Some new semi-exponential operators

Abstract

In the theory of approximation, linear operators play an important role. The exponential-type operators were introduced four decades ago, since then no new exponential-type operator was introduced by researchers, although several generalizations of existing exponential-type operators were proposed and studied. Very recently, the concept of semi-exponential operators was introduced and few semi-exponential operators were captured from the exponential-type operators. It is more difficult to obtain semi-exponential operators than the corresponding exponential-type operators. In this paper, we extend the studies and define semi-exponential Bernstein, semi-exponential Baskakov operators, semi-exponential Ismail–May operators related to $2x^{3/2}$ or $x^{3}$. Furthermore, we present a new derivation for the semi-exponential Post–Widder operators. In some examples, open problems are indicated.

1 Introduction

The exponential-type operators are important in the field of approximation theory. They were firstly considered by Ismail and May [4] in 1978. The exponential-type operators preserve the linear functions. Many generalizations of exponential-type operators are available in the literature. Tyliba and Wachnicki [7] extended the definition of Ismail and May [4] by proposing a more general family of operators. For a non-negative real number $\beta$, they introduced the operators $L_{\lambda }^{\beta }$. For $\beta >0$, they are not of exponential type but similar to exponential-type operators. Recently, Herzog [3] further extended the studies and termed such operators as semi-exponential type operators. Actually, an operator of the form

$$\begin{aligned} (L_{\lambda }f)(x)=\int _{I}W_{\beta }^{L}\left( \lambda ,x,t\right) f\left( t\right) dt \end{aligned}$$

is called a semi-exponential operator if its kernel $W_{\beta }^{L}\left( \lambda ,x,t\right)$ satisfies the differential equation

$$\begin{aligned} \frac{\partial }{\partial x}W_{\beta }^{L}\left( \lambda ,x,t\right) =\left( \frac{\lambda (t-x)}{p(x)}-\beta \right) W_{\beta }^{L}\left( \lambda ,x,t\right) . \end{aligned}$$

(1)

In particular, for $\beta >0$, one has $L_{\lambda }^{\beta }e_{1}\ne e_{1}$, where $e_{r}\left( t\right) =t^{r}$ $\left( r=0,1,2,\ldots \right)$. In the case $\beta =0$, the operator $L_{\lambda }^{\beta =0}$ is simply the exponential-type operator studied by Ismail and May [4]. A collection of such operators may be found in the recent book [2,  Ch. 1].

Choosing different functions $p\left( x\right)$ several exponential-type operators were captured in Ismail and May [4]. It is difficult to construct new exponential-type operators or the corresponding semi-exponential operators by just taking different functions $p\left( x\right)$. The essential obstacle is to fulfill the normalization condition

$$\begin{aligned} \int _{I}W_{\beta }^{L}\left( \lambda ,x,t\right) dt=1, \end{aligned}$$

which means that $L_{\lambda }^{\beta }$ preserves constant functions. Tyliba and Wachnicki [7] captured the semi-exponential operators of Weierstrass and Szász–Mirakyan operators, Herzog [3] got success to define the semi-exponential Post–Widder operators. We represent below the tabular form of known semi-exponential type operators available till date:

No.	Exponential operator	$p\left( x\right)$
(1)	Gauss–Weierstrass operators $(W_{n}f)\left( x\right)$	1
–	$(W_{n}f)\left( x\right) =\sqrt{\frac{n}{2\pi }}\int _{-\infty }^{\infty }\exp \left( \frac{-n(t-x)^{2}}{2}\right) f\left( t\right) dt$	Exponential
–	$(W_{n}^{\beta }f)\left( x\right) =\sqrt{\frac{n}{2\pi }}\int _{-\infty }^{\infty }\exp \left( \frac{-n(t-x-\beta /n)^{2}}{2}\right) f\left( t\right) dt$	Semi-exponential
(2)	Post–Widder operators $(P_{n}f)\left( x\right)$	$x^{2}$
–	$(P_{n}f)\left( x\right) =\frac{n^{n}}{\varGamma (n)x^{n}}\int _{0}^{\infty }e^{-nt/x}t^{n-1}f\left( t\right) dt$	Exponential
–	$(P_{n}^{\beta }f)\left( x\right) =\frac{n^{n}}{x^{n}e^{\beta x}} \sum _{k=0}^{\infty }\frac{\left( n\beta \right) ^{k}}{k!\varGamma \left( n+k\right) }\int _{0}^{\infty }e^{-nt/x}t^{n+k-1}f\left( t\right) dt$	Semi-exponential
(3)	Szász–Mirakyan operators $(S_{n}f)\left( x\right)$	x
–	$(S_{n}f)\left( x\right) =\sum _{k=0}^{\infty }e^{-nx}\frac{\left( nx\right) ^{k}}{k!}f\left( \frac{k}{n}\right)$	Exponential
–	$(S_{n}^{\beta }f)\left( x\right) =\sum _{k=0}^{\infty }e^{-(n+\beta )x} \frac{\left( \left( n+\beta \right) x\right) ^{k}}{k!}f\left( \frac{k}{n} \right)$	Semi-exponential

As pointed out earlier, one can obtain the exponential-type operator as the special $\beta =0$ from semi-exponential operators, but the converse is not analogous. Here we capture some more semi-exponential operators viz. semi-exponential Bernstein polynomials, semi-exponential Baskakov operators, etc.

