Partially Explore the Differences and Similarities between Riemann-Liouville Integral and Mellin Transform

Abstract

At present many researchers devote themselves to studying the relationship between continuous fractal functions and their fractional integral. But little attention is paid to the relationship between Mellin transform and fractional integral. This paper aims to partially explore the differences and similarities between Riemann-Liouville integral and Mellin transform, then a 1-dimensional continuous and unbounded variational function defined on the closed interval $[0,1]$ needs to be constructed. Through describing the image of the constructed function and its transformed function and proving the relevant properties, we obtain that Box dimension of its Riemann–Liouville integral of arbitrary order and its Mellin transformed function are also one. The smoothness of its Riemann–Liouville integral can only be improved, and its Mellin transformed function is differentiable.

1. Introduction

In the paper [1], Xiao draws a conclusion that the cardinality of fractal function set is the same as that of real number set. So We strongly believe that fractal functions have the same status as real numbers, which is worth spending a lot of time to discuss. As we all know, at the end of last century, researchers began to discuss the fractional calculus of continuous functions and the fractal dimension of fractal curves. Then, in [2,3], the precise relationship between fractals and fractional integrals has been studied. Through further research, researchers can discuss the relationship between fractal dimension of fractal or fractal functions and any order of fractional integrals. Xiao [4] has concluded upper box dimension of continuous fractal functions is equal to or more than their Riemann-Liouville integral, and Gao [5] has also concluded upper box dimension of continuous fractal functions is equal to or more than their Weyl integral. Both of them partly answer Fractal Calculus Conjecture [4]. In addition, Noreen Azhar [6] recently gives solution of fuzzy fractional order differential equations by fractional Mellin transform method. But little work about the relationship of Riemann-Liouville integral and Mellin transform has been investigated. In the present paper, we aim to discuss about the differences and similarities between Riemann-Liouville integral and Mellin transform by a certain function. The conclusion is that box dimension of its Riemann–Liouville integral of any order and its Mellin transformed function have also been proved to be 1. Compared with the constructed function, the smoothness of its Riemann–Liouville integral can only be improved, while its Mellin transformed function is differentiable. We give the definitions of upper box dimension, lower box dimension, box dimension and function of unbound variation as below.

Definitition 1

([7]). Let $\mathcal{F}$ be a bounded and nonempty set in $R^n$, then let $N_{\delta }\left(\mathcal{F}\right)$ be the minimum number of sets covering $\mathcal{F}$ completely whose diameter is at most δ. The upper and lower box dimensions of the set $\mathcal{F}$ respectively are defined as

$${\overline{dim}}_B\mathcal{F}=\underset{\delta \to 0}{\overline{lim}}\frac{log{N}_{\delta }\left(\mathcal{F}\right)}{-log\delta }$$

$${\underline{dim}}_B\mathcal{F}=\underset{\overline{\delta \to 0}}{lim}\frac{log{N}_{\delta }\left(\mathcal{F}\right)}{-log\delta }.$$

If those are equal, the value of either of them is regarded as the box dimension of the set $\mathcal{F}$

$$di{m}_B\mathcal{F}=\underset{\delta \to 0}{lim}\frac{log{N}_{\delta }\left(\mathcal{F}\right)}{-log\delta }.$$

Definitition 2

([8]). Let $g\left(t\right)$ be a bounded function defined on interval $\mathrm{I}=[0,1]$ and ${\left\{t_i\right\}}_{i=0}^n$ be an arbitrary point sequence satisfying $0=t_0<t_1<t_2<$⋯$<t_n=1$. If

$${\mathrm{V}}_g:=\underset{(t_0,t_1,...,t_n)}{sup}\sum _{i=1}^n|g\left(t_i\right)-g\left(t_{i-1}\right)|$$

is finite, that is, $g\left(t\right)$ is bounded variational on $\mathrm{I}$.

2. Constructing a Continuous and Unbounded Variational Function

After giving the above background and definition, the main work will be carried out. At the beginning of this section, a one-dimensional unbounded variational function defined on a closed interval $\mathrm{I}=[0,1]$ is constructed. Then, some properties mentioned of this function including continuity, unbounded variation and one-dimension are proved.

