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\begin{document}
\title{Pseudo-derivative and Pseudo-integral of fractional order}
\author[A. Zohri]{$^1$Ali Zohri}
\address{$^1${Department} of Mathematics, Payame Noor University, I. R. of IRAN}
\email{zohri\_a@pnu.ac.ir}
\author[Sh. Jamshidzadeh]{$^2$Shabnam Jamshidzadeh}
\address{$^2${Department} of Mathematics, Payame Noor University, I. R. of IRAN}
\email{jamshidzadeh\_shabnam@yahoo.com}
%\subjclass[2000]{Primary 46H25; Secondary 46H20}
\keywords{Pseudo-addition, Pseudo-moltiplication, Fractional pseudo-derivative, Fractional pseudo-integral}
\dedicatory{}
\smallskip
%................................................................................................................
\begin{abstract}
Integral and differential of fractional order are important notion that is often used in dealing with Frechet geometry. Based on pseudo-operations given by monotone and continuous function $g$, we study pseudo-derivative and pseudo-integral of fractional order through this paper.
\end{abstract} \maketitle
%................................................................................................................
\section{Introduction}
%................................................................................................................
Motivation for the research presented here lies both in capability of the pseudo-analysis, generalization of the classical analysis, to extends the range of the possible applications. The pseudo-analysis, see [4,7,8,9], is based, instead of the usual field of real numbers, on a semiring acting on the real interval $[a,b]\subset [{-\infty},{+\infty}]$, denoting the corresponding operations as $\oplus$ (pseudo-addition) and $\odot$ (pseudo-multiplication) of the following form:
\[x\oplus y=g^{-1}(g(x)+g(y))~~~,~~~x\odot y=g^{-1}(g(x)g(y)),\]
where $g$ is a strictly monotone and continuous generating function.\\
In recent years fractional calculus [1,2,5,6] has received much interest due to its new applications. A recent set of applicatons in mechanical engineering and an electrical engineering control theory approach may be found in [1,3].
The fractional calculus is a field of mathematic study that grows out of the traditional definitions of the calculus integral and derivative operaters in much the same way fractional exponents is an outgrowth of exponents with integer value.
In this paper, in order to broaden the area of possible applications of fractional derivative and fractional integral, we shall give some notions and prove theorem from g-calculus which are analogous to the classical theorems of the usual calculus of fractional derivative and fractional integral.
\section{Preliminaries}
In this section we collect the necessary Definition and Basic notions.\\
Let $[a,b]$ be a closed subinterval of $[-\infty,+\infty]$ (in some cases semiclosed subintervals will be considered) and let $\preceq$ be a total order on $[a,b]$. A semiring is the structure $([a,b],\oplus,\odot)$ when the following holds:
\begin{itemize}
\item
$\oplus$ is pseudo-addition, i.e., a function $\oplus: [a,b]\times[a, b]\longrightarrow [a, b]$ which is commutative, non-decreasing (with respect to $\preceq$), associative and with a zero element, denoted by 0;
\item
$\odot$ is pseudo-multiplication, i.e., a function $\odot:[a,b]\times [a,b]\longrightarrow [a,b]$which is commutative, positively non-decreasing ($x\preceq y$ implies $x\odot z\preceq y\odot z$,$z\in [a,b]_{+}=\{x:x\in [a,b],0\preceq x\}$), associative and for which there exists a unit element denoted by 1;
\item
$0\odot x=0$
\item
$x\odot (y\oplus z)=(x\odot y)\oplus(x\odot z)$
\end{itemize}
\begin{defn}$[9]$
The pseudo-operations are defined by a monotone and continuous function $g:[a,b]\longrightarrow[0,\infty]$, i.e., pseudo-operations are given with
\begin{equation}
x\oplus y=g^{-1}(g(x)+g(y))~~~~,~~~~x\odot y=g^{-1}(g(x)g(y)).
