Fractional Calculus of Zero

Zero in any differentiation or integration order can be expressed as (Zero summation formula).
$$D_{x}^{\alpha} (0)=\sum_{k=1}^{\infty}C_k\frac{x^{-(\alpha+k)}}{\Gamma(1-(\alpha+k))}$$
Where the summation terms are the values of Reciprocal gamma function at its roots for $\alpha = 0$ and $C_k$ is constant (See Reciprocal gamma function graph).
The zeros of Reciprocal gamma function are at $n$ where $-n \in \Bbb N$.
So for the first-order differentiation, it can be expressed as:
$$D_x^1f(x)=f'(x)+\sum_{k=1}^{\infty}C_k\frac{x^{-(1+k)}}{\Gamma(1-(1+k))}=f'(x)$$
This is the same first order differentiation and there is no need to set the constants to zero because this integer order of differentiation jumps from one zero of Reciprocal gamma function to another so all terms become zeros all time.
Now for the integration.
$$D_x^{-1}f(x)=g(x)+\sum_{k=1}^{\infty}C_k\frac{x^{-(-1+k)}}{\Gamma(1-(-1+k))}=g(x)+C_1 \frac{x^0}{\Gamma(1)}+\sum_{k=2}^{\infty}C_k\frac{x^{-(-1+k)}}{\Gamma(1-(-1+k))}$$
We can see a constant that need to be determined and that is the so-called integral constant which is usually set to zero as default.
So if we need to evaluate higher integration order, more constants comes out from zero summation formula need to be evaluated.
For example
$$D_x^{-3}f(x)=g(x)+\sum_{k=1}^{\infty}C_k\frac{x^{-(-3+k)}}{\Gamma(1-(-3+k))}=g(x)+C_1 \frac{x^2}{\Gamma(3)}+C_2 \frac{x}{\Gamma(2)}+C_3 \frac{x^0}{\Gamma(1)}+\sum_{k=4}^{\infty}C_k\frac{x^{-(-1+k)}}{\Gamma(1-(-1+k))}$$
And so on, the higher integration order the more constants and polynomial terms joining the integration result, but for a reason we set all these constants to zero.
Now let us see what happen if one wants to evaluate the half-order differentiation for a function.
$$D_x^{1/2}f(x)=g(x)+\sum_{k=1}^{\infty}C_k\frac{x^{-(\frac12+k)}}{\Gamma\left(1-(\frac12+k)\right)}$$
and if the function contains a constant
$$D_x^{1/2}\left(f(x)+C_0\right)=g(x)+C_0\frac{x^{-\frac12}}{\Gamma(\frac12)} +\sum_{k=1}^{\infty}C_k\frac{x^{-(\frac12+k)}}{\Gamma \left(1-(\frac12+k)\right)}$$
The zero summation formula is now a polynomial that diverges, making the half-order differentiation undefined at all, so for this reason and the same as integral constant reason we should set all these constants to zero.
Hence $$D_x^{1/2}(0)=0$$