For any given positive integer n, we let $A_k=(\frac{1}{2k},\frac{1}{2k-1}](k=1,2,...,n)$, $C_{2n}=[0,\frac{1}{2n}]$, and $B_k=(\frac{1}{2k-1},\frac{1}{2(k-1)}](k=2,3,...,n-1)$ so we get that $\mathrm{I}=({\displaystyle \bigcup _{k=1}^n}{A}_k)\cup ({\displaystyle \bigcup _{k=2}^n}{B}_k)\cup C_{2n}$. Then, let

$$f_n\left(x\right)=\left\{\begin{array}{ccc}1+(1-2k)x& \hfill & x\in A_k\hfill \\ -1+(1+2k)x& \hfill & x\in B_k\hfill \\ x& \hfill & x\in C_{2n}.\hfill \end{array}\right.$$

(1)

Given different n, we will get different functions. The following Figure 1 is the function image of $f_n$ when n = 10,000

Let $n\to \infty$, we get the limit function as follows:

$$\mathcal{M}\left(x\right)=\underset{n\to \infty }{lim}{f}_n\left(x\right),$$

where $n\in N^{*}$ and $n\ge 2$.

Property 1.

$\mathcal{M}\left(x\right)$ is a bounded continuous function on $\mathrm{I}$.

Proof. 

For any $\epsilon >0$ and $x\in \mathrm{I}$, $\exists N=\frac{1}{\epsilon }$, when $n\ge m>N$, by the definition of function $f_n\left(x\right)$ we get

$$|f_n\left(x\right)-f_m\left(x\right)|<\frac{1}{N}=\epsilon .$$

By the Cauchy criterion of uniformly convergent series of function terms, we have

$$f_n\left(x\right)\Rightarrow \mathcal{M}\left(x\right)(n\to \infty ).$$

Then we get $\mathcal{M}\left(x\right)$ is continuous because $f_n\left(x\right)$ is continuous on $\mathrm{I}$ for any $n\in N^{*}$ [9]. By the definition of function $f_n\left(x\right)$, it is also easy to know $\mathcal{M}\left(\mathrm{I}\right)\subseteq [0,1]$. □

Property 2.

$\mathcal{M}\left(x\right)$ is unbounded variational on $\mathrm{I}$.

Proof. 

$\mathcal{M}\left(x\right)$ is monotonous on any subinterval $(\frac{1}{n+1},\frac{1}{n}]$. Moreover, whenever n is odd or even, its amplitude is greater or equal to $\frac{1}{n+2}$. We get the variation of $\mathcal{M}\left(x\right)$ is greater than ${\displaystyle \sum _{n=1}^{\infty }}\frac{1}{n+2}=\infty$. The proof of the property has done. □

Lemma 1

([7]). Let $g:[0,1]\to \mathbb{R}$ be continuous. Set $0<\delta <1$, then let n be the minimum integer that equal to or more than $1/\delta$. If $N_{\delta }$ is the number of δ-mesh squares intersecting graph g, that is

$${\delta }^{-1}\sum _{m=0}^{n-1}{R}_g[m\delta ,(m+1)\delta ]\le N_{\delta }\le 2n+{\delta }^{-1}\sum _{m=0}^{n-1}{R}_g[m\delta ,(m+1)\delta ],$$

(2)

where $R_g[m\delta ,(m+1)\delta ]$ is the amplitude of g in interval $[m\delta ,(m+1\left)\delta \right]$,

Proof. 

$R_g[m\delta ,(m+1)\delta ]/\delta$ and $2+R_g[m\delta ,(m+1)\delta ]/\delta$ are respectively at least number and at most number of $\delta$-mesh squares intersecting graph f defined on interval $[m\delta ,(m+1\left)\delta \right]$. Sum the whole intervals, and the above inequality is proved. □

Property 3.

Not only Hausdorff dimension but box dimension of $\mathcal{M}\left(x\right)$ on $\mathrm{I}$ is equal to 1.

Proof. 

Obviously the topological dimension of any continuous function is more than or equal to 1. That is

$${\underline{dim}}_B\Gamma (\mathcal{M}\left(x\right),\mathrm{I})\ge 1,$$

(3)

where ($\Gamma (\mathcal{M},\mathrm{I})$ means the image of the $\mathcal{M}\left(x\right)$ defined on $\mathrm{I}$,

Let $0<\delta <\frac{1}{4}$, and there exists constant c satisfying $n^2+1<c<{(n+1)}^2$, subject to ${\delta }^{-1}<c<{\delta }^{-1}+1$. That is, ${\delta }^{-1}$ and $n^2$ are infinite of the same order. It is easy to know that the amplitude of the straight line with slope k is $R_f=k\delta$ in the interval with length $\delta$. In addition, we certainly know $\mathrm{I}=({\displaystyle \bigcup _{k=1}^n}{A}_k)\cup ({\displaystyle \bigcup _{k=2}^n}{B}_k)\cup C_{2n}$ which is mentioned in (1).

