On the Laplace-Type Transform and Its Applications

Abstract

Using the Laplace transform and the Gamma function, we obtain the Laplace-type transform, with the property of mapping a function to a functional sequence, which cannot be realized by the Laplace transform. In addition, we construct a backward difference as a generalization of the backward difference operator ∇. By connecting it to the Laplace-type transform, we deduce a method for solving difference equations and, relying on classical orthogonal polynomials, for obtaining combinatorial identities. A table of some elementary functions and their images is at the end of the text.

1. Introduction and Preliminaries

Very often, if we interpret continuity problems as discrete, we can solve them more easily. For example, complicated definite integrals of continuous functions are calculated numerically by being reduced to finite sums. The opposite is also true, i.e., the only way to solve some discrete problems is to view them in the light of continuity. For this purpose, we develop a specific transform and its inverse, enabling us to map the continuous functions into sequences and vice versa.

It is well-known that the integral defines the Gamma function

$$\Gamma \left(\alpha \right)={\int }_0^{\infty }{e}^{-x}{x}^{\alpha -1}\mathrm{d}x(\alpha >0).$$

(1)

We can regard it as the Mellin transform of the function $e^{-x}$, i.e., $\Gamma \left(\alpha \right)=\mathcal{M}\left\{e^{-x}\right\}\left(\alpha \right)$. On the other hand, for $s>0$, after introducing the substitution $x=st$, we find

$$\Gamma \left(\alpha \right)={\int }_0^{\infty }{s}^{\alpha }{e}^{-st}{t}^{\alpha -1}\mathrm{d}t\Rightarrow {\displaystyle \frac{1}{s^{\alpha }}}={\int }_0^{\infty }{e}^{-st}{\displaystyle \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}}\mathrm{d}t=\mathcal{L}\left\{{\displaystyle \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}}\right\}\left(s\right).$$

Thereby, we have determined the Laplace transform of the function $t^{\alpha -1}/\Gamma \left(\alpha \right)$.

However, if apart from $t^{\alpha -1}$ there appears a real-valued piecewise continuous function $f\left(t\right)$ over $(0,\infty )$, and of exponential order r, i.e., there exist positive numbers M and r such that the inequality $\left|f\left(t\right)\right|<M{e}^{rt}$ holds, according to and in keeping with [1] (p. 744, Equation (2.1)), we introduce a new integral transform $\mathcal{G}$ of the function $f\left(t\right)$ and name it the Laplace-type transform, i.e.

$$\mathcal{G}\left\{f\left(t\right)\right\}\left(s\right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_0^{\infty }{e}^{-st}{t}^{\alpha -1}f\left(t\right)\mathrm{d}t(s>r,\alpha >0).$$

(2)

Under the given conditions, this integral is convergent, since

$${\int }_0^{\infty }{e}^{-st}{\displaystyle \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}}f\left(t\right)\mathrm{d}t<M{\int }_0^{\infty }{e}^{-(s-r)t}{\displaystyle \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}}\mathrm{d}t={\displaystyle \frac{M}{{(s-r)}^{\alpha }}}.$$

By setting $\alpha -1=n$ in (2) for $n\in \mathbb{N}\cup \left\{0\right\}$, the Laplace-type transform maps the function $f\left(t\right)$ to a sequence of functions ${\left\{{\phi }_n\left(s\right)\right\}}_{n\in {\mathbb{N}}_0}$, given by

$${\phi }_n\left(s\right)={\int }_0^{\infty }{e}^{-st}{\displaystyle \frac{t^n}{n!}}f\left(t\right)\mathrm{d}t(s>0),$$

(3)

i.e., ${\phi }_n\left(s\right)\fallingdotseq f\left(t\right)$. For $s=1$, we write ${\phi }_n$ instead of ${\phi }_n\left(1\right)$ so that in this case, the Laplace-type transform maps the function $f\left(t\right)$ to a sequence of numbers ${\left\{{\phi }_n\right\}}_{n\in {\mathbb{N}}_0}$, i.e., ${\phi }_n\fallingdotseq f\left(t\right)$, and later, within Applications, it will enable us to solve difference equations through differential ones.

Example 1.

The Hurwitz zeta function $\zeta (s,a)$ initially defined for $\sigma >1$ ($s=\sigma +i\tau$) by the series having its integral representation

$$\zeta (s,a)=\sum _{k=0}^{\infty }{\displaystyle \frac{1}{{(k+a)}^s}},$$

(4)

where a is a fixed real number, $0<a⩽1$.

Multiplying the equality (4) by $\Gamma \left(s\right)$,

$$\zeta (s,a)\Gamma \left(s\right)=\sum _{k=0}^{+\infty }{\displaystyle \frac{1}{{(k+a)}^s}}{\int }_0^{\infty }{x}^{s-1}{e}^{-x}\mathrm{d}x,$$

and introducing the substitution $x=(k+a)t$, then interchanging the sum and integral, dividing by $\Gamma \left(z\right)$, we get the integral representation of the Hurwitz zeta function

$$\zeta (s,a)={\int }_0^{\infty }{t}^{s-1}\sum _{k=0}^{\infty }{e}^{-(k+a)t}\mathrm{d}t={\displaystyle \frac{1}{\Gamma \left(s\right)}}{\int }_0^{\infty }{\displaystyle \frac{t^{s-1}{e}^{-at}}{1-e^{-t}}}\mathrm{d}t.$$

(5)

This integral can be regarded as a Mellin transform of the function ${\displaystyle \frac{e^{-ax}}{1-e^{-x}}}$.

Referring to (2) for ${\displaystyle f\left(t\right)=\frac{1}{1-e^{-t}}}$, $\alpha =s$, $s=a$, and taking into account (5), we have

$$\mathcal{G}\left\{f\left(t\right)\right\}\left(s\right)=\mathcal{G}\left\{{\displaystyle \frac{1}{1-e^{-t}}}\right\}\left(\alpha \right)=\zeta (\alpha ,a).$$

(6)

However, taking ${\displaystyle f\left(t\right)=\frac{1}{e^t-1}}$, we find

$$\mathcal{G}\left\{{\displaystyle \frac{1}{e^t-1}}\right\}\left(s\right)={\displaystyle \frac{1}{\Gamma \left(s\right)}}{\int }_0^{\infty }{\displaystyle \frac{e^{-at}}{e^t-1}}{t}^{s-1}\mathrm{d}t={\displaystyle \frac{1}{\Gamma \left(s\right)}}{\int }_0^{\infty }{\displaystyle \frac{e^{-(a+1)t}}{1-e^{-t}}}{t}^{s-1}\mathrm{d}t=\zeta (s,a+1).$$

On the other hand, the generating function for the Bernoulli polynomials is

$${\displaystyle \frac{t{e}^{at}}{e^t-1}}=\sum _{k=0}^{\infty }{B}_k\left(a\right){\displaystyle \frac{t^k}{k!}},$$

which implies

$$\zeta (s,a+1)={\displaystyle \frac{1}{\Gamma \left(s\right)}}{\int }_0^{\infty }{\displaystyle \frac{t{e}^{at}}{e^t-1}}{e}^{-2at}{t}^{s-2}\mathrm{d}t={\displaystyle \frac{1}{\Gamma \left(s\right)}}{\int }_0^{\infty }{e}^{-2at}{t}^{s-2}\sum _{k=0}^{\infty }{B}_k\left(a\right){\displaystyle \frac{t^k}{k!}}\mathrm{d}t.$$

(7)

Interchanging summation and integration, we rewrite (7) in the form of

$$\zeta (s,a+1)=\sum _{k=0}^{\infty }{\displaystyle \frac{B_k\left(a\right)}{\Gamma \left(s\right)}}{\int }_0^{\infty }{e}^{-2at}{\displaystyle \frac{t^k}{k!}}{t}^{s-2}\mathrm{d}t=\sum _{k=0}^{\infty }{\displaystyle \frac{B_k\left(a\right)}{2^{s+k-1}\Gamma \left(s\right)}}{\int }_0^{\infty }{e}^{-at}{\displaystyle \frac{t^k}{k!}}{t}^{s-2}\mathrm{d}t.$$

Looking up in Appendix A for the image of $f\left(t\right)=t^{s-2}$, we find

$$\zeta (s,a+1)=\sum _{k=0}^{\infty }{\displaystyle \frac{B_k\left(a\right)}{\Gamma \left(s\right)}}{\displaystyle \frac{\Gamma (s+k-1)}{{\left(2a\right)}^{s+k-1}k!}}={\displaystyle \frac{\Gamma (s-1)}{{\left(2a\right)}^{s-1}\Gamma \left(s\right)}}+\sum _{k=1}^{\infty }{\displaystyle \frac{B_k\left(a\right)}{\Gamma \left(s\right)}}{\displaystyle \frac{\Gamma (s+k-1)}{{\left(2a\right)}^{s+k-1}k!}}.$$

Using the relation $B_k\left(a\right)=-k\zeta (1-k,a)$, $k\in \mathbb{N}$, one obtains the representation

$$\zeta (s,a+1)={\displaystyle \frac{1}{{\left(2a\right)}^{s-1}(s-1)}}-\sum _{k=1}^{\infty }{\displaystyle \frac{\zeta (1-k,a)}{{\left(2a\right)}^{s+k-1}}}{\displaystyle \frac{{\left(s\right)}_{k-1}}{k!}},{\left(s\right)}_{k-1}={\displaystyle \frac{\Gamma (s+k-1)}{\Gamma \left(s\right)}},$$

with ${\left(x\right)}_n$ being the Pochhammer symbol.

