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

A regular perturbation admits an asymptotic expansion about the limiting problem on the relevant scale without losing its solutions or conditions. A singular perturbation changes the limiting structure, so some solutions require a different scale: a differential equation may lose an order, or a polynomial may lose its degree.

The two finite roots approach $s=\pm \sqrt{2}$. Set $x=s+ϵa+O({ϵ}^2)$. Substitution into the polynomial gives $s^2-2=0$ at leading order and $s^3+2sa=0$ at the next order. Since $s^2=2$, $a=-1$ for either sign. Each branch is a regular perturbation of a simple algebraic root, withThe limiting quadratic has only two roots, whereas the original equation has three. The missing root is a divergent algebraic root in a singular perturbation. The dominant balance for algebraic roots sets $x={ϵ}^{-1}X$, giving $X^3+X^2=2{ϵ}^2$. Its nonzero leading root is $X=-1$. Put $X=-1+a{ϵ}^2+O({ϵ}^4)$; the left side is $a{ϵ}^2+O({ϵ}^4)$, so $a=2$. Hence the second nonzero term occurs at order $ϵ$, not order one:For small nonzero real $ϵ$ these three branches are real and distinct; the divergent branch changes sign with $ϵ$. The two bounded branches are regular even though the full root problem is singular.

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.

For small $x$, use $u=\cosh t$ to write the defining oscillatory integral asThe first integral is $-\mathrm{Ci}(x)$, where the cosine integral satisfies $\mathrm{Ci}(x)={\gamma }_E+\log x+O(x^2)$. Its logarithm follows by integrating the $1/x$ singularity of its derivative; its conventional constant is the Euler--Mascheroni constant. The primary normalization is given by dlmf.nist.gov/6.6.E6 .

The second integrand has an integrable inverse-square-root endpoint singularity and an $O(u^{-3})$ tail. The dominated convergence theorem therefore gives its limitSplitting at $u=1/x$ and using $|1-\cos (xu)|\le \min (x^2{u}^2/2,2)$ also bounds its error by $O(x^2|\log x|)$. ThusIn particular, the leading term is $(2/\pi )\log x$.

For large $x$, the phase $\cosh t$ has its only stationary point at the endpoint $t=0$. Its expansion is $1+t^2/2+O(t^4)$, so the contributing width is $t=O(x^{-1/2})$. The endpoint stationary phase method and the Fresnel integral yieldA smooth cutoff near the endpoint justifies this local replacement; the remaining nonstationary tail can be integrated by parts, with no stationary contribution at infinity. Taking the real part gives the Bessel function of the second kind asymptoticThe additive error formulation remains meaningful near zeros of the leading oscillation. Both limits were derived from the given integral, rather than identifying a special function and merely quoting its asymptotics.