ABSTRACT
The Riemann-Liouville fractional integral is usually defined by
D−αf(t) = 0D−αt f(t) = 1
Γ(α)
(t− x)α−1f(x)dx, Reα> 0. (6.2.1)
Clearly, D−α is a linear integral operator. A simple change of variable (t− x)α = u in (6.2.1) allows us to prove the
following result
D[D−αf(t)] =D−α[Df(t)] + f(0) tα−1
Γ(α) . (6.2.2)
Clearly, the integral in (6.2.1) is a convolution, and hence, the Laplace transform of (6.2.1) gives
L {D−αf(t)} = L {f(t)∗g(t)}=L {f(t)}L {g(t)}, (6.2.3) = s−αf¯(s), α> 0. (6.2.4)
where g(t) = tα−1
Γ(α) and g¯(s) = s−α.
