Denote the nth derivative D^n and the n-fold integral D^(-n). Then D^(-1) f(t) = integral_0^t f(ξ) d ξ. Now, if the equation D^(-n) f(t) = 1/((n - 1)!) integral_0^t (t - ξ)^(n - 1) f(ξ) d ξ for the multiple integral is true for n, then D^(-(n + 1)) f(t) | = | D^(-1)[1/((n - 1)!) integral_0^t (t - ξ)^(n - 1) f(ξ) d ξ] | = | integral_0^t[1/((n - 1)!) integral_0^x (x - ξ)^(n - 1) f(ξ) d ξ] d x. Interchanging the order of integration gives D^(-(n + 1)) f(t) = 1/(n!) integral_0^t (t - ξ)^n f(ξ) d ξ.

fractional calculus | Riemann-Liouville operator | semi-integral