= Laplace endpoint expansion with a logarithmic denominator
{c}
{title2=$I_0(x)\sim[x(\log x)^2]^{-1}[1-2\gamma_E/\log x+\cdots]$}
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: $\int_0^\infty e^{-v}\log v\,dv=-\gamma_E$ and $\int_0^\infty e^{-v}(\log v)^2dv=\gamma_E^2+\pi^2/6$. Other peaks or singularities must be treated separately before applying this endpoint result to a whole integral.
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