= Logarithmic integral representation of the natural logarithm
{title2=$\int_0^1(t^x-1)/\ln t\,dt=\ln(1+x)\quad(x>-1)$}
The <natural logarithm> has the parameter-integral representation
$$
\boxed{\int_0^1\frac{t^x-1}{\ln t}\,dt=\ln(1+x),\qquad x>-1.}
$$
For $0<t<1$, 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> $c\leq x\leq d$ 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 $I'(x)=\int_0^1t^xdt=1/(1+x)$. Since $I(0)=0$, the <fundamental theorem of calculus> gives $I(x)=\ln(1+x)$. At $x=-1$, the original <improper integral> diverges as $-\int_0dt/[t|\ln t|]$, and it also diverges for $x<-1$.
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