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Laplace’s formula : an approach by nonstandard analysis

OZAKI Ryushi
[email protected]
Abstract

Using nonstandard analysis (NSA), the proof of the Laplace’s formula is given. The usage of NSA reduces the intricacy of taking limit, and the crude line of the proof would be clearly seen, compared to the done with the rigorous classical calculus. We use very elementary tools of NSA.

Keywords Laplace’s formula  ⋅\cdot Laplace’s method  ⋅\cdot maximum term method  ⋅\cdot nonstandard analysis

1 Introduction

Laplace’s formula concerns the asymptotic behaviour of integral ∫abφ​(x)​exp⁡(n​h​(x))​𝑑x\displaystyle\int_{a}^{b}\varphi(x)\exp(n\,h(x))dx when nn goes to infinity, where hh takes maximum at only one point on the integral interval. Various application of the formula is well known. For example, if we take a=0,b=∞,φ​(x)≡1,a=0,\ b=\infty,\ \varphi(x)\equiv 1, and h​(x)=log⁡(x)−xh(x)=\log(x)-x, asymptotic behaviour of Γ​(n+1)/nn+1\Gamma(n+1)/n^{n+1} as n→∞n\to\infty, hence Stirling’s formula is obtained.

Before showing things correctly, let us make a rough observation. For the exact formulation and proof, see 2. Suppose h​(x)h(x) takes its only maximum at ξ0∈(a,b){\xi_{0}}\in(a,b). And now consider the integral

exp⁡(−n​h​(ξ0))​∫abφ​(x)​exp⁡(n​h​(x))​𝑑x=∫abφ​(x)​exp⁡[n​(h​(x)−h​(ξ0))]​𝑑x.\displaystyle\exp(-n\,h({\xi_{0}}))\int_{a}^{b}\varphi(x)\exp(n\,h(x))dx=\int_{a}^{b}\varphi(x)\exp\left[n(h(x)-h({\xi_{0}}))\right]dx. (1)

For some ε,α>0\varepsilon,\alpha>0 we have |x−ξ0|>ε⟹h​(x)−h​(ξ0)<−α|x-{\xi_{0}}|>\varepsilon\implies h(x)-h({\xi_{0}})<-\alpha, hence the term exp⁡[n​(h​(x)−h​(ξ0))]\exp\left[n(h(x)-h({\xi_{0}}))\right] decays as n→∞n\to\infty. So the major portion of the integral is

∫ξ0−εξ0+εφ​(x)​exp⁡[n​(h​(x)−h​(ξ0))]​𝑑x.\displaystyle\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\varphi(x)\exp[n\left(h(x)-h({\xi_{0}})\right)]\,dx.

For small ε\varepsilon, φ​(x)\varphi(x) can be approximated by φ​(ξ0)\varphi({\xi_{0}}) if φ\varphi is continuous. Hence we obtain

φ(ξ0)∫ξ0−εξ0+εexp[n(h(x)−h(ξ0)]dx.\displaystyle\varphi({\xi_{0}})\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\exp[n\left(h(x)-h({\xi_{0}})\right]\,dx.

Now suppose h′′​(ξ0)<0h^{\prime\prime}({\xi_{0}})<0. Then h​(x)−h​(ξ0)h(x)-h({\xi_{0}}) is well approximated by h′′​(ξ0)2​(x−ξ0)2\displaystyle\frac{h^{\prime\prime}({\xi_{0}})}{2}(x-{\xi_{0}})^{2}. Thus, by change of variable, we obtain

φ​(ξ0)​2−n​h′′​(ξ0)​∫|z|≤Rn,εe−z2​𝑑z,\displaystyle\varphi({\xi_{0}})\sqrt{\frac{2}{-n\,h^{\prime\prime}({\xi_{0}})}}\int_{|z|\leq R_{n,\varepsilon}}e^{-z^{2}}dz,

where

Rn,ε=ε​−n​h′′​(ξ0)2.\displaystyle R_{n,\varepsilon}=\varepsilon\sqrt{\frac{-n\,h^{\prime\prime}({\xi_{0}})}{2}}.

Taking nn so large to cancell the smallness of ε\varepsilon, Rn,εR_{n,\varepsilon} is effectively approximated by ∞\infty. Hence we obtain

φ​(ξ0)​2−n​h′′​(ξ0)​∫ℝe−z2​𝑑z≒φ​(ξ0)​2​π−n​h′′​(ξ0).\displaystyle\varphi({\xi_{0}})\sqrt{\frac{2}{-n\,h^{\prime\prime}({\xi_{0}})}}\int_{\mathbb{R}}e^{-z^{2}}dz\fallingdotseq\varphi({\xi_{0}})\sqrt{\frac{2\pi}{-n\,h^{\prime\prime}({\xi_{0}})}}.

Finally we obtain:

∫abφ​(x)​exp⁡(n​h​(x))​𝑑x∼φ​(ξ0)​exp⁡(n​h​(ξ0))​2​π−n​h′′​(ξ0)(n→∞).\displaystyle\int_{a}^{b}\varphi(x)\exp(n\,h(x))dx\sim\varphi({\xi_{0}})\exp(n\,h({\xi_{0}}))\sqrt{\frac{2\pi}{-n\,h^{\prime\prime}({\xi_{0}})}}\qquad(n\to\infty).

Of course, above explanation cannot be accepted as rigorous proof, though it captures the essense of the phenomenon. The most significant defects in the above explanation is (i) the arbitrariness of nn between infinite and some big natural and (ii) vagueness of the size of ε\varepsilon.

