= Moving Gumbel cutoff in an inverse-power tail
{title2=$\int_a^\infty x^{-r}e^{-\lambda e^{-x}}dx\sim\frac{L^{1-r}}{r-1}-\gamma L^{-r}\quad(L=\log\lambda)$}
For fixed $a>0$ and $r>1$, the factor $\exp(-\lambda e^{-x})$ is a translated <Gumbel distribution> cumulative function, with transition near $x=L$. Integration by parts transforms the <integral> to an exponentially small boundary term plus $(r-1)^{-1}\int_0^{\lambda e^{-a}}e^{-u}(L-\log u)^{1-r}du$. Expanding the latter against its finite logarithmic moments gives the displayed two terms because $\int_0^\infty e^{-u}\log u\,du=-\gamma$, with $\gamma$ <Euler's constant>. The smooth cutoff shifts the first correction even though replacing it by a step gives the leading tail.
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