\right)\theta \left(x\right)$ can be then determined by

$f_{\nu }\left(x\right)={\delta }_{\nu }\left(x\right)\ast f\left(x\right),$ (3.6)

with the consequence

${\delta }_{\mu }\left(x\right)\ast f_{\nu }\left(x\right)={\delta }_{\mu }\left(x\right)\ast {\delta }_{\nu }\left(x\right)\ast f\left(x\right)={\delta }_{\mu +\nu }\left(x\right)\ast f\left(x\right)=f_{\mu +\nu }\left(x\right),$ (3.7)

that means associativity as it has to be for successive convolutions using a notation of the form of multiplications.

Sumudu transformation and fractional integration of order $\nu$ do not commute. If we make first the fractional integration of order $\nu$ of the powers $x^n,\left(n=0,1,2,\cdots \right)$ with symbol $\stackrel{\text{Fract}\left(\nu \right)}{\to }$ and then with the result the Sumudu transformation we obtain

$x^n\stackrel{\text{Fract}\left(\nu \right)}{\to }\frac{n!}{\left(n+\nu \right)!}{x}^{n+\nu }\stackrel{\text{Sumudu}}{\to }n!x^{n+\nu }\mathrm{,}$ (3.8)

and if we make first a Sumudu transformation of the powers $x^n$ and then a fractional integration of order $\nu$ of the result we find

$x^n\stackrel{\text{Sumudu}}{\to }n!x^n\stackrel{\text{Fract}\left(\nu \right)}{\to }\frac{n{!}^2}{\left(n+\nu \right)!}{x}^{n+\nu }.$ (3.9)

In comparison, two successive fractional integrations of order $\mu$ and $\nu$ are commuting and result in a fractional integration of order $\mu +\nu$ and, therefore, form a one-parameter group with the identity operation belonging to $\nu =0$ .

By comparison of the Laplace transforms in the tables of Bateman and Erdélyi [9] (pp. 120-204, here, in particular, p. 121, (8) and (9)) with the Sumudu transforms (2.6) one finds that essentially the variable of the Laplace transform is substituted by its inverse in the Sumudu transform. In case of negative values

$\nu =-\left|\nu \right|$ we have in front of $\hat{f}\left(x\right)$ the powers $\frac{1}{{\left|x\right|}^{\nu }}$ which are singular

functions of x at $x=0$ where the singularities in the sense of generalized functions have to be determined more precisely.

We now consider a certain limiting transition from fractional integration to Sumudu transformation and write for this purpose (3.2) first in the way

$f_{\left(n+1\right)}\left(z\right)=\frac{1}{n\mathrm{<mo>\; !\; </mo>}}{\displaystyle {\int }_0^z}\text{d}y{\left(z-y\right)}^{n}f\left(y\right)=\frac{z^n}{n\mathrm{<mo>\; !\; </mo>}}{\displaystyle {\int }_0^z}\text{d}y{\left(1-\frac{y}{z}\right)}^{n}f\left(y\right)\mathrm{,}$ (3.10)

and substitute $z=nx$ with $x>0$ that leads to the following relation

$\frac{n!}{n^n{x}^{n+1}}{f}_{\left(n+1\right)}\left(nx\right)=\frac{1}{x}{\displaystyle {\int }_0^{nx}}\text{d}y{\left(1-\frac{y}{nx}\right)}^{n}f\left(y\right).$ (3.11)

If we now make the limiting transition $n\to +\infty$ using the definition of the exponential functions ${\text{e}}^z$ by the limiting process

$\underset{n\to \infty }{\lim }{\left(1+\frac{z}{n}\right)}^n={\text{e}}^z\mathrm{,}$ (3.12)

and known already to Euler we obtain from the right-hand side of (3.11) and in agreement with (1.5)

$\underset{n\to \infty }{\lim }\frac{1}{x}{\displaystyle {\int }_0^{nx}}\text{d}y{\left(1-\frac{y}{nx}\right)}^{n}f\left(y\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\exp \left(-\frac{y}{x}\right)f\left(y\right)\equiv \hat{f}\left(x\right)\mathrm{.}$ (3.13)

