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Appendix B: Complex Analysis and Asymptotics

Complex analysis turns local formulae into global data, while asymptotic analysis turns exact objects into controlled approximations. Both operations are convention-sensitive. A logarithm without a sheet, a saddle contribution without an oriented contour, or a period without a lifted cycle is not yet a reproducible quantity.

This appendix is a working desk for the recurring ingredients used throughout the book. It supplies the shortest safe derivations and points to the chapters where the analytic hypotheses are developed in depth.

A holomorphic germ [f]z[f]_{z_*} can be continued along a path γ:[0,1]X\gamma:[0,1]\to X by a chain of overlapping analytic elements. The endpoint germ can depend on the homotopy class of γ\gamma in XX. The monodromy theorem guarantees path independence when the relevant domain is simply connected and continuation exists along every path under consideration. Simple connectivity is sufficient, not necessary: a function can have trivial monodromy on a multiply connected domain.

The house default for elementary powers is

π<Argzπ,Logz=logz+iArgz.-\pi<\operatorname{Arg}z\leq\pi, \qquad \operatorname{Log}z = \log|z|+\ii\operatorname{Arg}z.

After continuation around a closed path avoiding the origin,

Logγz=Logz+2πiwind(γ,0).\operatorname{Log}^{\gamma}z = \operatorname{Log}z +2\pi\ii\,\operatorname{wind}(\gamma,0).

Consequently,

(zα)γ=e2πiαwind(γ,0)zα.\left(z^\alpha\right)^\gamma = \ee^{ 2\pi\ii\alpha\, \operatorname{wind}(\gamma,0) } z^\alpha.

The upper and lower boundary values on the principal cut are, for x>0x>0,

Log(x±i0)=logx±πi.\operatorname{Log}(-x\pm\ii0) = \log x\pm\pi\ii.

Thus z=exp[Logz/2]\sqrt z=\exp[\operatorname{Log}z/2] changes sign after one positive turn around zero. The sign change belongs to the two-sheeted surface; the negative-real cut is only one planar drawing of that surface.

Suppose ff is meromorphic in the region swept between two oriented contours Γ0\Gamma_0 and Γ1\Gamma_1 with the same endpoints. If C=Γ1Γ0C=\Gamma_1-\Gamma_0 is the resulting closed chain, then

Γ1f(z) ⁣dzΓ0f(z) ⁣dz=2πiawind(C,a)resz=af(z).\int_{\Gamma_1}f(z)\,\dd z - \int_{\Gamma_0}f(z)\,\dd z = 2\pi\ii \sum_{a} \operatorname{wind}(C,a) \operatorname*{res}_{z=a}f(z).

The sum runs over poles in the swept region. If the integrand is multivalued, this formula is applied on a covering surface or after the boundary values on every cut have been specified. A contour deformation in the plane that silently changes sheets is not a Cauchy deformation.

Object being continuedMinimum data to recordQuick invariant check
Logz\operatorname{Log}z or zαz^\alphaBase value, cut drawing, path, windingDifferentiate to 1/z1/z and check the endpoint multiplier
Frobenius solutionExponent lift, leading coefficient, pathLocal monodromy eigenvalue and Wronskian
Fundamental matrixOrdered basis, base point, side of actionProduct law for two named loops
Contour integralOriented contour, poles, cut boundary valuesResidue difference under a test deformation
Square-root differentialCover, sheet, lifted pathDeck involution reverses the square root

The book conventions fix the default elementary branches. The monodromy problem page shows how these choices enter normalized ODE bases, while the working toolkit collects the residue, winding, and Wronskian checks used in calculations.

The Gamma function is meromorphic, single-valued, and zero-free, but logarithms and asymptotic powers used to evaluate it are not. Its simple poles and the three most useful functional relations are recorded together as

Γ(z+1)=zΓ(z),Γ(z)Γ(1z)=πsin(πz),Γ(z)Γ(z+12)=212zπΓ(2z),resz=nΓ(z)=(1)nn!,nZ0.\begin{aligned} \Gamma(z+1)&=z\Gamma(z),\\ \Gamma(z)\Gamma(1-z) &= \frac{\pi}{\sin(\pi z)},\\ \Gamma(z)\Gamma\left(z+\frac12\right) &= 2^{1-2z}\sqrt{\pi}\,\Gamma(2z),\\ \operatorname*{res}_{z=-n}\Gamma(z) &= \frac{(-1)^n}{n!}, \qquad n\in\mathbb Z_{\geq0}. \end{aligned}

These are meromorphic identities. Reflection is often the safest way to move an argument away from the negative real axis before applying a large-zz expansion.

