= Solution
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 $r\log r\to0$; the distant horizontal segment vanishes exponentially. The two vertical edges therefore give the exact representation
$$
g(x)=i\int_0^\infty\left(\log s+\frac{i\pi}{2}\right)e^{-xs}\,ds-i e^{ix}\int_0^\infty\log(1+is)e^{-xs}\,ds.
$$
This 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 gives
$$
i\int_0^\infty\left(\log s+\frac{i\pi}{2}\right)e^{-xs}\,ds
=\frac{-\pi/2-i(\log x+\gamma)}{x}.
$$
For the second edge, the Taylor series at $s=0$ is $\log(1+is)=\sum_{n\ge1}(-1)^{n+1}(is)^n/n$. <Watson's lemma> integrates it term by term as an <asymptotic expansion>, using $\int_0^\infty s^ne^{-xs}\,ds=n!/x^{n+1}$. Consequently
$$
\boxed{g(x)\sim\frac{-\pi/2-i(\log x+\gamma)}x-e^{ix}\sum_{k=0}^\infty\frac{k!}{(ix)^{k+2}}.}
$$
Here $\gamma$ is the <Euler--Mascheroni constant>. The first smooth-endpoint terms are
$$
g(x)\sim\frac{-\pi/2-i(\log x+\gamma)}x
+e^{ix}\left(\frac1{x^2}-\frac{i}{x^3}-\frac2{x^4}+\frac{6i}{x^5}+\cdots\right).
$$
For every fixed $N\ge1$, retaining $k=0,\ldots,N-1$ 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.
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