2 New semi-exponential operators

In this section, we establish some new exponential-type operators. In all listed cases it is possible to solve the differential equation (1) in the form $W_{\beta }^{L}\left( n,x,t\right) =A_{L}\left( n,t,\beta \right) y$, but it is difficult to find the normalization, i.e., the factor $A_{L}\left( n,t,\beta \right)$ of the solution y such that

$$\begin{aligned} \int _{I}W_{\beta }^{L}\left( n,x,t\right) dt=1 \end{aligned}$$

or, in the discrete case,

$$\begin{aligned} \sum _{k=0}^{\infty }W_{\beta }^{L}\left( n,x,k/n\right) =1, \end{aligned}$$

respectively. Below we list some instances of $p\left( x\right)$, which were considered for well-known exponential-type operators.

2.1 Semi-exponential Bernstein operators

If we take $p\left( x\right) =x\left( 1-x\right)$, then for a kernel $W_{\beta }^{B}\left( n,x,k/n\right) =A_{B}\left( n,k,\beta \right) y$, we have

$$\begin{aligned} y^{\prime }=\frac{k-nx}{x(1-x)}y-\beta y, \end{aligned}$$

where the derivative of y is with respect to the variable x. We conclude that

$$\begin{aligned} \frac{y^{\prime }}{y}= & {} k\left( \frac{1}{1-x}+\frac{1}{x}\right) -\frac{n}{ 1-x}-\beta , \\ \log y= & {} \log (1-x)^{n-k}+\log x^{k}-\beta x, \end{aligned}$$

implying

$$\begin{aligned} y=x^{k}\left( 1-x\right) ^{n-k}e^{-\beta x}. \end{aligned}$$

In order to have normalization

$$\begin{aligned} \sum _{k=0}^{\infty }W_{\beta }^{B}\left( n,x,k/n\right) =\sum _{k=0}^{\infty }A_{B}\left( n,k,\beta \right) x^{k}(1-x)^{n-k}e^{-\beta x}=1. \end{aligned}$$

we evaluate $A_{B}\left( n,k,\beta \right)$ from the equation

$$\begin{aligned} \sum _{k=0}^{\infty }A_{B}\left( n,k,\beta \right) \left( \frac{x}{1-x} \right) ^{k}=e^{\beta x}\left( 1-x\right) ^{-n}. \end{aligned}$$

For $0\le x<1$, put $z=x/\left( 1-x\right)$. Then $x=z/\left( 1+z\right)$, and for any positive integer n, the generating function of the sequence $\left( A_{B}\left( n,k,\beta \right) \right) _{k=0}^{\infty }$

$$\begin{aligned} \sum _{k=0}^{\infty }A_{B}\left( n,k,\beta \right) z^{k}=e^{\beta z/\left( 1+z\right) }\left( 1+z\right) ^{n} \end{aligned}$$

is analytic, for $\left| z\right| <1$, with an essential singularity at $z=-1$. Hence, it can be developed as a power series in the disk $\left| z\right| <1$. The series

$$\begin{aligned} e^{\beta z/\left( 1+z\right) }\left( 1+z\right) ^{n} = \sum _{j=0}^{\infty }\frac{\left( \beta z\right) ^{j}}{j!}\left( 1+z\right) ^{n-j} \end{aligned}$$

is convergent for all complex z different from $-1$. It follows that, for $\left| z\right| <1$,

$$\begin{aligned} e^{\beta z/\left( 1+z\right) }\left( 1+z\right) ^{n} =\sum _{j=0}^{\infty }\frac{\left( \beta z\right) ^{j}}{j!}\sum _{\ell =0}^{\infty }{\left( {\begin{array}{c}n-j\\ \ell \end{array}}\right) }z^{\ell }=\sum _{k=0}^{\infty }z^{k}\sum _{j+\ell =k}{\left( {\begin{array}{c}n-j\\ \ell \end{array}}\right) }\frac{\beta ^{j}}{j!}, \end{aligned}$$

