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For fixed $0<\gamma <1$, the denominator's first squared term has no zero on the interval and is at least $(1-\gamma {)}^2$. Expand the integrand uniformly in ${\epsilon }^2$ and integrate:The asymptotic expansion is uniform for $\gamma$ in compact subsets below $1$, including the continuous limit $\gamma \to 0$. It is not uniform through $\gamma =1$, where a narrow endpoint peak forms. That transition occurs when $1-\gamma =O(\epsilon )$, as the next part shows.

The exponent $\Phi (t)=\frac{i}{2}(t-t^{-1})$ has saddles $t=\pm i$, with $\Phi (-i)=1$ and $\Phi (i)=-1$. Since the integrand is analytic off the origin, deform the contour to the unit circle and rotate $t=-i{e}^{i\theta }$. This gives the exact representationThe dominant saddle at $\theta =0$ is a steepest-descent direction for the method of steepest descent: $\cos \theta =1-{\theta }^2/2+{\theta }^4/24-\cdots$. Set $\theta =s/\sqrt{x}$ and expand the analytic amplitude against $e^{-s^2/2}$. Integrating its Gaussian moments gives the leading factor $e^x/\sqrt{2\pi x}$; the rest of the contour is exponentially smaller than this dominant contribution outside any fixed neighbourhood of the maximum.

To obtain every coefficient efficiently, the integral satisfies $x^2{I}_0^{\prime \prime }+x{I}_0^{\prime }-x^2{I}_0=0$, by differentiating and integrating the derivative of $\sin \theta e^{x\cos \theta }$. Substitute $I_0=e^x{x}^{-1/2}\sum a_n{x}^{-n}/\sqrt{2\pi }$ to get $a_0=1$ and $a_n=(2n-1{)}^2{a}_{n-1}/(8n)$. Therefore the full asymptotic expansion isHere $(-1)!!=1$. For each fixed truncation at $n=N$, the relative remainder is $O(x^{-N-1})$ as $x\to +\infty$. The infinite series is generally divergent, and this all-orders expansion does not purport to resolve exponentially small saddle contributions or their Stokes phenomenon.

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

Put $q={\mu }^2-1/4$ and $S=\log y$. The differential equation becomes the Riccati equation $S^{\prime \prime }+(S^{\prime }{)}^2=1/4+q/x^2$. The inverse-power exponential hierarchy begins withso choose $S_0=\epsilon x/2$, $S_1=0$ with $\epsilon =\pm 1$; additive constants specify the overall normalization. At order $x^{-2}$, $2S_0^{\prime }{S}_2^{\prime }=q/x^2$, and at subsequent ordersThese equations are formally equivalent coefficient by coefficient; the order-zero transport term happens to be constant, so its chosen zero value is not interpreted as a literal denominator in a successive-ratio hypothesis. The ansatz $S_j=c_j{x}^{-(j-1)}$ is consistent for $j\ge 2$, withEmpty sums are zero. In particular $c_3=-q$, which agrees with direct expansion of the Riccati equation.

Exponentiating the inverse-power series gives $y_{\epsilon }\sim e^{\epsilon x/2}{\sum }_{j\ge 0}{A}_j^{\epsilon }{x}^{-j}$ with $A_0=1$. Substitute this amplitude series into $u^{\prime \prime }+\epsilon u^{\prime }=q{x}^{-2}u$ to obtainFormal matching alone does not prove existence of actual solutions with these asymptotics. Here actual solutions are obtained from the Modified Bessel differential equation and the large-argument asymptotic expansion of a modified Bessel function: substituting $y=\sqrt{x}w(x/2)$ yields solutions $\sqrt{\pi }\sqrt{x}{I}_{\mu }(x/2)$ and $\sqrt{x/\pi }{K}_{\mu }(x/2)$, using the Modified Bessel function of the first kind and Modified Bessel function of the second kind, whose positive-real-axis asymptotic expansions have precisely these normalized growing and decaying series. They are linearly independent. The series need not converge; it is an asymptotic expansion with a remainder after each fixed truncation. When $q=j(j+1)$ the amplitude recurrence terminates.

