Reading guide · Proof index

Gaussian regularization and quadratic stationary phase

Written 4 October 2026. This is newly written exposition from the freely accessible sources named below. It is not a revision obtained by deleting references from an inherited lesson. The exact elementary programme proofs used are listed in F0. The parameter Morse argument is now written in its companion, with specific prerequisite completions. The existing full stationary-phase revision is retained for proof-by-proof reconciliation; its correct free-source arguments are to be preserved. The complete lesson has not yet been cleared.

F0. The exact earlier programme proofs

The proofs below use the following elementary results. This list declares actual dependencies; it does not grant an exception to the requirement to prove every used result inside the programme.

ID Result used Uses below Current programme binding
F0-CALC Fundamental theorem of one-variable calculus, product and chain rules, elementary real and complex exponential differentiation, and smoothness of the positive square root Q1–Q9 Exact combined chain in exponential-proof-chain.json: differential rules and roots, compact FTC/Taylor, real and complex exponentials, their modulus and derivative rules, and the right-half-plane root phase are proved. Global integration remains separate
F0-INT Construction, linearity and norm bounds for the used compact and absolutely convergent improper integrals, their tails, and iteration under the explicit product majorants in these proofs Q1–Q9 Exact chain in global-integral-proof-chain.json, including full compact-rectangle Fubini and its omitted steps, global tails, compact-uniform limits and the interchanges in Q3–Q6. Change of variables remains separate
F0-COV Change of variables for orthogonal linear maps, positive dilations, and planar polar coordinates, with their Jacobians Q2, Q4, Q6 Exact chain in change-of-variables-proof-chain.json: full compact Jordan substitution, determinant–volume scaling, global affine changes, compact chart substitution and the polar exhaustion needed for the Gaussian are proved
F0-COMP Compactness of a closed bounded Euclidean set; attainment of extrema and uniform continuity for continuous functions on it Q1, Q5, Q9 Exact transitive proof chain in topology-proof-chain.json, including P1/P6/P8 and existing compactness, extrema and uniform-continuity proofs; closed relative to its explicit axioms
F0-ALG Determinant multiplication, the cofactor inverse formula, and their smooth dependence on matrix entries Q5, Q6, Q8 Exact algebra and differentiation chain in differential-proof-chain.json, with specific omissions supplied by P7/P9/P10; closed relative to its declared axioms

Finite sums, multi-indices, Euclidean scalar products and the determinant are used with their usual algebraic definitions. An exact earlier linear-algebra binding for the determinant identities is recorded in prerequisite-bindings.json and differential-proof-chain.json. All integrals in this module have the compact Riemann or absolutely convergent improper meaning proved in P17–P18 of global-integral-prerequisite-completions.md. In particular the global interchanges below satisfy the product-majorant hypotheses of P18.4. All of Q1–Q9 and E1 now have exact closed prerequisite chains, recorded in change-of-variables-proof-chain.json. This completes the proofs of this quadratic module; it does not complete the proofs of the full stationary-phase lesson.

Set

a^(ξ)=∫Rne−ix⋅ξa(x) dx,a(x)=(2π)−n∫Rneix⋅ξa^(ξ) dξ.(F) \widehat a(\xi)=\int_{\mathbb R^n}e^{-ix\cdot\xi}a(x)\,dx, \qquad a(x)=(2\pi)^{-n}\int_{\mathbb R^n}e^{ix\cdot\xi} \widehat a(\xi)\,d\xi. \tag{F}

The second equality will be proved in Q4, before it is used. A function belongs to S(Rn)\mathcal S(\mathbb R^n) when it is smooth and sup⁡x(1+∣x∣)m∣∂αa(x)∣<∞\sup_x(1+|x|)^m|\partial^\alpha a(x)|<\infty for every integer m≥0m\geq0 and every multi-index α\alpha.

Q1. The limit and differentiation rules actually used here

Suppose continuous functions ftf_t converge to f0f_0 uniformly on every compact set as t→0t\to0, and ∣ft(x)∣≤g(x)|f_t(x)|\leq g(x) for one integrable gg, including t=0t=0. Then ∫ft→∫f0\int f_t\to\int f_0.

