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?

One way to deal with these constants is to change from indefinite integration to definite integration.  For example, the Riemann-Liouville integral may be used:
$$D^{-\alpha}_af(x)=\frac1{\Gamma(\alpha)}\int_a^xf(t)(x-t)^{\alpha-1}\ dt$$
Now our constant of integration is controlled by $a$, and there will exist some $a$ such that
$$D^{-1}_a\frac d{dx}f(x)=f(x)$$