\end{equation}
\end{defn}
\begin{rem}
For any pseudo-addition ${\oplus}$ with a generater $g$ there exists ${\oplus_{\lambda}}$ and ${\odot_{\lambda}}$ which generated by $g^{\lambda}$ (the function $g$ of the power ${\lambda}$), such that
\begin{equation}
x{\oplus_{\lambda}} y=(g^{\lambda})^{-1}(g^{\lambda}(x)+g^{\lambda}(y))~~~~,~~~~x{\odot_{\lambda}} y=(g^{\lambda})^{-1}(g^{\lambda}(x)g^{\lambda}(y)).
\end{equation}
For more details see [3,6,7,8].
\end{rem}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{$g$-derivative and $g$-integral}
\begin{defn}$[8]$
Let function $f$ be defined on the interval $[c,d]$ and with values in $[a,b]$. If $f$ is differentiable on $(c,d)$, and has same monotonicity as the function $g$ then we define the $g$-derivative of $f$ at the point $x\in(c,d)$ as
\begin{equation}
\frac{d^{\oplus}f(x)}{dx}=g^{-1}\Big(\frac{d}{dx}g(f(x))\Big).
\end{equation}
\end{defn}
\begin{defn}$[8]$
Let $g$ be a generating function defined on the interval $[a,b]$ and $\oplus$ and $\odot$ are pseudo-operations given by (1). The pseudo-integral for a function $f: [c,d]\longrightarrow [a,b]$ reduces on the $g$-integral as follows
\begin{equation}
\int_{[c,d]}^{\oplus}f(x)\ud x=g^{-1}\int_{c}^{d}g(f(x))\ud x.
\end{equation}
\end{defn}
\begin{thm}$[8]$
Suppose that $f$ is continuous on $[c,d]$. Then we have
\begin{equation}
\frac{d^\oplus}{dx}\Big(\int^{\oplus}_{[c,x]}f\ud x\Big)=f(x)
\end{equation}
for each $x\in(c,d)$.
\end{thm}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{The Gamma Function}
The gamma function is a continuous extension to the factorial function, which is only defined for the nonnegative integers. While there are other continuous extensions to the factorial function, the gamma function is the only one that is convex for positive real numbers [1,2,4,5].\\
The definition of the gamma function is given by
\begin{equation}
\Gamma(x)=\int^{\infty}_{0}t^{x-1}e^{-t}\ud t,
\end{equation}
for all $x\in R$. so, we have
\[\Gamma(1)=\int^{\infty}_{0}e^{-t}\ud t=-e^{-t}]^{\infty}_{0}=0-(-1)=1\]
Also, we can show $\Gamma(2)=\Gamma(1)=1$.\\
The beauty of the gamma function can be found in its properties. One of them, this function is unique in that the value for any quantity is, by consequence of the form of the integral, equivalent to that quantity $x$ minus one times the gamma of the quantity minuse one.
\begin{equation}
\Gamma(x+1)=x\Gamma(x)~~~x>0,
\end{equation}
Also we have
\begin{equation}
\Gamma(x+1)=x!~~~or~~~\Gamma(x)=(x-1)!~~~x>0,
\end{equation}
\begin{ex}
for non integer value we have
\begin{align*}
(\frac{7}{2})!&=\Gamma(\frac{7}{2}+1)=\frac{7}{2}\Gamma(\frac{7}{2})
=\frac{7}{2}\times\frac{5}{2}\Gamma(\frac{5}{2})\\
&=\cdots =\frac{7}{2}\times\frac{5}{2}\times\frac{3}{2}\times\frac{1}{2}\Gamma(\frac{1}{2})=\frac{105}{16}\Gamma(\frac{1}{2})
\end{align*}
and
\[\Gamma(\frac{1}{2})=\int^{\infty}_{0}t^{-1/2}e^{-t}\ud t=2\int^{\infty}_{0}e^{-u^2}\ud u=\sqrt{\pi}\]
Hence, we have
\[(\frac{7}{2})!=\frac{105}{16}\Gamma(\frac{1}{2})=\frac{105\sqrt{\pi}}{16}.\]
\end{ex}
(see [5,6]).\\
By above we can define derivative and integral from fractional order.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Fractional derivative and Fractional integral}
Assuming a function $f(x)$ that is defined where $x>0$, form the definite integral from $0$ to $x$. we have
\[(If)(x)=\int^{x}_{0}f(t)\ud t\]
Repeating this process gives
\[(I^{2}f)(x)=\int^{x}_{0}\Big(\int^{t}_{0}f(s)\ud s\Big)\ud t\]
and this can be extended arbitrarily.\\
The Cauchy formula for repeated integration, namely
\begin{equation}
(I^{n}f)(x)=\frac{1}{(n-1)!}\int^{x}_{0}(x-t)^{n-1}f(t)\ud t,
\end{equation}
leads to a straightforward way to a generalization for real n.