To forge the above explanation into rigorous proof in the classical analysis, one must pay the cost of taking intricate limit as done in [3], which is somewhat difficult to follow. However, nonstandard analysis (NSA) can afford the rigorous proof of Laplace’s formula along the line of the above explanation, without any complication. For in NSA, we can treat “infinitely large natural number” and “infinitely small real number”, though it may feel contradictory at first.

In this note, we prove the Laplace’s formula using NSA, along the line of the above explanation. We need only elementary tools of NSA and basic calculus; we do not need neither advanced understanding of NSA, intricate limit, nor Big O notation. In sec 2, we show the exact statement of the problem and the proof. In sec 3, we state a generalizaiton of the formula and show the proof of them.

In what follows, basic familiarity with Nelson’s IST (Internal Set Theory [1]) is supposed. However, the reader with some knowledge on (ordinary) nonstandard analysis would follow the argument with little effort.

2 Exact Statement and the proof

Theorem 1 (Laplace’s theorem).

Let [a,b]⊆ℝ[a,b]\subseteq\mathbb{R} be finite or infinite interval, φ​(x)\varphi(x) and h​(x)h(x) be functions defined on the interval, and ξ0∈(a,b){\xi_{0}}\in(a,b), satisfying following conditions:

  1. (C1)

    φ​(x)​exp⁡(n​h​(x))\varphi(x)\exp(n\,h(x)) is absolutely integrable over [a,b][a,b], for n=0,1,2,…n=0,1,2,....

  2. (C2)

    There is a neighbourhood of ξ0{\xi_{0}} where h′′​(x)h^{\prime\prime}(x) exists and is continuous and h′′​(ξ0)<0h^{\prime\prime}({\xi_{0}})<0.

  3. (C3)

    For some r1>0r_{1}>0 and for any ρ∈[0,r1)\rho\in[0,r_{1}) we have

    ξ0≥x≥ξ0−ρ≥y≥a⟹h​(x)≥h​(y).\displaystyle{\xi_{0}}\geq x\geq{\xi_{0}}-\rho\geq y\geq a\implies h(x)\geq h(y).
  4. (C4)

    For some r2>0r_{2}>0 and for any ρ∈[0,r2)\rho\in[0,r_{2}) we have

    ξ0≤x≤ξ0+ρ≤y≤b⟹h​(x)≥h​(y).\displaystyle{\xi_{0}}\leq x\leq{\xi_{0}}+\rho\leq y\leq b\implies h(x)\geq h(y).
  5. (C5)

    φ​(x)\varphi(x) is continuous at x=ξ0x={\xi_{0}} and φ​(ξ0)≠0\varphi({\xi_{0}})\not=0.

Then the following asymptotic formula holds as n→∞n\to\infty.

∫abφ​(x)​exp⁡(n​h​(x))​𝑑x∼φ​(ξ0)​exp⁡(n​h​(ξ0))​−2​πn​h′′​(ξ0).\displaystyle\int_{a}^{b}\varphi(x)\exp(n\,h(x))dx\sim\varphi({\xi_{0}})\exp(n\,h({\xi_{0}}))\sqrt{-\frac{2\pi}{n\,h^{\prime\prime}({\xi_{0}})}}.
Proof.

It suffices to see

limn→∞n​∫abφ​(x)​exp⁡[n​(h​(x)−h​(ξ0))]​𝑑x=φ​(ξ0)​−2​πh′′​(ξ0).\displaystyle\lim_{n\to\infty}\sqrt{n}\int_{a}^{b}\varphi(x)\exp\left[n(h(x)-h({\xi_{0}}))\right]dx\enskip=\enskip\varphi({\xi_{0}})\sqrt{-\frac{2\pi}{h^{\prime\prime}({\xi_{0}})}}. (2)

We assume a,b,φ​(x),h​(x)a,b,\varphi(x),h(x) and ξ0{\xi_{0}} are standard. Then, desired relation is now equivalent to:

∀ν∈ℕ∞​(ν​∫abφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃φ​(ξ0)​−2​πh′′​(ξ0)).\displaystyle\forall\nu\in\mathbb{N}_{\infty}\enskip\left(\sqrt{\nu}\int_{a}^{b}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\enskip\simeq\enskip\varphi({\xi_{0}})\sqrt{-\frac{2\pi}{h^{\prime\prime}({\xi_{0}})}}\>\right).

Now, take arbitrary ν∈ℕ∞\nu\in\mathbb{N}_{\infty} and let ε:=ν−1/6\varepsilon:=\nu^{-1/6}. Since ε\varepsilon is positive infinitesimal, h′′​(x)h^{\prime\prime}(x) exists and is less than 0 if |x−ξ0|<ε|x-{\xi_{0}}|<\varepsilon, by condition (C2). And since h′′​(ξ0)h^{\prime\prime}({\xi_{0}}) is standard negative number, more precise estimation is given:

|x−ξ0|≤ε⟹0<−1 2​h′′​(ξ0)≤−h′′​(x)≤−3 2​h′′​(ξ0).\displaystyle|x-{\xi_{0}}|\leq\varepsilon\implies 0<-\frac{1}{\,2\,}h^{\prime\prime}({\xi_{0}})\leq-h^{\prime\prime}(x)\leq-\frac{3}{\,2\,}h^{\prime\prime}({\xi_{0}}). (3)

We separate the integral into three parts:

ν​∫abφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x\displaystyle\sqrt{\nu}\int_{a}^{b}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx =ν​∫aξ0−εφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x\displaystyle=\sqrt{\nu}\int_{a}^{{\xi_{0}}-\varepsilon}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx
+ν​∫ξ0−εξ0+εφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x\displaystyle+\sqrt{\nu}\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx
+ν​∫ξ0+εbφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x.\displaystyle+\sqrt{\nu}\int_{{\xi_{0}}+\varepsilon}^{b}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx.

