# On the Fractional Derivative Duality in Some Transforms

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## Abstract

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## 1. Introduction

## 2. Suitable Derivatives

#### 2.1. Derivative Requirements

#### 2.2. Liouville-Type Derivatives

**Definition 1.**

**Theorem 1.**

- Derivative in t$${D}_{t\pm}^{\alpha}{e}^{st}={s}^{\alpha}{e}^{st},\phantom{\rule{1.em}{0ex}}\pm Re\left(s\right)>0;$$
- Derivative in s$${D}_{s\pm}^{\alpha}{e}^{-st}={(-t)}^{\alpha}{e}^{-st},\phantom{\rule{1.em}{0ex}}\mp t>0;$$

#### 2.3. Hadamard-Type Derivatives

**Definition 2.**

**Theorem 2.**

- Derivative in $\tau $$${\mathfrak{D}}_{\tau \pm}^{\alpha}{\tau}^{v}={v}^{\alpha}{\tau}^{v},\phantom{\rule{1.em}{0ex}}\pm Re\left(v\right)>0;$$
- Derivative in v$${D}_{v\pm}^{\alpha}{\tau}^{-v}={(-ln\tau )}^{\alpha}{\tau}^{-v},\phantom{\rule{1.em}{0ex}}{\tau}^{\mp 1}>1;$$

#### 2.4. Discrete-Time Derivatives

#### 2.4.1. Fractional Nabla and Delta Derivatives

**Definition 3.**

**Theorem 3.**

**Definition 4.**

**Corollary 1.**

#### 2.4.2. Forward and Backward Derivatives Based on the Bilinear Transformation

**Definition 5.**

**Definition 6.**

## 3. Main Transforms and Integer-Order Derivatives

#### 3.1. Continuous-Time Laplace and Fourier Transforms

**Definition 7.**

- piecewise continuous,
- with bounded variation,
- locally integrable (in the sense that the function is absolutely integrable in any real interval $[a,b]$, so that ${\int}_{a}^{b}\left|f\left(t\right)\right|\phantom{\rule{0.166667em}{0ex}}\mathrm{d}t<\infty $),
- of exponential order,

**Remark 2.**

- If $f\left(t\right)$ is absolutely integrable and of finite duration, then the ROC is the entire s-plane since the Laplace transform is finite and $F\left(s\right)$ exists for any s.
- If $f\left(t\right)$ is right-handed (i.e., it exists with $t\ge {t}_{0}\in \mathbb{R}$) and $Re\left(s\right)=a\in {\mathcal{R}}_{c}$, then any s to the right of a is also in ${\mathcal{R}}_{c}$.
- If $f\left(t\right)$ is left-handed (i.e., exists with $t\le {t}_{0}\in \mathbb{R}$) and $Re\left(s\right)=a\in {\mathcal{R}}_{c}$, then any s to the left of a is also in ${\mathcal{R}}_{c}$.
- A function $f\left(t\right)$ is absolutely integrable (satisfying the Dirichlet conditions and having the Fourier transform) if and only if the ROC of the corresponding Laplace transform $F\left(s\right)$ includes the imaginary axis since $Re\left(s\right)=0$ and $s=i\omega $.
- A given complex variable function only can define univocally an LT if it has attached a suitable ROC.
- If $F\left(s\right)=\mathcal{L}\left[f\left(t\right)\right]$, then $\mathcal{L}\left[f(-t)\right]=F(-s)$.
- If the region of convergence of $F\left(s\right)$ includes the frontiers $b\le Re\left(s\right)\le a$, then $F\left(s\right)$ is completely defined in that region by the values at the lines $F(a+i\tau )\phantom{\rule{0.277778em}{0ex}}\mathrm{and}\phantom{\rule{0.277778em}{0ex}}F(b+i\tau ),$$\phantom{\rule{0.277778em}{0ex}}\tau \in \mathbb{R}$.
- $F\left(s\right)$ is bounded in the strip $a+\u03f5\le Re\left(s\right)\le b-\u03f5$, with $\u03f5>0$.

