Solution
= 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 $\alpha\geq2$, its support interval is empty. For $0<\alpha<2$,
$$
\int_\alpha^2\frac{dx}{x}=\log\frac2\alpha.
$$
Setting this equal to one gives the unique admissible value
$$
\boxed{\alpha=\frac2e.}
$$
On this interval $1/x>0$, so both requirements for a <probability density function> are met.