Given that entropy term $-p \ln{p}$ has support (0,1) and beta distribution has same support, was interested in understanding if Beta $( x|\alpha, \beta )$ might exactly fit to normalized pdf $f(x)=A x \ln{x}$ ... using numerical methods (see graph below) it appears that it comes close for values $\alpha = 1.6267$ and $\beta = 2.0769$
How would you approach deriving analytical values for $\alpha$ and $\beta$ which mostly closely match Beta distribution to $-x \ln{x}$?
Also, is there a notational shortcut for representing/discussing entropy term $-x \ln{x}$
The beta distribution has pdf $f(x; \alpha , \beta )=\frac{\Gamma (\alpha + \beta)}{\Gamma (\alpha)\Gamma (\beta)}x^{\alpha - 1} (1-x)^{\beta -1}$
One approach I have seen many times when attempting to approximate an arbitrary pdf by a beta pdf is the moment matching approach. Let $\mu$ and $\sigma^2$ denote the mean and variance of the arbitrary pdf, respectively. Noting that the mean and variance of the beta pdf equals $\alpha/(\alpha +\beta)$ and $\alpha \beta / ((\alpha +\beta)^2(\alpha+\beta+1))$, respectively, one only has to solve the system $$ \begin{align} \frac{\alpha}{\alpha +\beta}&=\mu,\\ \frac{\alpha \beta}{(\alpha +\beta)^2(\alpha+\beta+1)}&=\sigma^2, \end{align} $$ with respect to $\alpha$ and $\beta$.
For the pdf $-4 x \log(x)$, it holds $\mu=1/4$ and $\sigma^2=17/324$. Solving the system results in $$ \begin{align} \alpha &= 28/17 \approx 1.64706,\\ \beta &= 35/17 \approx 2.05882, \end{align} $$ which are actually quite close to the values you identified numerically.