I am a bit confused as to how the beta distribution can be defined at $0$. The formulation of the beta distribution is given as -
$f(x)=\frac{\Gamma(a+b)}{\Gamma(\alpha) \Gamma(\beta)} x^{\alpha-1}(1-x)^{\beta-1}$, for $0 \le x \le 1$
$=0$, otherwise
If I use $\alpha = \beta = 0.5$, at $x=0$ the beta distribution would have a term $0^{-0.5}$. Does this not give $\infty$ at $x=0,1$? Can we safely deal with distribution functions having such singularities in their domain?
Besides, what might the interpretation be of such infinities when $\text{Beta}(0.5, 0.5)$ is used as a Bayesian prior? The $\alpha$, $\beta$ in the prior distribution are often interpreted as number of successes and failures previously seen; hence form the basis of our prior belief. In this context, how can $\text{Beta}(0.5, 0.5)$ assign the highest probability density to the extremes $x=0,1$?
What I found till now:
There are a few resources where the domain has been restricted to $0 < x < 1$. But a clear indication of the motivation has not been provided. For example, the Wikipedia article on Beta distribution provides both closed and open domains as options.
Although in the same article, the section on mode, does point to the issue of a singularity at $x=0,1$ when $\alpha, \beta < 1$. It does not, however, speak of any appropriate method to deal with this, or interpret it.