From (C3) and (C4) ξ0{\xi_{0}} is the local maximum point of h​(x)h(x), and h′​(ξ0)=0h^{\prime}({\xi_{0}})=0 by (C2).

Now, for any x∈[a,ξ0−ε]x\in[a,{\xi_{0}}-\varepsilon], we have

h​(ξ0)−h​(x)\displaystyle h({\xi_{0}})-h(x) =h​(ξ0)−h​(ξ0−ε)+h​(ξ0−ε)−h​(x)\displaystyle=h({\xi_{0}})-h({\xi_{0}}-\varepsilon)+h({\xi_{0}}-\varepsilon)-h(x)
≥h​(ξ0)−h​(ξ0−ε)=−∫ξ0−εξ0h′​(ξ0)⏟=0−h′​(t)​d​t\displaystyle\geq h({\xi_{0}})-h({\xi_{0}}-\varepsilon)=-\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}}\underbrace{h^{\prime}({\xi_{0}})}_{=0}-h^{\prime}(t)\,dt
=∫ξ0−εξ0(∫tξ0(−h′′​(u))​𝑑u)​𝑑t≥−h′′​(ξ0) 2​∫ξ0−εξ0(∫tξ0𝑑u)​𝑑t\displaystyle=\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}}\left(\int_{t}^{{\xi_{0}}}\bigl{(}-h^{\prime\prime}(u)\bigr{)}du\right)dt\geq-\frac{h^{\prime\prime}({\xi_{0}})}{\,2\,}\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}}\left(\int_{t}^{{\xi_{0}}}du\right)dt
=−h′′​(ξ0) 4​ε2=−h′′​(ξ0) 4​ν−1/3(>0),\displaystyle=-\frac{h^{\prime\prime}({\xi_{0}})}{\,4\,}\varepsilon^{2}=-\frac{h^{\prime\prime}({\xi_{0}})}{\,4\,}\nu^{-1/3}\enskip(>0),

by condition (C3) and (3). Thus, ν​(h​(x)−h​(ξ0))≤h′′​(ξ0)4​ν2/3(<0)\displaystyle\nu(h(x)-h({\xi_{0}}))\leq\frac{h^{\prime\prime}({\xi_{0}})}{4}\nu^{2/3}\>(<0) and, since ν2/3\nu^{2/3} is infinitely large number,

ν​exp⁡[ν​(h​(x)−h​(ξ0))]\displaystyle\sqrt{\nu}\enskip\exp\left[\nu(h(x)-h({\xi_{0}}))\right] ≤ν3/6​exp⁡(h′′​(ξ0)4​ν4/6)\displaystyle\leq\nu^{3/6}\,\exp\left(\frac{h^{\prime\prime}({\xi_{0}})}{4}\nu^{4/6}\right)
=ν−1/6​(ν4/6​exp⁡(h′′​(ξ0)4​ν4/6))\displaystyle=\nu^{-1/6}\left(\nu^{4/6}\exp\left(\frac{h^{\prime\prime}({\xi_{0}})}{4}\nu^{4/6}\right)\right)
≤C​ν−1/6,\displaystyle\leq C\,\nu^{-1/6}, (4)
by some positive constant CC. Hence
≃0.\displaystyle\simeq 0.

case 1: a>−∞a>-\infty In this case, we simply have

∫aξ0−εν​φ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ 0.\displaystyle\int_{a}^{{\xi_{0}}-\varepsilon}\sqrt{\nu}\enskip\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>0.

case 2: a=−∞a=-\infty  For any standard k=1,2,3,…k=1,2,3,... we have:

∫ξ0−kξ0−εν​φ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ 0.\displaystyle\int_{{\xi_{0}}-k}^{{\xi_{0}}-\varepsilon}\sqrt{\nu}\enskip\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>0.

By Robinson’s extension theorem, some κ∈ℕ∞\kappa\in\mathbb{N}_{\infty} exists and

∫ξ0−κξ0−εν​φ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ 0.\displaystyle\int_{{\xi_{0}}-\kappa}^{{\xi_{0}}-\varepsilon}\sqrt{\nu}\enskip\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>0.

From the integrability of |φ​(x)||\varphi(x)| over [−∞,b][-\infty,b] and (4) we have:

|∫−∞ξ0−κν​φ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x|≤C​ν−1/6​∫−∞ξ0−κ|φ​(x)|​𝑑x≃ 0.\displaystyle\left\lvert\int_{-\infty}^{{\xi_{0}}-\kappa}\sqrt{\nu}\enskip\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\right\rvert\leq C\,\nu^{-1/6}\int_{-\infty}^{{\xi_{0}}-\kappa}|\varphi(x)|dx\>\simeq\>0.

Hence, regardless of whether aa is −∞-\infty or not, we have

∫aξ0−εν​φ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ 0.\displaystyle\int_{a}^{{\xi_{0}}-\varepsilon}\sqrt{\nu}\enskip\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>0. (5)

By similar argument, we have:

∫ξ0+εbν​φ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ 0.\displaystyle\int_{{\xi_{0}}+\varepsilon}^{b}\sqrt{\nu}\enskip\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>0. (6)

From (5)(6) we have:

ν​∫abφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ν​∫ξ0−εξ0+εφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x.\displaystyle\sqrt{\nu}\int_{a}^{b}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>\sqrt{\nu}\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx. (7)

For any x∈[ξ0−ε,ξ0+ε]x\in[{\xi_{0}}-\varepsilon,{\xi_{0}}+\varepsilon], we set

p​(x):=h​(x)−h​(ξ0),q​(x):=h′′​(ξ0) 2​(x−ξ0)2.\displaystyle p(x):=h(x)-h({\xi_{0}}),\enskip q(x):=\frac{h^{\prime\prime}({\xi_{0}})}{\,2\,}(x-{\xi_{0}})^{2}.