Remark 1.

We can regard the Laplace-type transform as convergent versions of Watson’s lemma, proved by G. N. Watson [2], and the generalized Laplace transforms [3]:

$$F\left(z\right)={\int }_0^b{e}^{-z{t}^m}{t}^{s-1}{(1-t/b)}^{\sigma -1}f\left(t\right)\mathrm{d}t(\Re z⩾x_0>0,\Re s>0,0<\Re \sigma ⩽1),$$

with $m\in N$ and $0<b⩽\infty$. So, by and letting $b\to \infty$, the factor ${(1-t/b)}^{\sigma -1}$ becomes 1, and for $m=1$, we arrive at (2).

2. Basic Properties

We shall show later how the Laplace-type transform helps us derive a method for solving difference equations and obtaining combinatorial identities. For this purpose, we need to investigate some of its properties.

Lemma 1.

Let $f\left(t\right)$ be piecewise continuous on $[0,\infty )$ of exponential order r and $F\left(s\right)$ denote its Laplace transform, i.e., $F\left(s\right)=\mathcal{L}\left\{f\right(t\left)\right\}\left(s\right)$. If ${\phi }_n\left(s\right)\fallingdotseq f\left(t\right)$, then ${\phi }_n\left(s\right)$ can be expressed through $F^{\left(n\right)}\left(s\right)$ as

$${\phi }_n\left(s\right)={\displaystyle \frac{{(-1)}^n}{n!}}{F}^{\left(n\right)}\left(s\right).$$

(8)

Proof. 

By differentiating n times $F\left(s\right)$, we have

$$F^{\left(n\right)}\left(s\right)={\displaystyle \frac{d^n}{d{s}^n}}F\left(s\right)={\displaystyle \frac{d^n}{d{s}^n}}\left({\int }_0^{\infty }{e}^{-st}f\left(t\right)dt\right)={(-1)}^n{\int }_0^{\infty }{e}^{-st}{t}^{n}f\left(t\right)dt.$$

By the hypotheses, for $s\ge x_0>r$, it is justified to interchange the derivative and integral sign in the last calculation (see [4]). Because of (3), there immediately follows (8). □

Lemma 2.

Let ${\displaystyle g\left(t\right)={\int }_0^{t}f\left(x\right)dx}$. If $f\left(t\right)\fallingdotseq {\phi }_n\left(s\right)$ and $g\left(t\right)\fallingdotseq {\psi }_n\left(s\right)$, then

$${\psi }_n\left(s\right)=\sum _{k=0}^n{\displaystyle \frac{{\phi }_k\left(s\right)}{s^{n-k+1}}}.$$

Proof. 

Let $F_f\left(s\right)$ and $F_g\left(s\right)$ denote the Laplace transform of $f\left(t\right)$ and $g\left(t\right)$ respectively. Considering that

$$F_g\left(s\right)={\displaystyle \frac{1}{s}}{F}_f\left(s\right),$$

and making use of (3), (8) and the Leibnitz rule, we find ${\psi }_n\left(s\right)\fallingdotseq g\left(t\right)$, where

$${\psi }_n\left(s\right)={\displaystyle \frac{{(-1)}^n}{n!}}{\left({\displaystyle \frac{1}{s}}{F}_f\left(s\right)\right)}^{\left(n\right)}=\sum _{k=0}^n{\displaystyle \frac{1}{s^{n-k+1}}}{\displaystyle \frac{{(-1)}^k}{k!}}{F}_f^{\left(k\right)}\left(s\right)=\sum _{k=0}^n{\displaystyle \frac{{\phi }_k\left(s\right)}{s^{n-k+1}}}.$$

Whereby we have proved the statement. □

The convolution of two functions, $f\left(t\right)$ and $g\left(t\right)$, defined for $t>0$, is of great importance in numerous different physical applications. The integral gives the convolution:

$$(f\ast g)\left(t\right)={\int }_0^{t}f(t-\tau )g\left(\tau \right)d\tau ,$$

which exists if f and g are piecewise continuous.

Lemma 3.

Let ${\phi }_n\left(s\right)\fallingdotseq f\left(t\right)$ and ${\psi }_n\left(s\right)\fallingdotseq g\left(t\right)$. The following equality holds

$$\sum _{k=0}^n{\phi }_{n-k}\left(s\right){\psi }_k\left(s\right)\fallingdotseq (f\ast g)\left(t\right).$$

Proof. 

Let $F_f\left(s\right)$, $F_g\left(s\right)$ and $F_{f\ast g}\left(s\right)$ denote the Laplace transform of $f\left(t\right)$, $g\left(t\right)$ and $(f\ast g)\left(t\right)$ respectively. It is well known that

$$F_{f\ast g}\left(s\right)=F_f\left(s\right)·F_g\left(s\right).$$

Using the result of Lemma 1, we obtain

$$h_n\left(s\right)={\displaystyle \frac{{(-1)}^n}{n!}}{\left(F_f\left(s\right)·F_g\left(s\right)\right)}^{\left(n\right)}\fallingdotseq (f\ast g)\left(t\right),$$

and by using the Leibnitz rule and (8), we have

$$h_n\left(s\right)=\sum _{k=0}^n{\displaystyle \frac{{(-1)}^{n-k}{F}_f^{(n-k)}\left(s\right)}{(n-k)!}}{\displaystyle \frac{{(-1)}^k{F}_g^{(n-k)}\left(s\right)}{k!}}=\sum _{k=0}^n{\phi }_{n-k}\left(s\right){\psi }_k\left(s\right).$$

And it ends the proof. □

3. The Laplace-Type Transform and Laguerre Polynomials

We recall that the Laguerre polynomials, the explicit representation of which is

$$L_n\left(x\right)=\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{x^k}{k!}}.$$

(9)

By introducing an inner product as follows

$$(L_m,L_n)={\int }_0^{\infty }{e}^{-x}{L}_m\left(x\right)L_n\left(x\right)dx.$$

We make them into an orthogonal system with respect to the weight function $e^{-x}$, whereby they form an orthogonal base of the Hilbert space $L_2(0,\infty ;e^{-x})$. To connect them to the Laplace-type transform, we replace x with $sx(s>0)$, and the above inner product becomes

$${\int }_0^{\infty }{e}^{-sx}s{L}_m\left(sx\right)L_n\left(sx\right)dx.$$

That prompts us to construct a system of Laguerre functions over $(0,\infty )$, i.e.,

$$L_n^{*}(x;s)={(-1)}^n\sqrt{s}{L}_n\left(sx\right),$$

(10)

and define their inner product by

$$(L_m^{*},L_n^{*})={\int }_0^{\infty }{e}^{-sx}{(-1)}^m\sqrt{s}{L}_m\left(sx\right){(-1)}^n\sqrt{s}{L}_n\left(sx\right)dx.$$

Since we easily find $(L_m^{*},L_n^{*})=0,m\ne n$, $(L_n^{*},L_n^{*})=1$, we conclude the system $L_n^{*}\left(x\right)$ ($n\in {\mathbb{N}}_0$) form an orthonormal basis of the Hilbert space with respect to the weight function $e^{-sx}$. A function $f\left(x\right)$ of the exponential order r belongs to $L_2(0,\infty ;e^{-sx})$, because if there exist positive constants M and r such that $\left|f\left(x\right)\right|<M{e}^{rx}$, then

$${\parallel f\parallel }^2={\int }_0^{+\infty }{e}^{-sx}{\left|f\left(x\right)\right|}^{2}dx<M^2{\int }_0^{+\infty }{e}^{-(s-2r)x}dx={\displaystyle \frac{M^2}{s-2r}}(s>2r),$$

and can be expanded into the Fourier series over the functions $L_n^{*}(x;s)$, so that we have

$$f\left(x\right)=\sum _{k=0}^{\infty }{a}_k\left(s\right)L_k^{*}(x;s),a_n\left(s\right)={(-1)}^n{\int }_0^{\infty }{e}^{-st}f\left(t\right)\sqrt{s}{L}_n\left(st\right)dt.$$