The Mellin transform iswhere this integral converges. Rapid decay at infinity makes the integral over $[1,\infty )$ an entire function of $s$, but alone gives no control near zero. With the additional expansion, the integral initially converges absolutely for $\mathrm{Re}s>-{\sigma }_1$.

For $\mathrm{Re}s>1$, expand $(e^{\pi y}-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-\pi ry}$. Absolute convergence justifies termwise integration, using the Gamma integral, and gives the scaled Bose integral

There is a missing hypothesis in the general continuation claim: one needs ${\sigma }_j\to +\infty$. The original PDF, like the TeX, only says that the sequence increases. Under the intended additional hypothesis, split the integral at one and subtract $r$ terms of the asymptotic expansion:Continuity of $g_r$ at zero makes it bounded on $[0,1]$. The last integral is holomorphic on $\mathrm{Re}s>-{\sigma }_{r+1}$: on every compact subset, its integrand and all its $s$ derivatives are dominated by an integrable power of $y$ times a power of $|\log y|$. The expressions for successive $r$ agree on their common initial domain and hence everywhere they overlap by the identity theorem. As ${\sigma }_{r+1}\to \infty$, these half-planes cover $\mathbb{C}$. Taking $r\ge j$ with ${\sigma }_{r+1}>{\sigma }_j$ isolates the term $c_j/(s+{\sigma }_j)$, while all other terms are holomorphic near $-{\sigma }_j$. Thus the corrected claim isThis is meromorphic continuation of a Mellin transform from an asymptotic expansion.

To show why the correction matters, set ${\sigma }_j=1-1/j$, $c_j=2^{-j}$, andwhere $\eta$ is continuous, equals one for $0<y\le 1$ and vanishes for $y\ge 2$. On $[0,1]$, the remainder after division by $y^{{\sigma }_{r+1}}$ is a uniformly convergent series of nonnegative powers of $y$, so it extends continuously to zero with value $c_{r+1}$. Away from zero, define $g_r$ by the required remainder quotient; it is continuous on the rest of $[0,\infty )$ as well. Thus all the printed hypotheses hold. Its Mellin transform equalsThe series continues meromorphically on $\mathrm{Re}s>-1$ and has genuine poles at $-1+1/j$, accumulating at $-1$. A meromorphic function on $\mathbb{C}$ cannot have such an accumulation of poles. This is the accumulating asymptotic exponents obstruct Mellin continuation counterexample.

For the Gamma function, the Gamma function recurrence givesAt $s=-j$ the numerator is $\Gamma (1)=1$ and the product of the nonzero denominator factors is $(-1{)}^{j}j!$. Therefore the residues of the Gamma function are

The generating function of the Bernoulli numbers gives the convergent expansion near zeroThe subtraction proof above applies to this expansion, including its zero coefficients, which give no pole. At $s=1-n$, for $n\ge 1$, the residue of its transform is $B_n{\pi }^{n-1}/n!$, while the residue of ${\pi }^{-s}\Gamma (s)$ is ${\pi }^{n-1}(-1{)}^{n-1}/(n-1)!$. Their quotient defines a holomorphic continuation of the Riemann zeta function at this point, including when the first residue is zero. Dividing gives the Bernoulli formula for zeta values at nonpositive integers:In particular the generating-series convention is $B_1=-1/2$, so the formula includes $\zeta (0)=-1/2$.

Assume $m^2>0$ and take $\lambda \to 0^{+}$. The free Gaussian integral gives $Z_0=\sqrt{2\pi }/|m|$, and the normalized free Gaussian measure has covariance $C=m^{-2}$. Expanding only the interaction exponential gives the asymptotic expansionIn this zero-dimensional scalar field theory, the path integral has become an ordinary integral, but its combinatorics are precisely those of sextic scalar field theory.

