3.2.2. Riemann–Liouville Derivative Definition

This definition shows that the Riemann–Liouville derivative corresponds to the integerorder derivative of the Riemann–Liouville integral of *x*(*t*).

$${}^{RL}D\_t^n(\mathbf{x}(t)) = \frac{d}{dt} \left[ I\_t^{1-n}(\mathbf{x}) \right] = \frac{d}{dt} [h\_{1-n}(t) \* \mathbf{x}(t)] \tag{17}$$

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Using the Laplace transform, we obtain

$$\begin{aligned} \left(L\left\{ ^{RL}D\_t^n(\mathbf{x})\right\} \right) &= \mathbf{s}\left(L\left\{ I\_t^{1-n}(\mathbf{x}) \right\} \right) - \mathbf{g}(0) = \mathbf{s}\left(\frac{1}{s^{1-n}}\mathbf{X}(\mathbf{s})\right) - \mathbf{g}(0) = \mathbf{s}^n\mathbf{X}(\mathbf{s}) - \mathbf{g}(0) \\ \text{with } \mathbf{g}(0) &= \left\{ \_{-\infty}I\_t^{1-n}(\mathbf{x}) \right\}\_{t=0} \end{aligned}$$

Since *g*(0) does not have a physical and direct interpretation, the Riemann–Liouville derivative is generally not used to integrate system (6) or (7).

It is important to note that the two derivative definitions only require integer-order differentiation *<sup>d</sup> dt* and fractional-order integration *I*1−*n* . So, contrary to a common belief, the basic operation of fractional calculus is not fractional differentiation but fractional integration.