From the left-hand side of (3.11) in the limiting transition $n\to \infty$ using the well-known Stirling approximation of the Gamma function for $z\to +\infty$

$\Gamma \left(z\right)=\sqrt{2\text{π}}{\text{e}}^{-z}{z}^{z-\frac{1}{2}}\left\{1+\frac{1}{12z}+\cdots \right\}=\left(z-1\right)\mathrm{<mo>\; !\; </mo>,}$ (3.14)

inserting this for $\left(n-1\right)\mathrm{<mo>\; !\; </mo>}$ we find

$\underset{n\to \infty }{\lim }\frac{n!}{n^n{x}^{n+1}}{f}_{\left(n+1\right)}\left(nx\right)=\underset{n\to \infty }{\lim }\sqrt{\frac{2\text{π}n}{x^2}}\frac{f_{\left(n+1\right)}\left(nx\right)}{{\left(\text{e}x\right)}^n}\mathrm{,}\left(x=\left|x\right|>0\right)\mathrm{.}$ (3.15)

Thus we may write the Sumudu transformation $\hat{f}\left(x\right)$ as the following limit of the fractional integration if we substitute in addition $n\to \nu \in ℝ$

$\hat{f}\left(x\right)=\sqrt{\frac{2\text{π}}{x^2}}\underset{\nu \to \infty }{\lim }\frac{\sqrt{\nu }{f}_{\left(\nu +1\right)}\left(\nu x\right)}{{\left(\text{e}x\right)}^{\nu }}\mathrm{,}\left(x=\left|x\right|>0\right)\mathrm{.}$ (3.16)

The right-hand side of this relation is one of the ways to represent the Sumudu transformation by a limiting procedure from fractional integration.

4. Examples for Sumudu Transformations

We provide now a few simple examples for Sumudu transformations. The transformation of the “basic” monomial functions $f\left(x\right)=x^n$ was already considered in (1.18) and (1.19) and leads to $\hat{f}\left(x\right)=n!x^n$ . It illustrates the action of

the operator $\Gamma \left(\frac{\partial }{\partial x}x\right)$ in (1.12) in ideal way.

The most striking genuine example for a Sumudu transformation establishes the connection between the Exponential and the Geometric series, clearly, with a well-known result. One finds [2]

$f\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^n}{n!}={\text{e}}^{-x},F\left(t\right)=\frac{1}{1+t},$

$\hat{f}\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}{\left(-x\right)}^n=\frac{1}{x}F\left(\frac{1}{x}\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y{\text{e}}^{-y}\exp \left(-\frac{y}{x}\right)=\frac{1}{1+x}.$ (4.1)

For this example inserting $\hat{f}\left(x\right)=\frac{1}{1+x}$ on the right-hand side of the formula

(1.17) it is easy to demonstrate the inverse Sumudu transformation from the Geometric to the Exponential series according to

$f\left(x\right)=\frac{1}{\text{i}2\text{π}}{\displaystyle {\int }_{c-\text{i}\infty }^{c+\text{i}\infty }}\text{d}s\frac{1}{s}\frac{1}{1+\frac{1}{s}}{\text{e}}^{xs}=\frac{1}{\text{i}2\text{π}}{\displaystyle {\int }_C}\text{d}s\frac{{\text{e}}^{xs}}{s+1}={\text{e}}^{-x}\mathrm{.}$ (4.2)

Here we applied the residue theorem for the Cauchy integral over a closed contour C in the complex s-plane including the only singularity of the integrand at $s=-1$ . The contour C consists for parameter $x>0$ of the line through $s=c$ parallel to the imaginary axis closed by a semicircle $s=c+r{\text{e}}^{\text{i}\phi }$ with

$\frac{\text{π}}{2}\le \phi \le \frac{\text{3π}}{2}$ or $\cos \left(\phi \right)\le 0$ (i.e. to the left of the line) and followed by $r\to \infty$

where the integral over the semicircle then vanishes. The possible real values c in the integral are then only restricted by $c>-1$ for the purpose to include the singularity.

