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Put $L=\log \lambda$ and integrate by parts using $x^{-2}=-d(x^{-1})/dx$. The exponential cutoff has derivative $\lambda e^{-x}{e}^{-\lambda e^{-x}}$, so the endpoint at infinity vanishes and the lower endpoint givesWith $u=\lambda e^{-x}$ this becomesThe mass of the transformed integral lies at $u=O(1)$, whereTo control this asymptotic expansion, split at $u=e^{-L/2}$ and $u=e^{L/2}$. On the central interval the geometric-series remainder is bounded by $2(\log u{)}^2/L^3$, whose integral against $e^{-u}$ is finite. The lower tail has length $e^{-L/2}$ and denominator at least $L$; the upper tail is exponentially small in $e^{L/2}$, with the original denominator at least one. The omitted tails of the logarithmic moments are likewise negligible. Thus extending the moments to $(0,\infty )$ is justified.

The Gamma function gives ${\int }_0^{\infty }{e}^{-u}du=1$, and the first logarithmic moment is $-\gamma$, with $\gamma$ Euler's constant. ThereforeThis is the moving Gumbel cutoff in an inverse-power tail: replacing the original cutoff by a sharp step at $x=\log \lambda$ gives the first term, but its smooth transition supplies the Euler-constant correction.