Simply using the Gamma function to remove the discrete nature of the factorial function (recalling that $\Gamma(n+1)=n!)$) gives us a natural candidate for fractional applications of the integral operater.
\begin{equation}
(I^{a}f)(x)=\frac{1}{\Gamma(a)}\int^{x}_{0}(x-t)^{a-1}f(t)\ud t,~~~~~a>0
\end{equation}
This is in fact a well-defined operater.\\
Let us assume that $f(x)$ is a monomial of the form $f(x)=x^{k}$. we have
\[(If)(x)=\frac{x^{k+1}}{k+1}\]
\[(I^{2}f)(x)=\frac{x^{k+2}}{(k+1)(k+2)}\]
\[\vdots\]
\[(I^{n}f)(x)=\frac{k!}{(n+k)!}x^{n+k}\]
Which, after replacing the factorials with the Gamma function, leads us to
\begin{equation}
(I^{a}f)(x)=\frac{\Gamma(k+1)}{\Gamma(a+k+1)}x^{a+k}.
\end{equation}
at the same way, for fractional derivative of function $f$ we have
\begin{equation}
\frac{d^a}{dx^{a}}x^{k}=\frac{\Gamma(k+1)}{\Gamma(k-a+1)}x^{k-a}~~~~x>0~~,~~~a>0
\end{equation}
for more details see $[4,6]$.\\
Another important issue is relation between fractional derivative and fractional integral where we have used the fundamental theorem of the usual calculus.
\begin{thm} $[6]$
Let $a>0$, we choose the integer value of $p$ where $p>a$, then we have
\begin{equation}
\frac{d^a}{dx^{a}}f(x)=\frac{d^p}{dx^{p}}(I^{p-a}f)(x).
\end{equation}
\end{thm}
\begin{ex}
For $k=1$ and $a=1/2$, we obtain the half-derivative of the function $f(x)=x$, where $p=1$ as
\begin{align*}
\frac{d^{1/2}}{dx^{1/2}}x=\frac{d}{dx}(I^{1-1/2}x)=\frac{d}{dx}(I^{1/2}x)&=\frac{d}{dx}\Big(\frac{\Gamma(1+1)}{\Gamma(1/2+1+1)}x^{1/2+1}\Big)\\
&=\frac{d}{dx}\Big(\frac{\Gamma(2)}{\Gamma(5/2)}x^{3/2}\Big)\\
&=\frac{d}{dx}\Big(\frac{1}{\frac{3}{2}\times\frac{1}{2}\Gamma(\frac{1}{2})}x^{3/2}\Big)\\
&=\frac{d}{dx}\Big(\frac{4}{3\sqrt{\pi}}x^{3/2}\Big)\\
&=\frac{4}{3\sqrt{\pi}}
\times\frac{3}{2}x^{1/2}=\frac{2\sqrt{x}}{\sqrt{\pi}}.
\end{align*}
\end{ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Main Results}
Now we define pseudo-derivative of fractional order as generalization of both fractional order and pseudo-derivative.
%.................................................................................................................................................
\subsection{Pseudo-derivative of fractional order}
\begin{defn}
Let the function $f(x)=x^{k}$ be defined on the interval $[c,d]$ and with values in $[a,b]$. If $f$ is differentiable on $(c,d)$, and has same monotonicity as the function $g$ then we define the fractional order of pseudo-derivative for function $f$ at the point $x\in(c,d)$ as
\begin{equation}
\frac{d^{(a)\oplus}f(x)}{dx^a}=g^{-1}\Big(\frac{d^a}{dx^a}g(f(x))\Big),
\end{equation}
where $x>0$ and $a>0$.
