Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 79 / 1 / c

Take the principal complex logarithm and deform the interval upwards, where $e^{ixt}$ decays for $x>0$. The small indentation at zero contributes nothing because $r\log r\to 0$; the distant horizontal segment vanishes exponentially. The two vertical edges therefore give the exact representationThis is an endpoint steepest descent contour, with distinct contributions from the singular endpoint and the smooth endpoint. Setting $u=xs$ in the first integral and using the given logarithmic moment givesFor the second edge, the Taylor series at $s=0$ is $\log (1+is)={\sum }_{n\ge 1}(-1{)}^{n+1}(is{)}^n/n$. Watson's lemma integrates it term by term as an asymptotic expansion, using ${\int }_0^{\infty }{s}^n{e}^{-xs}ds=n!/x^{n+1}$. ConsequentlyHere $\gamma$ is the Euler--Mascheroni constant. The first smooth-endpoint terms areFor every fixed $N\ge 1$, retaining $k=0,\dots ,N-1$ leaves $O(x^{-N-2})$. The factorial series is an asymptotic series, not a convergent sum. Treating the logarithmic endpoint by an ordinary smooth-endpoint integration formula would lose the leading logarithm; this is the logarithmic endpoint Fourier asymptotics mechanism.