(11)

Also, the Bessel inequality holds

$$\sum _{k=0}^{\infty }{a}_k^2\left(s\right)\le {\parallel f\parallel }^2.$$

(12)

To prove the expansion (11), we form its nth partial sum, and after rearrangement obtain

$$S_n(x,f;s)={\int }_0^{\infty }{e}^{-st}f\left(t\right)\left[\sum _{k=0}^n{L}_k^{*}(t;s)L_k^{*}(x;s)\right]dt,$$

where applying the Christoffel–Darboux formula

$$\sum _{k=0}^n{L}_k^{*}(t;s)L_k^{*}(x,s)=(n+1){\displaystyle \frac{L_{n+1}^{*}(x;s)L_n^{*}(t;s)-L_{n+1}^{*}(t;s)L_n^{*}(x;s)}{t-x}}=(n+1){\displaystyle \frac{{\mathsf{\Phi }}_n(x,t;s)}{t-x}},$$

we have

$$S_n(x,f;s)=(n+1){\int }_0^{\infty }{e}^{-st}f\left(t\right){\displaystyle \frac{{\mathsf{\Phi }}_n(x,t;s)}{t-x}}dt.$$

Now, dealing with the difference

$$S_n(x,f;s)-f\left(x\right)=(n+1){\int }_0^{\infty }{e}^{-st}{\displaystyle \frac{f\left(x\right)-f\left(t\right)}{x-t}}{\mathsf{\Phi }}_n(x,t;s)dt,$$

we can make it arbitrarily small (see [5]).

Theorem 1.

The sequence $\left\{a_n\left(s\right)\right\}$ can be expressed in terms of the sequence $\left\{{\phi }_n\left(s\right)\right\}$ defined by (3), and the other way around:

$$a_n\left(s\right)=\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)s^{k+{\displaystyle \frac{1}{2}}}{\phi }_k\left(s\right),{\phi }_n\left(s\right)={\displaystyle \frac{\sqrt{s}}{s^{n+1}}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)a_k\left(s\right).$$

(13)

Proof. 

We prove the first equality of (13). Relying on (11) and then referring to (3), we obtain

$$\begin{array}{cc}\hfill a_n\left(s\right)& ={(-1)}^n\sqrt{s}{\int }_0^{+\infty }{e}^{-sx}f\left(x\right)\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\displaystyle \frac{{\left(sx\right)}^k}{k!}}dx\hfill \\ \hfill & =\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)s^{k+{\displaystyle \frac{1}{2}}}{\int }_0^{+\infty }{e}^{-sx}{\displaystyle \frac{x^k}{k!}}f\left(x\right)dx=\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)s^{k+{\displaystyle \frac{1}{2}}}{\phi }_k\left(s\right).\hfill \end{array}$$

To derive the second equality, we consider the first equality in (13) as the binomial transform of the sequence $\left\{b_n\left(s\right)\right\}$, i.e.,

$$a_n\left(s\right)=\sum _{k=0}^n{(-1)}^{n-k}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)b_k\left(s\right),b_k\left(s\right)=s^{k+{\displaystyle \frac{1}{2}}}{\phi }_k\left(s\right).$$

So its inverse transform is

$$b_n\left(s\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)a_k\left(s\right)\Rightarrow s^{n+\frac{1}{2}}{\phi }_n\left(s\right)=\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)a_k\left(s\right)\iff {\phi }_n\left(s\right)={\displaystyle \frac{\sqrt{s}}{s^{n+1}}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)a_k\left(s\right).$$

Whereby we complete the proof. □

4. The Inverse Laplace-Type Transform

Let $S_c$ be the space of sequences of functions ${\left\{t_n\left(s\right)\right\}}_{n\in {\mathbb{N}}_0}$ such that

$$\sum _{k=0}^{+\infty }{\left\{{\Delta }_n^k{\left(s^n{t}_n\left(s\right)\right)}_{n=0}\right\}}^2<+\infty ,$$

(14)

where ${\Delta }_n$ is the forward difference operator

$${\Delta }_n{t}_n\left(s\right)=t_{n+1}\left(s\right)-t_n\left(s\right),{\Delta }_n^k{t}_n\left(s\right)={\Delta }_n\left({\Delta }_n^{k-1}{t}_n\left(s\right)\right)=\sum _{j=0}^k{(-1)}^{k-j}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)t_{n+j}\left(s\right).$$

We shall show that the sequence ${\left\{{\phi }_n\left(s\right)\right\}}_{n\in {\mathbb{N}}_0}$ defined by (3) satisfies (14), i.e., belongs to the space $S_c$.

Taking account of (3), (9) and (11), we find

$$\begin{array}{cc}\hfill {\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}& =\sum _{j=0}^k{(-1)}^{k-j}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)s^j{\phi }_j\left(s\right)={(-1)}^k\sum _{j=0}^k{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)s^j{\int }_0^{\infty }{e}^{-sx}{\displaystyle \frac{x^j}{j!}}f\left(x\right)dx,\hfill \\ \hfill & ={(-1)}^k{\int }_0^{\infty }{e}^{-sx}f\left(x\right)\sum _{j=0}^k{\displaystyle \frac{{(-1)}^j}{j!}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){\left(sx\right)}^{j}dx,\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^k}{\sqrt{s}}}{\int }_0^{\infty }{e}^{-sx}f\left(x\right)\sqrt{s}{L}_k\left(sx\right)dx={\displaystyle \frac{a_k\left(s\right)}{\sqrt{s}}}.\hfill \end{array}$$

Hence, making use of the Bessel inequality (12), we finally have

$$\sum _{k=0}^{\infty }{\left\{{\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}\right\}}^2={\displaystyle \frac{1}{s}}\sum _{k=0}^{\infty }{a}_k^2\left(s\right)<+\infty .$$

Bearing in mind that and the relations (13) between orthogonal Laguerre functions (10) and the Laplace-type transform, we define a mapping $\mathcal{R}:S_c\mapsto L_2(0,+\infty )$ as follows

$$f\left(x\right)=\mathcal{R}\left\{{\phi }_n\right\}\left(x\right)=\sum _{k=0}^{\infty }{\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}\sqrt{s}{L}_k^{*}(x;s)$$

(15)

Theorem 2.

The transforms $\mathcal{G}$ and $\mathcal{R}$ are inverse to each other; in other words, if we obtain ${\phi }_n\left(s\right)$ by applying $\mathcal{G}\left\{f\right(t\left)\right\}\left(s\right)$, then $f\left(t\right)$ is obtained by $\mathcal{R}\left\{{\phi }_n\left(s\right)\right\}\left(t\right)$, and the other way round, if (15) holds, then $f\left(t\right)\risingdotseq {\phi }_n\left(s\right)$.

Proof. 

Let ${\phi }_n\left(s\right)=\mathcal{G}\left\{f\left(t\right)\right\}\left(s\right)$. Then, the first relation in (13) yields

$$a_k\left(s\right)=\sqrt{s}\sum _{j=0}^k{(-1)}^{k-j}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)s^j{\phi }_j\left(s\right)=\sqrt{s}{\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}.$$

Thus, we prove the first direction as $a_k\left(s\right)$ are coefficients in the expansion of $f\left(x\right)$ in terms of Laguerre polynomials:

$$f\left(x\right)=\sum _{k=0}^{\infty }{a}_k\left(s\right)L_k^{*}(x;s)=\sum _{k=0}^{\infty }{\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}\sqrt{s}{L}_k^{*}(x;s)=\mathcal{R}\left\{{\phi }_n\right\}\left(x\right).$$

Conversely, let $\mathcal{R}\left\{{\phi }_n\right\}\left(x\right)=f\left(x\right)$. So, by (3), but relying on (15) and referring to (10), we have

$$\begin{array}{cc}\hfill \mathcal{G}\left\{f\right(t\left)\right\}\left(s\right)& ={\displaystyle \frac{1}{n!}}{\int }_0^{\infty }{e}^{-sx}{x}^{n}f\left(x\right)dx={\int }_0^{\infty }{e}^{-sx}{\displaystyle \frac{x^n}{n!}}\sum _{k=0}^{\infty }{\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}\sqrt{s}{L}_k^{*}(x;s)dx\hfill \\ \hfill & =s\sum _{k=0}^{\infty }{(-1)}^k{\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}{\int }_0^{\infty }{e}^{-sx}{\displaystyle \frac{x^n}{n!}}{L}_k\left(sx\right)dx\hfill \\ \hfill & ={\displaystyle \frac{1}{s^n}}\sum _{k=0}^{\infty }{(-1)}^k{\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}{\int }_0^{\infty }{e}^{-sx}{\displaystyle \frac{{\left(sx\right)}^n}{n!}}{L}_k\left(sx\right)d\left(sx\right).\hfill \end{array}$$