The symbol \sim is Poincaré notation. For a power scale it means that, for every fixed NN,

F(z)n=0N1cnzn=O(zN)F(z) - \sum_{n=0}^{N-1}c_nz^{-n} = O(z^{-N})

in the stated limit, uniformly when a closed subsector is specified. It does not assert convergence of the infinite sum.

Fix δ>0\delta>0. On the slit plane C(,0]\mathbb C\setminus(-\infty,0], normalize the zero-free Gamma function by

LogcontΓ(1)=0,\operatorname{Log}_{\mathrm{cont}}\Gamma(1)=0,

and continue that logarithm from the positive real axis. Thus LogcontΓ(x)=logΓ(x)\operatorname{Log}_{\mathrm{cont}}\Gamma(x)=\log\Gamma(x) for x>0x>0; an arbitrary additive 2πim2\pi\ii m is not allowed. Uniformly on closed subsectors

argzπδ,|\arg z|\leq\pi-\delta,

Stirling’s logarithmic expansion is

LogcontΓ(z)(z12)Logzz+12log(2π)+k1B2k2k(2k1)z2k1.\begin{aligned} \operatorname{Log}_{\mathrm{cont}}\Gamma(z) \sim{}& \left(z-\frac12\right)\operatorname{Log}z -z+\frac12\log(2\pi)\\ &+ \sum_{k\geq1} \frac{B_{2k}} {2k(2k-1)z^{2k-1}}. \end{aligned}

If the sum is stopped at k=N1k=N-1, the Poincaré remainder is Oδ(z(2N1))O_\delta(|z|^{-(2N-1)}) for fixed NN. This is not a convergent-series claim. Near the excluded negative axis, use recurrence or reflection and continue every logarithm consistently.

For fixed aa and bb, the corresponding ratio card is

Γ(z+a)Γ(z+b)zab[1+(ab)(a+b1)2z+O(z2)],\frac{\Gamma(z+a)}{\Gamma(z+b)} \sim z^{a-b} \left[ 1+ \frac{(a-b)(a+b-1)}{2z} +O(z^{-2}) \right],

in the same type of sector. The power zabz^{a-b} uses the same branch as the Stirling expansion. Ratios should be simplified before separate numerical evaluation of two large Gamma functions; otherwise avoidable overflow and phase cancellation can dominate the result.

On vertical lines, a second form of Stirling’s estimate is often more useful. Uniformly for real xx in a bounded interval,

Γ(x+iy)2πyx1/2eπy/2,y.|\Gamma(x+\ii y)| \sim \sqrt{2\pi}\, |y|^{x-1/2}\ee^{-\pi|y|/2}, \qquad |y|\longrightarrow\infty.

This estimate controls Mellin–Barnes tails and makes the exponential damping in imaginary Gamma arguments visible before numerical quadrature.

The hypergeometric connection benchmark shows these identities inside an exact connection matrix. The Airy–Weber–Mathieu examples use the same Stirling sector in quantum-period calculations.

Consider

I(κ)=Γeκϕ(z)a(z) ⁣dz,κ.I(\kappa) = \int_\Gamma \ee^{-\kappa\phi(z)} a(z)\,\dd z, \qquad |\kappa|\longrightarrow\infty.

Assume that ϕ\phi and aa are holomorphic near the saddle and that the contour can be deformed locally onto the stated descent arc without crossing an endpoint, pole, or branch cut. A nondegenerate saddle zsz_s satisfies

ϕ(zs)=0,ϕ(zs)0.\phi'(z_s)=0, \qquad \phi''(z_s)\neq0.

Locally there is a holomorphic Morse coordinate uu for which

ϕ(z)ϕ(zs)=u22, ⁣dz ⁣duu=0=1ϕ(zs).\phi(z)-\phi(z_s)=\frac{u^2}{2}, \qquad \left.\frac{\dd z}{\dd u}\right|_{u=0} = \frac{1}{\sqrt{\phi''(z_s)}}.

