**Proof.** This is Theorem VIII in [31].

Since our work involves analytic functions defined on the complex plane, we find it useful to define the domain

$$\mathbb{D} := \mathbb{C} \backslash \mathbb{R}\_0^-,\tag{11}$$

namely the complex plane slit along a branch cut from the origin. This is used as a domain for various complex power functions and other related functions.

Our first main result is a fractional identity between the Mittag–Leffler functions with two and three parameters as defined by Equations (6) and (7). This is motivated by previous work (e.g., [32]), in which gamma functions that appear in infinite power series can be interpreted as arising from fractional differintegrals. After submitting the paper, we realised that this result was previously proved in [33]. However, our original proof is preserved below.

**Proposition 1.** *For any <sup>α</sup>*, *<sup>β</sup>*, *<sup>ρ</sup>* <sup>∈</sup> <sup>C</sup> *with* Re(*α*), Re(*β*) <sup>&</sup>gt; <sup>0</sup>*, we have:*

$$E\_{a,\beta}^{\rho}(z) = \frac{1}{\Gamma(\rho)} \, ^{RL}\_0 D\_z^{\rho-1} \left[ z^{\rho-1} E\_{a,\beta}(z) \right], \qquad z \in \mathbb{D}. \tag{12}$$

**Proof.** First, Lemma 1 tells us that a quotient of gamma functions can very often be interpreted as arising from a fractional differintegral of a power function. In this case, the expression <sup>Γ</sup>(*ρ*+*n*) *<sup>n</sup>*! appearing in the coefficients of the series in Equation (7) gives rise to the following:

$$\begin{split} E\_{a,\beta}^{\rho}(z) &= \sum\_{n=0}^{\infty} \frac{\Gamma(\rho+n)z^{n}}{\Gamma(\rho)\Gamma(n\alpha+\beta)n!} = \sum\_{n=0}^{\infty} \frac{1}{\Gamma(n\alpha+\beta)\Gamma(\rho)} \cdot \frac{\Gamma(\rho+n)}{\Gamma(n+1)} z^{n} \\ &= \sum\_{n=0}^{\infty} \frac{1}{\Gamma(n\alpha+\beta)\Gamma(\rho)} \cdot \,\_0^{RL}D\_{z}^{\rho-1} \left[ z^{n+\rho-1} \right] = \frac{1}{\Gamma(\rho)} \sum\_{n=0}^{\infty} \,\_0^{RL}D\_{z}^{\rho-1} \left[ \frac{z^{n+\rho-1}}{\Gamma(n\alpha+\beta)} \right] .\end{split}$$

Since the series here is uniformly convergent, we can use the result of Lemma 2 to swap the summation with the fractional differintegration. This gives:

$$\begin{split} E\_{a,\beta}^{\rho}(z) &= \frac{1}{\Gamma(\rho)} \sum\_{n=0}^{\infty} \, ^{RL}\_{0}D\_{z}^{\rho-1} \left[ \frac{z^{n+\rho-1}}{\Gamma(n\alpha+\beta)} \right] = \frac{1}{\Gamma(\rho)} \, ^{RL}\_{0}D\_{z}^{\rho-1} \left[ \sum\_{n=0}^{\infty} \frac{z^{n+\rho-1}}{\Gamma(n\alpha+\beta)} \right] \\ &= \frac{1}{\Gamma(\rho)} \, ^{RL}\_{0}D\_{z}^{\rho-1} \left[ z^{\rho-1} \sum\_{n=0}^{\infty} \frac{z^{n}}{\Gamma(n\alpha+\beta)} \right] = \frac{1}{\Gamma(\rho)} \, ^{RL}\_{0}D\_{z}^{\rho-1} \left[ z^{\rho-1} E\_{a,\beta}(z) \right], \end{split}$$

and we have the result desired.

Note that, by setting *ρ* = 1 in Equation (12), we recover the trivial identity *E*<sup>1</sup> *<sup>α</sup>*,*β*(*z*) = *Eα*,*β*(*z*).

