Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 67 / 1

For fixed $\mu \ne 0$, a large positive root must lie close to a pole of the tangent function: away from its poles, $\tan x$ cannot balance the unbounded right side. Label the adjacent poles by $a_n=(n+1/2)\pi$ and write $x_n=a_n+{\delta }_n$. The Taylor series of $-\cot \delta$ givesSeek the asymptotic expansion $\delta =A/a_n+B/a_n^3+\cdots$. The coefficients of $a_n$ and $a_n^{-1}$ giveThus the large roots near tangent poles satisfyThe first correction already supplies the requested dependence on $\mu$. For $\mu >0$ the root approaches the pole from below; for $\mu <0$ it approaches from above. This labels roots by their nearby poles, avoiding an irrelevant finite shift in the enumeration of positive roots.

No positivity restriction on $\mu$ is required. The expansion requires $\mu \ne 0$ fixed and $|\mu |a_n\gg 1$, so the displacement is small. It is not uniform as $\mu \to 0$. At $\mu =0$ the roots are exactly $n\pi$, a different leading sequence; these cannot be recovered by setting $\mu =0$ in the pole expansion. If $\mu$ varies with $n$, its size must be checked against the small-displacement and successive-term conditions rather than using the fixed-parameter remainder blindly.

Watson's lemma applies to Laplace-type endpoint integrals such as ${\int }_0^b{e}^{-\lambda t}f(t)dt$ as $\lambda \to +\infty$, when $f$ has a local algebraic expansion $f(t)\sim {\sum }_j{c}_j{t}^{p+j-1}$ with $p>0$ and suitable integrability or exponential-growth control away from the endpoint. Each term contributes $c_j\Gamma (p+j){\lambda }^{-p-j}$. The exponential selects a neighborhood of $t=0$.

Laplace method instead treats ${\int }_a^{b}g(t)e^{\lambda \varphi (t)}dt$ by locating the dominant maxima of a real phase $\varphi$. A smooth nondegenerate interior maximum produces a neighborhood governed by a Gaussian function of width ${\lambda }^{-1/2}$; endpoint or degenerate maxima require their corresponding local scaling. Contributions away from a strict dominant maximum must be smaller. These methods overlap after suitable changes of variable, but their usual starting forms emphasize an endpoint amplitude expansion and an exponential maximum, respectively.

For the Gamma function, put $t=xu$. ThenThe phase has its unique maximum at $u=1$, with value $-1$ and second derivative $-1$. Set $u=1+s/\sqrt{x}$. Near this maximum,Combining the amplitude and exponential expansions in Laplace method givesLocalization near $u=1$ justifies extending the limits for the Gaussian integral; the distant endpoint is not being expanded uniformly by this local series. Odd terms integrate to zero. The moments of the Gaussian integral satisfy $I_1=I_0$, $I_2=3I_0$, $I_3=15I_0$, so the relative correction isThe two terms of the Stirling formula, obtained by the Stirling correction by Gaussian moments, are therefore