where the binomial coefficient is to be read as ${\left( {\begin{array}{c}n-j\\ 0\end{array}}\right) =1}$ and $\left( {\begin{array}{c}n-j\\ \ell \end{array}}\right) =\left( {\ell }! \right) ^{-1} \prod _{\nu =0}^{\ell -1}\left( n-j-\nu \right)$, for $\ell \in {\mathbb {N}}$. We have

$$\begin{aligned} \left( B_{n}^{\beta }f\right) \left( x\right) =e^{-\beta x}\sum _{k=0}^{\infty }A_{B}\left( n,k,\beta \right) x^{k}(1-x)^{n-k}f\left( \frac{k}{n}\right) { \qquad }\left( 0\le x<\frac{1}{2}\right) , \end{aligned}$$

where

$$\begin{aligned} A_{B}\left( n,k,\beta \right) =\sum _{j=0}^{k}{\left( {\begin{array}{c}n-j\\ k-j\end{array}}\right) }\frac{\beta ^{j}}{j!}. \end{aligned}$$

Thus the semi-exponential Bernstein polynomials $B_{n}^{\beta }$ map a function f on $\left[ 0,+\infty \right)$ to a function $B_{n}^{\beta }f$ defined on $\left[ 0,1/2\right)$, whenever the sum is convergent. It can be shown that the operators $B_{n}^{\beta }$ apply to all polynomials. In the special case $\beta =0$ we have $j=0$ and $\ell =k$ such that $A_{B}\left( n,k,\beta \right) ={\left( {\begin{array}{c}n\\ k\end{array}}\right) }$. Hence, the sum defining $B_{n}^{\beta =0}f$ is finite, and we get the Bernstein polynomials.

The operators $B_{n}^{\beta }$ can be rewritten in the alternative form

$$\begin{aligned} \left( B_{n}^{\beta }f\right) \left( x\right) =e^{-\beta x}\sum _{j=0}^{\infty }\frac{\beta ^{j}}{j!}x^{j}(1-x)^{n-j}\sum _{k=0}^{ \infty }{\left( {\begin{array}{c}n-j\\ k\end{array}}\right) }\left( \frac{x}{1-x}\right) ^{k}f\left( \frac{j+k}{n} \right) \end{aligned}$$

$\left( 0\le x<\frac{1}{2}\right)$. The latter representation immediately reveals the special case $\beta =0$.

2.2 Semi-exponential Baskakov operators

If we take $p\left( x\right) =x\left( 1+x\right)$, then for a kernel $W_{\beta }^{V}\left( n,x,k/n\right) =A_{V}\left( n,k,\beta \right) y$, we have

$$\begin{aligned} y^{\prime }=\frac{k-nx}{x(1+x)}y-\beta y, \end{aligned}$$

where the derivative of y is with respect to the variable x. We conclude that

$$\begin{aligned} \frac{y^{\prime }}{y}= & {} k\left( \frac{1}{x}-\frac{1}{1+x}\right) -\frac{n}{ 1+x}-\beta , \\ \log y= & {} -k\log \left( 1+x\right) +k\log x-n\log \left( 1+x\right) -\beta x, \end{aligned}$$

implying

$$\begin{aligned} y=\frac{x^{k}}{\left( 1+x\right) ^{n+k}}e^{-\beta x}. \end{aligned}$$

In order to have normalization

$$\begin{aligned} \sum _{k=0}^{\infty }W_{\beta }^{V}\left( n,x,k/n\right) =\sum _{k=0}^{\infty }A_{V}\left( n,k,\beta \right) \frac{x^{k}}{\left( 1+x\right) ^{n+k}} e^{-\beta x}=1. \end{aligned}$$

Put, for $x\ge 0$, $z=x/\left( 1+x\right)$. Then $x=z/\left( 1-z\right)$. We obtain

$$\begin{aligned} \sum _{k=0}^{\infty }A_{V}\left( n,k,\beta \right) z^{k}= & {} e^{\beta z/\left( 1-z\right) }\left( 1-z\right) ^{-n} \\= & {} \sum _{j=0}^{\infty }\frac{\left( \beta z\right) ^{j}}{j!}\left( 1-z\right) ^{-n-j} \\= & {} \sum _{j=0}^{\infty }\frac{\left( \beta z\right) ^{j}}{j!}\sum _{\ell =0}^{\infty }{\left( {\begin{array}{c}{n+j-1+\ell }\\ {\ell }\end{array}}\right) }z^{\ell } \\= & {} \sum _{k=0}^{\infty }z^{k}\sum _{j+\ell =k}{\left( {\begin{array}{c}{n+k-1}\\ {\ell }\end{array}}\right) }\frac{ \beta ^{j}}{j!}. \end{aligned}$$