Obviously p​(x),q​(x)≤0p(x),q(x)\leq 0. So

|exp⁡(ν​p​(x))−exp⁡(ν​q​(x))|≤1ν​|∫q​(x)p​(x)|exp⁡(ν​t)|​𝑑t|≤|p​(x)−q​(x)|ν.\displaystyle\left\lvert\exp(\nu\,p(x))-\exp(\nu\,q(x))\right\rvert\leq\frac{1}{\nu}\left\lvert\int_{q(x)}^{p(x)}|\exp(\nu\ t)|\,dt\right\rvert\leq\frac{|p(x)-q(x)|}{\nu}. (8)

Since

p​(x)\displaystyle p(x) =h​(x)−h​(ξ0)=∫ξ0xdd​t​(t−x)​h′​(t)​𝑑t=∫ξ0x(x−t)​h′′​(t)​𝑑t\displaystyle=h(x)-h({\xi_{0}})=\int_{{\xi_{0}}}^{x}\frac{d}{dt}(t-x)\,h^{\prime}(t)\,dt=\int_{{\xi_{0}}}^{x}(x-t)h^{\prime\prime}(t)\,dt

and

q​(x)\displaystyle q(x) =h′′​(ξ0) 2​(x−ξ0)2=∫ξ0x(x−t)​h′′​(ξ0)​𝑑t,\displaystyle=\frac{h^{\prime\prime}({\xi_{0}})}{\,2\,}(x-{\xi_{0}})^{2}=\int_{{\xi_{0}}}^{x}(x-t)h^{\prime\prime}({\xi_{0}})\,dt, (9)

we have

|p​(x)−q​(x)|\displaystyle|p(x)-q(x)| ≤|∫ξ0x|x−t|​|h′′​(t)−h′′​(ξ0)|​𝑑t|≤1 2​|h′′​(ξ0)|​|∫ξ0x|x−t|​𝑑t|\displaystyle\leq\left\lvert\int_{{\xi_{0}}}^{x}|x-t|\,|h^{\prime\prime}(t)-h^{\prime\prime}({\xi_{0}})|\,dt\right\rvert\leq\frac{1}{\,2\,}|h^{\prime\prime}({\xi_{0}})|\left\lvert\int_{{\xi_{0}}}^{x}|x-t|\,dt\right\rvert
=1 4​|h′′​(ξ0)|​|x−ξ0|2≤1 4​|h′′​(ξ0)|​ε2=1 4​|h′′​(ξ0)|​ν−1/3,\displaystyle=\frac{1}{\,4\,}|h^{\prime\prime}({\xi_{0}})|\,|x-{\xi_{0}}|^{2}\leq\frac{1}{\,4\,}|h^{\prime\prime}({\xi_{0}})|\varepsilon^{2}=\frac{1}{\,4\,}|h^{\prime\prime}({\xi_{0}})|\nu^{-1/3},

by the estimate (3). Substituting this to (8), we obtain

|exp⁡(ν​p​(x))−exp⁡(ν​q​(x))|≤1 4​|h′′​(ξ0)|​ν−8/6.\displaystyle\left\lvert\exp(\nu\,p(x))-\exp(\nu\,q(x))\right\rvert\leq\frac{1}{\,4\,}|h^{\prime\prime}({\xi_{0}})|\nu^{-8/6}.

Hence

ν​exp⁡(ν​p​(x))≃ν​exp⁡(ν​q​(x)).\displaystyle\sqrt{\nu}\>\exp(\nu\,p(x))\>\simeq\>\sqrt{\nu}\>\exp(\nu\,q(x)).

From the continuity of φ​(x)\varphi(x) at x=ξ0x={\xi_{0}}, we have φ​(x)≃φ​(ξ0)\varphi(x)\simeq\varphi({\xi_{0}}). Thus,

∫ξ0−εξ0+εν​φ​(x)​exp⁡(ν​p​(x))​𝑑x\displaystyle\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\sqrt{\nu}\>\varphi(x)\>\exp(\nu\>p(x))\,dx\enskip ≃∫ξ0−εξ0+εν​φ​(ξ0)​exp⁡(ν​q​(x))​𝑑x\displaystyle\simeq\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\sqrt{\nu}\>\varphi({\xi_{0}})\>\exp(\nu\>q(x))\,dx
=φ​(ξ0)​ν​∫ξ0−εξ0+εexp⁡[ν​(h′′​(ξ0)2​(x−ξ0)2)]​𝑑x\displaystyle=\varphi({\xi_{0}})\>\sqrt{\nu}\>\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\exp\left[\nu\left(\frac{h^{\prime\prime}({\xi_{0}})}{2}(x-{\xi_{0}})^{2}\right)\right]\,dx
By letting z:=(x−ξ0)​−ν​h′′​(ξ0)2\displaystyle z:=(x-{\xi_{0}})\sqrt{-\nu\frac{h^{\prime\prime}({\xi_{0}})}{2}},
=φ​(ξ0)​ν​−2ν​h′′​(ξ0)​∫|z|≤Re−z2​𝑑z\displaystyle=\varphi({\xi_{0}})\>\sqrt{\nu}\>\sqrt{-\frac{2}{\nu\>h^{\prime\prime}({\xi_{0}})}}\int_{|z|\leq R}e^{-z^{2}}\,dz
=φ​(ξ0)​−2h′′​(ξ0)​∫|z|≤Re−z2​𝑑z,\displaystyle=\varphi({\xi_{0}})\sqrt{-\frac{2}{h^{\prime\prime}({\xi_{0}})}}\int_{|z|\leq R}e^{-z^{2}}\,dz, (*)

