Laplace's Method for Exponential Integrals

Derivation

Laplace's Method for Exponential Integrals

D-192 Home PU-205 Threads chance · energy Depends on Watson's Lemma

Statement

Let \( I(\lambda)=\int_a^b g(t)\,e^{\lambda \varphi(t)}\,dt \) with \(\lambda\to+\infty\), where \(\varphi\) attains a strict interior maximum at a single point \(t_0\in(a,b)\) with \(\varphi'(t_0)=0\) and \(\varphi''(t_0)<0\), and \(g(t_0)\neq 0\). Then the leading asymptotics is \[ I(\lambda)\sim g(t_0)\,e^{\lambda\varphi(t_0)}\sqrt{\frac{2\pi}{\lambda\,\lvert\varphi''(t_0)\rvert}}\,,\qquad \lambda\to+\infty. \]

Why it matters

An enormous class of physical quantities is an integral of an exponentiated "action" or "free-energy" function weighted by a large parameter: the partition function \(Z=\int e^{-\beta E(t)}\,dt\) at low temperature, the density of states, reaction rates in Arrhenius/Kramers theory, the Stirling expansion of \(n!\), and the semiclassical (large-\(\lambda\)) limit of quantum amplitudes. Laplace's method says such an integral is controlled entirely by the immediate neighbourhood of the single point where the exponent is largest, and returns not just the exponential order but the leading prefactor.

It is the real-axis prototype for the more general method of steepest descent (stationary phase) used for oscillatory and complex integrals, and it makes precise the recurring physical statement that in a large-\(\lambda\) limit "the maximum dominates and fluctuations about it are Gaussian." Both threads meet here: the energy thread supplies the exponent \(\varphi\) as a free energy, and the chance thread reads \(\sqrt{2\pi/\lambda\lvert\varphi''\rvert}\) as the width of the Gaussian of fluctuations.

Assumptions

\(\varphi\) has a unique global maximum on \([a,b]\), attained at an interior point \(t_0\). If the maximum is attained at several points, each contributes additively and the amplitudes add; if it sits at an endpoint the Gaussian is cut in half (factor \(\tfrac12\)) and one instead expands to first order, giving a different \(\lambda^{-1}\) power.

\(\varphi'(t_0)=0\) and \(\varphi''(t_0)<0\) (a non-degenerate maximum). If \(\varphi''(t_0)=0\) the quadratic term vanishes and the width is set by the first non-zero derivative \(\varphi^{(2m)}\), changing the power of \(\lambda\) from \(\lambda^{-1/2}\) to \(\lambda^{-1/(2m)}\) and the constant to a Gamma-function.

\(g\) is continuous at \(t_0\) with \(g(t_0)\neq 0\), and \(g,\varphi\) are smooth enough near \(t_0\) (here \(C^2\), \(C^3\) for the correction). If \(g(t_0)=0\) the leading term vanishes and one must carry the expansion of \(g\) to its first non-vanishing order, lowering the overall power of \(\lambda\).

The integral converges and the tails are exponentially subdominant: \(\varphi(t)\le \varphi(t_0)-\delta\) outside any fixed neighbourhood of \(t_0\), for some \(\delta>0\). If \(\varphi\) approaches its supremum only asymptotically (e.g. on an infinite interval with \(\varphi\to\varphi(t_0)\) at infinity), the "localisation" step fails and the error is no longer exponentially small in \(\lambda\).