Thus, the semi-exponential Baskakov operators can be defined by

$$\begin{aligned} \left( V_{n}^{\beta }f\right) \left( x\right)= & {} \sum _{k=0}^{\infty }b_{n,k}^{\beta }\left( x\right) f\left( \frac{k}{n}\right) \\= & {} \sum _{k=0}^{\infty }A_{V}\left( n,k,\beta \right) \frac{x^{k}}{(1+x)^{n+k} }e^{-\beta x}f\left( \frac{k}{n}\right) , \end{aligned}$$

where

$$\begin{aligned} A_{V}\left( n,k,\beta \right) =\sum _{j+\ell =k}{\left( {\begin{array}{c}{n+k-1}\\ {\ell }\end{array}}\right) } \frac{\beta ^{j}}{j!}=\sum _{j+\ell =k}\frac{\left( n+j\right) _{\ell }}{k!}{ \left( {\begin{array}{c}k\\ j\end{array}}\right) }\beta ^{j}. \end{aligned}$$

In special case $\beta =0$ we have $j=0$ and $\ell =k$ such that we get the Baskakov operators.

2.3 Semi-exponential Ismail–May operators related to $2x^{3/2}$

If we take $p\left( x\right) =2x^{3/2}$, then for a kernel $W_{\beta }^{U}\left( n,x,t\right) =A_{U}\left( n,t,\beta \right) y$, we have

$$\begin{aligned} y^{\prime }=\frac{n(t-x)}{2x^{3/2}}y-\beta y, \end{aligned}$$

where the derivative of y is with respect to the variable x. We conclude that

$$\begin{aligned} \frac{y^{\prime }}{y}= & {} \frac{nt}{2x^{3/2}}-\frac{n}{2\sqrt{x}}-\beta , \\ \log y= & {} \frac{-nt}{\sqrt{x}}-n\sqrt{x}-\beta x, \end{aligned}$$

implying

$$\begin{aligned} y=\exp \left( \frac{-nt}{\sqrt{x}}-n\sqrt{x}-\beta x\right) . \end{aligned}$$

Our target is to obtain $A_{U}\left( n,t,\beta \right)$ in order to have normalization

$$\begin{aligned} \int _{0}^{\infty }A_{U}\left( n,t,\beta \right) ydt=1. \end{aligned}$$

If we put, for abbreviation, $s=n/\sqrt{x}$, the normalization condition takes the form

$$\begin{aligned} \int _{0}^{\infty }A_{U}\left( n,t,\beta \right) e^{-st}dt=\exp \left( \frac{ n^{2}}{s}+\beta \frac{n^{2}}{s^{2}}\right) { \qquad }\left( s>0\right) . \end{aligned}$$

Since

$$\begin{aligned} \exp \left( \frac{n^{2}}{s}+\beta \frac{n^{2}}{s^{2}}\right) =\sum _{k=0}^{\infty }\left( \frac{n}{s}\right) ^{k}\sum _{\begin{array}{c} i,j\ge 0, \\ i+2j=k \end{array}}\frac{n^{i}\beta ^{j}}{i!j!}{ \qquad }\left( s\ne 0\right) \end{aligned}$$

we obtain

$$\begin{aligned} A_{U}\left( n,t,\beta \right) =\delta \left( t\right) +\sum _{k=1}^{\infty } \frac{n^{k}t^{k-1}}{\varGamma \left( k\right) }\sum _{\begin{array}{c} i,j\ge 0, \\ i+2j=k \end{array}}\frac{n^{i}\beta ^{j}}{i!j!}{ \qquad }\left( s>0\right) , \end{aligned}$$

where $\delta \left( t\right)$ denotes Dirac’s delta function. Hence, the operators are defined by

$$\begin{aligned} \left( U_{n}^{\beta }f\right) \left( x\right) =e^{-n\sqrt{x}-\beta x}f\left( 0\right) +e^{-n\sqrt{x}-\beta x}\int _{0}^{\infty }{\hat{A}}_{U}\left( n,t,\beta \right) \exp \left( -\frac{nt}{\sqrt{x}}\right) f\left( t\right) dt \end{aligned}$$

with

$$\begin{aligned} {\hat{A}}_{U}\left( n,t,\beta \right) =\sum _{k=0}^{\infty }\frac{\left( nt\right) ^{k}}{k!}\sum _{\begin{array}{c} i,j\ge 0, \\ i+2j=k+1 \end{array}}\frac{ n^{i+1}\beta ^{j}}{i!j!}{ \qquad }\left( s>0\right) . \end{aligned}$$