where

R:=ε​ν​−h′′​(ξ0)2=ν1/3​−h′′​(ξ0)2.\displaystyle R:=\varepsilon\sqrt{\nu\frac{-h^{\prime\prime}({\xi_{0}})}{2}}=\nu^{1/3}\sqrt{-\frac{h^{\prime\prime}({\xi_{0}})}{2}}.

Since RR is infinitely large, the integral in (* ‣ 2) is infinitely close to ∫−∞∞e−z2​𝑑z=π\displaystyle\int_{-\infty}^{\infty}e^{-z^{2}}\,dz=\sqrt{\pi}. Therefore,

(​* ‣ 2​)≃φ​(ξ0)​−2h′′​(ξ0)​π=φ​(ξ)​−2​πh′′​(ξ0).\displaystyle\eqref{ex:007}\simeq\varphi({\xi_{0}})\sqrt{-\frac{2}{h^{\prime\prime}({\xi_{0}})}}\sqrt{\pi}=\varphi(\xi)\sqrt{-\frac{2\pi}{h^{\prime\prime}({\xi_{0}})}}.

By (7) and above estimation, we obtain

ν​∫abφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃φ​(ξ0)​−2​πh′′​(ξ0).\displaystyle\sqrt{\nu}\int_{a}^{b}\varphi(x)\>\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>\varphi({\xi_{0}})\sqrt{-\frac{2\pi}{h^{\prime\prime}({\xi_{0}})}}.

Since ν\nu was arbitrary, (2) is shown for any standard a,b,φ​(x),h​(x)a,b,\varphi(x),h(x) and ξ0{\xi_{0}}. By transfer, (2) is true for any (possibly nonstandard) a,b,φ​(x),h​(x)a,b,\varphi(x),h(x) and ξ0{\xi_{0}}. This completes the proof. ∎

3 A Generalization

The approach used in proving Theorem 1 can be extended to show the genelalized formula.

Theorem 2 (generalized Laplace’s theorem).

Let [a,b]⊆ℝ[a,b]\subseteq\mathbb{R} be finite or infinite interval, φ​(x)\varphi(x) and h​(x)h(x) be functions defined on the interval, and ξ0∈(a,b){\xi_{0}}\in(a,b). And let mm be the natural number ≥1\geq 1. Suppose these data satisfies (C1)(C3)(C4)(C5) of Theorem 1, and (C2’) of the following:

  1. (C2’)

    h(k)​(ξ0)=0​(1≤k<2​m),h(2​m)​(ξ0)<0.h^{(k)}({\xi_{0}})=0\enskip(1\leq k<2m),\enskip h^{(2m)}({\xi_{0}})<0.

Then the following asymptotic formula holds as n→∞n\to\infty.

∫abφ​(x)​exp⁡(n​h​(x))​𝑑x∼φ​(ξ0)​exp⁡(n​h​(ξ0))​Γ​(12​m)m​(−(2​m)!n​h(2​m)​(ξ0))12​m.\displaystyle\int_{a}^{b}\varphi(x)\exp(n\,h(x))dx\sim\varphi({\xi_{0}})\exp(n\,h({\xi_{0}}))\frac{\Gamma\left(\frac{1}{2m}\right)}{m}\left(-\frac{(2m)!}{n\,h^{(2m)}({\xi_{0}})}\right)^{\frac{1}{2m}}.
Proof.

Since the following is paralell to the proof of Theorem1, sometimes we omit the detail. We assume a,b,φ​(x),h​(x)a,b,\varphi(x),h(x) and ξ0{\xi_{0}} are standard. So, desired relation is now equivalent to :

∀ν∈ℕ∞​(ν12​m​∫abφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃φ​(ξ0)​Γ​(12​m)m​(−(2​m)!h(2​m)​(ξ0))12​m).\displaystyle\forall\nu\in\mathbb{N}_{\infty}\left(\nu^{\frac{1}{2m}}\int_{a}^{b}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\enskip\simeq\enskip\varphi({\xi_{0}})\frac{\Gamma\left(\frac{1}{2m}\right)}{m}\left(-\frac{(2m)!}{h^{(2m)}({\xi_{0}})}\right)^{\frac{1}{2m}}\right). (10)

Take arbitrary ν∈ℕ∞\nu\in\mathbb{N}_{\infty} and let ε:=ν−1/6​m2\varepsilon:=\nu^{-1/6m^{2}}. Since h(2​m)​(ξ0)h^{(2m)}({\xi_{0}}) is standard negative number, following estimation holds:

|x−ξ0|≤ε⟹0<−1 2​h(2​m)​(ξ0)≤−h(2​m)​(x)≤−3 2​h(2​m)​(ξ0).\displaystyle|x-{\xi_{0}}|\leq\varepsilon\implies 0<-\frac{1}{\,2\,}h^{(2m)}({\xi_{0}})\leq-h^{(2m)}(x)\leq-\frac{3}{\,2\,}h^{(2m)}({\xi_{0}}). (11)

As before, we separate the integral into three parts:

ν12​m​∫abφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x\displaystyle\nu^{\frac{1}{2m}}\int_{a}^{b}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx =ν12​m​∫aξ0−εφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x\displaystyle=\nu^{\frac{1}{2m}}\int_{a}^{{\xi_{0}}-\varepsilon}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx
+ν12​m​∫ξ0−εξ0+εφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x\displaystyle+\nu^{\frac{1}{2m}}\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx
+ν12​m​∫ξ0+εbφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x.\displaystyle+\nu^{\frac{1}{2m}}\int_{{\xi_{0}}+\varepsilon}^{b}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx.