**Corollary 1.** *For any <sup>α</sup>*, *<sup>ρ</sup>* <sup>∈</sup> <sup>C</sup> *with* Re(*α*) <sup>&</sup>gt; <sup>0</sup>*, we have:*

$$E\_{a,1}^{\rho}(z) = \frac{1}{\Gamma(\rho)} \, \, ^{RL}\_0 D\_z^{\rho-1} \left[ z^{\rho-1} E\_a(z) \right], \qquad z \in \mathbb{D}. \tag{13}$$

**Proof.** This follows immediately by setting *β* = 1 in Proposition 1.

**Remark 1.** *By exactly the same argument as in Proposition 1, we can show that*

$$E\_{a,\beta}^{\rho}(\gamma z) = \frac{1}{\Gamma(\rho)} \, ^{RL}\_0 D\_z^{\rho - 1} \left[ z^{\rho - 1} E\_{a,\beta}(\gamma z) \right], \qquad z \in \mathbb{D}, \tag{14}$$

*for any <sup>α</sup>*, *<sup>β</sup>*, *<sup>γ</sup>*, *<sup>ρ</sup>* <sup>∈</sup> <sup>C</sup> *with* Re(*α*), Re(*β*) <sup>&</sup>gt; <sup>0</sup>*. The proof is as above with an extra factor of <sup>γ</sup><sup>n</sup> included in each term of the sum.*

We have now established a relation between the Mittag–Leffler functions of two and three parameters, and hence a relation between the Mittag–Leffler function of one parameter and the function *E<sup>ρ</sup> <sup>α</sup>*,1(*z*). However, in order to find a connection with the AB model, we need to consider not the function *Eα*(*x*) but rather the function *E<sup>α</sup>* <sup>−</sup>*<sup>α</sup>* <sup>1</sup>−*<sup>α</sup> <sup>x</sup>α* , which appears in the kernel of the definition in Equation (9). To this end, we note the following result, which is seen in (Equation (7.1), [12]) but without reference to fractional calculus.

**Proposition 2.** *For any <sup>α</sup>*, *<sup>β</sup>*, *<sup>γ</sup>* <sup>∈</sup> <sup>C</sup> *with* Re(*α*), Re(*β*) <sup>&</sup>gt; <sup>0</sup>*, we have:*

$$E\_{\mathfrak{a},\mathfrak{\beta}}\left(\gamma z^{\mathfrak{a}}\right) = z^{1-\mathfrak{\beta}} \, ^{RL}\_{0}D\_{z}^{1-\mathfrak{\beta}}\left[E\_{\mathfrak{a}}\left(\gamma z^{\mathfrak{a}}\right)\right], \qquad z \in \mathbb{D}.\tag{15}$$

.

**Proof.** This time we start from the right-hand side of the desired identity, and use the definition in Equation (5) of the function *Eα*:

$$z^{1-\beta} \, \_0^{RL} D\_z^{1-\beta} \left[ E\_a \left( \gamma z^a \right) \right] = z^{1-\beta} \, \_0^{RL} D\_z^{1-\beta} \left[ \sum\_{n=0}^{\infty} \frac{\gamma^n z^{\alpha n}}{\Gamma(n\alpha + 1)} \right]$$

This series is uniformly convergent, thus, by Lemma 2, we can swap the summation and fractional differintegration provided that (at least in the case 0 < Re(*β*) < 1) the resulting series also converges uniformly. We swap the operations now and justify this assumption at the end.

$$\begin{split} z^{1-\beta} \, ^{RL}\_{0}D\_{z}^{1-\beta} \left[ E\_{\mathfrak{a}} \left( \gamma z^{\mathfrak{a}} \right) \right] &= z^{1-\beta} \sum\_{n=0}^{\infty} \, ^{RL}\_{0}D\_{z}^{1-\beta} \left[ \frac{\gamma^{n} z^{an}}{\Gamma(na+1)} \right] \\ &= z^{1-\beta} \sum\_{n=0}^{\infty} \frac{\Gamma(an+1)}{\Gamma(an+\beta)} \cdot \frac{\gamma^{n} z^{an+\beta-1}}{\Gamma(na+1)} \\ &= \sum\_{n=0}^{\infty} \frac{\gamma^{n} z^{an}}{\Gamma(an+\beta)} . \end{split}$$

This series converges uniformly, being precisely the series expression in Equation (6) for *Eα*,*<sup>β</sup>* (*γzα*). Thus, our swapping of operations above was justified, and the proof is complete.

The key point here is that the dependence of the Mittag–Leffler function in Equation (7) on the parameters *ρ* and *β* can be encoded by fractional differintegrals. Proposition 1 enables us to interpret the parameter *ρ* as merely the order of a differintegral, and Proposition 2 enables us to do the same with *β*.

By combining the results of Proposition 1 and Proposition 2, it is possible to obtain a composite expression for the three-parameter Mittag–Leffler function in Equation (7) in terms of fractional-type integrals, as described by the following theorem.