The square root is not chosen independently: its sign and phase are fixed by the oriented contour through the saddle. If that contour maps to the oriented real uu-axis and κ>0\kappa>0, then

Is(κ)eκϕ(zs)a(zs) ⁣dz ⁣du02πκ.I_s(\kappa) \sim \ee^{-\kappa\phi(z_s)} a(z_s) \left.\frac{\dd z}{\dd u}\right|_{0} \sqrt{\frac{2\pi}{\kappa}}.

Writing κϕ(zs)=κϕ(zs)eiβ\kappa\phi''(z_s)=|\kappa\phi''(z_s)|\ee^{\ii\beta} makes the local directions explicit. For the exponential eκϕ\ee^{-\kappa\phi},

descent directionsarg(zzs)=β/2(modπ)ascent directionsarg(zzs)=(πβ)/2(modπ)\begin{array}{c|c} \text{descent directions} & \arg(z-z_s)=-\beta/2\pmod{\pi} \\ \text{ascent directions} & \arg(z-z_s)=(\pi-\beta)/2\pmod{\pi} \end{array}

because Im[κ(ϕϕs)]\operatorname{Im}[\kappa(\phi-\phi_s)] is constant on both sets, while the sign of the real part alternates.

Local descent and ascent rays through a Gaussian saddle.

For ϕ(z)=z2/2\phi(z)=z^2/2 and positive κ\kappa, the real axis is the descent thimble: Re(κϕ)\operatorname{Re}(\kappa\phi) increases away from the saddle and eκϕ\ee^{-\kappa\phi} decays. The dashed imaginary axis is the ascent direction, along which the same exponential grows. If arg[κϕ(zs)]=β\arg[\kappa\phi''(z_s)]=\beta, all four rays rotate by β/2-\beta/2.

Leading Stirling from a non-Gaussian integral

Section titled “Leading Stirling from a non-Gaussian integral”

The Gaussian lemma becomes useful only after a non-Gaussian exponent has been reduced near its saddle. For real z>0z>0, Euler’s integral and the scaling t=zst=zs give

Γ(z)=0tz1et ⁣dt=zz0ez(slogs) ⁣dss.\begin{aligned} \Gamma(z) &= \int_0^\infty t^{z-1}\ee^{-t}\,\dd t \\ &= z^z \int_0^\infty \ee^{-z(s-\log s)} \frac{\dd s}{s}. \end{aligned}

Here

ϕ(s)=slogs,a(s)=1s.\phi(s)=s-\log s, \qquad a(s)=\frac1s.

The positive contour passes through the unique saddle s=1s=1, where

ϕ(1)=1,ϕ(1)=1,a(1)=1.\phi(1)=1, \qquad \phi''(1)=1, \qquad a(1)=1.

The oriented real contour selects the positive Gaussian root, so

Γ(z)2πzz1/2ez.\Gamma(z) \sim \sqrt{2\pi}\, z^{z-1/2}\ee^{-z}.

The original integral is not Gaussian; only its local normal form is. Expanding ϕ\phi and aa to higher order generates the coefficients 1/(12z),1/(288z2),1/(12z),1/(288z^2),\ldots. For complex zz, rotating the contour requires the branch and sector data already recorded in the Gamma section.

Local saddle data do not decide the global contour. In relative homology, an admissible contour decomposes schematically as

[Γ]=sns[Js],nsZ,[\Gamma] = \sum_s n_s[\mathcal J_s], \qquad n_s\in\mathbb Z,

where Js\mathcal J_s is a downward thimble and nsn_s is its intersection number with the dual upward cycle. Endpoints, poles, and branch points are part of the relative problem. A saddle with ns=0n_s=0 does not contribute even if its exponential is large.

Thimble decompositions can jump when saddles are connected by a constant-phase trajectory. A necessary phase-alignment condition for such a connection is

Im[κ(ϕ(zs)ϕ(zt))]=0.\operatorname{Im} \left[ \kappa\bigl(\phi(z_s)-\phi(z_t)\bigr) \right] =0.

It is not sufficient: an actual global connecting flow must exist, and the relevant intersection data must change.

Equality of exponential magnitudes instead tests the real part. Because the names “Stokes line” and “anti-Stokes line” are interchanged in the literature, the book states the relevant real or imaginary condition rather than relying on the name.

Coalescing saddles, a saddle meeting an endpoint, or a saddle meeting a pole invalidate the isolated Gaussian approximation. They require a uniform local model—typically Airy, parabolic-cylinder, or another canonical integral—and a new scaling limit.

