 Discrete prize-count optimization for a power-valued contest

For prize scale $w(x)=x^{-\alpha }$, total effort is proportional to $h(x)=x^{1-\alpha }(1-x)$ on the feasible grid $x=m/n$. If $\alpha \ge 1$, one prize is optimal. For $0<\alpha <1$, $h$ increases up to $x_{\ast }=(1-\alpha )/(2-\alpha )$ and then decreases, so the optimal feasible integer is among $\lfloor n{x}_{\ast }\rfloor$ and $\lfloor n{x}_{\ast }\rfloor +1$. Remove infeasible candidates and compare their objective values. No definition at $x=0$ is needed.