Derivation

\[ I(\lambda)=e^{\lambda\varphi(t_0)}\int_a^b g(t)\,e^{\lambda\,[\varphi(t)-\varphi(t_0)]}\,dt \]

Factor out the value at the maximum; \(\psi(t)\equiv\varphi(t)-\varphi(t_0)\le 0\) with equality only at \(t_0\). This isolates the exponentially large scale from an integral whose integrand is \(O(1)\) at \(t_0\) and decays elsewhere. A

\[ I(\lambda)=e^{\lambda\varphi(t_0)}\!\left(\int_{t_0-\eta}^{t_0+\eta}+\int_{[a,b]\setminus(t_0-\eta,t_0+\eta)}\right)g(t)\,e^{\lambda\psi(t)}\,dt \]

Split at a small but fixed \(\eta>0\). On the outer region \(\psi\le-\delta<0\), so that piece is bounded by \(\big(\int|g|\big)\,e^{-\lambda\delta}\), exponentially smaller than any power of \(\lambda\): the integral is localised at \(t_0\). B

\[ \psi(t)=\tfrac12\varphi''(t_0)\,(t-t_0)^2+O\big((t-t_0)^3\big),\qquad g(t)=g(t_0)+O(t-t_0) \]

Taylor-expand about \(t_0\) inside the inner window, using \(\varphi'(t_0)=0\). Since \(\eta\) is fixed and small, the cubic remainder is uniformly controlled on the window. Write \(\varphi''(t_0)=-\lvert\varphi''(t_0)\rvert\). B

\[ I(\lambda)\approx e^{\lambda\varphi(t_0)}\,g(t_0)\int_{t_0-\eta}^{t_0+\eta}\exp\!\Big(-\tfrac12\lambda\lvert\varphi''(t_0)\rvert\,(t-t_0)^2\Big)dt \]

Replace \(\psi\) by its quadratic and \(g\) by \(g(t_0)\) on the window. The neglected cubic multiplies the exponent by \(1+O(\lambda(t-t_0)^3)\); the Gaussian confines \(t-t_0\sim\lambda^{-1/2}\), so this correction is \(O(\lambda^{-1/2})\), i.e. relatively small. C

\[ u=\sqrt{\lambda\lvert\varphi''(t_0)\rvert}\,(t-t_0),\qquad dt=\frac{du}{\sqrt{\lambda\lvert\varphi''(t_0)\rvert}} \]

Rescale to the natural fluctuation variable. The window \(\lvert t-t_0\rvert<\eta\) maps to \(\lvert u\rvert<\eta\sqrt{\lambda\lvert\varphi''\rvert}\to\infty\) as \(\lambda\to\infty\), so the limits extend to \(\pm\infty\) with exponentially small error. B

\[ I(\lambda)\approx e^{\lambda\varphi(t_0)}\,\frac{g(t_0)}{\sqrt{\lambda\lvert\varphi''(t_0)\rvert}}\int_{-\infty}^{\infty}e^{-u^2/2}\,du \]

Substitute; the constant \(\big(\lambda\lvert\varphi''\rvert\big)^{-1/2}\) comes out of the integral. This is a rearrangement of symbols only — no numbers yet. A

\[ \int_{-\infty}^{\infty}e^{-u^2/2}\,du=\sqrt{2\pi}\quad\Longrightarrow\quad I(\lambda)\sim g(t_0)\,e^{\lambda\varphi(t_0)}\sqrt{\frac{2\pi}{\lambda\,\lvert\varphi''(t_0)\rvert}} \]

Evaluate the standard Gaussian integral. Watson's lemma applied to the same window and to the subleading Taylor terms guarantees this is genuinely the leading term of a full asymptotic series in powers of \(\lambda^{-1}\). A

\[ I(\lambda)= g(t_0)\,e^{\lambda\varphi(t_0)}\sqrt{\frac{2\pi}{\lambda\lvert\varphi''(t_0)\rvert}}\left[\,1+\frac{c_1}{\lambda}+O\!\left(\frac1{\lambda^2}\right)\right] \]

Carrying the cubic and quartic Taylor terms and the linear term of \(g\) through the Gaussian moments (Watson's lemma) yields the first correction with \(c_1=\dfrac{1}{\lvert\varphi''\rvert}\!\left[\dfrac{g''(t_0)}{2g(t_0)}-\dfrac{\varphi'''(t_0)\,g'(t_0)}{2g(t_0)\varphi''(t_0)}+\dfrac{5\varphi'''(t_0)^2}{24\varphi''(t_0)^2}-\dfrac{\varphi^{(4)}(t_0)}{8\varphi''(t_0)}\right]\). Odd Gaussian moments vanish, so only \(\lambda^{-1}\) corrections survive. C

