Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 76 / 1 / b / Solution

The integrand is an entire function, so the defining contour is immaterial. A rectangular contour, closed far to the right where the Gaussian function decays, gives the exact horizontal-tail identityThe constant is the half-line Gaussian integral. Expand the endpoint amplitude $e^{-s^2}=1-s^2+\cdots$ in this Watson lemma. Alternatively, repeated integration by parts gives the same expansion, including purely imaginary $z$, because every derivative of $e^{-s^2}$ is integrable on the positive real axis. Thus, on each fixed ray in the stated quadrant,The remainder is measured on the exponential endpoint scale. When that scale grows, this formula does not resolve terms exponentially small relative to it.

Write $z=r{e}^{i\theta }$. The endpoint term has modulus $e^{-r^2\cos 2\theta }/(2r)$. For $0\le \theta <\pi /4$ it is exponentially small, and at $\theta =\pi /4$ it is oscillatory of order $r^{-1}$. For $\pi /4<\theta \le \pi /2$ it dominates the constant. The quadrant asymptotics of the error function therefore give the requested leading asymptotic expansionsThe boundary ray $\theta =\pi /4$ belongs to the constant-dominated case because of the extra factor $1/r$. On the imaginary axis this gives $f(ir)\sim i{e}^{r^2}/(2r)$, consistent with $f(ir)=i{\int }_0^r{e}^{s^2}ds$. The constant/exponential dominance exchange is an anti-Stokes line phenomenon; it should not be confused with dropping the constant everywhere in the quadrant.