A further interesting simple example establishes a connection between the Bessel function ${\text{J}}_0\left(2\sqrt{x}\right)$ and the Exponential function ${\text{e}}^{-x}$ in the following way [2]

$f\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^n}{n{!}^2}={\text{J}}_0\left(2\sqrt{x}\right)\mathrm{,}F\left(t\right)=\frac{1}{t}\exp \left(-\frac{1}{t}\right)\mathrm{,}$

$\hat{f}\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^n}{n\mathrm{<mo>\; !\; </mo>}}=\frac{1}{x}F\left(\frac{1}{x}\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y{\text{J}}_0\left(2\sqrt{y}\right)\exp \left(-\frac{y}{x}\right)={\text{e}}^{-x}\mathrm{.}$ (4.3)

In this example it is merely the integral transform from the Bessel function ${\text{J}}_0\left(2\sqrt{x}\right)$ to the exponential function ${\text{e}}^{-x}$ which is less known (but not unknown) in comparison to the other results. Clearly, by substitution $x\to \text{i}x$ one may obtain similar formulae for the Bessel function ${\text{I}}_0\left(2\sqrt{x}\right)$ .

Related to (4.1) one finds

$f\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^n}{n!\left(n+1\right)}=\frac{1}{x}{\displaystyle {\int }_0^x}\text{d}t{\text{e}}^{-t}=\frac{1-{\text{e}}^{-x}}{x},$

$\hat{f}\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^n}{n+1}=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\frac{1-{\text{e}}^{-y}}{y}\exp \left(-\frac{y}{x}\right)=\frac{\log \left(1+x\right)}{x},$ (4.4)

and related to (4.3)

$f\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^n}{n\mathrm{<mo>\; !\; </mo>}\left(n+\mu \right)\mathrm{<mo>\; !\; </mo>}}=\frac{{\text{J}}_{\mu }\left(2\sqrt{x}\right)}{{\left(\sqrt{x}\right)}^{\mu }}\mathrm{,}$

$\hat{f}\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^n}{\left(n+\mu \right)\mathrm{<mo>\; !\; </mo>}}=\frac{1}{x}F\left(\frac{1}{x}\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\frac{{\text{J}}_{\mu }\left(2\sqrt{y}\right)}{{\left(\sqrt{y}\right)}^{\mu }}\exp \left(-\frac{y}{x}\right)\mathrm{.}$ (4.5)

For integer $\mu =m=\left(0,1,2,\cdots \right)$ one finds from this relation

$\begin{array}{l}\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\frac{{\text{J}}_m\left(2\sqrt{y}\right)}{{\left(\sqrt{y}\right)}^m}\exp \left(-\frac{y}{x}\right)\\ ={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^n}{\left(n+m\right)\mathrm{<mo>\; !\; </mo>}}={\displaystyle \underset{k=m}{\overset{\infty }{\sum }}}\frac{{\left(-x\right)}^{k-m}}{k\mathrm{<mo>\; !\; </mo>}}=\frac{{\left(-1\right)}^m}{x^m}\left(\exp \left(-x\right)-{\displaystyle \underset{k=0}{\overset{m-1}{\sum }}}\frac{{\left(-x\right)}^k}{k\mathrm{<mo>\; !\; </mo>}}\right)\mathrm{.}\end{array}$ (4.6)

It is here again the integral over modified Bessel functions multiplied with an exponential function which is less known than the involved Taylor series. We mention that the n-th derivative of the Bessel function ${\text{J}}_0\left(2\sqrt{x}\right)$ leads to the

entire function ${\left(-1\right)}^n\frac{{\text{J}}_n\left(2\sqrt{x}\right)}{{\left(\sqrt{x}\right)}^n}$ according to the more general formulae

$\frac{{\partial }^n}{\partial u^n}\frac{{\text{J}}_{\mu }\left(2\sqrt{u}\right)}{{\left(\sqrt{u}\right)}^{\mu }}={\left(-1\right)}^n\frac{{\text{J}}_{\mu +n}\left(2\sqrt{u}\right)}{{\left(\sqrt{u}\right)}^{\mu +n}}\mathrm{,}\frac{{\partial }^n}{\partial u^n}\frac{{\text{I}}_{\mu }\left(2\sqrt{u}\right)}{{\left(\sqrt{u}\right)}^{\mu }}=\frac{{\text{I}}_{\mu +n}\left(2\sqrt{u}\right)}{{\left(\sqrt{u}\right)}^{\mu +n}}\mathrm{.}$ (4.7)