Proof. Given δ>0\delta>0, choose a cube KK so that ∫Kcg<δ\int_{K^c}g<\delta. The difference of the integrals on KcK^c is at most 2δ2\delta. On KK it is at most ∣K∣sup⁡K∣ft−f0∣|K|\sup_K|f_t-f_0|, which tends to zero. Taking a limit superior and then letting δ\delta decrease to zero proves the claim.

For a parameter derivative, suppose f(t,x)f(t,x) and ∂tf(t,x)\partial_t f(t,x) are continuous for tt in a neighbourhood of t0t_0, the integral at t0t_0 exists absolutely, and ∣∂tf(t,x)∣≤g(x)|\partial_t f(t,x)|\leq g(x) there. The fundamental theorem of calculus gives

f(t0+s,x)−f(t0,x)s=∫01∂tf(t0+us,x) du. \frac{f(t_0+s,x)-f(t_0,x)}s =\int_0^1\partial_t f(t_0+us,x)\,du.

These quotients are bounded by gg and converge uniformly on compact sets to ∂tf(t0,x)\partial_t f(t_0,x). The first part proves differentiation under the integral. It also proves continuity of that derivative when the same local bound holds. Repeating the argument proves any finite order for which the corresponding bounds are available. This proof uses F0, not a general unproved appeal to dominated convergence. □\square

Q2. A complex Gaussian without an analytic-continuation import

For z∈Cz\in\mathbb C with Re⁡z>0\operatorname{Re}z>0 and real η\eta,

∫Rexp⁡(−zx2/2+iηx) dx=2π z−1/2exp⁡(−η2/(2z)).(G) \int_{\mathbb R}\exp(-zx^2/2+i\eta x)\,dx =\sqrt{2\pi}\,z^{-1/2}\exp(-\eta^2/(2z)). \tag{G}

Here z\sqrt z is the square root with positive real part.

Proof. The integral and all derivatives used below converge absolutely: on a compact subset of the right half-plane their absolute values are bounded by a polynomial in ∣x∣|x| times e−cx2e^{-c x^2}, for some c>0c>0. Such bounds are integrable: differentiation shows that rme−cr2/2r^m e^{-cr^2/2}, for r≥0r\geq0, has a finite maximum, so the remaining factor is bounded by Ce−cr2/2C e^{-cr^2/2}, and hence by Ce−cr/2C e^{-cr/2} for r≥1r\geq1. The latter has a finite integral by the fundamental theorem. Q1 therefore applies.

First let J=∫e−x2/2 dx>0J=\int e^{-x^2/2}\,dx>0. Squaring, using absolute iteration and then planar polar coordinates, gives

J2=∫R2e−(x2+y2)/2 dx dy=∫02π∫0∞e−r2/2r dr dθ=2π. J^2=\int_{\mathbb R^2}e^{-(x^2+y^2)/2}\,dx\,dy =\int_0^{2\pi}\int_0^\infty e^{-r^2/2}r\,dr\,d\theta=2\pi.

Thus J=2πJ=\sqrt{2\pi}. P21.5 in change-of-variables-prerequisite-completions.md proves the polar Jacobian, the compact-sector substitution, and the limits at the origin, missing ray and infinity used in this step.

For fixed zz join 1 to zz by zt=1+t(z−1)z_t=1+t(z-1), 0≤t≤10\leq t\leq1. Its real part is positive. Put J(t)=∫e−ztx2/2 dxJ(t)=\int e^{-z_t x^2/2}\,dx. Integration of ∂x(xe−ztx2/2)\partial_x(xe^{-z_t x^2/2}) over [−R,R][-R,R], followed by R→∞R\to\infty, gives ∫x2e−ztx2/2 dx=J(t)/zt\int x^2e^{-z_t x^2/2}\,dx=J(t)/z_t, because the boundary terms vanish exponentially. Consequently

J′(t)=−zt′2ztJ(t). J'(t)=-\frac{z'_t}{2z_t}J(t).

