Calculus 0.5

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

\displaystyle F(x)=I_nf(x) = \frac{1}{(n-1)!}\int_0^x (x-t)^{n-1} f(t)\,dt  \ \ \ \ \ (1)

is the {n}th antiderivative of {f}, with initial conditions {F(0)=\dots = F^{(n-1)}(0)=0}. Indeed, differentiating under the integral sign {(n-1)} times, we get

\displaystyle  F^{(n-1)}(x) = \int_0^x f(t)\,dt

and one more derivative returns {f}.

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

\displaystyle I_\alpha f (x) = \frac{1}{\Gamma(\alpha)}\int_0^x (x-t)^{\alpha-1} f(t)\,dt

As a special case, the half-integral of {f} is

\displaystyle I_{1/2}f (x) = \pi^{-1/2}\int_0^x \frac{f(t)}{\sqrt{x-t}}\,dt

Of course, for this integral to be worthy of its name, applying it twice we should get the ordinary antiderivative of {f}. Does this work? It suffices to check the case of Dirac delta {f=\delta_a}. When {x>a}, {I_{1/2}f (x) = \pi^{-1/2} (x-a)^{-1/2}}; for {x<a} we have {I_{1/2}f (x)=0}. The task reduces to showing that

\displaystyle \int_a^x \frac{1}{\sqrt{x-t}\,\sqrt{t-a}}\,dt = \pi

whenever {x>a}. The substitution {t=\frac{x+a}{2}+\frac{x-a}{2}\,s} takes care of that:

\displaystyle  \int_a^x \frac{1}{\sqrt{x-t}\,\sqrt{t-a}}\,dt = \int_{-1}^1 \frac{1}{\sqrt{1-s^2}}\,ds =2\sin^{-1}1 = \pi

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 {x=0}, which is spread across the half-axis due to the non-local nature of fractional calculus.

Cosine, its half-integral, and integral (sine).
Cosine, its half-integral, and integral (sine).

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 {1-\cos x}.

Two more half-steps of integration
Two more half-steps of integration

For another example, consider the piecewise constant function {f=\chi_{[0,2]}-\chi_{[2,4]}}, pictured below in red with its half-integral in blue.

Half-integral of piecewise constant function
Half-integral of piecewise constant function

The antiderivative {F=\int_0^x f(t)\,dt} is a piecewise linear hat function. The half-integral of {f} is {I_{1/2}f}, which is also the half-derivative of {F}. The fractional derivative is non-local: its value at a particular point {x} is influenced by the values of {F} on the entire domain (and the boundary condition at {0} exerts some influence too).

Hat function, its half-derivative and derivative
Hat function, its half-derivative and derivative

The final plot shows {F}, {F^{(1/4)}}, {F^{(1/2)}}, {F^{(3/4)}}, and {F'=f}.

Same in increments of 1/4
Same in increments of 1/4

1 thought on “Calculus 0.5”

  1. To make the name more intuitive, we can reformulate it as \frac{1}{\Gamma(\alpha+1)}\int_0^x f(x-t) dt^\alpha. It looks similar to “integrating over t \alpha times”. Interestingly this forces us to integrate f “backwards”. For \alpha=1 the difference \delta=t_{i+1}-t_{i} is uniform so it does not matter whether we are integrating f(x) or f(x-t). But it does matter for other \alpha‘s. For example when \alpha=1/2, \delta'=\sqrt{t_{i+1}}-\sqrt{t_{i}}=\frac{\delta}{\sqrt{t_{i+1}}+\sqrt{t_{i}}}. Now the larger t is, the smaller \delta' 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 (x-t)^{\alpha-1} term in the original formula.

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