\end{defn}
\begin{ex}
Taking $g(x)=\sqrt{x}$, for function $f(x)=x$, using the ralation (13) for $p=1$, the pseudo half-derivative (means $a=\frac{1}{2}$) is given as
\begin{align*}
\frac{d^{(1/2)\oplus}}{dx^{1/2}}x&=g^{-1}\Big(\frac{d^{1/2}}{dx^{1/2}}g(x)\Big)=g^{-1}\Big(\frac{d^{1/2}}{dx^{1/2}}x^{1/2}\Big)\\
&=g^{-1}\frac{d}{dx}(I^{(1-1/2)}x^{1/2})\\
&=g^{-1}\Big(\frac{d}{dx}(\frac{\Gamma(1/2+1)}{\Gamma(1/2+1/2+1)}x^{1/2+1/2})\Big)\\
&=g^{-1}\Big(\frac{d}{dx}(\frac{\Gamma(3/2)}{\Gamma(2)}x)\Big)\\
&=g^{-1}\Big(\frac{d}{dx}(\frac{\sqrt{\pi}}{2}x)\Big)\\
&=g^{-1}(\frac{\sqrt{\pi}}{2})=(\frac{\sqrt{\pi}}{2})^{2}\\
&=\frac{\pi}{4}.
\end{align*}
\end{ex}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Pseudo-integral of fractional order}
\begin{defn}
Let $g$ be a generating and $\oplus$ and $\odot$ pseudo-operations given by (1). The pseudo-integral of fractional order for a function $f: [c,d]\longrightarrow [a,b]$ reduces on the fractional $g$-integral as follows
\end{defn}
\begin{equation}
(I^{(a)\oplus}f)(x)=g^{-1}\Big((I^{a}g(f))(x)\Big).
\end{equation}
where $f(x)=x^{k}$ and $x>0$ and $a>0$.
\begin{ex}
Let $g(x)=\sqrt{x}$ be continuous and strictly monotone function and $f(x)=x$ be identity function, then we have $g^{-1}(x)=x^{2}$, for pseudo integration of order $a=\frac{1}{2}$ we have
\begin{align*}
(I^{\oplus(a)}f)(x)&=g^{-1}\Big((I^{a}g(f))(x)\Big)\\
&=g^{-1}\Big((I^{1/2}(x^{1/2}))\Big)\\
&=g^{-1}\Big(\frac{\Gamma(k+1)}{\Gamma(a+k+1)}x^{a+k}\Big)\\
&=g^{-1}\Big(\frac{\Gamma(3/2)}{\Gamma(2)}x\Big)=g^{-1}\Big(\Gamma(3/2)x\Big)\\
&=g^{-1}\Big(1/2\Gamma(1/2)x\Big)=g^{-1}\Big(\frac{\sqrt{\pi}}{2}x\Big)\\
&=\Big(\frac{\sqrt{\pi}}{2}x\Big)^{2},
\end{align*}
(where $a=\frac{1}{2}$,~~~$k=\frac{1}{2}$,~~~$p=1$).
\end{ex}
\begin{thm}
If generating function $g$ is monotone bijection, then
\begin{equation}
\Gamma^{\oplus}(x\oplus 1)=x\odot\Gamma^{\oplus}(x).
\end{equation}
\end{thm}
\begin{proof}
\begin{align*}
\Gamma^{\oplus}(x\oplus 1)&=\Gamma^{\oplus}\Big(g^{-1}(g(x)+g(1))\Big)\\
&=g^{-1}\Big(\Gamma(g(g^{-1}(g(x)+g(1)))\big)\\
&=g^{-1}\Big(\Gamma(g(x)+g(1))\Big)\\
&=g^{-1}\Big(\Gamma(\sqrt{x}+1)\Big),
\end{align*}
on the other side we have
\begin{align*}
x\odot\Gamma^{\oplus}(x)&=g^{-1}\Big(g(x)g(\Gamma^{\oplus}(x)\Big)\\
&=g^{-1}\Big(g(x)g(g^{-1}(\Gamma(g(x))))\Big)\\
&=g^{-1}\Big(\sqrt{x}\Gamma(\sqrt{x})\Big).
