 Laplace endpoint expansion with a logarithmic denominator

A cut-off endpoint integral with amplitude $(\log u{)}^{-2}$ has a logarithmic rather than a Taylor endpoint expansion. Set $v=xu$ and expand $(\log x-\log v{)}^{-2}$ against $e^{-v}dv/x$. The derivatives of the gamma function at one supply the logarithmic moments: ${\int }_0^{\infty }{e}^{-v}\log vdv=-{\gamma }_E$ and ${\int }_0^{\infty }{e}^{-v}(\log v{)}^{2}dv={\gamma }_E^2+{\pi }^2/6$. Other peaks or singularities must be treated separately before applying this endpoint result to a whole integral.