This can be generalized by fractional differentiation from integer n to arbitrary positive real $\nu$ . The corresponding more general formulae for fractional integration of order $\nu$ are [15] (chap. 13.1. formula (63) in the table) or [18] (p. 142, Eq. 9)

$\frac{1}{\left(\nu -1\right)\mathrm{<mo>\; !\; </mo>}}{\displaystyle {\int }_0^x}\text{d}y{\left(x-y\right)}^{\nu -1}\left\{\begin{array}{c}{\left(\sqrt{y}\right)}^{\mu }{\text{J}}_{\mu }\left(2\sqrt{y}\right)\\ {\left(\sqrt{y}\right)}^{\mu }{\text{I}}_{\mu }\left(2\sqrt{y}\right)\end{array}\right\}=\left\{\begin{array}{c}{\left(\sqrt{x}\right)}^{\mu +\nu }{\text{J}}_{\mu +\nu }\left(2\sqrt{x}\right)\\ {\left(\sqrt{x}\right)}^{\mu +\nu }{\text{I}}_{\mu +\nu }\left(2\sqrt{x}\right)\end{array}\right\}\mathrm{.}$ (4.8)

The integrals (1.3) cannot be extended from positive to negative n since $\left(-n\right)!\to \infty ,\left(n=1,2,\cdots \right)$ . Nevertheless, the Sumudu transform can be calculated for some functions without a Taylor series at $x=0$ and it is interesting to see in an example what happens. For this purpose we calculate the Sumudu transform of the function $f\left(x\right)=\theta \left(x-x_0\right)$ where $\theta \left(x\right)$ is the Heaviside step function $\left(\theta \left(x\right)=0,\left(x<0\right),\theta \left(x\right)=1,\left(x>0\right)\right)$ . We find

$f\left(x\right)=\theta \left(x-x_0\right),\left(x_0>0\right),$

$\begin{array}{c}\hat{f}\left(x\right)=\frac{1}{x}{\displaystyle {\int }_0^{+\infty }}\text{d}y\theta \left(y-x_0\right)\exp \left(-\frac{y}{x}\right)=\frac{1}{x}{\displaystyle {\int }_{x_0}^{+\infty }}\text{d}y\exp \left(-\frac{y}{x}\right)\\ =\exp \left(-\frac{x_0}{x}\right)={\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^n}{n!}{\left(\frac{x_0}{x}\right)}^n+1.\end{array}$ (4.9)

The Sumudu transform $\hat{f}\left(x\right)$ is singular at $x=0$ with the Laurent series written in (4.9). If we formally write the inverse Sumudu transformation $f\left(x\right)$ to $\hat{f}(x)$

$f\left(x\right)={\displaystyle \underset{n=1}{\overset{\infty }{\sum }}}\frac{{\left(-1\right)}^n}{n!\left(-n\right)!}{\left(\frac{x_0}{x}\right)}^n+\frac{1}{0!},$ (4.10)

adding for negative powers $n<0$ of x in the denominator the factor $\left(-n\right)\mathrm{<mo>\; !\; </mo>}$

then for $n=\pm 1,\pm 2,\cdots$ we have $\frac{1}{n!\left(-n\right)!}=0$ and the series (4.10) becomes undetermined in the neighborhood of $x=0$ due to the singularities.

The Sumudu transformation is an interesting addition to other methods of the analysis of functions but often one possesses alternative possibilities to evaluate infinite series.