The chosen root s(t)=zts(t)=\sqrt{z_t} is differentiable and satisfies 2s(t)s′(t)=zt′2s(t)s'(t)=z'_t. One can see this without a complex-analysis theorem by writing zt=rteiθtz_t=r_t e^{i\theta_t}, with −π/2<θt<π/2-\pi/2<\theta_t<\pi/2, and setting s(t)=rteiθt/2s(t)=\sqrt{r_t}e^{i\theta_t/2}. P16.3 in exponential-prerequisite-completions.md proves the smooth angle, the unique positive-real-part root and this differentiation rule. The product rule now gives (sJ)′=0(sJ)'=0; hence J(1)=2π z−1/2J(1)=\sqrt{2\pi}\,z^{-1/2}.

Finally, with zz fixed write the left side of (G) as F(η)F(\eta). The integral of ∂xe−zx2/2+iηx\partial_x e^{-zx^2/2+i\eta x} vanishes, so ∫xe−zx2/2+iηx dx=(iη/z)F(η)\int xe^{-zx^2/2+i\eta x}\,dx=(i\eta/z)F(\eta). Thus F′(η)=−(η/z)F(η)F'(\eta)=-(\eta/z)F(\eta), and ∂η(eη2/(2z)F(η))=0\partial_\eta(e^{\eta^2/(2z)}F(\eta))=0. The value at zero already computed proves (G). □\square

Q3. Schwartz estimates and the signs in the Fourier rules

If a∈Sa\in\mathcal S, then a^∈S\widehat a\in\mathcal S, and

∂ξαa^=(−ix)αa^,∂xβa^=(iξ)βa^.(D) \partial_\xi^\alpha\widehat a=\widehat{(-ix)^\alpha a},\qquad \widehat{\partial_x^\beta a}=(i\xi)^\beta\widehat a. \tag{D}

These maps are continuous for the defining Schwartz seminorms.

Proof. Polynomial multiples of derivatives of aa are integrable. For example use the rectangular shells between [−2j,2j]n[-2^j,2^j]^n and [−2j+1,2j+1]n[-2^{j+1},2^{j+1}]^n. P18.3 proves their integral bounds directly from finite rectangle partitions. Decay (1+∣x∣)−m(1+|x|)^{-m}, m>nm>n, reduces the bound to a convergent geometric series. Q1 proves every differentiation in the first formula. For the second, integrate first in a single coordinate on [−R,R][-R,R]. Its boundary terms tend to zero by rapid decrease; the remaining integrals converge absolutely. Iteration proves the multi-index formula. Applying these identities together shows that each ξβ∂ξαa^\xi^\beta\partial_\xi^\alpha\widehat a is the Fourier transform of a finite linear combination of polynomial multiples of derivatives of aa. Its supremum is bounded by their L1L^1 norms. The rectangular-shell estimate bounds each such norm by finitely many input seminorms, proving both rapid decrease and continuity.

A useful quantitative consequence is, for integers M>N+n/2M>N+n/2,

∫∣ξ∣2N∣a^(ξ)∣ dξ≤Cn,N,M ∥(1−Δ)Ma∥L1,Cn,N,M=∫∣ξ∣2N(1+∣ξ∣2)M dξ<∞.(W) \int |\xi|^{2N}|\widehat a(\xi)|\,d\xi \leq C_{n,N,M}\,\|(1-\Delta)^M a\|_{L^1},\qquad C_{n,N,M}=\int\frac{|\xi|^{2N}}{(1+|\xi|^2)^M}\,d\xi<\infty. \tag{W}

Indeed (D) gives (1+∣ξ∣2)Ma^=(1−Δ)Ma^(1+|\xi|^2)^M\widehat a=\widehat{(1-\Delta)^M a}; take absolute values and integrate. The same rectangular-shell bound proves finiteness of Cn,N,MC_{n,N,M}, with its stated threshold unchanged. □\square

Q4. Fourier inversion proved before stationary phase

For every a∈Sa\in\mathcal S, the inversion identity in (F) holds.