After substituting $t=sx$ in the last integral, it becomes

$${\int }_0^{\infty }{e}^{-t}{\displaystyle \frac{t^n}{n!}}{L}_k\left(t\right)dt={\int }_0^{\infty }{e}^{-t}{\displaystyle \frac{t^n}{n!}}\left(\sum _{j=0}^k{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){\displaystyle \frac{t^j}{j!}}\right)dt={\displaystyle \frac{1}{n!}}\sum _{j=0}^k{\displaystyle \frac{{(-1)}^j}{j!}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){\int }_0^{\infty }{e}^{-t}{t}^{n+j}dt.$$

The right-hand side integral is actually $\Gamma (n+j+1)=(n+j)!$, and we now calculate the left-most integral

$$\begin{array}{cc}\hfill {\int }_0^{\infty }{e}^{-t}{\displaystyle \frac{t^n}{n!}}{L}_k\left(t\right)dt& ={\displaystyle \frac{1}{n!}}\sum _{j=0}^k{\displaystyle \frac{{(-1)}^j}{j!}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)(n+j)!=\sum _{j=0}^k{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)\hfill \\ \hfill & ={(-1)}^k\sum _{j=0}^k{(-1)}^{k-j}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)={(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right),k⩽n,\hfill \end{array}$$

where the sum in the last row presents the binomial transform (see [6]). So, one obtains

$$\mathcal{G}\left\{f\left(t\right)\right\}\left(s\right)={\displaystyle \frac{1}{n!}}{\int }_0^{\infty }{e}^{-sx}{x}^{n}f\left(x\right)dx={\displaystyle \frac{1}{s^n}}\sum _{k=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right){\Delta }_n^k{\left(s^n{\phi }_n\left(s\right)\right)}_{n=0}={\displaystyle \frac{1}{s^n}}{s}^n{\phi }_n\left(s\right)={\phi }_n\left(s\right).$$

Here, we have applied the inverse binomial transform ([6]) and proved the second direction. Thereby, we complete the proof. □

We make use of the formula for the inverse Laplace transform (see [7])

$$f\left(t\right)={\displaystyle \frac{1}{2\pi i}}{\int }_{c-i\infty }^{c+i\infty }{e}^{st}F\left(s\right)ds=\sum _{j=0}^k\underset{s=s_j}{\mathrm{Res}}{e}^{st}F\left(s\right),$$

(16)

to prove

Lemma 4.

The original $f\left(t\right)$ of the Laplace-type transform $\mathcal{R}$ can be determined by the equality

$$f\left(t\right)={\displaystyle \frac{n!}{t^n}}\sum _{j=0}^k\underset{s=s_j}{\mathrm{Res}}{e}^{st}{\phi }_n\left(s\right).$$

Proof. 

Considering that $n!f\left(t\right)\risingdotseq n!{\phi }_n\left(s\right)$, and relying on Lemma 1, we have already seen that $n!{\phi }_n\left(s\right)=\mathcal{L}\left\{t^{n}f\left(t\right)\right\}\left(s\right)$, whence we have ${\mathcal{L}}^{-1}\{n!{\phi }_n\left(s\right)\}\left(t\right)=t^{n}f\left(t\right)$. Referring to (16), we find

$$t^{n}f\left(t\right)={\displaystyle \frac{1}{2\pi i}}{\int }_{c-i\infty }^{c+i\infty }{e}^{st}n!{\phi }_n\left(s\right)ds=n!\sum _{j=0}^k\underset{s=s_j}{\mathrm{Res}}{e}^{st}{\phi }_n\left(s\right).$$

However, taking into account Theorem 2, we conclude that $f\left(t\right)=\mathcal{R}\left\{{\phi }_n\left(s\right)\right\}\left(t\right)$, i.e.,

$$\mathcal{R}\left\{{\phi }_n\left(s\right)\right\}\left(t\right)={\displaystyle \frac{n!}{t^n}}\sum _{j=0}^k\underset{s=s_j}{\mathrm{Res}}{e}^{st}{\phi }_n\left(s\right).$$

Thereby, we have proved the lemma. □

5. The Laplace-Type Transform and a Backward Difference Operator

For a sequence $\left\{{\phi }_n\left(s\right)\right\}$, the backward shift operator $E^{-1}$ is defined by ${\phi }_{n-1}\left(s\right)=E^{-1}{\phi }_n\left(s\right)$. We define the generalized backward difference operator ${\nabla }_s$ as

$${\nabla }_s{\phi }_n\left(s\right)=s{\phi }_n\left(s\right)-{\phi }_{n-1}\left(s\right),$$

(17)

which for $s=1$ reduces to the backward difference operator $\nabla {\phi }_n={\phi }_n-{\phi }_{n-1}$, with ${\phi }_n={\phi }_n\left(1\right)$.

By writing (17) in the form of

$${\nabla }_s{\phi }_n\left(s\right)=s{\phi }_n\left(s\right)-E^{-1}{\phi }_n\left(s\right)=(sI-E^{-1}){\phi }_n\left(s\right),$$

with I as identity operator, i.e., $I_s{\phi }_n\left(s\right)={\phi }_n\left(s\right)$, for the power ${\nabla }_s^p{\phi }_n\left(s\right)$ ($p\in \mathbb{N}$), we have

$$\begin{array}{cc}\hfill {\nabla }_s^p{\phi }_n\left(s\right)& ={(sI-E^{-1})}^p{\phi }_n\left(s\right)=\sum _{k=0}^p{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)s^{p-k}{\left(E^{-1}\right)}^k{\phi }_n\left(s\right)\hfill \\ \hfill & =\sum _{k=0}^p{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)s^{p-k}{E}^{-k}{\phi }_n\left(s\right)=\sum _{k=0}^p{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{p}{k}}\right)s^{p-k}{\phi }_{n-k}\left(s\right).\hfill \end{array}$$

Using this and starting with ${\phi }_{n-1}\left(s\right)=s{\phi }_n\left(s\right)-{\nabla }_s{\phi }_n\left(s\right)$, by the method of mathematical induction, we prove that ${\phi }_{n-p}\left(s\right)$ is expressed over the powers of ${\nabla }_s{\phi }_n\left(s\right)$, that is

$${\phi }_{n-p}\left(s\right)=s^p{\phi }_n\left(s\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{p}{1}}\right)s^{p-1}{\nabla }_s{\phi }_n\left(s\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{p}{2}}\right)s^{p-2}{\nabla }_s^2{\phi }_n\left(s\right)+\cdots +{(-1)}^p{\nabla }_s^p{\phi }_n\left(s\right),$$

(18)

and we notice that ${\nabla }_s^p{\phi }_n\left(s\right)$ for $s=1$ becomes ${\nabla }^p{\phi }_n$.

Theorem 3.

Let $f\left(t\right)\risingdotseq {\phi }_n\left(s\right)$. Then, there holds

$$f^{\left(p\right)}\left(t\right)\risingdotseq \left\{\begin{array}{c}{\nabla }_s^p{\phi }_n\left(s\right),p⩽n\hfill \\ {\displaystyle {\nabla }_s^p{\phi }_n-{(-1)}^n\sum _{j=1}^{p-n}\left(\genfrac{}{}{0pt}{}{p-j}{n}\right)s^{p-j}{f}^{(j-1)}\left(0\right),n⩽p-1.}\hfill \end{array}\right.$$

In other words, pth derivative of a function f is mapped by the Laplace-type transform $\mathcal{G}$ to pth backward difference of the ${\phi }_n$ defined by (3), if $p⩽n$.

Proof. 

