What is the fractional constant of integration?

Question

In fractional calculus, one usually tends to ignore the constant of "differ-integration" if you will, but when I attempted a problem with some fractional calculus, I found the result was somewhat off, which led me to the belief that I was missing the "simple" constant of integration.

After trying to put it to paper, I found this problem to be very annoying and confusing, as I came up with a few things.

Let $D^n_xf(x)$ represent the $n$th derivative of $f$ with respect to $x$.

$$D^n_xx^k=\frac{\Gamma(k+1)}{\Gamma(k+1-n)}x^{k-n}$$

For $k+1-n\in\mathbb Z$ and $k+1-n<0$, we get

$$D^n_xx^k=\frac{\Gamma(k+1)}{\pm\infty}x^{k-n}=0$$

So from this, we get

$$D^n_x0=\frac{\Gamma(k+1-n)}{\Gamma(k+1)}x^{k+n}$$

where $k\in\mathbb Z$ and $k\le0$.

I also run into other problems, like $D^n_xD^k_x0=D^{n+k}_x0$, and everything starts to fall apart. Is there a good defined way for representing the constant of differ-integration?

Answer 1

You are attacking something complicated. Consider the functions of one real variable $x$, in my notations $a,b,c \in \mathbb{R}^{*+}$, $n,m \in \mathbb{N}^*$, $\ \ C_k,C$ are real constants, $u,v,w$ are differentiable/smooth functions of $x$.

As I Wrote $$u \mapsto D^a u$$

is a linear operator. That is : $$D^a (u+v) =D^a u+D^a v, \qquad \qquad D^a (C u) = C D^au$$

When you write $D^{-a}u$, you are in fact considering the solutions of the linear equation $D^a v = u$ where $v$ is the unknown, and what you call the "integration constant" means how to define this set of solutions.

As every linear equation, once we have a solution $D^a v = u$, we can find all the other solutions from the kernel of $D^a$ : if $D^a w = 0$, then $v+w$ is also a solution of $D^a v = u$, and all the solutions are of this form.

You know that $D^1 x^0 = 0$, and from $D^{n+m} =D^{n}D^{m}$ you get that the kernel of $D^n$ are the degree $n-1$ polynomials. This is why you can write : $$D^n v = u \quad \implies \quad v = J^n u + \sum_{k=0}^{n-1} C_k x^k$$

Where "$J^n$ is the integrate $n$ times linear operator", i.e. $J^1 u (x) = \int_0^x u(t) dt$ and by induction $J^n = J^1 J^{n-1}$. Clearly $D^n J^n u = u$, so that $J^n$ is the right-inverse of $D^n$, but not the left-inverse, and there is no such, since the kernel of $D^n$ is non-empty, and hence not invertible.

Then you have to prove that with $\gamma(a,c) = \frac{\Gamma(c+1)}{\Gamma(c+1-a)}$ and when everything is well-defined (see below) : $$D^a x^c = \gamma(a,c) x^{c-a}, \qquad\qquad D^{a}D^{b} x^c = D^{a+b} x^c$$

How do you explain then that $D^1 x^0 = 0$ ? In fact $D^1 x^0 = \gamma(1,0) x^{-1}$ and this is $= 0$ only because $\gamma(1,0) =0$.

(Since every differentiable function can be written as a linear combination of powers of $x$, we can look only at those powers of $x$. )

And the kernel of $D^a$ is obtained from the set of values $c$ for which $\gamma(a,c) = 0$. It is a theorem that the $\Gamma$ function is nowhere zero, and has poles of order $1$ at negative integers. hence $$\gamma(a,c) = \frac{\Gamma(c+1)}{\Gamma(c+1-a)} = 0 \Leftrightarrow \Gamma(c+1)\ne \infty, \Gamma(c+1-a)= \infty \Leftrightarrow \ -(c+1) \not\in \mathbb{N}, \ -(c+1-a) \in \mathbb{N}$$

And this is the answer to your question. If $a$ is not an integer, then the kernel of $D^a$ is $$D^a w = 0 \qquad \Leftrightarrow \qquad w = \sum_{k=1}^\infty C_k x^{a-k} $$

And when $n$ is an integer $$D^n w = 0 \qquad \Leftrightarrow \qquad w = \sum_{k=1}^n C_k x^{n-k} $$

Now there is another problem : what happens if $\gamma(a,c) = \infty$ ? Because yes it can happen, when $-(c+1) \in \mathbb{N}, \ -(c+1-a) \not\in \mathbb{N}$.

This is how you get your problem, when trying to relate the kernel of $D^1$ with the kernel of $D^{1/2}$. Indeed $D^{1/2} u = 0 $ doesn't imply in general that $ D^1 u = D^{1/2}(D^{1/2} u) = 0$.

This is why you can't write that $D^{a+b} = D^a D^b$.

Answer 2

It's an undefined operation. Constants of integration are related to the singularities of the differential form.

I find the the best way to understand them is through Cauchy integration and as a generalization the Laplace or Fourier transforms.

For integer powers of the derivative you could define the constants of integration like so:

$$c(s)=c_0+c_1 s+c_2s^2+c_3s^3...$$ $$\frac{1}{2 \pi i} \oint\limits_{\gamma} s^a\exp(sx)c(s) ds$$

Where $\gamma$ circulates around s=0.

With Cauchy integration this turns all the $1 \over s$ terms into 1 and everything else to 0 for integer a.

This is because for integer a $s^a$ integrates to $s^{a+1}\over a+1$, and for a=-1, $\ln(s)$. If you start at s=1 and follow the output as you circulate around s=0, when you return to s=1, the output will be the same as when you started. But only for integer a. For $\ln(s)$ and fractional a, it's not the same.

So for a = -3 you would have:

$$\left(c_0+c_1s+c_2s^2 \right)\left(1+xs+\frac{x^2}{2}s^2 \right) \frac{1}{s^3} \to c_0 \frac{x^2}{2} +c_1x+c_2$$

This heavily implies that c(s) is a holomorphic function.

For non-integer a, the Cauchy integral won't do, because only ln(s) is scale independent and independent of starting position. It'd be best to use something like an adaptation of a Hankel contour integral or inverse Laplace/Fourier transform.

But the general implications would be that. There are an infinite number of constants. And that both the fractional derivative and it's inverse has such constants.

Where a is fractional:

$$s^a \exp(sx) c(s)$$

All terms are roots, all terms stick around.

So really $D^aD^b \ne D^{a+b}$, because $D^a$ isn't 1 to 1. And algebraic properties only truly work for 1 to 1 operations. But if you have it such that all constants are known, conserved, and fully recoverable like as some sort of hidden information, for example $Dc \to c \delta$ instead of $Dc \to 0$, then it becomes a 1 to 1 transformation.

But if $D^aD^b=D^{a+b}$ is all you want, then you could just treat it like eigenvalue eigenvector operations and completely ignore the constants.

$$D^a\exp(cx)=c^a \exp(cx)$$

However, unless you have the constants, you can't define all $D^a f(x)$. For example, $D^{-1} 1=0^{-1} \exp(0x)$ only really makes sense as $\lim\limits_{z \to 0} z^{-1}(\exp(zx)-1)+c_0=x+c_0$

I'd also like to point out that technically from an algebraic stand point $c^a$ is undefined for fractional a and any number c, too. That's because $c^a$ isn't 1 to 1 for numbers, either, but it is for complex paths as inputs. You can actually find many contradictory proofs that result from the fact that c^a isn't 1 to 1.

In general: $c^ac^b \ne c^{a+b}$