For the subinterval $(\frac{1}{2k},\frac{1}{2k-1}]k=1,2,...,n$,the length and the slope of it are equal to $\frac{1}{2k(2k-1)}$ and $2k-1$. Then we calculate the number of $\delta$-mesh squares intersecting $\Gamma (\mathcal{M},({\displaystyle \bigcup _{k=1}^n}{A}_k))$ is less than

$$\begin{array}{cc}\hfill N_1& \le \sum _{k=1}^n(2k-1)(1+\frac{1}{(2k-1)2k\delta })\hfill \\ \hfill & =\sum _{k=1}^n(2k-1)+{\delta }^{-1}\sum _{k=1}^n\frac{1}{2k}.\hfill \end{array}$$

(4)

Similarly, let $N_2$ and $N_3$ be respectively the number of all $\delta$-mesh squares intersecting $\Gamma (\mathcal{M},({\displaystyle \bigcup _{k=2}^n}{B}_k))$ and $\Gamma (\mathcal{M},C_{2n})$ respectively. We can also obtain that

$$N_2\le \sum _{k=2}^n(2k-1)+{\delta }^{-1}\sum _{k=2}^n\frac{1}{2(k-1)}$$

(5)

and

$$N_3\le 1+{\delta }^{-1}\frac{1}{2n}.$$

(6)

Combining (4)–(6), we can get that the number of the whole $\delta$-mesh squares intersecting $\Gamma (\mathcal{M},\mathrm{I})$ is

$$\begin{array}{cc}\hfill N_{\mathcal{M},\delta }& =N_1+N_2+N_3\hfill \\ \hfill & \le {\delta }^{-1}\sum _{k=1}^n\frac{1}{k}+2n^2\hfill \\ \hfill & ={\delta }^{-1}(C+log\left(n\right))+2n^2,\hfill \end{array}$$

(7)

where C is Euler constant.

So by the Definition 1 and Lemma 1, we can get

$$\begin{array}{cc}\hfill {\overline{dim}}_B\Gamma (\mathcal{M},\mathrm{I})& =\underset{\delta \to 0}{\overline{lim}}\frac{log{N}_{\mathcal{M},\delta }}{-log\delta }\hfill \\ \hfill & \le \underset{\delta \to 0}{\overline{lim}}\frac{log({\delta }^{-1}(C+log\left(n\right))+2n^2)}{log\left({\delta }^{-1}\right)}=1.\hfill \end{array}$$

(8)

Thus, gathering (3) and (8), we draw the conclusion, that is

$$di{m}_B\Gamma (\mathcal{M},\mathrm{I})=1.$$

Hausdorff dimension of $\Gamma (\mathcal{M},\mathrm{I})$ is always equal to or less than its Box dimension. So

$$1\le di{m}_H\Gamma (\mathcal{M},\mathrm{I})\le di{m}_B\Gamma (\mathcal{M},\mathrm{I})=1.$$

(9)

That is

$$di{m}_H\Gamma (\mathcal{M},\mathrm{I})=1.$$

□

Remark 1.

The function $\mathcal{M}\left(x\right)$ has only countable non-differentiable points on I, and all of them make up the set {0,1,$\frac{1}{2}$,$\frac{1}{3}$,...}. Besides, the function $\mathcal{M}\left(x\right)$ has fractal property only at the point 0 and 0 is the only one unbounded variation point [10]. For any $0<c<1$, $\Gamma (\mathcal{M},[0,c\left]\right)$ is similar to $\Gamma (\mathcal{M},\mathrm{I})$. That is to say, $\Gamma (\mathcal{M},\mathrm{I})$ is a self similar graph.

3. Riemann-Liouville Integral of $\mathcal{M}\left(\mathbf{x}\right)$

After fully studying the characteristics of fractal function $\mathcal{M}\left(x\right)$, we next study the functional properties of function $\mathcal{M}\left(x\right)$ after Riemann-Liouville integral, and compare the similarities and differences between them.