Proof. For ε>0\varepsilon>0 define

Aε(x)=(2π)−n∫eix⋅ξe−ε∣ξ∣2/2a^(ξ) dξ. A_\varepsilon(x)=(2\pi)^{-n}\int e^{ix\cdot\xi} e^{-\varepsilon|\xi|^2/2}\widehat a(\xi)\,d\xi.

Insert the definition of a^\widehat a. The double absolute integral is bounded by ∥a∥1∫e−ε∣ξ∣2/2 dξ<∞\|a\|_1\int e^{-\varepsilon|\xi|^2/2}\,d\xi<\infty, so F0-INT permits interchange. Applying (G) in every coordinate yields

Aε(x)=∫Kε(x−y)a(y) dy,Kε(v)=(2πε)−n/2e−∣v∣2/(2ε). A_\varepsilon(x)=\int K_\varepsilon(x-y)a(y)\,dy,\qquad K_\varepsilon(v)=(2\pi\varepsilon)^{-n/2} e^{-|v|^2/(2\varepsilon)}.

By positive dilation and Q2, ∫Kε=1\int K_\varepsilon=1. Since aa has bounded gradient, the fundamental theorem along a line segment gives ∣a(x−v)−a(x)∣≤∥∇a∥∞∣v∣|a(x-v)-a(x)|\leq\|\nabla a\|_\infty|v|, with the Euclidean norm for the gradient. Therefore

∣Aε(x)−a(x)∣≤ε ∥∇a∥∞∫∣w∣K1(w) dw⟶0. |A_\varepsilon(x)-a(x)| \leq\sqrt\varepsilon\,\|\nabla a\|_\infty \int |w|K_1(w)\,dw\longrightarrow0.

On the Fourier side the multiplier converges uniformly on compact sets to 1 and is bounded by 1. Q3 supplies integrability of a^\widehat a, so Q1 shows that Aε(x)A_\varepsilon(x) tends to the right side of (F). This proves inversion without using stationary phase. Applying the same argument to ∂αa\partial^\alpha a also gives every derivative of (F). □\square

Q5. Diagonalizing a real symmetric matrix

A real symmetric n×nn\times n matrix HH has an orthonormal basis of real eigenvectors. If HH is invertible, all its eigenvalues are nonzero. Define sgn⁡H\operatorname{sgn}H to be their number of positive entries minus their number of negative entries.

Proof. In dimension 0 the empty orthonormal basis proves the assertion. For n≥1n\geq1, the function v↦vTHvv\mapsto v^THv attains a maximum on the unit sphere, by F0-COMP. Let vv be a maximizing unit vector. For w⊥vw\perp v, differentiate this function along p(t)=(v+tw)/1+t2∣w∣2p(t)=(v+tw)/\sqrt{1+t^2|w|^2} at t=0t=0. The product, chain and positive-root derivative rules give p′(0)=wp'(0)=w. The scalar Fermat theorem at this maximum therefore gives 2wTHv=02w^THv=0. Set λ=vTHv\lambda=v^THv. The vector w=Hv−λvw=Hv-\lambda v is perpendicular to vv, and the preceding identity gives ∣w∣2=wTHv=0|w|^2=w^THv=0. Thus Hv=λvHv=\lambda v. If w⊥vw\perp v, then vTHw=(Hv)Tw=0v^THw=(Hv)^Tw=0; hence v⊥v^\perp is invariant under HH. P9.4 in differential-prerequisite-completions.md constructs an orthonormal basis of this complement and proves that the restriction is a symmetric (n−1)(n-1)-dimensional matrix. Apply induction to that matrix, starting with dimension 0 or 1. This gives an orthogonal matrix UU with UTHU=diag⁡(λ1,…,λn)U^THU=\operatorname{diag}(\lambda_1,\ldots,\lambda_n). If any λj=0\lambda_j=0, its eigenvector lies in the kernel, and conversely. Since UTU=IU^TU=I, determinant multiplication gives (det⁡U)2=1(\det U)^2=1 and det⁡(UTHU)=det⁡H\det(U^THU)=\det H. In the permutation formula for a diagonal matrix only the identity permutation can have a nonzero term; its term is ∏jλj\prod_j\lambda_j. Hence det⁡H=∏jλj\det H=\prod_j\lambda_j. □\square

