= Solution
For $0<x<1$,
$$
\frac{x^b-1}{\ln x}=\int_0^b x^s\,ds.
$$
In particular its <absolute value> is bounded by $|b|x^{\min(b,0)}$, proving convergence when $b>-1$; the apparent endpoint singularity at one is removable, with value $b$. On any compact parameter interval with lower bound $b_0>-1$, the differentiated integrand is $x^b$, dominated by $x^{b_0}$. The <dominated convergence theorem> therefore justifies <differentiation under the integral sign>:
$$
\boxed{J'(b)=\int_0^1x^b\,dx=\frac1{b+1}}.
$$
Hence $J(b)=\ln(b+1)+C$. Since the integrand is identically zero at $b=0$, $J(0)=0$ fixes $C=0$. Thus
$$
\boxed{J(b)=\ln(b+1),\qquad b>-1}.
$$
This proves the <logarithmic integral representation of the natural logarithm>, including its parameter range.
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