Thus, the semi-exponential operator, related to $2x^{3/2}$, takes the form

$$\begin{aligned} \left( U_{n}^{\beta }f\right) \left( x\right)= & {} e^{-n\sqrt{x}-\beta x}f\left( 0\right) +e^{-n\sqrt{x}-\beta x}\sum _{k=0}^{\infty }\frac{n^{k}}{k!}\left( \sum _{\begin{array}{c} i,j\ge 0, \\ i+2j=k+1 \end{array}}\frac{n^{i+1}\beta ^{j}}{i!j!} \right) \\&\times \int _{0}^{\infty }t^{k}\exp \left( -\frac{nt}{\sqrt{x}}\right) f\left( t\right) dt. \end{aligned}$$

In the special case $\beta =0$, the definition reduces to the Ismail–May operator of exponential type

$$\begin{aligned} \left( U_{n}^{\beta =0}f\right) \left( x\right)= & {} e^{-n\sqrt{x}}f\left( 0\right) +e^{-n\sqrt{x}}\sum _{k=0}^{\infty }\frac{n^{k}}{k!}\frac{n^{k+2}}{ \left( k+1\right) !}\int _{0}^{\infty }t^{k}\exp \left( -\frac{nt}{\sqrt{x}} \right) f\left( t\right) dt \\= & {} e^{-n\sqrt{x}}\left\{ f\left( 0\right) +n\int _{0}^{\infty }e^{-nt/\sqrt{x} }t^{-1/2}I_{1}\left( 2n\sqrt{t}\right) f\left( t\right) dt\right\} , \end{aligned}$$

where $I_{1}\left( x\right)$ is modified Bessel function of the first kind. Further results on the operators $U_{n}^{\beta =0}$ can be found in [1].

2.4 Semi-exponential Post–Widder operators

Although the semi-exponential Post–Widder operators were captured in [3,  Eq. (10)], using Laplace transform, we provide an alternative approach that is shorter. We proceed as follows.

If we take $p(x)=x^{2}$, then for a kernel $W_{\beta }^{P}\left( n,x,t\right) =A_{P}\left( n,t,\beta \right) y$, we have

$$\begin{aligned} y^{\prime }= & {} \frac{n(t-x)}{x^{2}}y-\beta y, \\ \frac{y^{\prime }}{y}= & {} ntx^{-2}-nx^{-1}-\beta , \\ \log y= & {} \frac{-nt}{x}-n\log x-{\beta x,} \\ y= & {} e^{{-nt}/{x}}x^{-n}e^{-\beta x}. \end{aligned}$$

For normalization, we look for a function $A_{P}\left( n,t,\beta \right)$ such that

$$\begin{aligned} \int _{0}^{\infty }W_{\beta }^{P}\left( n,x,t\right) dt=\int _{0}^{\infty }A_{P}\left( n,t,\beta \right) e^{{-nt}/{x}}x^{-n}e^{-\beta x}dt=1. \end{aligned}$$

Putting

$$\begin{aligned} A_{P}\left( n,t,\beta \right) =\sum _{k=0}^{\infty }a_{k}t^{k+\alpha } \end{aligned}$$

we have to choose coefficients $a_{k}$ such that

$$\begin{aligned} \sum _{k=0}^{\infty }a_{k}\int _{0}^{\infty }t^{k+\alpha }e^{{-nt}/{x} }dt=x^{n}e^{\beta x}. \end{aligned}$$

This is equivalent to

$$\begin{aligned} \sum _{k=0}^{\infty }a_{k}\varGamma \left( k+\alpha +1\right) \left( \frac{x}{n} \right) ^{k+\alpha +1}=\sum _{k=0}^{\infty }\frac{\beta ^{k}}{k!}x^{k+n}. \end{aligned}$$

It follows that $\alpha =n-1$ and

$$\begin{aligned} a_{k}=n^{k+n}\frac{\beta ^{k}}{k!\varGamma \left( k+n\right) }. \end{aligned}$$

Hence, $A_{P}\left( n,t,\beta \right)$ is given by

$$\begin{aligned} A_{P}\left( n,t,\beta \right) =n^{n}\sum _{k=0}^{\infty }\frac{\left( n\beta \right) ^{k}}{k!\varGamma \left( k+n\right) }t^{k+n-1}. \end{aligned}$$

Thus, semi-exponential Post–Widder operators take the form

$$\begin{aligned} \left( P_{n}^{\beta }f\right) \left( x\right) =\frac{n^{n}}{e^{\beta x}x^{n}} \sum _{k=0}^{\infty }\frac{(n\beta )^{k}}{k!}\frac{1}{\varGamma (n+k)} \int _{0}^{\infty }t^{n+k-1}e^{-nt/x}f(t)dt. \end{aligned}$$