Result

\[ \boxed{\,I(\lambda)=\int_a^b g(t)\,e^{\lambda\varphi(t)}\,dt\;\sim\;g(t_0)\,e^{\lambda\varphi(t_0)}\sqrt{\dfrac{2\pi}{\lambda\,\lvert\varphi''(t_0)\rvert}}\,}\]

Reading. The whole integral is set by the single dominant point \(t_0\): an exponential factor \(e^{\lambda\varphi(t_0)}\) fixing the order of magnitude, the value \(g(t_0)\) of the slowly-varying weight there, and a Gaussian "volume of fluctuations" \(\sqrt{2\pi/\lambda\lvert\varphi''\rvert}\) whose width \(\sim(\lambda\lvert\varphi''\rvert)^{-1/2}\) shrinks as \(\lambda\) grows or as the peak sharpens. Flatter peaks (small \(\lvert\varphi''\rvert\)) contribute more; sharper peaks contribute less.

Units check. Let \([t]=T\), \([\varphi]=\Phi\); then \(\lambda\varphi\) must be dimensionless, so \([\lambda]=\Phi^{-1}\). Now \([\varphi'']=\Phi\,T^{-2}\), hence \(\sqrt{2\pi/\lambda\lvert\varphi''\rvert}\) has units \(\sqrt{\Phi\cdot\Phi^{-1}T^{2}}=T\). Multiplying by \([g(t_0)]\) gives \([g]\cdot T\), which matches \([\,g\,dt\,]\) integrated. Dimensions balance.

Limiting cases

Sharp peak / large \(\lambda\): width \(\propto\lambda^{-1/2}\to 0\); the integral concentrates on \(t_0\) and \(g\) can be evaluated pointwise there — the "\(\delta\)-function" limit \(\lambda^{1/2}e^{-\lambda\varphi_0}I\to g(t_0)\sqrt{2\pi/\lvert\varphi''\rvert}\).
Endpoint maximum (\(\varphi'(a)\neq0\), \(\varphi\) decreasing): quadratic term absent, expand \(\psi\approx\varphi'(a)(t-a)\) and get \(I\sim -g(a)e^{\lambda\varphi(a)}/[\lambda\varphi'(a)]\), a \(\lambda^{-1}\) law, not \(\lambda^{-1/2}\).
Degenerate maximum (\(\varphi''=0\), \(\varphi^{(4)}<0\)): width scales as \(\lambda^{-1/4}\) and \(I\sim g(t_0)e^{\lambda\varphi_0}\,\Gamma(\tfrac14)\big(\tfrac{24}{\lambda\lvert\varphi^{(4)}\rvert}\big)^{1/4}/2\).
\(g(t_0)=0\): leading term vanishes; with \(g\approx\tfrac12 g''(t_0)(t-t_0)^2\) near \(t_0\) the result gains an extra factor \(\propto\lambda^{-1}\).
Two equal maxima \(t_0,t_1\): contributions add, \(I\sim e^{\lambda\varphi_0}\sum_i g(t_i)\sqrt{2\pi/\lambda\lvert\varphi''(t_i)\rvert}\).

Breaks when

The maximum is not isolated or moves with \(\lambda\). If \(\varphi\) has a flat plateau, or if \(g,\varphi\) depend on \(\lambda\) so that \(t_0\) drifts, the fixed-window localisation is invalid and the Gaussian width no longer scales as \(\lambda^{-1/2}\).
The exponent is imaginary (oscillatory), \(e^{i\lambda\varphi}\). Then \(\lvert e^{i\lambda\psi}\rvert=1\) everywhere, tails are not exponentially suppressed, and localisation fails; one needs stationary phase / steepest descent with complex contour deformation instead of real Laplace.
Non-uniform convergence at the boundary of validity. On an infinite interval where \(\varphi\to\varphi(t_0)\) only as \(t\to\infty\), or where \(g\) grows fast enough to fight the exponential decay, the outer integral is not \(o(e^{-\lambda\delta})\) and the estimate breaks.
Finite \(\lambda\) with slowly varying \(g\) that is not slowly varying. If \(g\) has structure on the scale \((\lambda\lvert\varphi''\rvert)^{-1/2}\) of the Gaussian (e.g. a narrow resonance near \(t_0\)), replacing \(g\) by \(g(t_0)\) is illegitimate.

