RiemannLiouville 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 $I^{k} x(t) = I^1\left(I^{k1}x(t)\right)$ for $k = 2,3,\ldots$. Coefficients $f_i(\alpha,t)$ are some known functions.
$\begingroup$
$\endgroup$
1

1$\begingroup$ There seems to be no such expansion, as this would imply that $(t  \tau)^{\alpha1}$ can be written as $\sum_k f_k(\alpha, t) (t  \tau)^{k1}$, 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 $(t  \tau)^{\alpha1}$ on $[0, t]$ by a polynomial. $\endgroup$– Mateusz KwaśnickiJul 26, 2020 at 19:22
Add a comment
