 Solution

We need $0<\alpha <2$. Indeed, if $\alpha <0$, the function is negative on $(\alpha ,0)$; if $\alpha =0$, the improper integral diverges at zero; and if $\alpha \ge 2$, its support interval is empty. For $0<\alpha <2$,Setting this equal to one gives the unique admissible valueOn this interval $1/x>0$, so both requirements for a probability density function are met.