Q6. Exact quadratic identity, with a justified regularization

Let HH be real, symmetric and invertible, h>0h>0, B=H−1B=H^{-1}, and a∈S(Rn)a\in\mathcal S(\mathbb R^n). Then

∫eixTHx/(2h)a(x) dx=CH(h)(2π)−n∫e−ihξTBξ/2a^(ξ) dξ,CH(h)=(2πh)n/2eiπsgn⁡H/4∣det⁡H∣.(QI) \int e^{ix^THx/(2h)}a(x)\,dx =C_H(h)(2\pi)^{-n}\int e^{-ih\xi^TB\xi/2} \widehat a(\xi)\,d\xi, \qquad C_H(h)=\frac{(2\pi h)^{n/2}e^{i\pi\operatorname{sgn}H/4}} {\sqrt{|\det H|}}. \tag{QI}

Both integrals displayed here are absolutely convergent because the amplitudes are Schwartz. The bare quadratic exponential is not treated as an absolutely integrable function.

Proof. Multiply the integrand on the left by e−ε∣x∣2/2e^{-\varepsilon|x|^2/2}, ε>0\varepsilon>0, and insert Q4. The double absolute integral is finite, bounded by (2π)−n∥a^∥1∫e−ε∣x∣2/2 dx(2\pi)^{-n}\|\widehat a\|_1\int e^{-\varepsilon|x|^2/2}\,dx. Thus its order can be exchanged. Diagonalize HH using Q5 and make the orthogonal change x=Uyx=Uy, η=UTξ\eta=U^T\xi. The inner integral is the product of (G) with zj=ε−iλj/hz_j=\varepsilon-i\lambda_j/h. Its value is

(2π)n/2∏j=1nzj−1/2exp⁡ ⁣(−∑j=1nηj22zj).(RG) (2\pi)^{n/2}\prod_{j=1}^n z_j^{-1/2} \exp\!\left(-\sum_{j=1}^n\frac{\eta_j^2}{2z_j}\right). \tag{RG}

Since Re⁡(1/zj)>0\operatorname{Re}(1/z_j)>0, the modulus of the exponential is at most 1, while ∣zj∣−1/2≤(h/∣λj∣)1/2|z_j|^{-1/2}\leq(h/|\lambda_j|)^{1/2}. This supplies an ε\varepsilon-independent integrable bound after multiplication by ∣a^(ξ)∣|\widehat a(\xi)|.

As ε↓0\varepsilon\downarrow0, with each root continued through the right half-plane,

zj−1/2⟶(h/∣λj∣)1/2eiπsgn⁡(λj)/4,zj−1⟶ih/λj. z_j^{-1/2}\longrightarrow (h/|\lambda_j|)^{1/2}e^{i\pi\operatorname{sgn}(\lambda_j)/4}, \qquad z_j^{-1}\longrightarrow ih/\lambda_j.

The convergence is uniform for ξ\xi in compact sets. Q1 therefore takes the limit on the Fourier side of (RG), giving the right side of (QI). On the original side it applies with bound ∣a∣|a|, removing the regularization. The product of the root phases is precisely eiπsgn⁡H/4e^{i\pi\operatorname{sgn}H/4}. □\square

The root limit and its π/4\pi/4 phase in this proof are established explicitly in P16.3, including both signs of λj\lambda_j.

