Is there a flaw in the theory of fractional calculus? Let's talk about the function $f(x)=x^n$.
It's derivative of $k^{th}$ order can be expressed by the formula:
$$\frac{d^k}{dx^k}x^n=\frac{n!}{(n-k)!}x^{n-k}$$
Similarly, the $k^{th}$ integral (integral operator applied $k$ times) can be expressed as:
$$\frac{n!}{(n+k)!}x^{n+k}$$
According the the Wikipedia article https://en.wikipedia.org/wiki/Fractional_calculus, we can replace the factorial with the Gamma function to get derivatives of fractional order.
So, applying the derivative of half order twice to $\frac{x^{n+1}}{n+1}+C$, should get us to $x^n$.
Applying the half-ordered derivative once gives:
$$\frac{d^{1/2}}{{dx^{1/2}}}\left(\frac{x^{n+1}}{n+1}+Cx^0\right)=\frac{1}{n+1}\frac{\Pi(n+1)}{\Pi(n+1/2)}x^{n+1/2}+C\frac{1}{\Pi(-1/2)}x^{-1/2}$$
where $\Pi(x)$ is the generalization of the factorial function, and $\Pi(x)=\Gamma(1+x)$
Again, applying the half-ordered derivative gives:
$$\frac{1}{n+1}\frac{\Pi(n+1)}{\Pi(n)}x^n+\frac{C}{\Pi(-1)}x^{-1}=x^n$$
which works fine because $\frac{C}{\Pi(-1)}\rightarrow 0$. So, the derivative works good but that's not the case with fractional-ordered integration.
Applying the half-ordered integral operator twice to $x^n$ should give us $\frac{x^{n+1}}{n+1}+C$. Applying the half-ordered integral once means finding a function whose half-ordered derivative is $x^n$. So, applying it once gives:
$$\frac{\Pi(n)}{\Pi{(n+1/2)}}x^{n+1/2}+C\frac{1}{\Pi(-1/2)}x^{-1/2}$$
Again, applying the half ordered derivative to this function should give a function whose half-ordered derivative is this function. So, again applying the half-integral operator gives:
$$\frac{x^{n+1}}{n+1}+C+C'\frac{1}{\Pi(-1/2)}x^{-1/2}\neq \frac{x^{n+1}}{n+1}+C$$
where $C'$ is another constant. So, why does this additional term containing $C'$ get introduced? Is the theory of fractional derivatives flawed? Is there any way to get a single constant $C$ in the end by applying the half-integral operator two times?

I learned about the existence of fractional calculus from this very recommendable site. 
In "integer" differential calculus you don't recover the function either.
If $f(x)=x^n$, $f'(x)=nx^{n-1}$ And integrating $\int nx^{n-1}=x^n+\mathbf C$ Even though initially we had $f(x)=x^n+C'$, we do the math but recover $f(x)=x^n+C$, but without any indication about whether or not $C$ and $C'$ are the same.
My reasoning seems the "tu quoque" fallacy, but it is not, simply says that fractional calculus have the same problems as "integer" calculus has. Maybe the solution is the same.
It can be put more dramatically by means of another example, this one almost paraphrasing yours: derive twice this $f(x)=x^3+1$. $f''(x)=6x$, turning back we get, $f'(x)=3x^2+C$, first and now, $f(x)=x^3+Cx+C_1$
In "integer" calculus we solve this stablishing the important thing is the fundamental theorem of calculus. In it, the result is independent of the primitive chosen. Maybe we can have the same for fractional calculus. I can figure to do this by the "method" of putting "by hand" the boundary conditions to recover the function, in the manner we use the contour conditions in differential equations. So is, I don't think we have for fractional calculus the nice notation the integral symbols offer, but the limits of integration are this: boundary or initial conditions to recover the solution. I'd do this:
$$f(x)=\frac{d^{-1/2}}{{dx^{-1/2}}}\left(\frac{\Pi(n)}{\Pi{(n+1/2)}}x^{n+1/2}+C\frac{1}{\Pi(-1/2)}x^{-1/2}\right)=$$
$$=\frac{x^{n+1}}{n+1}+C+C'\frac{1}{\Pi(-1/2)}x^{-1/2}\;\mathrm {with}\;f(0)=C\implies$$
$$f(x)=\frac{x^{n+1}}{n+1}+C$$
And indeed you did too without notice it!! If not, how did you put $C$ in the first "half integral"? the more general "half primitive" is:
$$\frac{d^{-1/2}}{{dx^{-1/2}}}x^n=\frac{\Pi(n)}{\Pi{(n+1/2)}}x^{n+1/2}+\mathbf A\frac{1}{\Pi(-1/2)}x^{-1/2}$$
So, is, a "half primitive" with some constant $A$ to determine. You imposed some condition on the "half-primitive" to get the original function with $C$, the original constant (in this case, the condition is an implicit and trivial relation)
