What is the fractional constant of integration? 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?
 A: 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 $D^{a+b} = D^a D^b$.