The large-order recurrence page applies the same saddle logic to recurrence integrals, while the instanton and large-order page tracks the corresponding exponential scales in transseries.

Borel–Laplace summation in the house convention

Section titled “Borel–Laplace summation in the house convention”

Let

f^()=a0+n1ann.\widehat f(\hbar) = a_0+\sum_{n\geq1}a_n\hbar^n.

Gevrey-1 growth means that some C,A>0C,A>0 obey

anCAnΓ(n+1).|a_n|\leq CA^n\Gamma(n+1).

The book removes the constant term and uses the shifted Borel transform

Bf^(ξ):=n1anΓ(n)ξn1.\mathcal B\widehat f(\xi) := \sum_{n\geq1} \frac{a_n}{\Gamma(n)} \xi^{n-1}.

Suppose this germ continues along argξ=θ\arg\xi=\theta and, for large t>0t>0, obeys a bound for some CB>0C_B>0 and σ0\sigma\geq0,

Bf^(eiθt)CBeσt.\left| \mathcal B\widehat f(\ee^{\ii\theta}t) \right| \leq C_B\ee^{\sigma t}.

Its directional Laplace transform is

Sθf^():=a0+0eiθeξ/Bf^(ξ) ⁣dξ,\mathcal S_\theta\widehat f(\hbar) := a_0+ \int_0^{\ee^{\ii\theta}\infty} \ee^{-\xi/\hbar} \mathcal B\widehat f(\xi)\,\dd\xi,

where

Re(eiθ)>σ\operatorname{Re} \left( \frac{\ee^{\ii\theta}}{\hbar} \right)>\sigma

is the corresponding tangent-domain condition. The weaker inequality >0>0 suffices only for exponential type zero. There is no factor 1/1/\hbar in this shifted convention. Indeed,

0eiθeξ/ξn1Γ(n) ⁣dξ=n.\int_0^{\ee^{\ii\theta}\infty} \ee^{-\xi/\hbar} \frac{\xi^{n-1}}{\Gamma(n)} \,\dd\xi = \hbar^n.

Many sources instead use

B0f^(ξ)=n0ann!ξn\mathcal B_0\widehat f(\xi) = \sum_{n\geq0} \frac{a_n}{n!}\xi^n

and invert with

10eiθeξ/B0f^(ξ) ⁣dξ.\frac1\hbar \int_0^{\ee^{\ii\theta}\infty} \ee^{-\xi/\hbar} \mathcal B_0\widehat f(\xi)\,\dd\xi.

The conversion is

Bf^=ξB0f^.\mathcal B\widehat f = \partial_\xi\mathcal B_0\widehat f.

Mixing the shifted transform with the unshifted inverse shifts every power by one. Testing the monomial n\hbar^n detects the error immediately.

The factorial series

F^()=n0n!n+1\widehat F(\hbar) = \sum_{n\geq0}n!\,\hbar^{n+1}

has shifted Borel transform

BF^(ξ)=11ξ.\mathcal B\widehat F(\xi) = \frac{1}{1-\xi}.

The positive ray meets its pole. Let S0+\mathcal S_{0+} approach the ray from the upper half-plane and S0\mathcal S_{0-} from the lower half-plane, both oriented from zero to infinity. For >0\hbar>0,

S0+F^S0F^=2πie1/.\mathcal S_{0+}\widehat F - \mathcal S_{0-}\widehat F = 2\pi\ii\,\ee^{-1/\hbar}.

More generally, set Discθ=Sθ+Sθ\operatorname{Disc}_\theta=\mathcal S_{\theta+}-\mathcal S_{\theta-}. If a simple pole at AA is the only obstruction between the two lateral contours, the connecting arcs contribute no boundary term, and its Borel residue is rr, then

Discθf^=2πireA/.\operatorname{Disc}_\theta\widehat f = -2\pi\ii\,r\,\ee^{-A/\hbar}.

The minus sign comes from the clockwise contour obtained by upper-forward minus lower-forward. In the factorial example r=1r=-1, which gives the positive jump above.

This residue calculation exhibits the information invisible to the formal power series: the two sums differ by an exponentially flat term. For a cut or several singularities, the discontinuity is computed from the continued Borel transform, not from a finite Taylor list.