**Theorem 1.** *For any <sup>α</sup>*, *<sup>β</sup>*, *<sup>γ</sup>*, *<sup>ρ</sup>* <sup>∈</sup> <sup>C</sup> *with* Re(*α*), Re(*β*) <sup>&</sup>gt; <sup>0</sup> *and* Re(*ρ*) <sup>&</sup>lt; <sup>1</sup>*, we have:*

$$E\_{a,\beta}^{\rho}(\gamma z^a) = \frac{a\sin(\pi\rho)}{\pi} \int\_0^z (z^a - u^a)^{-\rho} u^{a\rho - \beta} \, \_0^{\mathbb{R}I} D\_u^{1-\beta} \left[ \mathcal{E}\_a \left( \gamma u^a \right) \right] \, \mathrm{d}u, \qquad z \in \mathbb{D}.\tag{16}$$

**Proof.** For Re(*ρ*) < 1, the fractional differintegrals appearing in Proposition 1 and Corollary 1 are *integrals* (because Re(*ρ* − 1) < 0), and so Equation (14) can be rewritten as follows:

$$E\_{a,\emptyset}^{\rho}(\gamma z) = \frac{1}{\Gamma(\rho)} \, ^{RL}\_0 I\_z^{1-\rho} \left[ z^{\rho-1} E\_{a,\emptyset}(\gamma z) \right] = \frac{1}{\Gamma(\rho)\Gamma(1-\rho)} \int\_0^z (z-y)^{-\rho} y^{\rho-1} E\_{a,\emptyset}(\gamma y) \, \mathrm{d}y \, \mathrm{d}z$$

Making the change of variables *u* = *y*1/*α*, and using the reflection formula for the gamma function:

$$\begin{split} E\_{a,\beta}^{\rho}(\gamma z) &= \frac{\sin(\pi \rho)}{\pi} \int\_0^{z^{1/a}} (z - \mu^a)^{-\rho} \, \mu^{a(\rho - 1)} E\_{a,\beta}(\gamma \mu^a) a \mu^{a - 1} \, \mathrm{d}\mu \\ &= \frac{a \sin(\pi \rho)}{\pi} \int\_0^{z^{1/a}} (z - \mu^a)^{-\rho} \, \mu^{a\rho - 1} E\_{a,\beta}(\gamma \mu^a) \, \mathrm{d}\mu. \end{split}$$

Now, we can apply the result of Proposition 2 to the two-parameter Mittag–Leffler function appearing in the integrand of this expression:

$$\begin{split} \mathcal{E}\_{a,\beta}^{\rho}(\gamma z) &= \frac{a\sin(\pi\rho)}{\pi} \int\_{0}^{z^{1/a}} (z - u^{a})^{-\rho} \, u^{a\rho - 1} \Big[ u^{1 - \beta^{\ }Rl} \, \_0^{L} D\_{u}^{1 - \beta^{\ }} \left[ E\_{\mathfrak{a}} \left( \gamma u^{a} \right) \right] \, \mathrm{d}u \\ &= \frac{a\sin(\pi\rho)}{\pi} \int\_{0}^{z^{1/a}} (z - u^{a})^{-\rho} \, u^{a\rho - \beta^{\ }Rl} \, \_0^{L} D\_{u}^{1 - \beta^{\ }} \left[ E\_{\mathfrak{a}} \left( \gamma u^{a} \right) \right] \, \mathrm{d}u. \end{split}$$

Substituting *z<sup>α</sup>* for *z*:

$$E\_{a,\beta}^{\rho}(\gamma z^a) = \frac{a\sin(\pi\rho)}{\pi} \int\_0^z (z^a - u^a)^{-\rho} u^{a\rho - \beta} \, \_0^{RL} D\_u^{1-\beta} \left[ E\_a \left( \gamma u^a \right) \right] \, \mathrm{d}u.$$

Note that this can almost, but not quite, be expressed as a composition of two fractional differintegrals.

**Corollary 2.** *For any <sup>α</sup>*, *<sup>β</sup>*, *<sup>γ</sup>*, *<sup>ρ</sup>* <sup>∈</sup> <sup>C</sup> *with* Re(*α*) <sup>&</sup>gt; <sup>0</sup>*,* Re(*β*) <sup>&</sup>gt; <sup>1</sup>*, and* Re(*ρ*) <sup>&</sup>lt; <sup>1</sup>*, we have:*

$$E\_{a,\beta}^{\rho}(\gamma z^a) = \frac{a\sin(\pi\rho)}{\pi\Gamma(\beta - 1)} \int\_0^z (z^a - u^a)^{-\rho} u^{a\rho - \beta} \int\_0^u (u - t)^{\beta - 2} E\_a\left(\gamma t^a\right) \,\mathrm{d}t \,\mathrm{d}u, \qquad z \in \mathbb{D}.\tag{17}$$

