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For $0<x<1$,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:Hence $J(b)=\ln (b+1)+C$. Since the integrand is identically zero at $b=0$, $J(0)=0$ fixes $C=0$. ThusThis proves the logarithmic integral representation of the natural logarithm, including its parameter range.

At $x=0$, the integrand vanishes for $0<t<1$, so $I(0)=0$ andSolving the first-order linear differential equation from part (i) on $(-1,\infty )$ gives $g=C(x+1)$. The initial condition fixes $C=1$, soEquivalently, the defining exponent is the logarithmic integral representation of the natural logarithm, $I(x)=\ln (x+1)$.