Observing that $A_{P}\left( n,t,\beta \right) =n\left( nt/\beta \right) ^{\left( n-1\right) /2}I_{n-1}\left( 2\sqrt{n\beta t}\right)$, where $I_{n}$ denotes the modified Bessel function of the first kind, we obtain the alternative representation

$$\begin{aligned} \left( P_{n}^{\beta }f\right) \left( x\right) =\frac{n}{x^{n}e^{\beta x}} \int _{0}^{\infty }\left( \frac{nt}{\beta }\right) ^{\left( n-1\right) /2}e^{{ -nt}/{x}}I_{n-1}\left( 2\sqrt{n\beta t}\right) f\left( t\right) dt. \end{aligned}$$

2.5 Semi-exponential Ismail–May operators related to $x\left( 1+x\right) ^{2}$

If we take $p\left( x\right) =x\left( 1+x\right) ^{2}$, then for a kernel $W_{\beta }^{R}\left( n,x,k/n\right) =A_{R}\left( n,k,\beta \right) y$, we have

$$\begin{aligned} y^{\prime }= & {} \frac{k-nx}{x\left( 1+x\right) ^{2}}y-\beta y, \\ \frac{y^{\prime }}{y}= & {} k\left( \frac{1}{x}-\frac{1}{1+x}-\frac{1}{\left( 1+x\right) ^{2}}\right) -\frac{n}{\left( 1+x\right) ^{2}}-\beta , \\ \log y= & {} k\log x-k\log \left( 1+x\right) +\frac{n+k}{1+x}-\beta x, \end{aligned}$$

implying

$$\begin{aligned} y=\left( \frac{x}{1+x}\right) ^{k}\exp \left( \frac{n+k}{1+x}\right) e^{-\beta x}. \end{aligned}$$

If we put $y=x/\left( 1+x\right)$ the normalization condition reads

$$\begin{aligned} \sum _{k=0}^{\infty }A_{R}\left( n,k,\beta \right) \left( ye^{1-y}\right) ^{k}=\exp \left( \beta \frac{y}{1-y}-n\left( 1-y\right) \right) . \end{aligned}$$

Now we put $z=ye^{1-y}$, so we have the inverse $y=-W\left( -z/e\right)$, where W denotes the Lambert W function, i.e., the inverse of $z\mapsto ze^{z}$. Hence, $A_{R}\left( n,k,\beta \right)$ are the coefficients of the power series

$$\begin{aligned} \sum _{k=0}^{\infty }A_{R}\left( n,k,\beta \right) z^{k}=\exp \left( -\beta \frac{W\left( -z/e\right) }{1+W\left( -z/e\right) }-n\left( 1+W\left( -z/e\right) \right) \right) , \end{aligned}$$

which is convergent in a neighborhood of $z=0$. Following Ismail and May [4,  Eq. (3.13)] we take advantage of the identity [6,  p. 348]

$$\begin{aligned} e^{nw}=\sum _{k=0}^{\infty }\frac{n\left( n+k\right) ^{k-1}}{k!}\left( we^{-w}\right) ^{k}{ \qquad }\left( n\ne 0\right) , \end{aligned}$$

which is an easy consequence of the Lagrange expansion theorem. With $w=-W\left( -z/e\right)$ we obtain

$$\begin{aligned} e^{-nW\left( -z/e\right) }=\sum _{k=0}^{\infty }\frac{n\left( n+k\right) ^{k-1}}{k!}\left( -W\left( -z/e\right) e^{W\left( -z/e\right) }\right) ^{k}=\sum _{k=0}^{\infty }\frac{n\left( n+k\right) ^{k-1}}{k!}\left( \frac{z}{ e}\right) ^{k}. \end{aligned}$$

It follows

$$\begin{aligned} \sum _{k=0}^{\infty }A_{R}\left( n,k,\beta \right) z^{k}=\exp \left( -n-\beta \frac{W\left( -z/e\right) }{1+W\left( -z/e\right) }\right) \sum _{k=0}^{\infty }\frac{n\left( n+k\right) ^{k-1}}{k!}\left( \frac{z}{e} \right) ^{k}, \end{aligned}$$

i.e., $A_{R}\left( n,k,\beta \right)$ is the coefficient of $z^{k}$ in the latter power series expansion. The semi-exponential operators related to $x(1+x)^{2}$, take the form

$$\begin{aligned} \left( R_{n}^{\beta }f\right) \left( x\right) =e^{-\beta x}\sum _{k=0}^{\infty }A_{R}\left( n,k,\beta \right) \left( \frac{x}{1+x} \right) ^{k}\exp \left( \frac{n+k}{1+x}\right) f\left( \frac{k}{n}\right) . \end{aligned}$$