Q7. Every coefficient and an explicit remainder

Put ΔB=∑j,kBjk∂j∂k\Delta_B=\sum_{j,k}B_{jk}\partial_j\partial_k. For every integer N≥1N\geq1,

∫eixTHx/(2h)a(x) dx=CH(h)[∑k=0N−1(ih/2)kk!(ΔBka)(0)+RN(h;H,a)],(SP) \int e^{ix^THx/(2h)}a(x)\,dx =C_H(h)\left[\sum_{k=0}^{N-1}\frac{(ih/2)^k}{k!} (\Delta_B^ka)(0)+R_N(h;H,a)\right], \tag{SP}

where

∣RN(h;H,a)∣≤hN2NN!(2π)n∫∣ξTBξ∣N∣a^(ξ)∣ dξ≤hN∥B∥NCn,N,M2NN!(2π)n∥(1−Δ)Ma∥1(R) |R_N(h;H,a)|\leq \frac{h^N}{2^NN!(2\pi)^n} \int |\xi^TB\xi|^N|\widehat a(\xi)|\,d\xi \leq\frac{h^N\|B\|^N C_{n,N,M}}{2^NN!(2\pi)^n} \|(1-\Delta)^M a\|_1 \tag{R}

for any integer M>N+n/2M>N+n/2. Here ∥B∥\|B\| is its Euclidean operator norm. The error in the unnormalized integral is ∣CH(h)RN∣|C_H(h)R_N|, of order hn/2+Nh^{n/2+N} for fixed H,aH,a.

Proof. Repeatedly apply the fundamental theorem of calculus to the function t↦eitt\mapsto e^{it}, or inductively integrate its remainder once. This gives, for real tt,

eit−∑k=0N−1(it)kk!=(it)N(N−1)!∫01(1−s)N−1eist ds,∣eit−∑k<N(it)kk!∣≤∣t∣NN!.(T) e^{it}-\sum_{k=0}^{N-1}\frac{(it)^k}{k!} =\frac{(it)^N}{(N-1)!}\int_0^1(1-s)^{N-1}e^{ist}\,ds, \quad \left|e^{it}-\sum_{k<N}\frac{(it)^k}{k!}\right| \leq\frac{|t|^N}{N!}. \tag{T}

For completeness, the inductive integral form starts with f(1)−f(0)=∫01f′(s)dsf(1)-f(0)=\int_0^1 f'(s)ds. Integration by parts changes ∫01(1−s)N−1f(N)(s)/(N−1)! ds\int_0^1 (1-s)^{N-1}f^{(N)}(s)/(N-1)!\,ds into −f(N−1)(0)/(N−1)!+∫01(1−s)N−2f(N−1)(s)/(N−2)! ds-f^{(N-1)}(0)/(N-1)!+ \int_0^1(1-s)^{N-2}f^{(N-1)}(s)/(N-2)!\,ds; descending to the first identity proves the formula for f(s)=eitsf(s)=e^{its}.

Apply (T) with t=−hξTBξ/2t=-h\xi^TB\xi/2 in (QI). The remainder integrates absolutely by Q3, giving the first inequality (R). Moreover, (D) and Q4 give

(2π)−n∫(ξTBξ)ka^(ξ) dξ=(−1)k(ΔBka)(0). (2\pi)^{-n}\int(\xi^TB\xi)^k\widehat a(\xi)\,d\xi =(-1)^k(\Delta_B^ka)(0).

Thus (−ih/2)k(-ih/2)^k becomes (ih/2)k(ih/2)^k, establishing the sign in (SP). Finally ∣ξTBξ∣≤∥B∥∣ξ∣2|\xi^TB\xi|\leq\|B\||\xi|^2, and (W) proves the second inequality. □\square

Q8. What is uniform when the Hessian varies

Let pp range over an open parameter set, let H(p)H(p) be smooth, real symmetric and invertible, and let a(p,⋅)a(p,\cdot) be smooth with values in S\mathcal S: every parameter derivative has locally bounded Schwartz seminorms. On every compact parameter subset, the normalized integral in (QI), divided by CH(p)(h)C_{H(p)}(h), extends smoothly through h=0h=0 as a function of real hh. Its Taylor coefficients are the ones in (SP), and the normalized remainder is O(hN)O(h^N) after any fixed number of parameter derivatives.