LayerWhat has actually been constructed
Formal seriesA coefficient sequence
Borel germA convergent function near ξ=0\xi=0 after a Gevrey bound
Continued Borel transformAnalytic data along a specified path or lateral ray
Directional sumA Laplace integral with growth and decay control
Borel–Padé valueA finite numerical approximation, not automatically a certified sum

The rigorous hierarchy is developed on the exact-WKB summability page. The resurgent-singularity page continues from simple poles to logarithmic and branch-point discontinuities, while the Borel–Padé laboratory separates continuation error, quadrature error, and spectral root finding.

Quadratic differentials, covers, and homology

Section titled “Quadratic differentials, covers, and homology”

A meromorphic quadratic differential is locally

φ=Q(z)( ⁣dz)2.\varphi=Q(z)(\dd z)^2.

Under z=z(w)z=z(w) its coefficient transforms as

Q~(w)=Q(z(w))( ⁣dz ⁣dw)2.\widetilde Q(w) = Q(z(w)) \left( \frac{\dd z}{\dd w} \right)^2.

Its square root becomes a one-form on the normalized spectral cover

Σ^:λ2=φ.\widehat\Sigma: \qquad \lambda^2=\varphi.

Locally λ=y ⁣dz\lambda=y\,\dd z and y2=Q(z)y^2=Q(z). A zero or pole of φ\varphi of odd order is a branch point of the normalized double cover. The deck involution ι\iota reverses the one-form:

ιλ=λ.\iota^*\lambda=-\lambda.

Hence the same projected path on the opposite sheet has the opposite action.

On a simply connected chart of Σ^\widehat\Sigma away from critical points, choose a sheet and base point p0p_0. The distinguished local coordinate is

w(p)=p0pλ.w(p)=\int_{p_0}^{p}\lambda.

In the book’s Schrödinger normalization the leading WKB differential is

φ0=R0(z)( ⁣dz)2,R0=V0E.\varphi_0 = R_0(z)(\dd z)^2, \qquad R_0=V_0-E.

For exact-WKB phase θ\theta,

φθ=e2iθφ,wθ=eiθw.\varphi_\theta = \ee^{-2\ii\theta}\varphi, \qquad w_\theta = \ee^{-\ii\theta}w.

Horizontal trajectories satisfy

Imwθ=constant.\operatorname{Im}w_\theta = \text{constant}.

If

φ=c(zz0)m( ⁣dz)2[1+O(zz0)],m1,\varphi = c(z-z_0)^m(\dd z)^2 \left[1+O(z-z_0)\right], \qquad m\geq1,

then

ww02cm+2(zz0)(m+2)/2.w-w_0 \sim \frac{2\sqrt c}{m+2} (z-z_0)^{(m+2)/2}.

There are m+2m+2 horizontal prongs at that zero; a simple turning point has three.

Let SS be the lifted poles or other deleted points and set X^=Σ^S\widehat X=\widehat\Sigma\setminus S. Closed cycles lie in

H1(X^,Z).H_1(\widehat X,\mathbb Z).

With Πγ=γλ\Pi_\gamma=\oint_\gamma\lambda, oddness gives

Πιγ=Πγ.\Pi_{\iota_*\gamma} = -\Pi_\gamma.

Invariant cycles therefore have zero period. Over Q\mathbb Q, the period of an arbitrary cycle is determined by its anti-invariant projection

γQ:=12(γιγ),Πγ=ΠγQ.\gamma^-_{\mathbb Q} := \frac12 \left( \gamma-\iota_*\gamma \right), \qquad \Pi_\gamma=\Pi_{\gamma^-_{\mathbb Q}}.

That projection need not be integral. The natural integral charge lattice is

Γ:=ker ⁣(1+ι:H1(X^,Z)H1(X^,Z))={γ:ιγ=γ}.\begin{aligned} \Gamma^- &:= \ker\!\left( 1+\iota_*: H_1(\widehat X,\mathbb Z) \to H_1(\widehat X,\mathbb Z) \right) \\ &= \left\{ \gamma:\iota_*\gamma=-\gamma \right\}. \end{aligned}

Paths ending in a marked set PP lie in

H1(X^,P;Z).H_1(\widehat X,P;\mathbb Z).

The boundary map remembers the signed endpoints:

:H1(X^,P;Z)H0(P;Z).\partial: H_1(\widehat X,P;\mathbb Z) \longrightarrow H_0(P;\mathbb Z).

A closed WKB period and an open connection action therefore belong to different groups. Deforming either across a pole of λ\lambda changes the integral by the appropriate 2πi2\pi\ii times residue; compactification does not erase that obstruction.