In the special case $\beta =0$ we have

$$\begin{aligned} A_{R}\left( n,k,\beta =0\right) =\frac{n\left( n+k\right) ^{k-1}}{k!} e^{-\left( n+k\right) } \end{aligned}$$

and the operators reduce to

$$\begin{aligned} \left( R_{n}f\right) \left( x\right) =\exp \left( \frac{-nx}{1+x}\right) \sum _{k=0}^{\infty }\frac{n\left( n+k\right) ^{k-1}}{k!}\left( \frac{x}{1+x} \right) ^{k}\exp \left( \frac{-kx}{1+x}\right) f\left( \frac{k}{n}\right) \end{aligned}$$

[4,  Eq. (3.14)]. As Ismail and May remarked, the substitution $y=x/\left( 1+x\right)$ leads to the operators

$$\begin{aligned} (R_{n}^{*}f)(y)=e^{-ny}\sum _{k=0}^{\infty }\frac{n\left( n+k\right) ^{k-1}}{k!}\left( ye^{-y}\right) ^{k}f\left( \frac{k}{n+k}\right) { \qquad }\left( y\in \left( 0,1\right) \right) \end{aligned}$$

It may be considered as an open problem to find a closed form of the coefficients $A_{R}\left( n,k,\beta \right)$.

2.6 Semi-exponential Ismail–May operators related to $x^{3}$

If we take $p\left( x\right) =x^{3}$, then for a kernel $W_{\beta }^{Q}\left( n,x,t\right) =A_{Q}\left( n,t,\beta \right) y$, we have

$$\begin{aligned} y^{\prime }= & {} \frac{n\left( t-x\right) }{x^{3}}y-\beta y, \\ \frac{y^{\prime }}{y}= & {} n\left( \frac{t}{x^{3}}-\frac{1}{x^{2}}\right) -\beta , \\ \log y= & {} n\left( -\frac{t}{2x^{2}}+\frac{1}{x}\right) -\beta x. \end{aligned}$$

Thus

$$\begin{aligned} y=\exp \left( \frac{n}{x}-\frac{nt}{2x^{2}}-\beta x\right) . \end{aligned}$$

If we put $s=n/\left( 2x^{2}\right)$ the normalization condition reads

$$\begin{aligned} \int _{0}^{\infty }A_{Q}\left( n,t,\beta \right) e^{-st}dt=\exp \left( \beta \sqrt{\frac{n}{2s}}-\sqrt{2ns}\right) , \end{aligned}$$

such that $A_{Q}\left( n,t,\beta \right)$ is the inverse Laplace transform ${\mathcal {L}}^{-1}$ of $\exp \left( \beta \sqrt{n/\left( 2s\right) }-\sqrt{2ns} \right)$. We have

$$\begin{aligned} {\mathcal {L}}^{-1}\left\{ \exp \left( 1/\sqrt{s}\right) -1\right\} =\sum _{k=1}^{\infty }\frac{1}{k!}{\mathcal {L}}^{-1}\left\{ s^{-k/2}\right\} =\sum _{k=1}^{\infty }\frac{1}{k!\varGamma \left( k/2\right) }t^{k/2-1}, \end{aligned}$$

which implies

$$\begin{aligned} {\mathcal {L}}^{-1}\left\{ \exp \left( \beta \sqrt{\frac{n}{2s}}\right) \right\} =\delta \left( t\right) +\sum _{k=1}^{\infty }\frac{1}{k!\varGamma \left( k/2\right) }\left( \frac{n\beta ^{2}}{2}\right) ^{k/2}t^{k/2-1}, \end{aligned}$$

(2)

where $\delta \left( t\right)$ denotes Dirac’s delta function. It is well known that

$$\begin{aligned} {\mathcal {L}}^{-1}\left\{ \exp \left( -\sqrt{2ns}\right) \right\} =\sqrt{\frac{ n}{2\pi }}t^{-3/2}e^{-n/\left( 2t\right) }. \end{aligned}$$

(3)

We will take advantage of the convolution formula

$$\begin{aligned} {\mathcal {L}}^{-1}\left\{ {\mathcal {L}}\left\{ g\right\} {\mathcal {L}}\left\{ h\right\} \right\} =g*h, \end{aligned}$$

where

$$\begin{aligned} \left( g*h\right) \left( t\right) =\int _{0}^{t}g\left( t-u\right) h\left( u\right) du. \end{aligned}$$

Combining Eqs. (2) and (3)

$$\begin{aligned}&{\mathcal {L}}^{-1}\left\{ \exp \left( \beta \sqrt{n/\left( 2s\right) }-\sqrt{ 2ns}\right) \right\} \\&\quad =\sqrt{\frac{n}{2\pi }}\int _{0}^{t}u^{-3/2}e^{-n/\left( 2u\right) }\delta \left( t-u\right) du \\&\qquad +\sqrt{\frac{n}{2\pi }}\sum _{k=1}^{\infty }\frac{1}{k!\varGamma \left( k/2\right) }\left( \frac{n\beta ^{2}}{2}\right) ^{k/2}\int _{0}^{t}u^{-3/2}e^{-n/\left( 2u\right) }\left( t-u\right) ^{k/2-1}du. \end{aligned}$$