We use the Laplace transform of the pth derivative of the function f, which is

$$\mathcal{L}\left\{f^{\left(p\right)}\left(t\right)\right\}\left(s\right)=s^p\mathcal{L}\left\{f\left(t\right)\right\}\left(s\right)-\sum _{j=1}^p{s}^{p-j}{f}^{(j-1)}\left(0\right).$$

(19)

By virtue of (19) and (8), there follows

$$\begin{array}{cc}\hfill f^{\left(p\right)}\left(t\right)& \risingdotseq {\displaystyle \frac{{(-1)}^n}{n!}}\mathcal{L}{\left\{f^{\left(p\right)}\left(t\right)\right\}}^{\left(n\right)}\left(s\right)\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^n}{n!}}{\displaystyle \frac{d^n}{d{s}^n}}\left(s^p\mathcal{L}\left\{f\left(t\right)\right\}\left(s\right)\right)-{\displaystyle \frac{{(-1)}^n}{n!}}{\displaystyle \frac{d^n}{d{s}^n}}\sum _{j=1}^p{s}^{p-j}{f}^{(j-1)}\left(0\right).\hfill \end{array}$$

However, for $p⩽n$ there holds

$${\displaystyle \frac{d^n}{d{s}^n}}\sum _{j=1}^p{s}^{p-j}{f}^{(j-1)}\left(0\right)=0.$$

So, there remains to calculate the first part

$$\begin{array}{cc}\hfill f^{\left(p\right)}\left(t\right)& \risingdotseq {\displaystyle \frac{{(-1)}^n}{n!}}{\left(s^p\mathcal{L}\left\{f\left(t\right)\right\}\left(s\right)\right)}^{\left(n\right)}={\displaystyle \frac{{(-1)}^n}{n!}}\sum _{j=0}^n\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j}}\right){\left(s^p\right)}^{\left(j\right)}\mathcal{L}{\left\{f\left(t\right)\right\}}^{(n-j)}\left(s\right)\hfill \\ \hfill & =\sum _{j=0}^n{(-1)}^j{\displaystyle \frac{p^{\left(j\right)}}{j!}}{s}^{p-j}{\displaystyle \frac{{(-1)}^{n-j}}{(n-j)!}}\mathcal{L}{\left\{f\left(t\right)\right\}}^{(n-j)}\left(s\right)\hfill \\ \hfill & =\sum _{j=0}^p{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{p}{j}}\right)s^{p-j}{\phi }_{n-j}\left(s\right)={\nabla }_s^p{\phi }_n\left(s\right),\hfill \end{array}$$

where $p^{\left(j\right)}$ stands for the falling factorial, i.e., $p^{\left(j\right)}=p(p-1)\cdots (p-j+1)$. For $s=1$, we obtain $f^{\left(p\right)}\left(t\right)\risingdotseq {\nabla }^p{\phi }_n$.

Otherwise, for $n⩽p-1$, we have

$$\begin{array}{cc}\hfill \mathcal{G}\left\{f^{\left(p\right)}\left(t\right)\right\}\left(s\right)& ={\displaystyle \frac{{(-1)}^n}{n!}}{\displaystyle \frac{d^n}{d{s}^n}}\left(s^p\mathcal{L}\left\{f\left(t\right)\right\}\left(s\right)\right)-{\displaystyle \frac{{(-1)}^n}{n!}}{\displaystyle \frac{d^n}{d{s}^n}}\sum _{j=1}^p{s}^{p-j}{f}^{(j-1)}\left(0\right)\hfill \\ \hfill & =\sum _{j=0}^n{(-1)}^j{\displaystyle \frac{p^{\left(j\right)}}{j!}}{s}^{p-j}{\displaystyle \frac{{(-1)}^{n-j}}{(n-j)!}}\mathcal{L}{\left\{f\left(t\right)\right\}}^{(n-j)}\left(s\right)\hfill \\ \hfill & -{(-1)}^n\sum _{j=1}^{p-n}{\displaystyle \frac{{(p-j)}^{\left(n\right)}}{n!}}{s}^{p-j-n}{f}^{(j-1)}\left(0\right).\hfill \end{array}$$

The Gamma function (1) often fills in the values of the factorial, and can be analytically extended to the whole complex plane, except for the non-positive integers, where it has poles. However, the value of its reciprocal function is zero. So, this allows the summation index j in the first sum to run up to p because the values of $1/(n-j)!=1/\Gamma (n-j+1)$ for $j=n+1,\dots ,p$ are equal to zero! In that case, this sum is actually ${\nabla }_s^p{\phi }_n\left(s\right)$, so that we have

$$f^{\left(p\right)}\left(t\right)\risingdotseq {\nabla }_s^p{\phi }_n\left(s\right)-{(-1)}^n\sum _{j=1}^{p-n}\left({\displaystyle \genfrac{}{}{0pt}{}{p-j}{n}}\right)s^{p-j-n}{f}^{(j-1)}\left(0\right),$$

whereby we complete the proof. □

Lemma 5.

If $f\left(t\right)\risingdotseq {\phi }_n\left(s\right)$ and $r\in \mathbb{N}$, then $t^{r}f\left(t\right)\risingdotseq {\nabla }_s^p{\phi }_{n+r}\left(s\right)$ for $r+n⩾p$. If $r+n⩽p-1$ there holds

$$t^r{f}^{\left(p\right)}\left(t\right)\risingdotseq {(n+1)}_r\left({\nabla }_s^p{\phi }_{n+r}\left(s\right)-{(-1)}^{n+r}\sum _{j=1}^{p-n-r}\left({\displaystyle \genfrac{}{}{0pt}{}{p-j}{n+r}}\right)s^{p-j-n-r}{f}^{j-1}\left(0\right)\right).$$

where ${(n+1)}_r=(n+1)(n+2)(n+3)\cdots (n+r),r\in \mathbb{N}$. In the special case, for $p=0$, we have

$$t^{r}f\left(t\right)\risingdotseq {(n+1)}_r{\phi }_{n+r}\left(s\right).$$

Proof. 

Making use of one of the elemental properties of the Laplace transform

$$\mathcal{L}\left\{t^{r}g\left(t\right)\right\}\left(s\right)={(-1)}^r{\displaystyle \frac{d^r}{d{s}^r}}\mathcal{L}\left\{g\left(t\right)\right\}\left(s\right),$$

whence, after replacing $g\left(t\right)$ with $f^{\left(s\right)}\left(t\right)$ and applying (8), we obtain

$$\begin{array}{cc}\hfill t^r{f}^{\left(p\right)}\left(t\right)& \risingdotseq {\displaystyle \frac{{(-1)}^n}{n!}}{\displaystyle \frac{d^n}{d{s}^n}}\mathcal{L}\left\{t^r{f}^{\left(s\right)}\left(t\right)\right\}\left(s\right)={\displaystyle \frac{{(-1)}^n}{n!}}{\displaystyle \frac{d^n}{d{s}^n}}{(-1)}^r{\displaystyle \frac{d^r}{d{s}^r}}\mathcal{L}\left\{f^{\left(s\right)}\left(t\right)\right\}\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^{n+r}}{n!}}{\displaystyle \frac{d^{r+n}}{d{s}^{r+n}}}\mathcal{L}\left\{f^{\left(s\right)}\left(t\right)\right\}\left(s\right)={(n+1)}_r{\displaystyle \frac{{(-1)}^{n+r}}{(n+r)!}}{\displaystyle \frac{d^{r+n}}{d{s}^{r+n}}}\mathcal{L}\left\{f^{\left(s\right)}\left(t\right)\right\}\left(s\right).\hfill \end{array}$$

For $p=0$, on the basis of Lemma 1, one obtains the equality $t^{r}f\left(t\right)\risingdotseq {(n+1)}_r{\phi }_{n+r}\left(s\right)$. Let $p⩾1$. In view of (19), we find

$$t^r{f}^{\left(p\right)}\left(t\right)\risingdotseq {(n+1)}_r{\displaystyle \frac{{(-1)}^{n+r}}{(n+r)!}}\left({\displaystyle \frac{d^{r+n}}{d{s}^{r+n}}}\left(s^p\mathcal{L}\left\{f\left(t\right)\right\}\left(s\right)\right)-{\displaystyle \frac{d^{r+n}}{d{s}^{r+n}}}\sum _{j=1}^p{s}^{p-j}{f}^{(j-1)}\left(0\right)\right).$$

If $n+r⩾p$, the $(n+r)$th derivative of the second sum becomes zero, so there remains

$$\begin{array}{cc}\hfill t^r{f}^{\left(p\right)}\left(t\right)& \risingdotseq {(n+1)}_r{\displaystyle \frac{{(-1)}^{n+r}}{(n+r)!}}\sum _{j=0}^{r+n}\left({\displaystyle \genfrac{}{}{0pt}{}{r+n}{j}}\right){\left(s^p\right)}^{\left(j\right)}\mathcal{L}{\left\{f\left(t\right)\right\}}^{(r+n-j)}\left(s\right)\hfill \\ \hfill & ={(n+1)}_r\sum _{j=0}^{r+n}{(-1)}^j{\displaystyle \frac{{\left(s^p\right)}^{\left(j\right)}}{j!}}{(-1)}^{n+r-j}{\displaystyle \frac{\mathcal{L}{\left\{f\left(t\right)\right\}}^{(r+n-j)}\left(s\right)}{(r+n-j)!}}\hfill \\ \hfill & ={(n+1)}_r\sum _{j=0}^{r+n}{(-1)}^j{\displaystyle \frac{p^{\left(j\right)}}{j!}}{s}^{p-j}{\phi }_{r+n-j}\left(s\right)\hfill \\ \hfill & ={(n+1)}_r\sum _{j=0}^p{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{p}{j}}\right)s^{p-j}{\phi }_{r+n-j}\left(s\right)={\nabla }_s^p{\phi }_{r+n}\left(s\right).\hfill \end{array}$$

Setting $s=1$ gives rise to the equality $t^r{f}^{\left(s\right)}\left(t\right)\risingdotseq {(n+1)}_r{\nabla }^p{f}_{n+r}$.

