 Solution

The required range isNear $t=1$, the ratio $(t^x-1)/\ln t$ tends to $x$, so it causes no convergence problem. Near $t=0$, one useful identity isFor $x>-1$, this bounds its absolute value by $|x|t^{\min (x,0)}$, which is integrable on $(0,1)$. It also supplies the domination used in part (i).

For $x=-1$, the integrand is negative and behaves as $-1/[t|\ln t|]$ near zero; its improper integral diverges. For $x<-1$ its negative singularity is even stronger, and comparison with that same divergent integral again rules out convergence. Thus the original finite real-valued definition of $g$ has precisely the stated domain.