It is easy to integrate a function times using a single integral:

is the th antiderivative of , with initial conditions . Indeed, differentiating under the integral sign times, we get

and one more derivative returns .

Formula (1) makes sense for fractional orders of integration too, except that the factorial should be replaced with the gamma function.

As a special case, the half-integral of is

Of course, for this integral to be worthy of its name, applying it twice we should get the ordinary antiderivative of . Does this work? It suffices to check the case of Dirac delta . When , ; for we have . The task reduces to showing that

whenever . The substitution takes care of that:

Here is the graph of cosine (in red), its half-integral (also half-derivative of sine, in blue), and the sine. Notice that the half-integral is not quite periodic, although it is close. This is the effect of the boundary at , which is spread across the half-axis due to the non-local nature of fractional calculus.

Going further, we get half-integral of the sine, and then full integral of the sine, which with our initial condition turns out to be .

For another example, consider the piecewise constant function , pictured below in red with its half-integral in blue.

The antiderivative is a piecewise linear hat function. The half-integral of is , which is also the half-derivative of . The fractional derivative is non-local: its value at a particular point is influenced by the values of on the entire domain (and the boundary condition at exerts some influence too).

The final plot shows , , , , and .

To make the name more intuitive, we can reformulate it as . It looks similar to “integrating over times”. Interestingly this forces us to integrate “backwards”. For the difference is uniform so it does not matter whether we are integrating or . But it does matter for other ‘s. For example when , . Now the larger t is, the smaller becomes. For a symmetric function like a constant step inside a step function, this conforms with our intuition that the integral grows slower at larger values. For a non-symmetric one like the boundary between one step and another, this funny backward integration means the most recent points get the most credits, so the integral suddenly change shapes. I guess the same effect is achieved by the term in the original formula.