5. A modified Sumudu or a Kind of Gauss Transformation

In analogy to the integral (1.3) we now consider the sequence of integrals for non-negative integers $n=0,1,\cdots$ with real positive parameter x (identity $\sqrt{\text{π}}=\left(-\frac{1}{2}\right)!$ )

${\displaystyle {\int }_0^{+\infty }}\text{d}y{y}^n\exp \left(-\frac{y^2}{x^2}\right)=\frac{1}{2}\left(\frac{n-1}{2}\right)\mathrm{<mo>\; !\; </mo>}{x}^{n+1}=\frac{\sqrt{\text{π}}n\mathrm{<mo>\; !\; </mo>}}{\left(\frac{n}{2}\right)\mathrm{<mo>\; !\; </mo>}}{\left(\frac{x}{2}\right)}^{n+1}\mathrm{,}\left(x=\left|x\right|\right)\mathrm{,}$ (5.1)

which after simple substitutions is related to (1.3) and extends it practically to semi-integers n. We used in (5.1) the well-known formula for doubling of the argument of the Gamma function $\Gamma (z)$

$\Gamma \left(2z+1\right)=\frac{2^{2z}}{\sqrt{\text{π}}}\Gamma \left(z+\frac{1}{2}\right)\Gamma \left(z+1\right)=\frac{2^{2z}}{\sqrt{\text{π}}}\left(z-\frac{1}{2}\right)\mathrm{<mo>\; !\; </mo>}z\mathrm{<mo>\; !\; </mo>}=\left(2z\right)\mathrm{<mo>\; !\; </mo>,}$ (5.2)

with $z=\frac{n}{2}$ . Two special cases of the sequence of integrals (5.1), first, setting $n=2m$ leads to

${\displaystyle {\int }_0^{+\infty }}\text{d}y{y}^{2m}\exp \left(-\frac{y^2}{x^2}\right)=\frac{1}{2}\left(m-\frac{1}{2}\right)!x^{2m+1}=\sqrt{\text{π}}\frac{\left(2m\right)\mathrm{<mo>\; !\; </mo>}}{m\mathrm{<mo>\; !\; </mo>}}{\left(\frac{x}{2}\right)}^{2m+1}\mathrm{,}$ (5.3)

with $x\equiv \sqrt{x^2}>0$ for real $x^2$ and, second, setting $n=2m+1$ leads to

${\displaystyle {\int }_0^{+\infty }}\text{d}y{y}^{2m+1}\exp \left(-\frac{y^2}{x^2}\right)=\frac{1}{2}m\mathrm{<mo>\; !\; </mo>}{x}^{2m+2}=\sqrt{\text{π}}\frac{\left(2m+1\right)\mathrm{<mo>\; !\; </mo>}}{\left(m+\frac{1}{2}\right)\mathrm{<mo>\; !\; </mo>}}{\left(\frac{x}{2}\right)}^{2m+2}\mathrm{.}$ (5.4)

The integral (5.1) can be used to transform a function $f\left(x\right)$ with the Taylor series

$f\left(x\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{f^{\left(n\right)}\left(0\right)}{n!}{x}^n,$ (5.5)

in analogy to (1.5) into a function ${\hat{f}}^{\frac{1}{2}}\left(x\right)$ in the rank of its definition by the Taylor series as follows

$\begin{array}{c}{\hat{f}}^{\frac{1}{2}}\left(x\right)\equiv {\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{f^{\left(n\right)}\left(0\right)}{\left(\frac{n}{2}\right)!}{\left(\frac{x}{2}\right)}^n=\frac{1}{\sqrt{\text{π}}}{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{\left(\frac{n-1}{2}\right)!}{n!}{f}^{\left(n\right)}\left(0\right)x^n\\ =\frac{2}{\sqrt{\text{π}{x}^2}}{\displaystyle {\int }_0^{+\infty }}\text{d}y\left({\displaystyle \underset{n=0}{\overset{\infty }{\sum }}}\frac{f^{\left(n\right)}\left(0\right)}{n!}{y}^n\right)\exp \left(-\frac{y^2}{x^2}\right),\left(x>0\right).\end{array}$ (5.6)

This transformation can be compactly written in the integral form

${\hat{f}}^{\frac{1}{2}}\left(x\right)\equiv \frac{2}{\sqrt{\text{π}{x}^2}}{\displaystyle {\int }_0^{+\infty }}\text{d}yf\left(y\right)\exp \left(-\frac{y^2}{x^2}\right)\mathrm{,}\left(x>0\right)\mathrm{.}$ (5.7)