There is an integral factor-of-two trap. If an open lift β\beta has endpoints fixed by ι\iota, its anti-invariant closed lift obeys

βιβλ=2βλ.\int_{\beta-\iota_*\beta}\lambda = 2\int_\beta\lambda.

If γ\gamma is already anti-invariant, applying 1ι1-\iota_* produces 2γ2\gamma, not a new primitive representative.

If a connected degree-two cover of a compact genus-gg surface has rr simple branch points, Riemann–Hurwitz gives

2g(Σ^)2=2(2g2)+r.2g(\widehat\Sigma)-2 = 2(2g-2)+r.

Thus a four-branch-point cover of the sphere has genus one. An oriented symplectic basis is still extra data: the branch-cut drawing does not decide which representatives are AA and BB or how they are oriented. The book fixes their order and orientation by A,B=+1\langle A,B\rangle=+1.

The spectral-cover page derives the branch and genus rules. The cycle and residue toolkit develops absolute and relative periods, and the Stokes-graph page connects the phased foliation to sectorial WKB solutions.

TaskData that make it well posedFast audit
Continue a local solutionGerm, base point, punctured domain, pathRound trip around one generator
Simplify a Gamma productExact identity, excluded poles, logarithm branchCompare recurrence at a shifted argument
Evaluate a saddle integralLarge-parameter sector, oriented contour, contributing thimblesRotated Gaussian benchmark
Borel sum a formal tailTransform convention, continued germ, ray, lateral side, growth boundReconstruct one monomial
Integrate a WKB formNormalized cover, sheet, cycle or relative path, orientation, residuesApply the deck involution and one intersection check

Treating a branch cut as an intrinsic boundary. Moving a cut changes a planar representative, not the covering surface. Carry the lifted path and sheet label through the move.

Using Stirling’s series across its excluded ray. The negative axis is where the selected logarithm and exponentially improved terms require special care. Use recurrence or reflection first and state the sector of the remaining expansion.

Choosing the Gaussian square root without the contour. The Hessian determines two square roots, while the oriented thimble selects one. Reversing the contour reverses the saddle contribution.

Calling a finite Borel polynomial a Borel sum. A coefficient list gives a local approximation to a germ. Analytic continuation, lateral choice, Laplace growth, and numerical error remain separate questions.

Identifying a projected loop with a cycle on the cover. The sheet, orientation, deleted poles, and endpoint set can change the period even when the planar drawing looks unchanged.

At the base point z=1/2z_*=1/2, choose the principal values of

f(z)=zα(1z)β.f(z)=z^\alpha(1-z)^\beta.

Find the multiplier after a positive loop around 00 that does not enclose 11, and after a positive loop around 11 that does not enclose 00.

Solution

Around the first loop, Logz\operatorname{Log}z gains 2πi2\pi\ii while Log(1z)\operatorname{Log}(1-z) returns to its original branch. Therefore

fe2πiαf.f\longmapsto\ee^{2\pi\ii\alpha}f.

Around the second loop, 1z1-z winds once positively around zero, so

fe2πiβf.f\longmapsto\ee^{2\pi\ii\beta}f.

The multipliers commute because this scalar example has an Abelian one-dimensional monodromy representation. The corresponding matrix problem need not commute.

2. Check an exact Gamma ratio against Stirling

Section titled “2. Check an exact Gamma ratio against Stirling”

Use recurrence to evaluate

Γ(z+3/2)Γ(z+1/2)\frac{\Gamma(z+3/2)}{\Gamma(z+1/2)}

exactly. Then recover its first two large-zz terms from the ratio card.

Solution

Recurrence gives

Γ(z+32)=(z+12)Γ(z+12),\Gamma\left(z+\frac32\right) = \left(z+\frac12\right) \Gamma\left(z+\frac12\right),

hence the exact ratio is z+1/2z+1/2. In the asymptotic formula take a=3/2a=3/2 and b=1/2b=1/2. Then

ab=1,a+b1=1,a-b=1, \qquad a+b-1=1,

so

Γ(z+3/2)Γ(z+1/2)z(1+12z+O(z2))=z+12+O(z1).\frac{\Gamma(z+3/2)}{\Gamma(z+1/2)} \sim z\left(1+\frac{1}{2z}+O(z^{-2})\right) = z+\frac12+O(z^{-1}).

