Logarithmic integral representation of the natural logarithm (${\int }_0^1(t^x-1)/\ln tdt=\ln (1+x)(x>-1)$)

The natural logarithm has the parameter-integral representationFor $0<t<1$, the identity $(t^x-1)/\ln t={\int }_0^x{t}^{s}ds$ shows convergence: the absolute value is at most $|x|t^{\min (x,0)}$, which is integrable when $x>-1$. At $t=1$, the apparent singularity is removable, with limiting value $x$.

Write the left side as $I(x)$. On a compact parameter interval $c\le x\le d$ with $c>-1$, its differentiated integrand $t^x$ is bounded by the integrable function $t^c$. The dominated convergence theorem justifies differentiation under the integral sign, giving $I^{\prime }(x)={\int }_0^1{t}^{x}dt=1/(1+x)$. Since $I(0)=0$, the fundamental theorem of calculus gives $I(x)=\ln (1+x)$. At $x=-1$, the original improper integral diverges as $-{\int }_{0}dt/[t|\ln t|]$, and it also diverges for $x<-1$.