\end{align*}
from (7) we have \[\Gamma(\sqrt{x}+1)=\sqrt{x}\Gamma(\sqrt{x})\]
Hence,
\[g^{-1}\Big(\Gamma(\sqrt{x}+1)\Big)=g^{-1}\Big(\sqrt{x}\Gamma(\sqrt{x})\Big)\]
which completes the proof.
\end{proof}
\begin{thm}
Let $g(u)=u^{k}$ be a strictly monotone and continuous function and $\oplus_{1/k}$ and $\odot_{1/k}$ operations of the form (2), then for $g^{-1}(u)=u^{1/k}$ and $f(x)=x$ we have
\begin{equation}
(I^{\oplus_{1/k}(a)}f)(x)=\frac{\Gamma^{\oplus_{1/k}}(k\oplus_{1/k} 1)}{\Gamma^{\oplus}(a\oplus_{1/k}k\oplus_{1/k}1)}\odot_{1/k} x^{(a{\oplus_{1/k}}k)}.
\end{equation}
\end{thm}
\begin{proof}
\begin{equation*}
(I^{\oplus_{1/k}(a)}f)(x)=g^{-1/k}\Big(I^{a}x^{k}\Big)=g^{-1/k}\Big(\frac{\Gamma(k+1)}{\Gamma(a+k+1)}x^{a+k}\Big)
\end{equation*}
on the other side we have
\begin{align*}
&\frac{\Gamma^{\oplus_{1/k}}(k{\oplus_{1/k}} 1)}{\Gamma^{\oplus_{1/k}}(a{\oplus_{1/a}}k{\oplus_{1/k}}1)}{\odot_{1/k}} x^{(a{\oplus_{1/k}}k)}\\
&=\frac{\Gamma^{\oplus_{1/k}}\Big(g^{-1/k}(g^{1/k}(k)+g(1))\Big)}
{\Gamma^{\oplus_{1/k}}\Big(g^{-1/k}(g^{1/a}(a)+g^{1/k}(k)+g(1)\Big)}
{\odot_{1/k}} x^{\Big(g^{-1/k}(g^{1/k}(k)+g^{1/a}(a))\Big)}\\
&=\frac{g^{-1/k}\Big(\Gamma(g^{1/k}(k)+g(1))\Big)}{g^{-1/k}\Big(\Gamma(g^{1/a}(a)+g^{1/k}(k)+g(1)\Big)}{\odot_{1/k}}x^{g^{-1/k}(a+k)}\\
&=\frac{g^{-1/k}\Big(\Gamma(k+1)\Big)}{g^{-1/k}\Big(\Gamma(a+k+1)\Big)}{\odot_{1/k}}x^{g^{-1/k}(a+k)}\\
&=g^{-1/k}\Big(\frac{(\Gamma(k+1)}{(\Gamma(a+k+1)\Big)}g^{1/k}(x^{g^{-1/k}(a+k)})\Big)\\
&=g^{-1/k}\Big(\frac{\Gamma(k+1)}{\Gamma(a+k+1)}x^{a+k}\Big),
\end{align*}
which proves the claim.
\end{proof}
Of the special interest for this paper is to provide estimation for pseudo-analysis, counterpart of relation (13).
\begin{thm}
Suppose that $f$ is continuous on $[c,d]$ and is differentiable on $(c,d)$, for every $a>0$, we choose integer value $p$ which $p>a$. Then we have
\begin{equation}
\frac{d^{(a)\oplus}}{dx^{a}}f(x)=\frac{d^{(p)\oplus}}{dx^{p}}(I^{(p-a)\oplus}f)(x).