If $n+r⩽p-1$, there follows

$$\begin{array}{cc}\hfill t^r{f}^{\left(p\right)}\left(t\right)& \risingdotseq {(n+1)}_r\sum _{j=0}^{r+n}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{p}{j}}\right)s^{p-j}{\phi }_{r+n-j}\left(s\right)-{\displaystyle \frac{{(-1)}^{n+r}}{n!}}{\displaystyle \frac{d^{r+n}}{d{s}^{r+n}}}\sum _{j=1}^p{s}^{p-j}{f}^{(j-1)}\left(0\right)\hfill \\ \hfill & ={(n+1)}_r\sum _{j=0}^{r+n}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{p}{j}}\right)s^{p-j}{\phi }_{r+n-j}\left(s\right)\hfill \\ \hfill & -{(-1)}^{n+r}\sum _{j=1}^{p-r-n}{\displaystyle \frac{{(p-j)}^{(r+n)}}{n!}}{s}^{p-j-r-n}{f}^{(j-1)}\left(0\right)\hfill \\ \hfill & ={(n+1)}_r\sum _{j=0}^{r+n}{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{p}{j}}\right)s^{p-j}{\phi }_{r+n-j}\left(s\right)\hfill \\ \hfill & -{(-1)}^{n+r}{(n+1)}_r\sum _{j=1}^{p-r-n}{\displaystyle \frac{{(p-j)}^{(r+n)}}{(r+n)!}}{s}^{p-j-r-n}{f}^{(j-1)}\left(0\right).\hfill \end{array}$$

As in the proof of Theorem 3, we allow j in the first sum to run up to $r+n$, so that we obtain

$$t^r{f}^{\left(p\right)}\left(t\right)\risingdotseq {(n+1)}_r\left({\nabla }_s^p{\phi }_{r+n}\left(s\right)-{(-1)}^{n+r}\sum _{j=1}^{p-r-n}\left({\displaystyle \genfrac{}{}{0pt}{}{p-j}{r+n}}\right)s^{p-j-r-n}{f}^{(j-1)}\left(0\right)\right).$$

Hereby, we complete the proof. □

6. The Laplace-Type Transform and Fractional Derivative

For $m⩽\alpha <m+1$, the Riemann–Liouville definition of the fractional derivative of a function $f\left(t\right)$ is (see [8])

$$D^{\alpha }f={\displaystyle \frac{1}{\Delta (1-\alpha )}}{\displaystyle \frac{d^{m+1}}{d{t}^{m+1}}}{\int }_0^t{(t-\tau )}^{m-\alpha }f\left(\tau \right)d\tau ,$$

(20)

where D is a differential operator. By repeatedly performing integration by parts and differentiation, this gives

$$D^{\alpha }f=\sum _{k=0}^m{\displaystyle \frac{t^{k-\alpha }{f}^{\left(k\right)}\left(0\right)}{\Gamma (k-\alpha +1)}}+{\displaystyle \frac{1}{\Gamma (m-\alpha +1)}}{\int }_0^t{(t-\tau )}^{m-\alpha }{f}^{(m+1)}\left(\tau \right)d\tau .$$

Let $m=0$ and $g\left(t\right)$, denote the integral in (20), and $G\left(s\right)=\mathcal{L}\left\{g\right\}\left(s\right)$. Knowing that $\mathcal{L}\left\{g^{\prime }\right\}\left(s\right)=sG\left(s\right)-g(0+)$, and because $g(0+)=0$, we find

$$\mathcal{L}\left\{D^{\alpha }f\right\}\left(s\right)={\displaystyle \frac{sG\left(s\right)}{\Gamma (1-\alpha )}}={\displaystyle \frac{s}{\Gamma (1-\alpha )}}\mathcal{L}\{t^{-\alpha }\ast f\}\left(s\right)=s\mathcal{L}\left\{t^{-\alpha }\right\}\mathcal{L}\left\{f\right\}\left(s\right)=s^{\alpha }F\left(s\right).$$

Replacing $f\left(t\right)$ with $D^{\alpha }f\left(t\right)$ in (3), according to (8), we obtain

$$\begin{array}{cc}\hfill {\displaystyle \mathcal{G}\left\{D^{\alpha }f\right\}\left(s\right)=\frac{{(-1)}^n}{n!}{\left(s^{\alpha }F\left(s\right)\right)}^{\left(n\right)}}& {\displaystyle =\frac{{(-1)}^n}{n!}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){\left(s^{\alpha }\right)}^{\left(k\right)}{F}^{(n-k)}\left(s\right)}\hfill \\ \hfill & {\displaystyle =\frac{{(-1)}^n}{n!}\sum _{k=0}^n\left(\genfrac{}{}{0pt}{}{n}{k}\right){\alpha }^{\left(k\right)}{s}^{\alpha -k}{F}^{(n-k)}\left(s\right).}\hfill \end{array}$$

(21)

where ${\alpha }^{\left(k\right)}=\alpha (\alpha -1)\cdots (\alpha -k+1)$ is falling factorial. After rearranging (21), and relying on (8), we rewrite it as follows

$$\mathcal{G}\left\{D^{\alpha }f\right\}\left(s\right)=\sum _{k=0}^n{(-1)}^k{\displaystyle \frac{{\alpha }^{\left(k\right)}}{k!}}{s}^{\alpha -k}{\displaystyle \frac{{(-1)}^{n-k}}{(n-k)!}}{F}^{(n-k)}\left(s\right)=\sum _{k=0}^n{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{\alpha }{k}}\right)s^{\alpha -k}{\phi }_{n-k}\left(s\right).$$

So, the Laplace-type transform maps the fractional derivative $D^{\alpha }f$ to a sum in terms of the sequence $\left\{{\phi }_n\left(s\right)\right\}$.

7. Applications

We shall demonstrate how, by applying the Laplace-type transform, one can solve difference equations and obtain combinatorial identities based on orthogonal polynomials.

7.1. Solving Difference Equations

We can use the Laplace-type transform $\mathcal{G}$ and its inverse transform $\mathcal{R}$ to solve some difference equations.

Example 2.

Fibonacci numbers are solutions of the difference equation $F_n=F_{n-1}+F_{n-2}$. So, we deal first with the equation:

$${\phi }_n\left(s\right)-{\phi }_{n-1}\left(s\right)-{\phi }_{n-2}\left(s\right)=0,$$

and by using (18), it is transformed into

$${\phi }_n\left(s\right)-s{\phi }_n\left(s\right)+{\nabla }_s{\phi }_n\left(s\right)-s^2{\phi }_n\left(s\right)+2s{\nabla }_s{\phi }_n\left(s\right)-{\nabla }_s^2{\phi }_n\left(s\right)=0$$

or

$$(1-s-s^2){\phi }_n\left(s\right)+(1+2s){\nabla }_s{\phi }_n\left(s\right)-{\nabla }_s^2{\phi }_n\left(s\right)=0$$

where $Q_2\left(s\right)=1-s-s^2$. By setting $s=1$, it leads to the difference equation

$$F_n-3\nabla F_n+{\nabla }^2{F}_n=0,$$

which is the result of applying Theorem 3, i.e., $\mathcal{G}\left\{y\left(t\right)\right\}=F_n$, $\mathcal{G}\left\{y^{\prime }\right\}=\nabla F_n$, $\mathcal{G}\left\{y^{″}\right\}={\nabla }^2{F}_n$, to the differential equation

$$y^{″}-3y^{\prime }+y=0.$$

We shall solve Cauchy’s problem $y\left(0\right)=0y^{\prime }\left(0\right)=1$. At first, we find its general solution

$$y=C_1{e}^{{\displaystyle \frac{3+\sqrt{5}}{2}}t}+C_2{e}^{{\displaystyle \frac{3-\sqrt{5}}{2}}t},$$

and taking into account Cauchy’s conditions, we obtain $C_1={\displaystyle \frac{1}{\sqrt{5}}}$, $C_2=-{\displaystyle \frac{1}{\sqrt{5}}}$, and as a particular solution

$$y={\displaystyle \frac{1}{\sqrt{5}}}{e}^{{\displaystyle \frac{t}{2}}(3+\sqrt{5})}-{\displaystyle \frac{1}{\sqrt{5}}}{e}^{{\displaystyle \frac{t}{2}}(3-\sqrt{5})}.$$

However, applying the Laplace-type transform to this particular solution of the differential equation and using Lemma 1, we find

$$F_n={\displaystyle \frac{1}{\sqrt{5}}}{\left({\displaystyle \frac{\sqrt{5}+1}{2}}\right)}^n-{\displaystyle \frac{1}{\sqrt{5}}}{\left({\displaystyle \frac{1-\sqrt{5}}{2}}\right)}^n,n=0,1,2,\dots .$$

This is a particular solution of the Fibonacci difference equation, well-known as Binet’s formula.

