Riemann–Liouville integral

Remarks

The integral converges for almost every $x$, and for every $x$ when $f$ is continuous on $[a, b)$, because $(x - t)^{\alpha - 1}$ is integrable near $t = x$ for $\alpha > 0$. Example: for $\beta > -1$ and $a = 0$, the substitution $t = xs$ and the beta integral give $I_0^\alpha t^\beta = \frac{\Gamma(\beta + 1)}{\Gamma(\alpha + \beta + 1)} x^{\alpha + \beta}$. For $\beta = 0$ this is $x^\alpha / \Gamma(\alpha + 1)$. Non-example: for $\alpha = -\tfrac{1}{2}$ and $f = 1$, the integrand $(x - t)^{-3/2}$ is not integrable near $t = x$, so the formula defines nothing; negative orders need a derivative, as in fractional calculus. The operators satisfy the semigroup law $I_a^\alpha I_a^\beta f = I_a^{\alpha + \beta} f$ for $\alpha, \beta > 0$. For $\alpha = 1$, $I_a^1 f(x) = \int_a^x f(t) \, dt$.