Failure modes

Sign slip on the curvature: writing \(\sqrt{2\pi/\lambda\varphi''}\) instead of \(\lvert\varphi''\rvert\). At a maximum \(\varphi''<0\), so \(\varphi''\) itself gives an imaginary root — always take the absolute value.
Minimum instead of maximum: applying the formula at a point where \(\varphi''>0\). For \(e^{+\lambda\varphi}\) a minimum is exponentially suppressed, not dominant; for \(e^{-\lambda\varphi}\) it is the minimum that dominates.
Forgetting the \(\sqrt{\lambda}\): keeping only \(g(t_0)e^{\lambda\varphi_0}\) and dropping the prefactor, i.e. treating \(e^{\lambda\varphi}\) as an exact delta function of unit weight.
Expanding \(g\) but not re-Gaussianising: keeping the linear term \(g'(t_0)(t-t_0)\) and mistakenly getting a non-zero contribution — its Gaussian moment is odd and vanishes at leading order.
Endpoint treated as interior: using the \(\sqrt{2\pi}\) full-Gaussian normalisation when \(t_0\) coincides with \(a\) or \(b\), which double-counts the missing half-line.
Wrong sign of \(\lambda\to\infty\): applying the real Laplace result for \(\lambda\to-\infty\) or \(\lambda\) complex, where the dominant point changes or oscillation appears.

Discussion

Laplace's method is the mathematical face of a physical principle: when a large parameter multiplies an exponent, the system is overwhelmingly likely to be found near the exponent's maximum, and everything else is an exponentially rare fluctuation. In statistical mechanics with \(\lambda=\beta=1/k_BT\) (or the thermodynamic-limit parameter \(N\)), \(\varphi\) is minus a free energy: the partition function \(Z=\int e^{-N f(m)}\,dm\) is dominated by the equilibrium value \(m_0\) minimising \(f\), and the Gaussian width \((N f'')^{-1/2}\) is exactly the size of thermal fluctuations of the order parameter — the curvature \(f''\) is the inverse susceptibility. The method is thus the rigorous underpinning of the saddle-point (mean-field) evaluation of free energies.

The prefactor \(\sqrt{2\pi/\lambda\lvert\varphi''\rvert}\) carries real physics, not mere bookkeeping. In Kramers' rate theory the escape rate is a Laplace/steepest-descent ratio of two such factors — one at the barrier top, one at the well bottom — and the curvatures there set the attempt frequency. In the Stirling expansion \(n!=\int_0^\infty t^n e^{-t}dt\), writing the integrand as \(e^{n\ln t-t}\) and applying the method at \(t_0=n\) gives \(n!\sim\sqrt{2\pi n}\,(n/e)^n\) immediately, prefactor and all — a canonical demonstration that Laplace captures the constant \(\sqrt{2\pi}\) that naive "maximum-term" arguments miss.

The deeper structure is that Laplace's method is the leading order of a complete asymptotic (generally divergent, Borel-summable in good cases) expansion, and it is the real-line degeneration of the method of steepest descent on complex contours. For \(e^{i\lambda\varphi}\) the saddle points are the same stationary points, but the contour is deformed onto paths of steepest descent through them where the phase is stationary and the modulus decays; the same Gaussian bookkeeping yields \(\sqrt{2\pi/\lambda\lvert\varphi''\rvert}\,e^{\pm i\pi/4}\), the stationary-phase formula, and the \(e^{\pm i\pi/4}\) Maslov phase is the analytic continuation of the real Gaussian normalisation. This is precisely the WKB/semiclassical limit \(\hbar\to0\) of the Feynman path integral, where \(\lambda=1/\hbar\), \(\varphi=iS/\hbar\) is the classical action, \(t_0\) is the classical trajectory, and \(\varphi''\) is the Van Vleck determinant of second variations.