\end{equation}
\end{thm}
\begin{proof}
\begin{align*}
\frac{d^{(p)\oplus}}{dx^{p}}(I^{(p-a)\oplus}f)(x)&=
\frac{d^{(p)\oplus}}{dx^{p}}\Big(g^{-1}(I^{(p-a)}g(f(x)))\Big)\\
&=g^{-1}\Big(\frac{d^p}{dx^{p}}g(g^{-1}(I^{(p-a)}g(f(x)))\Big)\\
&=g^{-1}\Big(\frac{d^p}{dx^{p}}(I^{(p-a)}g(f(x))\Big)\\
&=g^{-1}\Big(\frac{d^a}{dx^{a}}g(f(x))\Big)\\
&=\frac{d^{(a)\oplus}}{dx^{a}}f(x).
\end{align*}
\end{proof}
\begin{ex}
Let $g(x)=\sqrt{x}$, for $f(x)=x$, pseudo half-derivative of $f$ is given as
\begin{align*}
\frac{d^{(1/2)\oplus}}{dx^{1/2}}x=\frac{d^\oplus}{dx}(I^{(1-1/2)\oplus}x)&=\frac{d^\oplus}{dx}\Big(g^{-1}(I^{1/2}g(x))\Big)\\
&=\frac{d^\oplus}{dx}\Big(g^{-1}(I^{1/2}x^{1/2})\Big)\\
&=\frac{d^\oplus}{dx}\Big(g^{-1}(\frac{\Gamma(1/2+1)}{\Gamma(1/2+1/2+1)}x^{1/2+1/2})\Big)\\
&=\frac{d^\oplus}{dx}\Big(g^{-1}(\frac{\Gamma(3/2)}{\Gamma(2)}x)\Big)\\
&=\frac{d^\oplus}{dx}\Big(g^{-1}(\frac{\sqrt{\pi}}{2}x)\Big)\\
&=g^{-1}\Big(\frac{d}{dx}g(\frac{\sqrt{\pi}}{2}x)^{2}\Big)\\
&=g^{-1}\Big(\frac{d}{dx}(\frac{\sqrt{\pi}}{2}x)\Big)\\
&=g^{-1}(\frac{\sqrt{\pi}}{2})=(\frac{\sqrt{\pi}}{2})^{2}\\
&=\frac{\pi}{4}.
\end{align*}
(where $p=1$~~,~~$k=\frac{1}{2}$~~,~~$a=\frac{1}{2}$).
\end{ex}
\begin{ex}
Let $g(x)=\sqrt{x}$, for $f(x)=x^{-1}$, pseudo half-derivative of $f$ is given as
\begin{align*}
\frac{d^{(1/2)\oplus}}{dx^{1/2}}x^{-1}&=\frac{d^\oplus}{dx}(I^{(1-1/2)\oplus}x^{-1})=\frac{d^\oplus}{dx}\Big(g^{-1}(I^{1/2}g(x^{-1}))\Big)\\
&=\frac{d^\oplus}{dx}\Big(g^{-1}(I^{1/2}x^{-1/2})\Big)\\
&=\frac{d^\oplus}{dx}\Big(g^{-1}\Big(\frac{\Gamma(-1/2+1)}{\Gamma(-1/2+1/2+1)}x^{-1/2+1/2}\Big)\Big)\\
&=\frac{d^\oplus}{dx}\Big(g^{-1}\Big(\frac{\Gamma(1/2)}{\Gamma(1)}\Big)\Big)\\
&=g^{-1}\Big(\frac{d}{dx}g(g^{-1}(\sqrt{\pi}))\Big)\\
&=g^{-1}(\frac{d}{dx}\sqrt{\pi})=g^{-1}(0)=0.
\end{align*}
\end{ex}
(where $p=1$~~,~~$k=\frac{-1}{2}$~~,~~$a=\frac{1}{2}$).
\section{Conclusion}
In this paper, we have introduced pseudo-derivative and pseudo-integral of fractional order. Using pseudo-analysis, the relations between the integral of fractional order and pseudo-integral of fractional order was shown. We plan to work further on equations of fractional order in the pseudo-analysis framework.
%...................................................................................................
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