Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 79 / 1

Use dominant balance for algebraic roots to separate the three scales. On writing $x=ϵX$, the equation becomes $X^3-X^2+2ϵX+2{ϵ}^3=0$. Its nonzero simple limiting root is $X=1$, so one root has $x_1\sim ϵ$.

For the two remaining roots put $x={ϵ}^{2}Y$. Division by ${ϵ}^5$ gives $ϵY^3-Y^2+2Y+2ϵ=0$. The nonzero limiting root is $Y=2$, giving $x_2\sim 2{ϵ}^2$. Finally $x={ϵ}^{3}Z$ gives, after division by ${ϵ}^6$,Its simple limiting root is $Z=-1$. These three distinct balances account for all roots of the cubic, andFor the root of smallest magnitude, set $Z=-1+bϵ+O({ϵ}^2)$. The order-$ϵ$ coefficient is $-1+2b$, so $b=1/2$. HenceEach rescaled root is simple at $ϵ=0$, justifying the regular perturbation of a simple algebraic root rather than treating the original triple root as an ordinary regular perturbation.

The complete integral is $\Gamma (1/2)=\sqrt{\pi }$. Therefore apply asymptotic expansion by repeated integration by parts to the tail $I(x)={\int }_x^{\infty }{t}^{-1/2}{e}^{-t}dt$:The last integral is bounded by $x^{-7/2}{e}^{-x}$. ThusIn particular the first three additive terms are $\sqrt{\pi }$, $-e^{-x}{x}^{-1/2}$ and $\frac{1}{2}{e}^{-x}{x}^{-3/2}$; the formula also supplies three successive terms of the exponentially small tail if the constant is counted separately. A power expansion alone would retain only $\sqrt{\pi }$ and miss this information.

Take the principal complex logarithm and deform the interval upwards, where $e^{ixt}$ decays for $x>0$. The small indentation at zero contributes nothing because $r\log r\to 0$; the distant horizontal segment vanishes exponentially. The two vertical edges therefore give the exact representationThis is an endpoint steepest descent contour, with distinct contributions from the singular endpoint and the smooth endpoint. Setting $u=xs$ in the first integral and using the given logarithmic moment givesFor the second edge, the Taylor series at $s=0$ is $\log (1+is)={\sum }_{n\ge 1}(-1{)}^{n+1}(is{)}^n/n$. Watson's lemma integrates it term by term as an asymptotic expansion, using ${\int }_0^{\infty }{s}^n{e}^{-xs}ds=n!/x^{n+1}$. ConsequentlyHere $\gamma$ is the Euler--Mascheroni constant. The first smooth-endpoint terms areFor every fixed $N\ge 1$, retaining $k=0,\dots ,N-1$ leaves $O(x^{-N-2})$. The factorial series is an asymptotic series, not a convergent sum. Treating the logarithmic endpoint by an ordinary smooth-endpoint integration formula would lose the leading logarithm; this is the logarithmic endpoint Fourier asymptotics mechanism.