Proof. Matrix inversion is smooth where the determinant is nonzero, by the cofactor formula, so B(p)B(p) and all derivatives are bounded on a compact parameter subset contained in the open set. For real hh define the normalized expression directly by

A(p,h)=(2π)−n∫e−ihξTB(p)ξ/2a(p,⋅)^(ξ) dξ. A(p,h)=(2\pi)^{-n}\int e^{-ih\xi^TB(p)\xi/2}\widehat {a(p,\cdot)}(\xi)\,d\xi.

Each hh-derivative multiplies the integrand by a power of the quadratic polynomial −iξTB(p)ξ/2-i\xi^TB(p)\xi/2. A parameter derivative either differentiates the Fourier amplitude or contributes a derivative of this polynomial multiplied by hh. Repeated product rules therefore give a finite sum of polynomial factors in ξ\xi, with bounded coefficients for pp in a compact set and ∣h∣≤1|h|\leq1, times a Fourier transform of a parameter derivative of aa. Q3 makes all these factors uniformly integrable. Q1 proves smoothness and permits all the derivatives just described.

At h=0h=0, Q4 and (D) give ∂hkA(p,0)=(i/2)kΔB(p)ka(p,0)\partial_h^k A(p,0)=(i/2)^k\Delta_{B(p)}^k a(p,0). Apply the integral Taylor formula proved in Q7 to h↦∂pαA(p,h)h\mapsto\partial_p^\alpha A(p,h). Its NNth hh-derivative is uniformly bounded by the preceding integral estimates. The remainder is bounded by CN,α∣h∣NC_{N,\alpha}|h|^N, uniformly on the compact parameter subset. This proves the asserted parameter estimate. It is not an estimate for unnormalized parameter derivatives of an additional oscillatory factor eiϕ(p)/he^{i\phi(p)/h}; that factor must be kept separate. □\square

Q9. The region with no stationary point

Suppose ϕ\phi is smooth and real on an open neighbourhood of the compact support of a∈Cc∞a\in C_c^\infty, and ∣∇ϕ∣≥c>0|\nabla\phi|\geq c>0 on such a neighbourhood. Set v=∇ϕ/∣∇ϕ∣2v=\nabla\phi/|\nabla\phi|^2 there and Ta=−div⁡(va)Ta=-\operatorname{div}(va). Then, for h>0h>0 and each integer N≥0N\geq0,

∫eiϕ(x)/ha(x) dx=(h/i)N∫eiϕ(x)/hTNa(x) dx,∣∫eiϕ/ha∣≤hN∥TNa∥1.(NS) \int e^{i\phi(x)/h}a(x)\,dx =(h/i)^N\int e^{i\phi(x)/h}T^Na(x)\,dx, \qquad \left|\int e^{i\phi/h}a\right|\leq h^N\|T^Na\|_1. \tag{NS}

The right side is finite and independent of hh.

Proof. The identity v⋅∇eiϕ/h=(i/h)eiϕ/hv\cdot\nabla e^{i\phi/h}=(i/h)e^{i\phi/h} follows by the chain rule. Integrating it against aa and using integration by parts in each coordinate gives (NS) for N=1N=1. Each TjaT^j a has support inside supp⁡a\operatorname{supp}a, because derivatives and multiplication do not enlarge support. Thus the same argument applies repeatedly. There are no boundary terms: one can choose the integration box outside the compact support. The final bound follows from the absolute-value bound on the integral and ∣eiϕ/h∣=1|e^{i\phi/h}|=1. For a family with a common compact support, a uniform lower bound on ∣∇ϕ∣|\nabla\phi|, and bounded derivatives of ϕ,a\phi,a to the required finite order, the same finite product-rule expansion bounds ∥TNa∥1\|T^Na\|_1 uniformly. A parameter derivative of the oscillatory factor costs a power of h−1h^{-1}; applying (NS) sufficiently many extra times still proves decay to every order for each fixed derivative order. □\square