**Theorem 4.**

#### 3.2. The Mellin Transform

**Definition 8.**

**Theorem 5.**

#### 3.3. On the Z and Discrete-Time Fourier Transforms

**Definition 9.**

**Definition 10.**

- If the signal is right (i.e., $x\left(n\right)=0,\phantom{\rule{0.277778em}{0ex}}n<{n}_{0}\in \mathbb{Z}$), then the ROC is the exterior of a circle centered at the origin (${r}_{+}=\infty $): $\left|z\right|>{r}_{-}$.
- If the signal is left (i.e., $x\left(n\right)=0,\phantom{\rule{0.277778em}{0ex}}n>{n}_{0}\in \mathbb{Z}$), then the ROC is the interior of a circle centered at the origin (${r}_{-}=0$): $\left|z\right|<{r}_{+}$.
- If the signal is a pulse (i.e., non null only on a finite set), then the ROC is the whole complex plane, possibly with the exception of the origin. In the ROC, the ZT defines an analytical function.

**Theorem 6.**

**Definition 11.**

**Theorem 7.**

## 4. Main Transforms and Non Integer-Order Derivatives

#### 4.1. Laplace Transform

- Right function case ($Re\left(s\right)>0$)$${F}_{-}^{\left(\alpha \right)}\left(s\right)=\underset{0}{\overset{\infty}{\int}}{(-t)}^{\alpha}f\left(t\right){e}^{-st}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}t,$$$${f}_{+}^{\left(\alpha \right)}\left(t\right)=\frac{1}{2\pi i}\underset{a-i\infty}{\overset{a+i\infty}{\int}}F\left(s\right){s}^{\alpha}{e}^{st}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}s,\phantom{\rule{4pt}{0ex}}a>0.$$
- Left function case ($Re\left(s\right)<0$)$${F}_{+}^{\left(\alpha \right)}\left(s\right)=\underset{-\infty}{\overset{0}{\int}}{(-t)}^{\alpha}f\left(t\right){e}^{-st}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}t,$$$${f}_{-}^{\left(\alpha \right)}\left(t\right)=\frac{1}{2\pi i}\underset{a-i\infty}{\overset{a+i\infty}{\int}}F\left(s\right){s}^{\alpha}{e}^{st}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}s,\phantom{\rule{4pt}{0ex}}a<0.$$
- Two-sided function case ($\left|Re\right(s\left)\right|<b$)The above relations suggest we introduce the two-sided fractional derivative [28]:$${D}_{\theta}^{\alpha}f\left(t\right):=\underset{h\to {0}^{+}}{lim}{h}^{-\alpha}\sum _{n=-\infty}^{+\infty}{(-1)}^{n}\frac{\mathsf{\Gamma}\left(\alpha +1\right)}{\mathsf{\Gamma}\left(\frac{\alpha +\theta}{2}-n+1\right)\mathsf{\Gamma}\left(\frac{\alpha -\theta}{2}+n+1\right)}f(t-nh),$$$${D}_{\theta}^{\alpha}{e}^{\pm i\omega t}={\left|\omega \right|}^{\alpha}{e}^{\mp i\theta \frac{\pi}{2}sgn\left(\omega \right)}{e}^{\pm i\omega t}.$$Assuming that $\left|\theta \right|\ne \left|\alpha \right|,$ then$${D}_{\theta}^{\alpha}F\left(i\omega \right)=\underset{-\infty}{\overset{\infty}{\int}}{\left|t\right|}^{\alpha}{e}^{-i\theta \frac{\pi}{2}sgn\left(t\right)}f\left(t\right){e}^{-i\omega t}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}t,$$$${D}_{\theta}^{\alpha}f\left(t\right)=\frac{1}{2\pi}\underset{-\infty}{\overset{\infty}{\int}}{\left|\omega \right|}^{\alpha}{e}^{i\theta \frac{\pi}{2}sgn\left(\omega \right)}F\left(i\omega \right){e}^{i\omega t}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\omega .$$In particular, we obtain$${D}_{0}^{\alpha}F\left(i\omega \right)=\underset{-\infty}{\overset{\infty}{\int}}{\left|t\right|}^{\alpha}f\left(t\right){e}^{-i\omega t}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}t,$$$${D}_{0}^{\alpha}f\left(t\right)=\frac{1}{2\pi}\underset{-\infty}{\overset{\infty}{\int}}{\left|\omega \right|}^{\alpha}F\left(i\omega \right){e}^{i\omega t}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\omega .$$