Example 3.

Consider the difference equation

$$a_0{f}_n+a_1{f}_{n-1}+\dots +a_p{f}_{n-p}=g_n,a_0,a_1,\dots ,a_p\in \mathbb{R}.$$

(22)

We regard it as a special case of the functional difference equation

$$a_0{\phi }_n\left(s\right)+a_1{\phi }_{n-1}\left(s\right)+\dots +a_p{\phi }_{n-p}\left(s\right)={\gamma }_n\left(s\right)$$

for $s=1$. Applying (18), we find

$$Q_p\left(s\right){\phi }_n\left(s\right)-{\displaystyle \frac{Q_p^{\prime }\left(s\right)}{1!}}{\nabla }_s{\phi }_n\left(s\right)+{\displaystyle \frac{Q_p^{″}\left(s\right)}{2!}}{\nabla }_s^2{\phi }_n\left(s\right)+\dots +{(-1)}^p{\displaystyle \frac{Q_p^{\left(s\right)}\left(s\right)}{p!}}{\nabla }_s^p{\phi }_n\left(s\right)={\gamma }_n\left(s\right),$$

where $Q_p\left(s\right)=a_0+a_{1}s+a_2{s}^2+\cdots +a_p{s}^p$. Taking $s=1$ gives rise to the difference equation

$$b_0{f}_n+b_1\nabla f_n+b_2{\nabla }^2{f}_n+\cdots +b_{p-1}{\nabla }^{p-1}+b_p{\nabla }^p{f}_n={\gamma }_n,$$

(23)

where ${\displaystyle b_k={(-1)}^k\frac{Q_p^{\left(k\right)}\left(1\right)}{k!},k=0,1,\dots ,p}$. Thus, (22) and (23) are equivalent. Relying on Theorem 3, using $\mathcal{G}\left\{y^{\left(k\right)}\left(t\right)\right\}={\nabla }^k{f}_n$, $k=0,1,\dots ,p$, we conclude that the differential equation:

$$L\left[y\right]=b_p{y}^{\left(s\right)}+b_{p-1}{y}^{(p-1)}+\cdots +b_2{y}^{″}+b_1{y}^{\prime }+b_{0}y=g\left(t\right)$$

gives rise to the difference Equation (23). We search for a solution of Cauchy’s problem $L\left[Y\right]=0$, $Y^{\left(k\right)}\left(0\right)=0$, $k=0,1,\dots ,n-2$, $Y^{(n-1)}\left(0\right)=1$, which is in [9] given by

$$y={\int }_0^{t}Y(t-\tau )g\left(\tau \right)d\tau .$$

If we now apply Lemma 3, so that $\mathcal{G}\left\{y\left(t\right)\right\}=f_n$, $\mathcal{G}\left\{Y\left(t\right)\right\}=Y_n$ and $\mathcal{G}\left\{g\left(t\right)\right\}=g_n$, we solve the difference Equation (22) in the form of

$$f_n=\sum _{k=0}^n{Y}_{n-k}{g}_k.$$

Example 4.

Consider the system of difference equations

$${\nabla }_s^2{X}_n=Y_n,{\nabla }_s^2{Y}_n=-2X_n.$$

Applying the $\mathcal{R}$ transform to this system, we come to the system of differential equations:

$${\displaystyle \frac{d^{2}x}{d{t}^2}}=y,{\displaystyle \frac{d^{2}y}{d{t}^2}}=-2x.$$

The general solution of this system is

$$x=e^t(C_{1}cost+C_{2}sint),y=e^t(C_{1}sint-C_{2}cost),$$

where $C_1$ and $C_2$ are arbitrary constants. Application of the Laplace-type transform to this solution yields the general solution of the above system of difference equations

$$X_n=(C_{1}cos\alpha +C_{2}sin\alpha )A\left(s\right),Y_n=(C_{1}sin\alpha +C_{2}cos\alpha )A\left(s\right),$$

where ${\displaystyle A=\frac{1}{{\left(\sqrt{s^{2}s+2}\right)}^{n+1}}}$, ${\displaystyle \alpha =(n+1)\mathrm{arctg}\frac{1}{s-1}}$.

Example 5.

By applying the Laplace-type transform to the integral equation (see [10])

$${\int }_0^x{(x-t)}^{\beta }f\left(t\right)dt=x^{\lambda }(\lambda ⩾0,\beta ⩾-1,\lambda ,\beta \in \mathbb{R}),$$

we shall map it to the discrete equation

$$\sum _{k=0}^n{\displaystyle \frac{\Gamma (n-k+\beta +1)}{s^{\beta -k}}}{\phi }_k\left(s\right)={\displaystyle \frac{\Gamma (n+\lambda +1)}{s^{\lambda }}}.$$

The solution of the integral equation is given by

$${\displaystyle f\left(x\right)=\frac{\Gamma (\lambda +1)}{\Gamma (\beta +1)\Gamma (\lambda -\beta )}{x}^{\lambda -\beta -1}(\lambda -\beta +k\ne 0,k\in {\mathbb{N}}_0).}$$

Applying the Laplace-type transform to it, the solution of the discrete equation is as follows:

$${\displaystyle {\phi }_k\left(s\right)=\frac{\Gamma (\lambda +1)\Gamma (n+\lambda -\beta +2)}{n!\Gamma (\beta +1)\Gamma (\lambda -\beta )}{s}^{n+\lambda -\beta -1}(\lambda -\beta +k\ne 0,k\in {\mathbb{N}}_0).}$$

Example 6.

The solution of the system of integral equations (see [10])

$$\left\{\begin{array}{c}{\displaystyle f_1\left(t\right)=1-2{\int }_0^t{e}^{2(t-\tau )}{f}_1\left(\tau \right)d\tau +{\int }_0^t{f}_2\left(\tau \right)d\tau ,}\hfill \\ {\displaystyle f_2\left(t\right)=4t-{\int }_0^t{f}_1\left(\tau \right)d\tau +4{\int }_0^t(t-\tau )f_2\left(\tau \right)d\tau }\hfill \end{array}\right.$$

(24)

is $f_1\left(t\right)=e^{-t}-t{e}^{-t}$, $f_2\left(t\right)={\displaystyle \frac{8}{9}}{e}^{2t}+{\displaystyle \frac{1}{3}}t{e}^{-t}-{\displaystyle \frac{8}{9}}{e}^{-t}$.

Denote $\mathcal{G}\left\{f_1\right\}\left(s\right)=\left\{{\phi }_n\left(s\right)\right\}$ and $\mathcal{G}\left\{f_2\right\}\left(s\right)=\left\{{\psi }_n\left(s\right)\right\}$. We apply the Laplace-type transform, and use Lemmas 2 and 3 to map (24) to the system of equations

$$\left\{\begin{array}{c}{\displaystyle {\phi }_n\left(s\right)=\frac{1}{s^{n+1}}-2\sum _{k=0}^n\frac{{\phi }_k\left(s\right)}{{(s-2)}^{n-k+1}}+\sum _{k=0}^n\frac{{\psi }_k\left(s\right)}{s^{n-k+1}},}\hfill \\ {\displaystyle {\psi }_n\left(s\right)=\frac{4(n+1)}{s^{n+2}}-\sum _{k=0}^n\frac{{\phi }_k\left(s\right)}{s^{n-k+1}}+4\sum _{k=0}^n\frac{(n-k+1){\psi }_k\left(s\right)}{s^{n-k+2}}.}\hfill \end{array}\right.$$

(25)

By applying the Laplace-type transform to $f_1\left(t\right)$ and $f_2\left(t\right)$, we find the solution of (25)

$$\left\{\begin{array}{c}{\displaystyle {\phi }_n\left(s\right)=\frac{1}{{(s+1)}^{n+1}}-\frac{n+1}{{(s+1)}^{n+2}},}\hfill \\ {\displaystyle {\psi }_n\left(s\right)=\frac{8}{9{(s-2)}^{n+1}}+\frac{n+1}{3{(s+1)}^{n+2}}-\frac{8}{9{(s+1)}^{n+1}}.}\hfill \end{array}\right.$$

Example 7.