Thus, semi-exponential operators related to $p\left( x\right) =x^{3}$ take the form

$$\begin{aligned} \left( Q_{n}^{\beta }f\right) \left( x\right) =e^{n/x-\beta x}\int _{0}^{\infty }A_{Q}\left( n,t,\beta \right) e^{-nt/\left( 2x^{2}\right) }f\left( t\right) dt, \end{aligned}$$

where

$$\begin{aligned} A_{Q}\left( n,t,\beta \right)= & {} \sqrt{\frac{n}{2\pi }}\left( t^{-3/2}e^{-n/\left( 2t\right) }+\sum _{k=1}^{\infty }\frac{1}{k!\varGamma \left( k/2\right) }\left( \frac{n\beta ^{2}}{2}\right) ^{k/2}\right. \\&\times \left. \int _{0}^{t}u^{-3/2}e^{-n/\left( 2u\right) }\left( t-u\right) ^{k/2-1}du\right) . \end{aligned}$$

In the special case $\beta =0$ we have

$$\begin{aligned} A_{Q}\left( n,t,\beta =0\right) =\sqrt{\frac{n}{2\pi }}t^{-3/2}e^{-n/\left( 2t\right) } \end{aligned}$$

and the operators reduce to

$$\begin{aligned} \left( Q_{n}^{\beta =0}f\right) \left( x\right) =\sqrt{\frac{n}{2\pi }} e^{n/x}\int _{0}^{\infty }t^{-3/2}\exp \left( -\frac{n}{2t}-\frac{nt}{2x^{2}} \right) f\left( t\right) dt \end{aligned}$$

[4,  Eq. (3.11)].

2.7 Semi-exponential Ismail–May operators related to $1+x^{2}$

If we take $p\left( x\right) =1+x^{2}$, then for a kernel $W_{\beta }^{T}\left( n,x,t\right) =A_{T}\left( n,t,\beta \right) y$, we have

$$\begin{aligned} y^{\prime }= & {} \frac{n(t-x)}{1+x^{2}}y-\beta y, \\ \frac{y^{\prime }}{y}= & {} \frac{nt}{1+x^{2}}-\frac{nx}{1+x^{2}}-\beta , \\ \log y= & {} nt\arctan x-\frac{n}{2}\log \left( 1+x^{2}\right) -\beta x, \end{aligned}$$

implying

$$\begin{aligned} y=e^{nt\arctan x-\beta x}\left( 1+x^{2}\right) ^{-n/2}. \end{aligned}$$

The operators related to $1+x^{2}$ take the form

$$\begin{aligned} \left( T_{n}^{\beta }f\right) \left( x\right) =\frac{e^{-\beta x}}{\left( 1+x^{2}\right) ^{n/2}}\int _{-\infty }^{\infty }A_{T}\left( n,t,\beta \right) e^{nt\arctan x}f(t)dt \end{aligned}$$

To have the normalization, we need

$$\begin{aligned} \int _{-\infty }^{\infty }A_{T}\left( n,t,\beta \right) e^{nt\arctan x}dt=e^{\beta x}\left( 1+x^{2}\right) ^{n/2}. \end{aligned}$$

If we put $s=n\arctan x,$ this is equivalent to

$$\begin{aligned} \int _{-\infty }^{\infty }A_{T}\left( n,t,\beta \right) e^{st}dt=\frac{ e^{\beta \tan \left( s/n\right) }}{\cos ^{n}\left( s/n\right) }. \end{aligned}$$

Using the identity [5,  Section 9, p. 46] (see [4,  Lemma 3.3])

$$\begin{aligned} \int _{-\infty }^{\infty }\left| \varGamma \left( \frac{\lambda +it}{2} \right) \right| ^{2}e^{st}dt=\frac{\pi \varGamma \left( \lambda \right) }{ 2^{\lambda -2}\cos ^{\lambda }s}{ \qquad }\left( \lambda >0, -\pi /2<s<\pi /2\right) \end{aligned}$$

Ismail and May [4,  Eq. (3.10)] obtained in the special case $\beta =0$,

$$\begin{aligned} A_{T}\left( n,t,\beta =0\right) =\frac{2^{n-2}n}{\pi \varGamma \left( n\right) } \left| \varGamma \left( n\frac{1+it}{2}\right) \right| ^{2}. \end{aligned}$$

The main target to find a closed expression for $A_{T}\left( n,t,\beta \right)$, for general $\beta >0$, may be considered as an open problem.