Common misconceptions. (i) "The maximum value of the integrand is the answer" — no; the answer is the maximum value times a fluctuation width, and the width contains the physics of the prefactor. (ii) "Laplace and stationary phase are different methods" — they are the same saddle-point idea, real vs. imaginary exponent, differing only in contour and in the \(e^{i\pi/4}\) phase. (iii) "It requires \(\lambda\) to be truly infinite" — it is an asymptotic statement; the relative error is \(O(1/\lambda)\), often excellent already for \(\lambda\gtrsim 5\text{--}10\).

Worked examples

\[ \text{Stirling: } n!=\Gamma(n+1)=\int_0^\infty t^{\,n}e^{-t}\,dt=\int_0^\infty e^{\,n\ln t - t}\,dt. \]

Write in Laplace form with large parameter \(n\): set \(t=n s\) so the exponent scales with \(n\). A

\[ t=ns:\quad n!=\int_0^\infty e^{\,n\ln(ns)-ns}\,n\,ds = n^{\,n+1}\!\int_0^\infty e^{\,n(\ln s - s)}\,ds,\quad \varphi(s)=\ln s - s. \]

Here \(g=n^{n+1}\) (constant in \(s\)) and \(\lambda=n\). A

\[ \varphi'(s)=\frac1s-1=0\Rightarrow s_0=1,\quad \varphi(1)=-1,\quad \varphi''(s)=-\frac1{s^2},\ \varphi''(1)=-1. \]

Non-degenerate interior maximum, \(\lvert\varphi''(s_0)\rvert=1\). A

\[ n!\sim n^{\,n+1}\,e^{-n}\sqrt{\frac{2\pi}{n\cdot 1}} = \sqrt{2\pi n}\,\Big(\frac n e\Big)^{n}. \]

Insert into the master formula with \(g(s_0)=n^{n+1}\), \(\varphi(s_0)=-1\). A

\[ n!\sim\sqrt{2\pi n}\left(\frac n e\right)^{n} \]

Check (numbers). For \(n=10\): \(\sqrt{20\pi}\approx 7.927\), \((10/e)^{10}=(3.6788)^{10}\approx 4.540\times10^{5}\), product \(\approx 3.599\times10^{6}\). True \(10!=3.6288\times10^{6}\); relative error \(\approx 0.83\%\), consistent with the \(1/(12n)\approx0.83\%\) first correction. Units: \(n!\) is dimensionless, as is the right side.

\[ \text{Low-}T\text{ partition function: } Z(\beta)=\int_{-\infty}^{\infty} e^{-\beta E(x)}\,dx,\quad E(x)=E_0+\tfrac12 k\,x^2+\alpha x^4, \]

Anharmonic well, large parameter \(\lambda=\beta=1/k_BT\to\infty\). Put \(\varphi(x)=-E(x)\) (maximum of \(\varphi\) = minimum of \(E\)). A

\[ \varphi'(x)=-(kx+4\alpha x^3),\quad x_0=0,\quad \varphi(x_0)=-E_0,\quad \varphi''(0)=-k. \]

The quartic does not shift the minimum (it is stationary at \(x=0\)); curvature at the bottom is \(-k\), so \(\lvert\varphi''\rvert=k\). B

\[ Z(\beta)\sim g(x_0)\,e^{\beta\varphi(x_0)}\sqrt{\frac{2\pi}{\beta k}} = e^{-\beta E_0}\sqrt{\frac{2\pi}{\beta k}},\qquad g\equiv 1. \]