Now, for any x∈[a,ξ0−ε]x\in[a,{\xi_{0}}-\varepsilon], we have

h​(ξ0)−h​(x)\displaystyle h({\xi_{0}})-h(x) ≥h​(ξ0)−h​(ξ0−ε)=−∫ξ0−εξ0h′​(ξ0)⏟=0−h′​(t1)​d​t1\displaystyle\geq h({\xi_{0}})-h({\xi_{0}}-\varepsilon)=-\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}}\underbrace{h^{\prime}({\xi_{0}})}_{=0}-h^{\prime}(t_{1})\,dt_{1}
=−∫ξ0−εξ0(∫t1ξ0(h′′​(ξ0)⏟=0−h′′​(t2))​dt2)​𝑑t1\displaystyle=-\int_{{\xi_{0}}-\varepsilon}^{\xi_{0}}\left(\int_{t_{1}}^{\xi_{0}}\left(\underbrace{h^{\prime\prime}({\xi_{0}})}_{=0}-h^{\prime\prime}(t_{2})\right)d_{t_{2}}\right)dt_{1}
=−∫ξ0−εξ0(∫t1ξ0(∫t2ξ0(h(3)​(ξ0)⏟=0−h(3)​(t3))​dt3)​dt2)​dt1\displaystyle=-\int_{{\xi_{0}}-\varepsilon}^{\xi_{0}}\left(\int_{t_{1}}^{\xi_{0}}\left(\int_{t_{2}}^{\xi_{0}}\left(\underbrace{h^{(3)}({\xi_{0}})}_{=0}-h^{(3)}(t_{3})\right)d_{t_{3}}\right)d_{t_{2}}\right)d_{t_{1}}
=…=−∫ξ0−εξ0(∫t1ξ0(⋯​∫t2​m−2ξ0(∫t2​m−1ξ0h(2​m)​(t2​m)​dt2​m)​dt2​m+1​⋯)​dt2)​dt1\displaystyle=...=-\int_{{\xi_{0}}-\varepsilon}^{\xi_{0}}\left(\int_{t_{1}}^{\xi_{0}}\left(\cdots\int_{t_{2m-2}}^{\xi_{0}}\left(\int_{t_{2m-1}}^{\xi_{0}}h^{(2m)}(t_{2m})\>d_{t_{2m}}\right)d_{t_{2m+1}}\cdots\right)d_{t_{2}}\right)d_{t_{1}}
≥−h(2​m)​(ξ0)2​∫ξ0−εξ0(∫t1ξ0(⋯​∫t2​m−2ξ0(∫t2​m−1ξ0dt2​m)​dt2​m+1​⋯)​dt2)​dt1\displaystyle\geq-\frac{h^{(2m)}({\xi_{0}})}{2}\int_{{\xi_{0}}-\varepsilon}^{\xi_{0}}\left(\int_{t_{1}}^{\xi_{0}}\left(\cdots\int_{t_{2m-2}}^{\xi_{0}}\left(\int_{t_{2m-1}}^{\xi_{0}}\>d_{t_{2m}}\right)d_{t_{2m+1}}\cdots\right)d_{t_{2}}\right)d_{t_{1}}
=−h(2​m)​(ξ0)2​1(2​m)!​ε2​m=−h(2​m)​(ξ0)2​1(2​m)!​ν−13​m,\displaystyle=-\frac{h^{(2m)}({\xi_{0}})}{2}\frac{1}{(2m)!}\varepsilon^{2m}=-\frac{h^{(2m)}({\xi_{0}})}{2}\frac{1}{(2m)!}\nu^{-\frac{1}{3m}},

by condition (C3) and (11). Thus, ν​(h​(x)−h​(ξ0))≤h(2​m)​(ξ0)2​(2​m)!​ν1−13​m(<0)\displaystyle\nu(h(x)-h({\xi_{0}}))\leq\frac{h^{(2m)}({\xi_{0}})}{2(2m)!}\nu^{1-\frac{1}{3m}}\>(<0) and, since ν1−13​m\nu^{1-\frac{1}{3m}} is infinitely large number,

ν12​m​exp⁡[ν​(h​(x)−h​(ξ0))]\displaystyle\nu^{\frac{1}{2m}}\enskip\exp\left[\nu(h(x)-h({\xi_{0}}))\right] ≤ν26​m​exp⁡(h(2​m)​(ξ0)2​(2​m)!​ν1−26​m)\displaystyle\leq\nu^{\frac{2}{6m}}\,\exp\left(\frac{h^{(2m)}({\xi_{0}})}{2(2m)!}\nu^{1-\frac{2}{6m}}\right)
=ν56​m−1​(ν1−26​m​exp⁡(h(2​m)​(ξ0)2​(2​m)!​ν1−26​m))\displaystyle=\nu^{\frac{5}{6m}-1}\left(\nu^{1-\frac{2}{6m}}\exp\left(\frac{h^{(2m)}({\xi_{0}})}{2(2m)!}\nu^{1-\frac{2}{6m}}\right)\right)
≤C​ν56​m−1,\displaystyle\leq C\,\nu^{\frac{5}{6m}-1}, (12)
by some positive constant CC. Hence
≃0.\displaystyle\simeq 0.

