Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 76 / 2

Use the Gaussian pole transition for a large gamma parameter. With $t=1+s/\sqrt{n}$ and $\lambda =n+i\mu \sqrt{n}+\nu +o(1)$, Taylor expansion givesThe upper semicircle is traversed from left to right, so $\int ds/s=-i\pi$ on the indentation. The minus sign in $dt/(1-t)$ therefore makes its contribution $+i\pi$. The remaining leading integral isAt $\mu =0$ its Cauchy principal value vanishes by oddness. Differentiating with respect to $\mu$ removes the pole, leaving the Fourier transform of a Gaussian:Integrating from zero givesThis is uniform for bounded real $\mu ,\nu$ (and extends analytically over bounded complex values where the local contour is continued consistently). Contributions away from the saddle are exponentially small after localization. The bounded shift $\nu$ enters at the next order, not the leading error function. For example, retaining the displayed cubic correction gives $\sqrt{2\pi /n}{e}^{-{\mu }^2/2}[\nu +(2+{\mu }^2)/3]$ as the next term. This also checks the constant and contour sign.

For positive integer $n$ and any positive integer $N$, the finite geometric series identitygives an exact decomposition along the indented contour. The polynomial terms have no pole and may be returned to the positive real axis; their Gamma function integrals areEquivalently, expansion about the endpoint $t=0$ in the Laplace transform integral gives the same coefficients. The resulting formal asymptotic expansion isThe ratio of consecutive terms is $r/\lambda$, so the infinite series is factorially divergent. It should be truncated, not treated as a convergent sum; the pole prescription controls the contribution beyond its algebraic orders. Although the source initially calls $n$ an integer, the ordinary lower endpoint integral requires $n>0$, as is already implicit in the large-positive-$n$ limit of part (i).

By Stirling's approximation, for fixed $a$,Apply this with $a=1/6,5/6$ and use $\Gamma (r+1)=r\Gamma (r)$. Their exponents add to one, and henceMore explicitly, the ratio is $1-5/(36r)+O(r^{-2})$. The exact ratio of neighboring terms isThus the Airy function expansion is factorially divergent, and optimal truncation retains about $|\sigma |$ terms.

The printed summation starts at $r=1$, but its leading $r=0$ term is missing. The Gamma reflection formula gives $\Gamma (1/6)\Gamma (5/6)=2\pi$, so $Y_0=1$. The correctly normalized dominant Airy function expansion starts at zero. This correction is essential for the exponentially small remainder in the next part; if the printed series is used literally, its remainder also contains the entire missing leading term.

Let $P(z)=1/(2\sqrt{\pi }{z}^{1/4})$, choose a consistent branch, and truncate the corrected series after $r=n-1$, with $n=|\sigma |+O(1)$. The leading factorial tail in part (iii) is represented by the same indented integral as in part (ii):Here the divergent sum means its lateral Borel summation, not ordinary summation. To see why the pole approximation controls this remainder, write $a_r={\sigma }^r{Y}_r$ and insert $\Gamma (r)={\int }_0^{\infty }{e}^{-v}{v}^{r-1}dv$ in the leading late terms. Summing the geometric tail creates $1/(1-v/\sigma )$; $v=\sigma t$ produces $I(\sigma ,n)$. The relative $O(r^{-1})$ corrections to the late coefficients give lower-order contributions on its saddle scale. More concretely, the Borel transform of the coefficient series isThe simple pole supplies the leading Stokes phenomenon; its weaker logarithmic singularity supplies the smaller corrections. This is the integral justification for retaining the factorial tail.

For $\arg \sigma =\varphi /\sqrt{|\sigma |}$, expand $\sigma =n+i\varphi \sqrt{n}+O(1)$. Part (i), with $\mu =\varphi$, now givesThis formula uses the upper-pole prescription and the upper Airy function switching ray: near $\arg z=2\pi /3$, take $\arg \sigma =3\arg z/2-\pi$ near zero. The coefficient of the subdominant exponential changes smoothly from zero to $i$ across this ray and has value $i/2$ on it. The angular transition width is $O(|\sigma {|}^{-1/2})=O(|z{|}^{-3/4})$. The conjugate lower ray has the conjugate contour sign, rather than the same $+i$ on both rays. This is error-function smoothing of a Stokes multiplier, not a discontinuity of the analytic Airy function. With the literal printed $r=1$ series, the missing $P(z)e^{\sigma /2}$ would invalidate this remainder formula.