# 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 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.

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 ${1-\cos x}$.

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.

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).

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