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The pole contour in a Gaussian saddle coalescing with a simple pole changes the multiplier smoothly. For the lower contour, ${\int }_{{\mathcal{C}}_{-}}{e}^{-u^2-i\mu u}du/u=i\pi \mathrm{erfc}(\mu /2)$. This follows by integrating $-i\sqrt{\pi }{e}^{-{\mu }^2/4}$ and adding the indentation contribution $i\pi$. The transition width is inverse square root of the large parameter; on a Stokes line the multiplier takes half its switched-on value.

Use the local coordinate $t=1/\lambda$. The differential system becomesFor the nonzero quadratic leading term, the coefficient of the local linear ordinary differential equation system has a pole of order four, so infinity is an irregular singular point of Poincaré rank three in this gauge. The matrix-valued function $A(\lambda )$ itself has a pole of order two; the extra factor $d\lambda /dt=-t^{-2}$ is essential when classifying the equation. If $A_0$ vanishes the degree and these orders decrease. The specified distinct eigenvalues $1,-1$ give the nondegenerate unramified case.

The leading exponential function factors have actions $q_{\pm }=\pm {\lambda }^3/3$, so their difference is $2{\lambda }^3/3$. Equal exponential magnitudes occur onThese are commonly called Stokes rays in the sectorial ODE convention. With the action-phase convention used by the wiki's Stokes line article they are anti-Stokes lines; the same-phase rays instead haveGiving both defining equations removes the nomenclature ambiguity; both sets contain six directions.

The standard construction with distinct leading eigenvalues uses six sectorial fundamental solutions per turn, with basic angular spacing $\pi /3$. They are fundamental matrices, each containing two independent column solutions. On the logarithm cover choose six consecutive overlapping sectors of width $\pi /3+2\delta$, with $\delta >\epsilon /2$ small. Their union has angular opening $2\pi +2\delta$, covering the requested $2\pi +\epsilon$. One often also writes a seventh index to compare the closing sector with the first after a full turn; that is a closing copy, not a seventh independent sector in the six-sector construction. The number is a generic sectorial fundamental solution normalization count, not a universal minimum for exceptional systems with vanishing Stokes matrices: the exactly diagonal system already has one global elementary fundamental matrix.

There is a missing qualification in the requested product identity. Polynomial coefficients give an entire invertible fundamental matrix, hence trivial actual monodromy about a large circle: the circle contracts through ordinary points. But the formal matrix generally has the form $\hat{G}{\lambda }^{\Theta }{e}^Q$, and a positive turn changes its logarithm branch by the formal monodromy $M_f=e^{2\pi i\Theta }$. This factor cannot be discarded.

To specify the printed product order, let $S_j$ act on column vectors of solution coefficients across the $j$th sector boundary, so adjacent normalized matrices satisfy $Y_{j+1}=Y_j{S}_j^{-1}$. Six transitions give $Y_7=Y_1(S_6\cdots S_1{)}^{-1}$. The normalized formal branch after the turn is ${\hat{Y}}_1{M}_f$, so sectorial uniqueness gives $Y_7=Y_1{M}_f$. ConsequentlyUsing Stokes matrices for right changes of fundamental matrices instead reverses/inverts the corresponding convention; the formal monodromy factor remains necessary. The formal monodromy correction to an entire-system Stokes product shows that the printed identity holds when $M_f=I$, including the corrected Lax specialization discussed below.

It does not follow from polynomiality alone. For example takeThe coefficient matching in question 2 gives $\Theta =\frac{1}{2}\mathrm{diag}(1,-1)$, hence $M_f=-I$ and the ordered Stokes matrix product is $-I$, not $I$, despite having no finite singularity. This is an explicit counterexample to the unqualified final clause.

Choose $N$ near the least term, so $\gamma =N+2/3=R+O(1)$ with $R=|z{|}^3$. On writing $\beta =3(\arg z-\pi /3)$, the parameter is $\lambda =R{e}^{i\beta }$. For $\beta =O(R^{-1/2})$,The supplied Gaussian pole transition for a large gamma parameter therefore gives $I\sim i\pi [1+\mathrm{erf}(\mu /\sqrt{2})]$. Substitution in solution gives the smoothing formula in solution once the amplitude identity in solution is used.

The origin of the error function can also be seen locally. Set $t=1+s/\sqrt{\gamma }$ in the Borel integral. ThenThe saddle has become a Gaussian integral meeting a simple pole. With the upper-pole contour, the leading integral is $-{\int }_{{\mathcal{C}}_{+}}{e}^{-s^2/2-i\mu s}ds/s$. Its value at $\mu =0$ is $i\pi$ from the indentation. Differentiating with respect to $\mu$ gives $i\sqrt{2\pi }{e}^{-{\mu }^2/2}$, so integration from zero gives $i\pi [1+\mathrm{erf}(\mu /\sqrt{2})]$. This independently fixes the sign and the half-jump on the Stokes line.