Leading Laplace term; the \(\alpha x^4\) term enters only the \(O(1/\beta)\) correction via step 8's \(c_1\). B

\[ Z(\beta)= e^{-\beta E_0}\sqrt{\frac{2\pi}{\beta k}}\left[\,1-\frac{3\alpha}{\beta k^2}+O\!\Big(\frac1{\beta^2}\Big)\right],\quad c_1=-\frac{\varphi^{(4)}(0)}{8\varphi''(0)}=-\frac{-24\alpha}{8(-k)}=-\frac{3\alpha}{k^2}. \]

Using \(\varphi^{(4)}(0)=-24\alpha\), \(\varphi''(0)=-k\), \(g'=g''=0\) in the correction formula of step 8. C

\[ Z(\beta)\sim e^{-\beta E_0}\sqrt{\frac{2\pi}{\beta k}}\left(1-\frac{3\alpha}{\beta k^2}\right) \]

Numbers. Take \(k=2\ \mathrm{J/m^2}\), \(\alpha=0.05\ \mathrm{J/m^4}\), \(E_0=0\), \(T\) with \(\beta=50\ \mathrm{J^{-1}}\). Leading: \(\sqrt{2\pi/(50\cdot2)}=\sqrt{0.0628}=0.2507\ \mathrm{m}\). Correction \(3\alpha/(\beta k^2)=0.15/(50\cdot4)=7.5\times10^{-4}\), so \(Z\approx0.2507(1-0.00075)=0.2505\ \mathrm{m}\). Units: \([Z]=\sqrt{1/([\beta][k])}=\sqrt{\mathrm{J}\cdot\mathrm{m^2/J}}=\mathrm{m}\), matching \(\int dx\).