#### 4.2. Mellin Transform

- Stretching case ($Re\left(v\right)>0$)$${G}_{-}^{\left(\alpha \right)}\left(v\right)=\underset{1}{\overset{\infty}{\int}}{(-ln\tau )}^{\alpha}g\left(\tau \right){\tau}^{-v-1}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\tau ,$$$${\mathfrak{D}}_{\tau +}^{\alpha}g\left(\tau \right)=\frac{1}{2\pi i}\underset{a-i\infty}{\overset{a+i\infty}{\int}}{v}^{\alpha}G\left(v\right){\tau}^{v}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}s,\phantom{\rule{4pt}{0ex}}a>0.$$
- Shrinking case ($Re\left(v\right)<0$)$${G}_{+}^{\left(\alpha \right)}\left(v\right)=\underset{0}{\overset{1}{\int}}{(ln\tau )}^{\alpha}g\left(\tau \right){\tau}^{-v-1}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\tau ,$$$${\mathfrak{D}}_{\tau -}^{\alpha}g\left(\tau \right)=\frac{1}{2\pi i}\underset{a-i\infty}{\overset{a+i\infty}{\int}}{v}^{\alpha}G\left(v\right){\tau}^{v}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}v,\phantom{\rule{4pt}{0ex}}a<0.$$
- Bilateral scale case ($\left|Re\right(v\left)\right|<b$)The above relations suggest we introduce the two-sided scale derivative by:$${\mathfrak{D}}_{\tau ,\theta}^{\alpha}g\left(\tau \right)=\underset{q\to {1}^{+}}{lim}ln{\left(q\right)}^{-\alpha}\sum _{n=-\infty}^{+\infty}{(-1)}^{n}\frac{\mathsf{\Gamma}\left(\alpha +1\right)}{\mathsf{\Gamma}\left(\frac{\alpha +\theta}{2}-n+1\right)\mathsf{\Gamma}\left(\frac{\alpha -\theta}{2}+n+1\right)}g\left(\tau {q}^{-n}\right),$$$${\mathfrak{D}}_{\tau ,\theta}^{\alpha}{\tau}^{i\omega}={\left|\omega \right|}^{\alpha}{e}^{-i\theta \frac{\pi}{2}sgn\left(\omega \right)}{\tau}^{i\omega}.$$Assuming that $\left|\theta \right|\ne \left|\alpha \right|,$ then$${D}_{\theta}^{\alpha}G\left(i\omega \right)=\underset{0}{\overset{\infty}{\int}}{|ln\tau |}^{\alpha}{e}^{i\theta \frac{\pi}{2}sgn(ln\tau )}g\left(\tau \right){\tau}^{-i\omega -1}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\tau ,$$$${\mathfrak{D}}_{\tau ,\theta}^{\alpha}g\left(\tau \right)=\frac{1}{2\pi}\underset{-\infty}{\overset{\infty}{\int}}{\left|\omega \right|}^{\alpha}{e}^{i\theta \frac{\pi}{2}sgn\left(\omega \right)}G\left(i\omega \right){\tau}^{i\omega t}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\omega .$$In particular, we obtain$${D}_{0}^{\alpha}G\left(i\omega \right)=\underset{0}{\overset{\infty}{\int}}{|ln\tau |}^{\alpha}g\left(\tau \right){\tau}^{-i\omega -1}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\tau ,$$$${\mathfrak{D}}_{0}^{\alpha}g\left(t\right)=\frac{1}{2\pi}\underset{-\infty}{\overset{\infty}{\int}}{\left|\omega \right|}^{\alpha}G\left(i\omega \right){e}^{i\omega t}\phantom{\rule{0.166667em}{0ex}}\mathrm{d}\omega .$$