First take $z>0$. Set $s=t^3-1$ in the tail integral. This puts it in the endpoint form needed for Watson's lemma:The binomial coefficients are $(-1{)}^r(2/3{)}_r/r!$, where $(a{)}_r=\Gamma (a+r)/\Gamma (a)$ is the rising factorial. Integrating each endpoint term givesAlso, scaling the complete positive-ray integral gives ${\int }_0^{\infty }{e}^{-z^3{t}^3}dt=\Gamma (1/3)/(3z)$. HenceThe endpoint term is exponentially small on this positive ray but must be retained for continuation to other sectors.

For complex $z$, use $s=zt$ so $G(z)=z^{-1}{\int }_0^z{e}^{-s^3}ds$. Put $C=\Gamma (1/3)/3$ and $\omega =e^{2\pi i/3}$. The complete steepest descent contours from the cubic saddle are the rays $\arg s=2\pi j/3$, whose integrals are $C{\omega }^j$. Deform the finite contour onto one of these rays followed by the endpoint contour. The cubic exponential integral Stokes sectors giveFor $0\le \arg z\le 2\pi$, take $j=0$ below $\pi /3$, $j=1$ between $\pi /3$ and $\pi$, $j=2$ between $\pi$ and $5\pi /3$, and $j=3$ above $5\pi /3$. The corresponding algebraic coefficients are $C,C\omega ,C{\omega }^2,C$. A negative-real $z$ provides a useful sign check: $G(z)=G(|z|e^{i\pi })$ is real and positive, so its two lateral algebraic coefficients must be complex conjugates.

The subdominant algebraic term switches across the Stokes linesThere $z^3$ is negative real and the endpoint exponential is dominant. The intervening anti-Stokes lines are $\arg z=\pi /6+j\pi /3$, where $\mathrm{Re}{z}^3=0$ and the exponential changes from growth to decay. The naming convention here distinguishes switching from equal exponential magnitude.

To resolve the jump near ${\theta }_s=\pi /3$, optimally truncate the endpoint series at $N$ with $N+2/3=R+O(1)$, where $R=|z{|}^3$. Let $P_N$ denote that truncated endpoint contribution. The Borel remainder for a cubic exponential integral derived in solution, together with the Gaussian pole transition for a large gamma parameter in solution, givesThis estimate is on the transition scale $\theta -\pi /3=O(R^{-1/2})$ after optimal truncation. The error-function smoothing of a Stokes multiplier has width $O(|z{|}^{-3/2})$. Before the line $S\to 0$ and the coefficient is $C$; after it $S\to 1$ and the coefficient is $C\omega$, agreeing with the sector expansions. On the line the leading coefficient is their average. The original $G$ is an entire function; only its asymptotic decomposition has these sector-dependent coefficients.

Put $w=1/z$ and $Y(w)=y(1/w)$. The equation becomesAt $w=0$ the regular-singular criterion requires $w(2/w)$ and $w^2(-w^{-n-4})$ to be analytic. This holds precisely for $n\le -2$. Thus the singularity at infinity is irregular, corresponding to essential exponential behaviour of solutions, exactly for $\boxed{n\ge -1}$. For $n\le -2$ it is regular singular, with powers and possibly logarithms rather than such essential exponentials.

For $y^{\prime \prime }=z^{3}y$, the Liouville–Green approximation uses $Q=z^3$, amplitude $Q^{-1/4}$ and phase integral $\int \sqrt{Q}dz=2z^{5/2}/5$. On a consistently chosen branch, the two leading solutions areThe derivative correction is smaller by order $z^{-5/2}$, and the two exponential signs give independent solutions.

Using the convention that Stokes lines have equal exponential phase, $\mathrm{Im}(z^{5/2})=0$, their rays are $\arg z=2\pi k/5$. The equal-magnitude, oscillatory rays, often called anti-Stokes lines, satisfy $\mathrm{Re}(z^{5/2})=0$, giving $\arg z=(2k+1)\pi /5$. If the alternative naming convention is used, these two names interchange; the stated equations specify the rays unambiguously.

To give the sector precisely, the solution recessive on the positive axis can be represented bywhere $K_{\nu }$ is the Modified Bessel function of the second kind. Substitution verifies the differential equation. Its asymptotic $K_{\nu }(w)\sim \sqrt{\pi /(2w)}{e}^{-w}$ is valid on closed subsectors of $|\arg w|<3\pi /2$ on the continued branch. ConsequentlyActual exponential decay occurs in the narrower sector $|\arg z|<\pi /5$; in the adjacent portions of the larger asymptotic sector the same continued exponential grows. This distinguishes the validity of the approximation to the positive-axis recessive solution from the directions where its modulus decays.

Set $z=x^2/\epsilon$. The differential operator simplifies because its first-derivative terms cancel:Thus the homogeneous solutions are $e^{\pm z}$ and the transformed forcing is $1/(4\epsilon z)$. For the decaying-sector variation of parameters solution choose the contour directionsThe first integrand decays at positive-real infinity and the second at positive-imaginary infinity. Join each endpoint to $x$ within the first quadrant, avoiding zero. This also fixes the exponentially small homogeneous contributions, which cannot be determined from an algebraic asymptotic expansion alone.