**Proof.** In this case, we have Re(1 − *β*) < 0 and so the fractional differintegral which appears in the integrand of Equation (16) is an *integral*. Thus, we can write

$${}^{RL}\_{0}I\_{\mu}^{\mathbb{R}-1} \left[ E\_{\mathfrak{a}} \left( \gamma \mu^{\mathfrak{a}} \right) \right] = \frac{1}{\Gamma(\beta - 1)} \int\_{0}^{\mathfrak{u}} (\mathfrak{u} - t)^{\beta - 2} E\_{\mathfrak{a}} \left( \gamma t^{\mathfrak{a}} \right) \, \mathrm{d}t \,\omega$$

and substitute this into Equation (16) to find:

$$E\_{a,\beta}^{\rho}(\gamma z^a) = \frac{a\sin(\pi\rho)}{\pi} \int\_0^z (z^a - u^a)^{-\rho} u^{a\rho - \beta} \left[ \frac{1}{\Gamma(\beta - 1)} \int\_0^u (u - t)^{\beta - 2} E\_a(\gamma t^a) \,\mathrm{d}t \right] \,\mathrm{d}u \,\rho$$

which rearranges to the required result.

**Theorem 2.** *The three-parameter Mittag–Leffler function in Equation* (7) *can be written as an integral transform of the one-parameter Mittag–Leffler function in Equation* (5) *in the following way:*

$$E\_{a,\beta}^{\rho}(\gamma z^a) = \frac{a \sin(\pi \rho)}{\pi \Gamma(\beta - 1)} \int\_0^z F\_{a,\beta,\rho}(t; z) E\_a \left(\gamma t^a\right) dt \qquad z \in \mathbb{D},\tag{18}$$

*where we assume <sup>α</sup>*, *<sup>β</sup>*, *<sup>γ</sup>*, *<sup>ρ</sup>* <sup>∈</sup> <sup>C</sup> *with* Re(*α*) <sup>&</sup>gt; <sup>0</sup>*,* Re(*β*) <sup>&</sup>gt; <sup>1</sup>*, and* Re(*ρ*) <sup>&</sup>lt; <sup>1</sup>*, and where the function <sup>F</sup> is defined as*

$$F\_{a,\beta,\rho}(t;z) := \int\_{t}^{z} (z^a - u^a)^{-\rho} \, u^{a\rho - \beta} (u - t)^{\beta - 2} \, \mathrm{d}u. \tag{19}$$

**Proof.** By Fubini's theorem, it is possible to swap the order of the integrals in Equation (17). We have 0 ≤ *u* ≤ *z* and 0 ≤ *t* ≤ *u*, which after swapping is equivalent to 0 ≤ *t* ≤ *z* and *t* ≤ *u* ≤ *z*. We have from Equation (17):

$$\begin{split} E\_{a,\beta}^{\rho}(\gamma z^{a}) &= \frac{a\sin(\pi\rho)}{\pi\Gamma(\beta-1)} \int\_{0}^{z} \int\_{0}^{u} (z^{a} - u^{a})^{-\rho} \, u^{a\rho - \beta} (u - t)^{\beta - 2} E\_{a} \left(\gamma t^{a}\right) \, \mathrm{d}t \, \mathrm{d}u \\ &= \frac{a\sin(\pi\rho)}{\pi\Gamma(\beta-1)} \int\_{0}^{z} \int\_{t}^{z} (z^{a} - u^{a})^{-\rho} \, u^{a\rho - \beta} (u - t)^{\beta - 2} E\_{a} \left(\gamma t^{a}\right) \, \mathrm{d}u \, \mathrm{d}t \\ &= \frac{a\sin(\pi\rho)}{\pi\Gamma(\beta-1)} \int\_{0}^{z} E\_{a} \left(\gamma t^{a}\right) \int\_{t}^{z} (z^{a} - u^{a})^{-\rho} \, u^{a\rho - \beta} (u - t)^{\beta - 2} \, \mathrm{d}u \, \mathrm{d}t \\ &= \frac{a\sin(\pi\rho)}{\pi\Gamma(\beta-1)} \int\_{0}^{z} E\_{a} \left(\gamma t^{a}\right) F\_{a,\beta,\rho}(t; z) \, \mathrm{d}t, \end{split}$$

as required.

Thus, using Riemann–Liouville fractional calculus, we have forged new connections between the Mittag–Leffler functions of one, two and three parameters. The connection between those of one and three parameters, in particular, may give rise to new formulae linking AB fractional calculus with Prabhakar fractional calculus, in a way more profound than simply writing one as a special case of the other.