Definitition 3

([11]). Let $0<\nu <1$ and $g\left(x\right)\in \mathrm{C}[0,1]$. The Riemann-Liouville integral of $g\left(x\right)$ is defined as

$${\mathcal{D}}^{-\nu }g\left(x\right)=\frac{1}{\Gamma \left(\nu \right)}{\int }_0^{x}g\left(t\right){(x-t)}^{\nu -1}dt.$$

Lemma 2

([12]). If $g\left(x\right)\in \mathrm{C}[0,1]$ and it’s bounded variational, ${\mathcal{D}}^{-\nu }g\left(x\right)\in \mathrm{C}[0,1]$ for $\nu \in (0,1)$ and it’s also bounded variational on $[0,1]$.

Property 4.

For any $0<\nu <1$, $di{m}_H\Gamma ({\mathcal{D}}^{-\nu }\mathcal{M},\mathrm{I})=di{m}_H\Gamma ({\mathcal{D}}^{-\nu }\mathcal{M},\mathrm{I})=1$ and ${\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)$ is bounded on $\mathrm{I}$.

Proof. 

For any $0<\nu <1$, by the definition of Gamma Function and $\mathcal{M}\left(x\right)$, we have

$$\begin{array}{cc}\hfill |{\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)|& =|\frac{1}{\Gamma \left(\nu \right)}{\int }_0^x\mathcal{M}\left(t\right){(x-t)}^{\nu -1}dt|\hfill \\ \hfill & \le Max\left|\mathcal{M}\left(x\right)\right|\frac{1}{\Gamma \left(\nu \right)}\frac{1}{\nu }{x}^{\nu }\hfill \\ \hfill & \le \frac{1}{\Gamma \left(\nu \right)\nu }.\hfill \end{array}$$

(10)

So ${\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)$ is bounded Let $0<\delta <1$, then $\mathcal{M}\left(x\right)$ is continuous and bounded variational on $[\delta ,1]$, so by the Lemma 2 it’s proved that ${\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)$ is bounded variational and continuous on $[\delta ,1]$. So there absolutely exists a positive number $\mathrm{C}$ so that $\frac{\mathrm{C}}{\delta }$ is the number $\delta$-mesh squares covering the graph of ${\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)$ defined on $[\delta ,1]$. the number of $\delta$-mesh squares covering ${\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)$ defined on $[0,\delta ]$ at most $\frac{1}{\delta }$. Thus, we get

$$\begin{array}{cc}\hfill {\overline{dim}}_B\Gamma ({\mathcal{D}}^{-\nu }\mathcal{M},\mathrm{I})& =\underset{\delta \to 0}{\overline{dim}}\frac{log{N}_{\delta }(\Gamma ({\mathcal{D}}^{-\nu }\mathcal{M},\mathrm{I}))}{-log\delta }\hfill \\ \hfill & \le \underset{\delta \to 0}{dim}\frac{log\frac{\mathrm{C}+1}{\delta }}{-log\delta }=1.\hfill \end{array}$$

(11)

Obviously the topological dimension of ${\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)$ is equal to or more than 1, that is,

$${\underline{dim}}_B\Gamma ({\mathcal{D}}^{-\nu }\mathcal{M},\mathrm{I})\ge 1.$$

(12)

Combining (11) and (12), we get the box dimension of $\Gamma ({\mathcal{D}}^{-\nu }\mathcal{M},\mathrm{I})$ is 1. Similar to (9), that is,

$$di{m}_H\Gamma ({\mathcal{D}}^{-\nu }\mathcal{M},\mathrm{I})=di{m}_B\Gamma ({\mathcal{D}}^{-\nu }\mathcal{M},\mathrm{I})=1.$$

□

Thus, both $\mathcal{M}\left(x\right)$ and ${\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)$ are 1-dimension and bounded on $\mathrm{I}$. But maybe when $\nu$ is big enough, ${\mathcal{D}}^{-\nu }\mathcal{M}\left(x\right)$ is bounded variational on $\mathrm{I}$.

4. Mellin Transform of $\mathcal{M}\left(\mathit{x}\right)$

Definitition 4

([11]). The Mellin transform of a function $\psi \left(t\right)$ of real variable $t\in {\mathbb{R}}^{+}$ is defined as

$$\left(\mathcal{M}\psi \right)\left(p\right)=\left(\mathcal{M}\right)\left[\psi \right]\left(p\right)={\psi }^{*}\left(s\right):={\int }_0^{\infty }{t}^{s-1}\psi \left(t\right)dt(s\in \mathbb{C}).$$

Considering that $\mathcal{M}\left(x\right)$ is defined on $\mathrm{I}=[0,1]$ and we want to do the Mellin transform to the function $\mathcal{M}\left(x\right)$, now we define $\mathcal{M}\left(x\right)=0$ on interval $(1,\infty )$. Then, we start to discuss the properties of $\mathcal{M}\left(x\right)$ on $[0,\infty )$ after Mellin transformation. The following Figure 2 is the function image of Mellin Transform of $\mathcal{M}\left(x\right)$ defined on $[-0.98,2]$.