Consider the integro-differential equation (see [10])

$$f^{″}\left(t\right)+{\int }_0^t{e}^{2(t-\tau )}{f}^{\prime }\left(\tau \right)d\tau =e^{2t},f\left(0\right)=0,f^{\prime }\left(0\right)=0.$$

Here is $p=2$, so we rely on Theorem 3 for $n⩾2$, leverage Lemma 2 as well as the convolution of $e^{2(t-\tau )}$ and $f^{\prime }\left(\tau \right)$, and we apply the Laplace-type transform to map the above equation to the equation

$${\nabla }_s^2{\phi }_n\left(s\right)+\sum _{k=0}^n{\displaystyle \frac{{\nabla }_s{\phi }_k\left(s\right)}{{(s-2)}^{n-k+1}}}={\displaystyle \frac{1}{{(s-2)}^{n+1}}}.$$

(26)

Applying the Laplace-type transform now to the solution $f\left(t\right)=t{e}^t-e^t+1$ of the integro-differential equation provides the solution of (26)

$${\phi }_n\left(s\right)={\displaystyle \frac{n+1}{{(s-1)}^{n+2}}}-{\displaystyle \frac{1}{{(s-1)}^{n+1}}}+{\displaystyle \frac{1}{s^{n+1}}}(n⩾2).$$

7.2. Some Identities Based on Orthogonal Polynomials

Relying on the classical orthogonal polynomials and applying the Laplace-type transform leads us to some interesting combinatorial identities.

Example 8.

The explicit representation of the standard Laguerre polynomials is

$$L_m\left(x\right)=\sum _{k=0}^m{\displaystyle \frac{{(-1)}^k}{k!}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)x^k.$$

Applying the Laplace-type transform, we map them to

$$L_m\left(x\right)\risingdotseq \sum _{k=0}^m{\displaystyle \frac{{(-1)}^{n+k}}{s^{n+k+1}}}\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+k}{k}}\right).$$

Taking this into account and applying the Laplace-type transform to the algebraic equality involving Laguerre polynomials (see [5])

$$\sum _{k=0}^m{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)L_k\left(x\right)={\displaystyle \frac{x^m}{m!}},$$

we obtain the combinatorial identity

$$\sum _{k=0}^m\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)\sum _{j=0}^k{\displaystyle \frac{{(-1)}^{k+j}}{s^j}}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+j}{j}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{m+n}{n}}\right){\displaystyle \frac{1}{s^m}}.$$

Example 9.

To apply the Laplace-type transform to Legendre polynomials

$$P_n\left(x\right)={\displaystyle \frac{1}{2^n}}\sum _{k=0}^{\left[{\displaystyle \frac{n}{2}}\right]}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2k}{n}}\right)x^{n-2k},x\in (-1,1),$$

we first have to map the interval $(-1,1)$ on to the interval $(0,+\infty )$. For that purpose, we make use of the function $x=1-2e^{-t}$, whereby

$$p_n\left(t\right)=P_n(1-2e^{-t})={\displaystyle \frac{1}{2^n}}\sum _{k=0}^{\left[{\displaystyle \frac{n}{2}}\right]}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2k}{n}}\right)\sum _{j=0}^{n-2k}\left({\displaystyle \genfrac{}{}{0pt}{}{n-2k}{j}}\right){(-2e^{-t})}^j.$$

We can rewrite this after interchanging the order of summation in the form of

$$p_n\left(t\right)=\sum _{j=0}^n{e}^{-jt}\sum _{k=0}^{\left[{\displaystyle \frac{n}{2}}\right]}{\displaystyle \frac{{(-1)}^{j+k}}{2^{n-j}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2k}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n-2k}{j}}\right)=\sum _{j=0}^n{A}_{n,j}{e}^{-jt},$$

where

$$A_{n,j}=\sum _{k=0}^{\left[{\displaystyle \frac{n}{2}}\right]}{\displaystyle \frac{{(-1)}^{j+k}}{2^{n-j}}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2n-2k}{n}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n-2k}{j}}\right).$$

(27)

By means of the function $x=1-2e^{-t}$ Bonnet’s recursion formula of Legendre polynomials

$$(m+1)P_{m+1}\left(x\right)=(2m+1)x{P}_m\left(x\right)-m{P}_{m-1}\left(x\right)$$

is transformed to

$$(m+1)p_{m+1}\left(t\right)=(2m+1)(1-2e^{-t})p_m\left(t\right)-m{p}_{m-1}\left(t\right).$$

So, we have

$$(m+1)\sum _{j=0}^{m+1}{A}_{m+1,j}{e}^{-jt}=(2m+1)(1-2e^{-t})\sum _{j=0}^m{A}_{m,j}{e}^{-jt}-m\sum _{j=0}^{m-1}{A}_{m-1,j}{e}^{-jt}.$$

After rearrangement and comparing corresponding coefficients, we obtain combinatorial identities. First of all, we obtain simple ones. The equalities $(m+1)A_{m+1,m+1}=-2(2m+1)A_{m,m}$ and $(m+1)A_{m+1,m}=(2m+1)(A_{m,m}-2A_{m,m-1})$ both yield

$${\displaystyle 1^{\circ }\left(\genfrac{}{}{0pt}{}{2m+2}{m+1}\right)=\frac{4m+2}{m+1}\left(\genfrac{}{}{0pt}{}{2m}{m}\right),}$$

but the equality $(m+1)A_{m+1,0}=(2m+1)A_{m,0}-m{A}_{m-1,0}$ yields

$$\begin{array}{c}{\displaystyle 2^{\circ }\sum _{k=0}^{\left[\frac{m+1}{2}\right]}{(-1)}^k\left(\genfrac{}{}{0pt}{}{m+1}{k}\right)\left(\genfrac{}{}{0pt}{}{2m+2-2k}{m+1}\right)=\frac{4m+2}{m+1}\sum _{k=0}^{\left[\frac{m}{2}\right]}{(-1)}^k\left(\genfrac{}{}{0pt}{}{m}{k}\right)\left(\genfrac{}{}{0pt}{}{2m-2k}{m}\right)}\hfill \\ \hfill -{\displaystyle \frac{4m}{m+1}}\sum _{k=0}^{\left[{\displaystyle \frac{m-1}{2}}\right]}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m-1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2m-2-2k}{m-1}}\right).\end{array}$$

More generally, the equality $(m+1)A_{m+1,j}=(2m+1)(A_{m,j}-2A_{m,j-1})-m{A}_{m-1,j}$, where $j=1,2,\dots ,m-1$, yields

$$\begin{array}{c}{3}^{\circ }\sum _{k=0}^{\left[{\displaystyle \frac{m+1}{2}}\right]}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m+1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2m+2-2k}{m+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m+1-2k}{j}}\right)\hfill \\ ={\displaystyle \frac{4m+2}{m+1}}\sum _{k=0}^{\left[{\displaystyle \frac{m}{2}}\right]}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2m-2k}{m}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m-2k+1}{j}}\right)\hfill \\ -{\displaystyle \frac{4m}{m+1}}\sum _{k=0}^{\left[{\displaystyle \frac{m-1}{2}}\right]}{(-1)}^k\left({\displaystyle \genfrac{}{}{0pt}{}{m-1}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{2m-2-2k}{m-1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{m-1-2k}{j}}\right).\hfill \end{array}$$

8. Concluding Remarks

Apart from results concerned with solving difference equations and obtaining combinatorial identities, the application of the Laplace-type transform yields many other results.

By choosing various functions $f\left(t\right)$ in the Laplace-type transform, one can achieve different results. For instance, in [11], taking $f\left(x\right)=E_{\alpha ,\beta }\left(x^{\alpha }z\right)$, they make use of the integral

$${\int }_0^{\infty }{e}^{-x}{x}^{\beta -1}{E}_{\alpha ,\beta }\left(x^{\alpha }z\right)\mathrm{d}x={\displaystyle \frac{1}{1-z}}$$

fundamental in the evaluation of the Laplace transforms of the function $E_{\alpha ,\beta }(-\lambda t^{\alpha })$, with $x=t,z=-\lambda$, where

$$E_{\alpha ,\beta }^{\gamma }\left(\tau \right)=\sum _{n=0}^{\infty }{\displaystyle \frac{{\left(\gamma \right)}_n}{\Gamma (\alpha n+\beta )}}{\displaystyle \frac{{\tau }^n}{n!}}(\Re \left(\alpha \right)>0,\beta ,\gamma \in \mathbb{C},{\left(\gamma \right)}_n=\gamma (\gamma -1)\cdots (\gamma -n+1))$$

stands for the generalized three-parameter Mittag–Leffler function ($\gamma =1$ in the above example) occurring in many interesting applied problems involving fractional differential equations.

Also, we can obtain significant results through the Laplace-type transform involving generalized hypergeometric functions [12]

$${\int }_0^{\infty }{e}^{-st}{x}^{\alpha -1}{}_{p}F_q\left[{\displaystyle \genfrac{}{}{0pt}{}{a_1,\dots ,a_p}{b_1,\dots ,b_q}};t\right]\mathrm{d}t,$$

where the hypergeometric function ${}_{p}F_q$ takes the place of $f\left(t\right)$ in (1).