# Expansion of the Riemann-Liouville integral

Riemann-Liouville integral operator of the $$\alpha$$-th order ($$\alpha > 0$$) is defined as $$I^{\alpha} x(t) := \frac{1}{\Gamma(\alpha)}\int\limits_{0}^{t} (t-\tau)^{\alpha - 1} x(\tau)d\tau$$ I would like to know are there some expansion formulas of this operator like: $$I^{\alpha} x(t) = f_0(\alpha,t) I^0 x(t) + f_1(\alpha,t) I^1 x(t) + \ldots$$ where $$I^0 x(t) = x(t)$$, $$I^1(t) = \int\limits_{0}^{t} x(\tau) d\tau$$ and in general $$I^{k} x(t) = I^1\left(I^{k-1}x(t)\right)$$ for $$k = 2,3,\ldots$$. Coefficients $$f_i(\alpha,t)$$ are some known functions.

• There seems to be no such expansion, as this would imply that $(t - \tau)^{\alpha-1}$ can be written as $\sum_k f_k(\alpha, t) (t - \tau)^{k-1}$, which is not the case. On the other hand, if $\alpha > 1$, one can get arbitrarily good approximations (say, with respect to the supremum norm on $[0, t]$) by using finitely many terms: simply approximate $(t - \tau)^{\alpha-1}$ on $[0, t]$ by a polynomial. Jul 26, 2020 at 19:22