The free generating functional is $⟨e^{J\varphi }{⟩}_0=e^{C{J}^2/2}$. Differentiating it proves Wick theorem: odd moments vanish, while each even moment is the sum over pairings of its factors. Each Wick contraction contributes $C$, soThus the Feynman rules are a quantum field theory propagator $C$ for each internal line and a factor $-\lambda$ for each six-valent interaction vertex. The factorials $V!$ and $(6!{)}^V$ remove vertex and half-edge labels; the remaining weight is $(-\lambda {)}^V{C}^I/S$, where $S$ is the Feynman-diagram symmetry factor, the order of the automorphism group of the resulting diagram, including its half-edge symmetries. A tadpole diagram has an additional interchange symmetry of the two ends of its loop.

There is one empty Vacuum Feynman diagram, contributing $1$. At one interaction vertex, the six half-edges form three loops. Its Feynman-diagram symmetry factor is $2^{3}3!=48$, so its contribution is $-\lambda C^3/48$.

At two interaction vertices let $r$ count the lines joining them. Each interaction vertex has $(6-r)/2$ self-loops, so $r$ can only be $0,2,4,6$. These four possibilities exhaust all Vacuum Feynman diagrams at this order. Their Feynman-diagram symmetry factors areHere the leading $2$ exchanges the two interaction vertices, $r!$ permutes the joining lines, and the remaining factors permute and reverse the self-loops. The $r=0$ graph is disconnected and must be included in $Z/Z_0$; only a connected generating functional would discard it.

Consequently,Indeed ${\sum }_r{S}_r^{-1}=77/7680$, independently agreeing with $11!!/[2(6!{)}^2]$ from Wick theorem.

This is an asymptotic expansion, rather than a convergent perturbation series: the coefficient of ${\lambda }^V$ contains $(6V-1)!!/[V!(6!{)}^V]$ and grows too rapidly for a nonzero radius of convergence. For any fixed truncation order, the Taylor expansion remainder bound for $e^{-x}$ on $x\ge 0$, followed by the finite Gaussian moment integral, proves the stated remainder estimate. For negative real $\lambda$ the original integral diverges.

The transition equation reduces to the Bessel differential equation because $d/dT=-Zd/dZ$ and $d^2/d{T}^2=Z^2{d}^2/d{Z}^2+Zd/dZ$. Thus $Y_{TT}+e^{-2T}Y=0$ becomes $Z^2{Y}_{ZZ}+Z{Y}_Z+Z^{2}Y=0$.

The Bessel function of the first kind has $J_0(Z)=1+O(Z^2)$ at zero, while the Bessel function of the second kind has $Y_0(Z)=(2/\pi )(\log (Z/2)+\gamma )+O(Z^2|\log Z|)$. Since $\log Z=-T$, these yield a constant and a linear function of $T$, explaining the late-time behavior. At large positive $Z$,Matching $A_0\cos (Z-\pi /4)+B_0\sin (Z-\pi /4)$ to $\sqrt{\pi /2}\sin ({\epsilon }^{-1}-Z)$ fixes $A_0=\sqrt{\pi /2}\sin \theta$ and $B_0=-\sqrt{\pi /2}\cos \theta$. The large- and small-argument asymptotic expansions thereby connect the oscillatory WKB approximation to the nonoscillatory late-time solution through one Bessel transition for an exponentially decaying oscillator.

The intended integral representation of the modified Bessel function isDifferentiation under the integral sign givesSinceintegration by parts and the vanishing endpoint term yield $x{I}_0^{\prime \prime }+I_0^{\prime }-x{I}_0=0$.

For large positive $x$, Laplace's method localizes the integral near the endpoint $\theta =0$. Put $\theta =u/\sqrt{x}$ and useExtending the exponentially small tail to infinity and evaluating the resulting Gaussian integrals givesand therefore

Now write $y=x^{-1/2}w$. Direct differentiation givesso substitution into the ordinary differential equation givesFor the Liouville–Green approximation, put $Q(x)=1-1/(4x^2)$. The two approximate solutions areso at leading order $y_{\pm }(x)\sim x^{-1/2}{e}^{\pm x}$. The growing solution agrees with the preceding asymptotic expansion, including its leading normalization $I_0(x)\sim e^x/\sqrt{2\pi x}$.