By similar argument in the proof of Theorem1, we obtain

∫aξ0−εν12​m​φ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ 0\displaystyle\int_{a}^{{\xi_{0}}-\varepsilon}\nu^{\frac{1}{2m}}\enskip\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>0 (13)
and
∫ξ0+εbν12​m​φ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ 0.\displaystyle\int_{{\xi_{0}}+\varepsilon}^{b}\nu^{\frac{1}{2m}}\enskip\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>0. (14)

From (13)(14) we have:

ν12​m​∫abφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃ν12​m​∫ξ0−εξ0+εφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x.\displaystyle\nu^{\frac{1}{2m}}\int_{a}^{b}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>\nu^{\frac{1}{2m}}\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\varphi(x)\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx. (15)

For any x∈[ξ0−ε,ξ0+ε]x\in[{\xi_{0}}-\varepsilon,{\xi_{0}}+\varepsilon], we set

p​(x):=h​(x)−h​(ξ0),q​(x):=h(2​m)​(ξ0)(2​m)!​(x−ξ0)2​m.\displaystyle p(x):=h(x)-h({\xi_{0}}),\enskip q(x):=\frac{h^{(2m)}({\xi_{0}})}{(2m)!}(x-{\xi_{0}})^{2m}.

Obviously p​(x),q​(x)≤0p(x),q(x)\leq 0. So again we have

|exp⁡(ν​p​(x))−exp⁡(ν​q​(x))|≤1ν​|∫q​(x)p​(x)|exp⁡(ν​t)|​𝑑t|≤|p​(x)−q​(x)|ν.\displaystyle\left\lvert\exp(\nu\,p(x))-\exp(\nu\,q(x))\right\rvert\leq\frac{1}{\nu}\left\lvert\int_{q(x)}^{p(x)}|\exp(\nu\ t)|\,dt\right\rvert\leq\frac{|p(x)-q(x)|}{\nu}. (16)

As is easily verified,

p​(x)\displaystyle p(x) =1(2​m−1)!​∫ξ0x(x−t)2​m−1​h(2​m)​(t)​𝑑t\displaystyle=\frac{1}{(2m-1)!}\int_{{\xi_{0}}}^{x}(x-t)^{2m-1}h^{(2m)}(t)dt

and

q​(x)\displaystyle q(x) =1(2​m−1)!​∫ξ0x(x−t)2​m−1​h(2​m)​(ξ0)​𝑑t.\displaystyle=\frac{1}{(2m-1)!}\int_{{\xi_{0}}}^{x}(x-t)^{2m-1}h^{(2m)}({\xi_{0}})\,dt. (17)

Thus we have

|p​(x)−q​(x)|\displaystyle|p(x)-q(x)| ≤1(2​m−1)!​|∫ξ0x|x−t|2​m−1​|h(2​m)​(t)−h(2​m)​(ξ0)|​𝑑t|\displaystyle\leq\frac{1}{(2m-1)!}\left\lvert\int_{{\xi_{0}}}^{x}|x-t|^{2m-1}\,\left\lvert h^{(2m)}(t)-h^{(2m)}({\xi_{0}})\right\rvert\,dt\right\rvert
≤12​(2​m−1)!​|h(2​m)​(ξ0)|​|∫ξ0x|x−t|2​m−1​𝑑t|\displaystyle\leq\frac{1}{2(2m-1)!}\left\lvert h^{(2m)}({\xi_{0}})\right\rvert\left\lvert\int_{{\xi_{0}}}^{x}|x-t|^{2m-1}\,dt\right\rvert
=12​(2​m)!​|h(2​m)​(ξ0)|​|x−ξ0|2​m≤12​(2​m)!​|h(2​m)​(ξ0)|​ε2​m=12​(2​m)!​|h(2​m)​(ξ0)|​ν−13​m,\displaystyle=\frac{1}{2(2m)!}\left\lvert h^{(2m)}({\xi_{0}})\right\rvert\,|x-{\xi_{0}}|^{2m}\leq\frac{1}{2(2m)!}\left\lvert h^{(2m)}({\xi_{0}})\right\rvert\varepsilon^{2m}=\frac{1}{2(2m)!}\left\lvert h^{(2m)}({\xi_{0}})\right\rvert\nu^{-\frac{1}{3m}},

by the estimate (11). Substituting this to (16), we obtain

|exp⁡(ν​p​(x))−exp⁡(ν​q​(x))|≤12​(2​m)!​|h(2​m)​(ξ0)|​ν−1−26​m.\displaystyle\left\lvert\exp(\nu\,p(x))-\exp(\nu\,q(x))\right\rvert\leq\frac{1}{2(2m)!}\left\lvert h^{(2m)}({\xi_{0}})\right\rvert\nu^{-1-\frac{2}{6m}}.

Hence

ν12​m​exp⁡(ν​p​(x))≃ν12​m​exp⁡(ν​q​(x)).\displaystyle\nu^{\frac{1}{2m}}\>\exp(\nu\,p(x))\>\simeq\>\nu^{\frac{1}{2m}}\>\exp(\nu\,q(x)).