#### 4.3. Z and Discrete-Time Fourier Transforms

- Right sequence case ($\left|z\right|>1$)$${\mathfrak{D}}_{z+}^{\alpha}X\left(z\right)=\sum _{k=0}^{\infty}{(-k)}^{\alpha}x\left(kh\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-k}$$$${\nabla}^{\alpha}x\left(kh\right)=\frac{1}{2\pi i}{\oint}_{\gamma}X\left(z\right){\left[{\displaystyle \frac{1-{z}^{-1}}{h}}\right]}^{\alpha}{z}^{k-1}\mathrm{d}z,$$
- Left sequence case ($\left|z\right|<1$)$${\mathfrak{D}}_{z-}^{\alpha}X\left(z\right)=\sum _{k=-\infty}^{-1}{(-k)}^{\alpha}x\left(kh\right)\phantom{\rule{0.166667em}{0ex}}{z}^{-k}$$$${\nabla}^{\alpha}x\left(kh\right)=\frac{1}{2\pi i}{\oint}_{\gamma}X\left(z\right){\left[{\displaystyle \frac{1-{z}^{-1}}{h}}\right]}^{\alpha}{z}^{k-1}\mathrm{d}z,$$
- Two-sided function case (${r}_{-}<\left|z\right|<{r}_{+}$)and considering again two real parameters α, the derivative order, and θ, the asymmetry parameter, such that $\alpha >-1$ if $\theta \ne \pm \alpha $, or $\alpha \in \mathbb{R}$ if $\theta =\pm \alpha ,$ we define a discrete-time two-sided derivative by
**Definition****12.**$${\mathsf{\Theta}}_{\theta}^{\alpha}x\left(kh\right):={h}^{-\alpha}\sum _{n=-\infty}^{+\infty}{(-1)}^{n}\frac{\mathsf{\Gamma}\left(\alpha +1\right)}{\mathsf{\Gamma}\left(\frac{\alpha +\theta}{2}-n+1\right)\mathsf{\Gamma}\left(\frac{\alpha -\theta}{2}+n+1\right)}x(kh-nh).$$**Theorem****8.**Let $x\left(kh\right)={e}^{i\omega hk}$. Then, [29]$${\mathsf{\Theta}}_{\theta}^{\alpha}{e}^{i\omega hk}={\left|2sin(\omega h/2)\right|}^{\alpha}{e}^{i\theta \frac{\pi}{2}sgn\left(\omega \right)}{e}^{i\omega hk}.$$The proof comes from the left side in (66) by letting $z={e}^{i\omega hk}$.Assuming that $\left|\theta \right|\ne \left|\alpha \right|,$ thenand**Theorem****9.**$${D}_{\theta}^{\alpha}X\left({e}^{i\omega}\right)=\sum _{k=-\infty}^{\infty}{\left|kh\right|}^{\alpha}{e}^{i\theta \frac{\pi}{2}sgn\left(k\right)}x\left(kh\right){e}^{-i\omega hk}$$$${\mathsf{\Theta}}_{\theta}^{\alpha}x\left(kh\right)=\frac{1}{2\pi h}\underset{-\pi /h}{\overset{\pi /h}{\int}}{\left|2sin(\omega h/2)\right|}^{\alpha}{e}^{i\theta \frac{\pi}{2}sgn\left(\omega \right)}X\left({e}^{i\omega}\right){e}^{i\omega hk}\mathrm{d}\omega .$$In particular, we obtain$${D}_{0}^{\alpha}X\left({e}^{i\omega}\right)=\sum _{k=-\infty}^{\infty}{\left|kh\right|}^{\alpha}x\left(kh\right){e}^{-i\omega hk}$$$${\mathsf{\Theta}}_{0}^{\alpha}x\left(kh\right)=\frac{1}{2\pi h}\underset{-\pi /h}{\overset{\pi /h}{\int}}{\left|2sin(\omega h/2)\right|}^{\alpha}{e}^{i\omega hk}\mathrm{d}\omega .$$

