Singularity of Riemann-Liouville Fractional Derivative

Question

I have recently started studying fractional order calculus and it seems that the definition of the fractional order differentiation, in the Riemann-Liouville (RL) sense, has singularities. To explain my point, let $\alpha \in (0,1)$ and $f(t)\in \mathbb{R}$ be a differentiable function. We have \begin{align} D^{\alpha} f(t) = \frac{1}{\Gamma(1-\alpha)}\frac{d}{dt}\int_{0}^t f(\tau)(t-\tau)^{-\alpha}\ d\tau. \end{align} Performing integration by parts and carrying out the differentiation, we further have \begin{align} D^{\alpha} f(t) = \frac{f(0)t^{-\alpha}}{\Gamma(1-\alpha)}+\frac{1}{\Gamma(1-\alpha)}\int_{0}^t \dot{f}(\tau)(t-\tau)^{-\alpha}\ d\tau. \end{align} The term $t^{-\alpha}$ brings about a singularity at $t=0$. So, can we say that (i) for a differentiable function the RL derivative of order $\alpha\in (0,1)$ is singular at $t=0$ unless $f(0) = 0$; (ii) if $f(0) = 0$, then the RL derivative corresponds to that of the Caputo?