There are two positive roots for $u>1$: the derivative of $x{e}^{1/x}$ is $e^{1/x}(1-1/x)$, so the function decreases to its minimum $e$ at $x=1$ and then increases. Negative $x$ cannot solve the equation. For the larger root, expansion of the exponential function in $1/x$ givesso series reversion yieldsFor the smaller root, set $y=1/x$ and take a logarithm. Then $y-\log y=u$, so successive substitution gives $y=u+\log u+O(\log u/u)$ andThese are the first two nonzero terms of the two asymptotic expansions. Equivalently, the roots are $-1/W_0(-e^{-u})$ and $-1/W_{-1}(-e^{-u})$, from the real branches of Lambert W.

Write the remainders after order $N$ as $r_N=o((x-x_0{)}^N)$ for $f$ and $s_N=o((x-x_0{)}^N)$ for $g$. ThenHence $h=f+g$ has the termwise sum of the two asymptotic expansions.

Watson lemma states that ifand the Laplace integral has suitable growth control away from zero, then

The Taylor series of the amplitude isTermwise application of Watson lemma therefore gives the asymptotic expansionIn particular, the leading asymptotic approximation is $I(x)\sim 2x^{-3}$.

A reliable transmission rate $R$ means that length-$n$ codes can carry $M_n$ messages with $n^{-1}{\log }_2{M}_n$ tending to at least $R$ while the probability measure of the decoding-error event tends to zero. By the Shannon second coding theorem, the supremum of reliable rates is the binary symmetric channel capacityFor small positive $p$, the binary entropy function has the asymptotic expansionso $C(p)$ starts at one bit per channel use and decreases rapidly from one as errors are introduced.

At $\theta =1$, the two saddles have receded to infinity and the ordinary quadratic saddle approximation is nonuniform. One should first use an asymptotic expansion of the phase and amplitude for large $z$:soThe phase varies on the scale $z=O(\lambda )$ rather than in an $O({\lambda }^{-1/2})$ neighbourhood of a finite saddle. Rescaling $z=\lambda Z$ in the positive and negative tails, retaining the corresponding large-$z$ amplitude, and matching these tail integrals to the finite part of the deformed contour produces the leading approximation. This is an endpoint at infinity problem; a uniform calculation may equivalently begin from the $\theta <1$ saddle representation and take the coalescing-at-infinity limit.

Complete the square: $z^2-2iz=(z-i{)}^2+1$. The pole of $(1-iz{)}^{-1}$ is at $z=-i$, so the contour may be shifted upward through the saddle point $z=i$ without crossing a singularity. Setting $z=t+i$ gives the exact steepest-descent representationNear the saddle,Odd terms integrate to zero, whileThe Watson lemma therefore yields the full asymptotic expansion

Write $\mu =GM$ and $s=\sqrt{b^2+r^2}$. The gravitational potential of the spherical isochrone model obeysThe spherical Poisson equation therefore givesIts small-radius asymptotic expansion isso the model has a finite-density core. At large radius,The $r^{-4}$ envelope has finite total mass $M$.

Since $y=x^{-1/2}u$, the two linearly independent solutions of the original equation have the asymptotic expansionsAs a check, the substitution $z=1/x$ converts the equation into the modified Bessel equation of order zero, whose solutions $I_0(z)$ and $K_0(z)$ have exactly these large-$z$ expansions.

For $t\to +\infty$ with $qE>0$, an asymptotic expansion givesso the longitudinal velocity tends to the speed of light. More generally, its limiting sign is the sign of $qE$.

For small $t$, the taylor series of the square root givesIn the nonrelativistic limit ${\mathcal{E}}_0\simeq m{c}^2$, this becomesThe transverse motion is then uniform:Eliminating $t$ gives the nonrelativistic parabolic trajectory

LetThe inequality $|1-e^{-x^2}|\le x^2$ and a split into $0<x<{\epsilon }^2$, ${\epsilon }^2<x<\epsilon$, and $\epsilon <x<1$ show thatThus replacing the exponential by one does not affect any term through $O(1)$.

FactorwherePartial fraction decomposition givesThe required asymptotic expansion follows fromThereforeThe two small roots reveal the same nested scales $x=O({\epsilon }^2)$ and $x=O(\epsilon )$ that a divide-and-conquer asymptotic expansion would match explicitly.