E1. An exact model that checks the normalization and the error

Take n=1n=1, H=1H=1, a(x)=e−x2/2a(x)=e^{-x^2/2}. Q2 gives

I(h)=2π (1−i/h)−1/2,I(h)2πh eiπ/4=(1+ih)−1/2.(EX) I(h)=\sqrt{2\pi}\,(1-i/h)^{-1/2},\qquad \frac{I(h)}{\sqrt{2\pi h}\,e^{i\pi/4}}=(1+ih)^{-1/2}. \tag{EX}

The roots in this equality have the branches specified in Q2: their arguments add to −π/2+arctan⁡h-\pi/2+\arctan h, which lies in (−π/2,0)(-\pi/2,0). Hence no sign ambiguity is hidden in the factorization.

Repeated integration by parts gives the Gaussian moments ∫ξ2Ne−ξ2/2dξ=(2N−1)!!2π\int\xi^{2N}e^{-\xi^2/2}d\xi=(2N-1)!!\sqrt{2\pi}: the boundary term in ∫ξ2N−1ξe−ξ2/2\int\xi^{2N-1}\xi e^{-\xi^2/2} vanishes and each step reduces the exponent by two; the base case is Q2. For the derivatives of the amplitude, use a′=−xaa'=-xa. Its (m−1)(m-1)st derivative at zero, for m≥2m\geq2, gives a(m)(0)=−(m−1)a(m−2)(0)a^{(m)}(0)=-(m-1)a^{(m-2)}(0). Together with a(0)=1a(0)=1, a′(0)=0a'(0)=0, induction gives a(2k)(0)=(−1)k(2k)!/(2kk!)a^{(2k)}(0)=(-1)^k(2k)!/(2^kk!). Consequently Q7 reads

(1+ih)−1/2=∑k=0N−1(−1)k(2kk)4k(ih)k+RN(h),∣RN(h)∣≤(2NN)4NhN.(ER) (1+ih)^{-1/2} =\sum_{k=0}^{N-1}(-1)^k\frac{\binom{2k}{k}}{4^k}(ih)^k+R_N(h), \qquad |R_N(h)|\leq\frac{\binom{2N}{N}}{4^N}h^N. \tag{ER}

For example the first three terms are 1−ih/2−3h2/81-ih/2-3h^2/8. The bound holds for every h>0h>0; convergence of an infinite Taylor series for large hh is neither asserted nor used.

Exact Gaussian model and the proved normalized remainder bound

Figure 1. The left panel samples the real and imaginary parts of the exact normalized integral (EX) and of 1−ih/2−3h2/81-ih/2-3h^2/8. The right panel samples ∣RN(h)∣/hN|R_N(h)|/h^N for N=1,2,3N=1,2,3; the horizontal dashed lines are the proved constants (2NN)/4N\binom{2N}{N}/4^N in (ER), not fitted estimates. The horizontal variable is 0.01≤h≤10.01\leq h\leq1. The sampled curves illustrate Q7 and E1; the proof of the bound is (T), (QI), and the moment calculation above. Reproducible source: figures/draw_quadratic_remainder.py. The human-source route for the Gaussian and quadratic calculation is Guillemin–Sternberg, author draft, §§14.1–14.5; the displayed example and figure are written for this reconstruction.

Free source reading and remaining work

The mathematical source versions actually consulted are:

The source PDFs are not redistributed here. No permission to copy their text is inferred from free access. All arguments above are written out here in new exposition, and the displayed normalization is verified directly by (G) and (EX), rather than copied without checking.

The F0 programme proofs use the verified free author edition of Jiří Lebl, Basic Analysis 6.3, with exact earlier proof locators and the used exercises completed in the licensed prerequisite companions. P21 supplies the change-of-variables steps instead of invoking a free citation in their place.

Next required mathematical work: integrate the smooth parameter inverse-function and Morse arguments now written in P2–P5 and M1; and reconcile the existing cutoff, invariant isolated-point and clean critical-manifold proofs, including parameter derivatives and normal density. Only an actual gap needs a replacement proof. The exact full lesson and all its transitive prerequisite files still require source and P514 review before any publication.