From the continuity of φ​(x)\varphi(x) at x=ξ0x={\xi_{0}}, we have φ​(x)≃φ​(ξ0)\varphi(x)\simeq\varphi({\xi_{0}}). Thus:

∫ξ0−εξ0+εν12​m​φ​(x)​exp⁡(ν​p​(x))​𝑑x\displaystyle\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\nu^{\frac{1}{2m}}\>\varphi(x)\>\exp(\nu\>p(x))\,dx\enskip ≃∫ξ0−εξ0+εν12​m​φ​(ξ0)​exp⁡(ν​q​(x))​𝑑x\displaystyle\simeq\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\nu^{\frac{1}{2m}}\>\varphi({\xi_{0}})\>\exp(\nu\>q(x))\,dx
=φ​(ξ0)​ν12​m​∫ξ0−εξ0+εexp⁡[ν​(h(2​m)​(ξ0)(2m)!)​(x−ξ0)2​m)]​𝑑x\displaystyle=\varphi({\xi_{0}})\>\nu^{\frac{1}{2m}}\>\int_{{\xi_{0}}-\varepsilon}^{{\xi_{0}}+\varepsilon}\exp\left[\nu\left(\frac{h^{(2m)}({\xi_{0}})}{(2m)!)}(x-{\xi_{0}})^{2m}\right)\right]\,dx
By letting z:=(x−ξ0)​(−ν​h(2​m)​(ξ0)(2​m)!)12​m\displaystyle z:=(x-{\xi_{0}})\left(-\nu\frac{h^{(2m)}({\xi_{0}})}{(2m)!}\right)^{\frac{1}{2m}},
=φ​(ξ0)​ν12​m​(ν​−h(2​m)​(ξ0)(2​m)!)−12​m​∫|z|≤R′e−z2​m​𝑑z\displaystyle=\varphi({\xi_{0}})\>\nu^{\frac{1}{2m}}\>\left(\nu\frac{-h^{(2m)}({\xi_{0}})}{(2m)!}\right)^{-\frac{1}{2m}}\int_{|z|\leq R^{\prime}}e^{-z^{2m}}\,dz
=φ​(ξ0)​(−h(2​m)​(ξ0)(2​m)!)−12​m​∫|z|≤R′e−z2​m​𝑑z,\displaystyle=\varphi({\xi_{0}})\>\left(\frac{-h^{(2m)}({\xi_{0}})}{(2m)!}\right)^{-\frac{1}{2m}}\int_{|z|\leq R^{\prime}}e^{-z^{2m}}\,dz, (**)

where

R′:=ε​(ν​−h(2​m)​(ξ0)(2​m)!)12​m=ν−16​m2+3​m6​m2​(−h(2​m)​(ξ0)(2m)!))12​m.\displaystyle R^{\prime}:=\varepsilon\left(\nu\frac{-h^{(2m)}({\xi_{0}})}{(2m)!}\right)^{\frac{1}{2m}}=\nu^{-\frac{1}{6m^{2}}+\frac{3m}{6m^{2}}}\left(-\frac{h^{(2m)}({\xi_{0}})}{(2m)!)}\right)^{\frac{1}{2m}}.

Since R′R^{\prime} is infinitely large, the integral in (** ‣ 3) is infinitely close to ∫−∞∞e−z2​m​𝑑z=1m​Γ​(12​m)\displaystyle\int_{-\infty}^{\infty}e^{-z^{2m}}\,dz=\frac{1}{m}\Gamma\left(\frac{1}{2m}\right). Therefore,

(​** ‣ 3​)≃φ​(ξ0)​Γ​(12​m)m​(−(2​m)!h(2​m)​(ξ0))12​m.\displaystyle\eqref{ex:007'}\simeq\varphi({\xi_{0}})\frac{\Gamma\left(\frac{1}{2m}\right)}{m}\left(-\frac{(2m)!}{h^{(2m)}({\xi_{0}})}\right)^{\frac{1}{2m}}.

By (15) and above estimation, we obtain

ν​∫abφ​(x)​exp⁡[ν​(h​(x)−h​(ξ0))]​𝑑x≃φ​(ξ0)​Γ​(12​m)m​(−(2​m)!h(2​m)​(ξ0))12​m.\displaystyle\sqrt{\nu}\int_{a}^{b}\varphi(x)\>\exp\left[\nu(h(x)-h({\xi_{0}}))\right]dx\>\simeq\>\varphi({\xi_{0}})\frac{\Gamma\left(\frac{1}{2m}\right)}{m}\left(-\frac{(2m)!}{h^{(2m)}({\xi_{0}})}\right)^{\frac{1}{2m}}.

Since ν\nu was arbitrary, (10) is shown for any standard a,b,φ​(x),h​(x)a,b,\varphi(x),h(x) and ξ0{\xi_{0}}. By transfer, (10) is true for any (possibly nonstandard) a,b,φ​(x),h​(x)a,b,\varphi(x),h(x) and ξ0{\xi_{0}}. This completes the proof. ∎

4 Note

The statement of the theorem here is taken and modified from the problem 201 of part two of [3]. To be more specific, the condition (C3) and (C4) are added to the premise. On the other hand, one condition is dropped from the original statement. It seems conditions (C3) and (C4) hold for typical cases of application of the formula.

Laplace’s theorem is already proved by nonstandard method in [2], based on wider point of view. The author of [2] uses general technique dealing with external numbers to show the theorem.

References

  • [1] Edward Nelson. Internal Set Theory: A new approach to nonstandard analysis. In Bull. Amer. Math. Soc. vol 83(6), pp.1165-1198,  1977.
  • [2] Imme van den Berg. Nonstandard Asymptotic Analysis. Springer,  1980.
  • [3] George Pólya and Gabor Szegö. Problems and Theorems in Analysis, Vol.1 Springer,  1970.