Property 5.

By the Definition 4, ${\mathcal{M}}^{*}\left(s\right)={\int }_0^{\infty }{t}^{s-1}\mathcal{M}\left(t\right)dt={\int }_0^1{t}^{s-1}\mathcal{M}\left(t\right)dt$ is bounded on any $[\mathrm{a},\mathrm{b}]$$(\mathrm{a}>0)$. $di{m}_H\Gamma ({\mathcal{M}}^{*}\left(s\right),[\mathrm{a},\mathrm{b}])=di{m}_B\Gamma ({\mathcal{M}}^{*}\left(s\right),[\mathrm{a},\mathrm{b}])=1.$

Proof. 

For any $s\in [\mathrm{a},\mathrm{b}]$$(\mathrm{a}>0)$, we have

$$\begin{array}{cc}\hfill {\mathcal{M}}^{*}\left(s\right)& ={\int }_0^{\infty }{t}^{s-1}\mathcal{M}\left(t\right)dt={\int }_0^1{t}^{s-1}\mathcal{M}\left(t\right)dt\hfill \\ \hfill & \le {\int }_0^1{t}^{s-1}dt\underset{t\in I}{Max}\left|\mathcal{M}\left(t\right)\right|=\frac{1}{s}\le \frac{1}{\mathrm{a}}.\hfill \end{array}$$

(13)

So ${\mathcal{M}}^{*}\left(s\right)$ is bounded on $[\mathrm{a},\mathrm{b}]$$(\mathrm{a}>0)$.

Let $0<\mathrm{a}\le s_1\le s_2\le \mathrm{b}.$ It holds that

$$\begin{array}{cc}\hfill {\mathcal{M}}^{*}\left(s_2\right)-{\mathcal{M}}^{*}\left(s_1\right)& ={\int }_0^1{t}^{s_2-1}\mathcal{M}\left(t\right)dt-{\int }_0^1{t}^{s_1-1}\mathcal{M}\left(t\right)dt\hfill \\ \hfill & ={\int }_0^1(t^{s_2-1}-t^{s_1-1})\mathcal{M}\left(t\right)dt\le 0.\hfill \end{array}$$

(14)

Thus, the function ${\mathcal{M}}^{*}\left(s\right)$ is monotonically decreasing and bounded on any interval $[\mathrm{a},\mathrm{b}]$. So ${\mathcal{M}}^{*}\left(s\right)$ is bounded variational on $[\mathrm{a},\mathrm{b}]$. We also get

$$di{m}_H\Gamma ({\mathcal{M}}^{*}\left(s\right),[\mathrm{a},\mathrm{b}])=di{m}_B\Gamma ({\mathcal{M}}^{*}\left(s\right),[\mathrm{a},\mathrm{b}])=1.$$

□

Property 6.

${\mathcal{M}}^{*}\left(s\right)={\int }_0^{\infty }{t}^{s-1}\mathcal{M}\left(t\right)dt={\int }_0^1{t}^{s-1}\mathcal{M}\left(t\right)dt$ is continuous on $(-1,\infty )$.

Proof. 

For any $s\in [\mathrm{a},\mathrm{b}]$$(-1<\mathrm{a}<\mathrm{b})$, the function $f(s,t)=t^{s-1}\mathcal{M}\left(t\right)$ is continuous on $[\mathrm{a},\mathrm{b}]\times (0,1]$. Furthermore, for any given positive number $\epsilon >0$, it does exist $\delta >0$$(\delta \to 0)$, for all $s\in [\mathrm{a},\mathrm{b}]$, we have

$$\begin{array}{cc}\hfill |{\mathcal{M}}^{*}\left(s\right)-{\int }_{\delta }^1{t}^{s-1}\mathcal{M}\left(t\right)dt|& =|{\int }_0^{\delta }{t}^{s-1}\mathcal{M}\left(t\right)dt|\le |{\int }_0^{\delta }{t}^{s-1}tdt|\hfill \\ \hfill & =|{\int }_0^{\delta }{t}^{s}dt|=\frac{1}{1+s}{\delta }^{s+1}\le \epsilon .\hfill \end{array}$$

(15)

So function ${\mathcal{M}}^{*}\left(s\right)$ converges uniformly on $[\mathrm{a},\mathrm{b}]$ [9]. Due to the arbitrariness of both $\mathrm{a}$ and $\mathrm{b}$, ${\mathcal{M}}^{*}\left(s\right)$ is continuous on $(-1,\infty )$. □

Remark 2.