After $s=t^2/\epsilon$, write $A(z)={\int }_{+\infty }^z{e}^{-s}ds/s$ and $B(z)={\int }_{-\infty +i0}^z{e}^{s}ds/s$. Their derivatives are $A^{\prime }=e^{-z}/z$ and $B^{\prime }=e^z/z$. Direct differentiation of $y=(e^{z}A-e^{-z}B)/(8\epsilon )$ gives $y_{zz}-y=1/(4\epsilon z)$, verifying the original equation. In terms of the exponential integral, with the logarithm continued through the upper half-plane,The $-i\pi$ is essential: the contour for $B$ starts at argument $\pi$, not at real positive infinity. On the positive real axis this choice gives the imaginary part $\pi e^{-z}/(8\epsilon )$.

Repeated endpoint integration by parts gives $e^z{E}_1(z)\sim z^{-1}(1-z^{-1}+2!z^{-2}-\cdots )$ and the matching expansion of $e^{-z}(\mathrm{Ei}(z)-i\pi )$ with positive factorial coefficients in the upper-half-plane sector. Their odd inverse-power corrections cancel. Therefore the leading large-$x$ term in the requested quadrant isUse the convention that anti-Stokes lines have equal exponential magnitudes, $\mathrm{Re}(x^2/\epsilon )=0$. They are $\arg x=\pi /4+m\pi /2$. The Stokes lines are $\arg x=m\pi /2$. Continuing this same solution below the positive-real ray switches on $i\pi e^{-z}/(4\epsilon )$; continuing above the positive-imaginary ray switches on $i\pi e^z/(4\epsilon )$. These exponentials remain decaying up to the neighbouring anti-Stokes lines. Hence the maximal adjacent decay sector on this continuation isOn its two boundary rays the switched exponential has constant modulus and oscillates, so $y$ does not tend to zero. Within the sector the leading algebraic term above persists. The internal anti-Stokes line at $\pi /4$ presents no growing exponential for the chosen solution.

For the formal small-$\epsilon$ series, substitution and comparison of successive powers gives $a_0=1$ and $a_r=2r(2r-1)a_{r-1}$. Thus all coefficients areThis is a factorially divergent asymptotic expansion. The ratio of consecutive term magnitudes is $(2r+2)(2r+1)/|z{|}^2$. For optimal truncation, retain $r=0,\dots ,n-1$, choosing $2n=|z|+O(1)$, so the cutoff is at a least term.

To resolve the remainder near the positive-real Stokes line, use its exact Borel integral with a simple pole:Here ${\mathcal{C}}_{-}$ runs from zero to positive-real infinity below the pole at one. Partial fractions reproduce the exponential-integral solution, including its positive imaginary part. The finite geometric identity for $(1-t^2{)}^{-1}$ gives the displayed exact remainder after integrating the first $n$ monomials.

Write $z=2n+i\mu \sqrt{n}+\nu$, with real bounded $\mu$ and bounded $\nu$. For a Gaussian saddle coalescing with a simple pole, put $t=1+u/\sqrt{n}$. The exponent becomes $-u^2-i\mu u+O(n^{-1/2})$ and $dt/(1-t^2)=-du/(2u)+O(n^{-1/2})du$. ThereforeThe integral is its Cauchy principal value plus $i\pi$. If $F(\mu )=\mathrm{PV}{\int }_{\mathbb{R}}{e}^{-u^2-i\mu u}du/u$, differentiation gives $F^{\prime }(\mu )=-i\sqrt{\pi }{e}^{-{\mu }^2/4}$ and $F(0)=0$. Consequently $F(\mu )=-i\pi \mathrm{erf}(\mu /2)$, and $J_n=-i\pi \mathrm{erfc}(\mu /2)/2+O(n^{-1/2})$.

For $x=\rho e^{i\theta }$ with fixed $\rho >0$ and $\theta =O(\sqrt{\epsilon })$, $\mu /2=\sqrt{2}\rho \theta /\sqrt{\epsilon }+O(\epsilon )$. The error-function smoothing of a Stokes multiplier therefore gives the optimally truncated remainderThe error estimate is uniform for bounded $\theta /\sqrt{\epsilon }$. On the Stokes line the multiplier is half its switched-on value; above it the exponentially small term switches off, and below it switches on. This term is invisible at every algebraic order. Near the line the remainder can exceed the least retained term by a square-root factor because a pole meets the saddle point.

The printed auxiliary Gaussian hint has inconsistent constants. Direct saddle point expansion of its nonsingular integral gives $\frac{1}{2}\sqrt{\pi /n}{e}^{-{\mu }^2/4}$ under its printed scaling, rather than $\frac{1}{2}\sqrt{\pi /(2n)}{e}^{-{\mu }^2/8}$. The derivation above uses the correct saddle point normalization and independently fixes both the smoothing width and the contour sign.