Problems

Evaluate \(\displaystyle\int_0^\pi e^{\lambda\cos t}\,dt\) as \(\lambda\to+\infty\) to leading order.
Solution
\(\varphi(t)=\cos t\) has its maximum on \([0,\pi]\) at the endpoint \(t_0=0\), where \(\varphi'(0)=-\sin0=0\) as well, so it is an interior-type (stationary) maximum sitting at the boundary. \(\varphi(0)=1\), \(\varphi''(0)=-\cos0=-1\). Because the peak is at an endpoint, only half the Gaussian lies in the range, giving a factor \(\tfrac12\): \[ \int_0^\pi e^{\lambda\cos t}dt\sim \tfrac12\,e^{\lambda}\sqrt{\frac{2\pi}{\lambda}} = e^{\lambda}\sqrt{\frac{\pi}{2\lambda}}. \] (This reproduces the large-argument asymptotics of \(\pi I_0(\lambda)=\int_0^\pi e^{\lambda\cos t}dt\), namely \(I_0(\lambda)\sim e^\lambda/\sqrt{2\pi\lambda}\).) For \(\lambda=20\): \(e^{20}\sqrt{\pi/40}=4.85\times10^{8}\times0.2802=1.36\times10^{8}\).
Find the leading asymptotics of \(\displaystyle\int_{-\infty}^{\infty} e^{-\lambda(x^2+x^4)}\,dx\) as \(\lambda\to+\infty\), and the first correction.
Solution
Maximum of \(\varphi=-(x^2+x^4)\) at \(x_0=0\); \(\varphi(0)=0\), \(\varphi''(0)=-2\Rightarrow\lvert\varphi''\rvert=2\). Leading: \(\sqrt{2\pi/(2\lambda)}=\sqrt{\pi/\lambda}\). Correction: \(\varphi^{(4)}(0)=-24\), so \(c_1=-\varphi^{(4)}/(8\varphi'')=-(-24)/(8\cdot(-2))=-3/2\); thus \[ \int_{-\infty}^\infty e^{-\lambda(x^2+x^4)}dx\sim\sqrt{\frac{\pi}{\lambda}}\left(1-\frac{3}{2\lambda}\right). \] For \(\lambda=10\): \(\sqrt{\pi/10}=0.5605\), factor \((1-0.15)=0.85\), estimate \(0.4764\); numerical integration gives \(\approx0.4816\) (2\% off, consistent with the truncated series).
Use Laplace's method to obtain the leading behaviour of \(\displaystyle\int_0^1 t^{2}\,e^{-\lambda t^2}\,dt\) as \(\lambda\to\infty\). Comment on why the naive formula must be modified.
Solution
The maximum of \(\varphi=-t^2\) is at the endpoint \(t_0=0\), where \(g(t)=t^2\) also vanishes: the standard interior formula gives \(0\), signalling a degenerate/endpoint case. Substitute \(u=\sqrt{\lambda}\,t\): \(\int_0^1 t^2 e^{-\lambda t^2}dt=\lambda^{-3/2}\int_0^{\sqrt\lambda}u^2e^{-u^2}du\to\lambda^{-3/2}\int_0^\infty u^2 e^{-u^2}du=\lambda^{-3/2}\cdot\tfrac{\sqrt\pi}{4}\). So \[ \int_0^1 t^2 e^{-\lambda t^2}dt\sim\frac{\sqrt\pi}{4}\,\lambda^{-3/2}. \] The power is \(\lambda^{-3/2}\), not \(\lambda^{-1/2}\), because \(g(t_0)=0\) (two extra powers of \(t\sim\lambda^{-1/2}\)) at an endpoint maximum. For \(\lambda=100\): \(\tfrac{\sqrt\pi}{4}\cdot10^{-3}=4.43\times10^{-4}\).
The exponent \(\varphi(t)=t-\tfrac13 t^3\) governs \(\displaystyle I(\lambda)=\int_0^\infty e^{\lambda(t-t^3/3)}\,dt\). Locate the dominant contribution and give \(I(\lambda)\) to leading order.
Solution
\(\varphi'(t)=1-t^2=0\Rightarrow t=1\) (the root in \((0,\infty)\)); \(\varphi(1)=1-\tfrac13=\tfrac23\), \(\varphi''(t)=-2t\), \(\varphi''(1)=-2\), \(\lvert\varphi''\rvert=2\). Interior non-degenerate maximum with \(g\equiv1\): \[ I(\lambda)\sim e^{2\lambda/3}\sqrt{\frac{2\pi}{2\lambda}}=e^{2\lambda/3}\sqrt{\frac{\pi}{\lambda}}. \] (As \(t\to\infty\), \(\varphi\to-\infty\), so tails are exponentially suppressed and localisation holds.) For \(\lambda=30\): \(e^{20}\sqrt{\pi/30}=4.85\times10^{8}\times0.3236=1.57\times10^{8}\).
A biased random walk has \(P(m)\propto e^{N h(m)}\) with \(h(m)=-\tfrac12(m-\mu)^2/\sigma^2\) plus a smooth weight \(g(m)=1+m^2\). Estimate the normalising integral \(\int_{-\infty}^\infty g(m)e^{Nh(m)}dm\) for large \(N\), and interpret the width.
Solution
\(h'(m)=-(m-\mu)/\sigma^2=0\Rightarrow m_0=\mu\); \(h(\mu)=0\), \(h''(\mu)=-1/\sigma^2\), \(\lvert h''\rvert=1/\sigma^2\). Weight at the peak \(g(\mu)=1+\mu^2\). Master formula with \(\lambda=N\): \[ \int_{-\infty}^\infty(1+m^2)e^{Nh(m)}dm\sim(1+\mu^2)\sqrt{\frac{2\pi}{N/\sigma^2}}=(1+\mu^2)\,\sigma\sqrt{\frac{2\pi}{N}}. \] The Gaussian width is \((N\lvert h''\rvert)^{-1/2}=\sigma/\sqrt N\): fluctuations of \(m\) about the mean \(\mu\) shrink as \(1/\sqrt N\), the central-limit scaling — the chance-thread reading of the curvature as an inverse variance. For \(N=400\), \(\sigma=2\), \(\mu=1\): \((1+1)\cdot2\sqrt{2\pi/400}=4\times0.1253=0.501\).