**Remark 4.**

## 5. Some Consequences

- Substitute the inverse LT (30) in (44) and note that the LT converges uniformly in the ROC to commute the integrations. Using the definition of the gamma function [31], we obtain$${F}_{-}^{\left(\alpha \right)}\left(s\right)=\frac{{(-1)}^{\alpha}\mathsf{\Gamma}(\alpha +1)}{2\pi i}{\int}_{a-i\infty}^{a+i\infty}F\left(u\right){(s-u)}^{-\alpha -1}\mathrm{d}u,\phantom{\rule{1.em}{0ex}}Re(s>0).$$This last formula is another way of expressing the left derivative in the complex plane, suitable for dealing with LT, but in agreement with previous formulations [30,32,33,34]. It is important to highlight an interesting fact: (73) is defined only in the right-handed complex plane. In general, we do not know what happens in the left half plane since it is out of the ROC. We profit on this fact to define there a branchcut line implicit in the definition. For this reason, we choose ${(-1)}^{\alpha}={e}^{i\alpha \pi}.$
- Let $f\left(t\right)=\epsilon \left(t\right)$ be the Heaviside unit step. Insert it into (44) and use again the definition of gamma function. Assume that $\alpha >0$. We arrive at$${F}_{-}^{\left(\alpha \right)}\left(s\right)={\displaystyle \frac{{(-1)}^{\alpha}\mathsf{\Gamma}(\alpha +1)}{{s}^{\alpha +1}}},\phantom{\rule{1.em}{0ex}}Re\left(s\right)>0.$$From this relation, we conclude that$$\mathcal{L}\left[\frac{{t}^{\alpha}}{\mathsf{\Gamma}(\alpha +1)}\epsilon \left(t\right)\right]={\displaystyle \frac{1}{{s}^{\alpha +1}}},\phantom{\rule{1.em}{0ex}}Re\left(s\right)>0,$$$${D}_{-}^{\alpha}{\displaystyle \frac{1}{s}}={\displaystyle \frac{{(-1)}^{\alpha}\mathsf{\Gamma}(\alpha +1)}{{s}^{\alpha +1}}},\phantom{\rule{1.em}{0ex}}Re\left(s\right)>0,$$$${D}_{-}^{\alpha}{s}^{-\beta}={\displaystyle \frac{{(-1)}^{\alpha}\mathsf{\Gamma}(\alpha +\beta )}{\mathsf{\Gamma}\left(\beta \right)}}{s}^{-\beta -\alpha},\phantom{\rule{1.em}{0ex}}Re\left(s\right)>0,$$

**Remark 5.**

## 6. Conclusions

## Author Contributions

## Funding

## Data Availability Statement

## Conflicts of Interest

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**MDPI and ACS Style**

Ortigueira, M.D.; Bengochea, G.
On the Fractional Derivative Duality in Some Transforms. *Mathematics* **2023**, *11*, 4464.
https://doi.org/10.3390/math11214464

**AMA Style**

Ortigueira MD, Bengochea G.
On the Fractional Derivative Duality in Some Transforms. *Mathematics*. 2023; 11(21):4464.
https://doi.org/10.3390/math11214464

**Chicago/Turabian Style**

Ortigueira, Manuel Duarte, and Gabriel Bengochea.
2023. "On the Fractional Derivative Duality in Some Transforms" *Mathematics* 11, no. 21: 4464.
https://doi.org/10.3390/math11214464