For $s=-1$, we carry out numerical experiments on ${\mathcal{M}}^{*}\left(s\right)$. Using the original integral definition, divide the interval $[0,1]$ into n segments equally. If $n=5000$, the value of ${\mathcal{M}}^{*}\left(s\right)$ is $5.4$. If n is equal to 8000, 12,000, 24,000, 48,000 respectively, the value of ${\mathcal{M}}^{*}\left(s\right)$ is equal to $5.7,6.4,6.9,7.24$ respectively. It’s easy to see that as the number n increases, the value of ${\mathcal{M}}^{*}\left(s\right)$ increases. So we guess ${\mathcal{M}}^{*}\left(s\right)$ is not convergent on the point $s=-1$. As for $s<-1$, combining the definition of Mellin transform and the constructed function, we obviously know that ${\mathcal{M}}^{*}\left(s\right)$ is meaningless.

Property 7.

${\mathcal{M}}^{*}\left(s\right)={\int }_0^{\infty }{t}^{s-1}\mathcal{M}\left(t\right)dt={\int }_0^1{t}^{s-1}\mathcal{M}\left(t\right)dt$ is differentiable on $(-1,\infty )$.

Proof. 

For any $s\in [\mathrm{a},\mathrm{b}]$$(-1<\mathrm{a}<\mathrm{b})$, $f(s,t)=t^{s-1}\mathcal{M}\left(t\right)$ and $f_s(s,t)=t^{s-1}logt\mathcal{M}\left(t\right)$ are continuous on $[\mathrm{a},\mathrm{b}]\times (0,1]$. In addition, in the Property 6, we proof that ${\mathcal{M}}^{*}\left(s\right)$ converges uniformly on $[\mathrm{a},\mathrm{b}]$. Therefore, for any given positive number $\epsilon >0$, there is always a real number $\delta \to 0$ independent of $\epsilon$, for all $s\in [\mathrm{a},\mathrm{b}]$$(-1<\mathrm{a}<\mathrm{b})$, we have

$$\begin{array}{cc}\hfill |{\left({\mathcal{M}}^{*}\left(s\right)\right)}^{{}^{\prime }}-{\int }_{\delta }^1{t}^{s-1}logt\mathcal{M}\left(t\right)dt|& =|{\int }_0^{\delta }{t}^{s-1}logt\mathcal{M}\left(t\right)dt|\le |{\int }_0^{\delta }{t}^{s}logtdt|\hfill \\ \hfill & =|(\frac{1}{1+s}{t}^{s+1}logt-\frac{1}{{(1+s)}^2}{t}^{1+s}){|}_0^{\delta }|\hfill \\ \hfill & =(\frac{1}{1+s}log\delta -\frac{1}{{(s+1)}^2}){\delta }^{1+s}\le \epsilon .\hfill \end{array}$$

(16)

So ${\left({\mathcal{M}}^{*}\left(s\right)\right)}^{{}^{\prime }}$ converges uniformly on $[\mathrm{a},\mathrm{b}]$ [9]. Due to the arbitrariness of $\mathrm{a}$ and $\mathrm{b}$, ${\mathcal{M}}^{*}\left(s\right)$ is differentiable on $(-1,\infty )$. At the same time, it is verified that for $0<\mathrm{a}<\mathrm{b}<\infty$, $di{m}_H\Gamma ({\mathcal{M}}^{*}\left(s\right),[\mathrm{a},\mathrm{b}])=di{m}_B\Gamma ({\mathcal{M}}^{*}\left(s\right),[\mathrm{a},\mathrm{b}])=1$. □

5. Conclusions

Through comparison of their properties of the constructed function, the box dimension of its Riemann–Liouville integral of any order and its Mellin transformed function have been proved to be 1. Both the two previous transformations can be regarded as integral transformations. Compared with $\mathcal{M}\left(x\right)$, the smoothness of its Riemann–Liouville fractional integral can only be improved [4], while its Mellin transformed function is differentiable.