= Solution
Use the principal <complex logarithm> in the upper half-plane. Deform the interval upward into a half-strip: the vertical side at zero is traversed upward, and the side at one downward. The top side vanishes exponentially; a small indentation at zero contributes $O(r|\log r|)$ and vanishes. Thus the <contour deformation> gives the exact representation, for $x>0$,
$$
f(x)=i\int_0^\infty\log(is)e^{-xs}\,ds
-i e^{ix}\int_0^\infty\log(1+is)e^{-xs}\,ds.
$$
Since $\log(is)=\log s+i\pi/2$, scaling $u=xs$ and using the integral defining <Euler's constant> gives
$$
i\int_0^\infty\log(is)e^{-xs}\,ds
=-\frac{i(\log x+\gamma)}x-\frac\pi{2x}.
$$
For the second endpoint, $\log(1+is)=is+s^2/2+O(s^3)$. <Watson's lemma>, or integration of these powers against $e^{-xs}$ with an exponentially small tail, yields
$$
-i e^{ix}\int_0^\infty\log(1+is)e^{-xs}\,ds
=\frac{e^{ix}}{x^2}-\frac{i e^{ix}}{x^3}+O(x^{-4}).
$$
In particular \b[the required first terms are]
$$
\boxed{f(x)=-\frac{i\log x}{x}-\frac{i\gamma+\pi/2}{x}
+\frac{e^{ix}}{x^2}+O(x^{-3}).}
$$
The two different endpoint scales come from the <logarithmic singularity> at zero and the simple zero of the amplitude at one.
Back to article page