The apparent remainder vanishes identically because recurrence already gave the exact polynomial.

For κ>0\kappa>0, consider

Iα(κ)=eiαReκz2/2 ⁣dz,I_\alpha(\kappa) = \int_{\ee^{\ii\alpha}\mathbb R} \ee^{-\kappa z^2/2}\,\dd z,

with the contour oriented by z=eiαxz=\ee^{\ii\alpha}x as x:+x:-\infty\to+\infty. Determine the absolute-convergence sectors and evaluate the integral for α<π/4|\alpha|<\pi/4.

Solution

Along the contour,

eκz2/2=exp[κx22cos(2α)].\left| \ee^{-\kappa z^2/2} \right| = \exp\left[ -\frac{\kappa x^2}{2}\cos(2\alpha) \right].

The integral converges absolutely with Gaussian decay when cos(2α)>0\cos(2\alpha)>0, namely in the sectors

αkπ<π4,kZ.|\alpha-k\pi|<\frac{\pi}{4}, \qquad k\in\mathbb Z.

On the boundary cos(2α)=0\cos(2\alpha)=0, the corresponding Fresnel integrals are conditionally convergent, but those rays are not exponentially decaying thimbles.

For α<π/4|\alpha|<\pi/4, rotate the contour to the real axis without leaving the decay sectors. No singularity is crossed and the connecting arcs vanish, so

Iα(κ)=2πκ.I_\alpha(\kappa) = \sqrt{\frac{2\pi}{\kappa}}.

At α=π\alpha=\pi the same geometric line has the opposite orientation and the answer changes sign. This is why the square-root phase cannot be detached from the oriented contour.

For

F^()=n0n!n+1,\widehat F(\hbar) = \sum_{n\geq0}n!\hbar^{n+1},

derive its shifted Borel transform and the difference between the upper and lower positive-ray sums for >0\hbar>0.

Solution

The coefficient of n+1\hbar^{n+1} is n!=Γ(n+1)n!=\Gamma(n+1), so

BF^(ξ)=n0ξn=11ξ.\mathcal B\widehat F(\xi) = \sum_{n\geq0}\xi^n = \frac{1}{1-\xi}.

The upper contour minus the lower contour is clockwise around the pole at ξ=1\xi=1. Since

resξ=1eξ/1ξ=e1/,\operatorname*{res}_{\xi=1} \frac{\ee^{-\xi/\hbar}}{1-\xi} = -\ee^{-1/\hbar},

the residue theorem gives

S0+F^S0F^=2πi(e1/)=2πie1/.\mathcal S_{0+}\widehat F - \mathcal S_{0-}\widehat F = -2\pi\ii\left(-\ee^{-1/\hbar}\right) = 2\pi\ii\ee^{-1/\hbar}.

The jump has zero formal power series at =0\hbar=0 in the positive decay sector.

Let

φ=(z21)( ⁣dz)2,\varphi = (z^2-1)(\dd z)^2,

and choose the sheet on which λ=+i1x2 ⁣dx\lambda=+\ii\sqrt{1-x^2}\,\dd x for 1<x<1-1<x<1. Let β\beta be the lift of the interval from 1-1 to 11, and set γ=βιβ\gamma=\beta-\iota_*\beta. Determine the phase θ\theta for which the projected interval is horizontal, show that γ\gamma is an anti-invariant closed cycle, and evaluate γλ\oint_\gamma\lambda.

Solution

Along the chosen lift,

eiθλ=eiθi1x2 ⁣dx.\ee^{-\ii\theta}\lambda = \ee^{-\ii\theta}\ii \sqrt{1-x^2}\,\dd x.

This is real when θ=π/2(modπ)\theta=\pi/2\pmod\pi, so the interval is horizontal for those phases. Its endpoints are branch points and hence are fixed by ι\iota. Therefore

γ=0,ιγ=γ.\partial\gamma = 0, \qquad \iota_*\gamma=-\gamma.

Finally,

γλ=2βλ=2i111x2 ⁣dx=iπ.\begin{aligned} \oint_\gamma\lambda &= 2\int_\beta\lambda \\ &= 2\ii\int_{-1}^{1}\sqrt{1-x^2}\,\dd x \\ &= \ii\pi. \end{aligned}

The middle line is the factor-of-two rule, and the last integral is twice the area of a unit semicircle. Reversing the chosen sheet or the cycle orientation reverses the answer.