Riemann–Liouville integral

The integral is a manner of generalization of the repeated antiderivative of f in the sense that for positive integer values of α, Iα f is an iterated antiderivative of f of order α.

The Riemann–Liouville integral is named for Bernhard Riemann and Joseph Liouville, the latter of whom was the first to consider the possibility of fractional calculus in 1832.

For a function f continuous on the interval [a,x], the Cauchy formula for n-fold repeated integration states that

Now, this formula can be generalized to any positive real number by replacing positive integer n with α, Therefore we obtain the definition of Riemann-Liouville fractional Integral by The Riemann–Liouville integral is defined by where Γ is the gamma function and a is an arbitrary but fixed base point.

The dependence on the base-point a is often suppressed, and represents a freedom in constant of integration.

Another notation, which emphasizes the base point, is[6] This also makes sense if a = −∞, with suitable restrictions on f. The fundamental relations hold the latter of which is a semigroup property.

Thus Iα defines a linear operator on L1(a,b): Fubini's theorem also shows that this operator is continuous with respect to the Banach space structure on L1, and that the following inequality holds: Here ‖ · ‖1 denotes the norm on L1(a,b).

Moreover, by estimating the maximal function of I, one can show that the limit Iα f → f holds pointwise almost everywhere.

The operator Iα is well-defined on the set of locally integrable function on the whole real line

It defines a bounded transformation on any of the Banach spaces of functions of exponential type

For f ∈ Xσ, the Laplace transform of Iα f takes the particularly simple form for Re(s) > σ.

Here F(s) denotes the Laplace transform of f, and this property expresses that Iα is a Fourier multiplier.

[8] The Caputo fractional derivative with base point x, is then: Another representation is: Let us assume that f(x) is a monomial of the form The first derivative is as usual Repeating this gives the more general result that which, after replacing the factorials with the gamma function, leads to For k = 1 and a = ⁠1/2⁠, we obtain the half-derivative of the function

as To demonstrate that this is, in fact, the "half derivative" (where H2f(x) = Df(x)), we repeat the process to get: (because

Indeed, given the convolution rule and shorthanding p(x) = xα − 1 for clarity, we find that which is what Cauchy gave us above.

Laplace transforms "work" on relatively few functions, but they are often useful for solving fractional differential equations.