Watson’s lemma and Laplace’s method

Jordan Bell
June 17, 2014

1 Watson’s lemma

Our proof of Watson’s lemma follows Miller.11 1 Peter D. Miller, Applied Asymptotic Analysis, p. 53, Proposition 2.1.

Theorem 1 (Watson’s lemma).

Suppose that T>0, that ϕ:RC belongs to L1([0,T]), that σ>-1, and that g(t)=t-σϕ(t) is C on some neighborhood of 0. Then F:(0,)C defined by




Take g to be C on some interval with left endpoint <0 and right endpoint s, 0<s<T. For p a nonnegative integer and λ>1, define


which satisfies, doing the change of variable τ=λt,

Fp(λ) = 0e-λttσ+p𝑑t-se-λttσ+p𝑑t
= λ-(σ+p+1)0e-ττσ+p𝑑τ-se-λttσ+p𝑑t
= λ-(σ+p+1)Γ(σ+p+1)-se-λttσ+p𝑑t.

Using the Cauchy-Schwarz inequality,

se-λttσ+p𝑑t = se-λt/2e-λt/2tσ+p𝑑t
= e-λs/2λ-1/2(se-λtt2σ+2p𝑑t)1/2
< e-λs/2(0e-tt2σ+2p𝑑t)1/2
= e-λs/2Γ(2σ+2p+1)1/2.

For any nonnegative integer m we have e-λs/2=om(λ-(σ+m+1)) as λ, hence, dealing with Γ(2σ+2p+1) merely as a constant depending on p,

Fp(λ)=λ-(σ+p+1)Γ(σ+p+1)+om,p(λ-(σ+m+1)) (1)

as λ.



One the one hand,


which shows that for any nonnegative integer n,


as λ.

One the other hand, for each nonnegative integer N, Taylor’s theorem tells us that the function rN:(r,s) defined by




where the supremum is over those τ strictly between 0 and t. Then for t(0,s),


Using the definition of rN,

0se-λtϕ(t)𝑑t = 0se-λttσn=0Ng(n)(0)n!tndt+0se-λttσrN(t)𝑑t
= n=0Ng(n)(0)n!Fn(λ)+0se-λttσrN(t)𝑑t

and using the inequality for rN(t),

|0se-λttσrN(t)𝑑t| 0se-λttσ|rN(t)|𝑑t
= sup0<τ<s|g(N+1)(τ)|1(N+1)!FN+1(λ)dt.

Using this and (1),


Putting together what we have shown, for any nonnegative integer N, as λ,

F(λ) = n=0Ng(n)(0)n!Fn(λ)+ON(λ-(σ+N+2))+ON(λ-(σ+N+2))
= n=0Ng(n)(0)n!λ-(σ+n+1)Γ(σ+n+1)+n=0Ng(n)(0)n!oN,n(λ-(σ+N+1))
= n=0Ng(n)(0)n!λ-(σ+n+1)Γ(σ+n+1)+oN(λ-(σ+N+1)),

which proves the claim. ∎

2 Laplace’s method for an interval

Theorem 2.

Suppose that a<b, that fC2([a,b],R), and that there is a unique x0[a,b] at which f is equal to its supremum over [a,b]. Suppose also that a<x0<b and that f′′(x0)<0. Then


as M.


We remark first that f(x0)=0 because f is equal to its supremum over [a,b] at this point, which is not a boundary point. The claim says that a ratio has limit 1 as M. We shall prove that the liminf and the limsup of this ratio are both 1, which will prove the claim. Let ϵ>0. Because f′′:[a,b] is continuous, there is some δ>0 such that |x-x0|<δ implies f′′(x)f′′(x0)-ϵ; we take δ small enough that (x0-δ,x0+δ)[a,b]. Writing


Taylor’s theorem tells us that for each x[a,b] there is some ξx strictly between x0 and x such that


Thus for |x-x0|<δ we have |ξx-x0|<δ, so


Using this inequality, which applies for any x(x0-δ,x0+δ), and because the integrand in the following integral is positive,

abeMf(x)𝑑x x0-δx0+δeMf(x)𝑑x
= eMf(x0)x0-δx0+δe-M-f′′(x0)+ϵ2(x-x0)2𝑑x.

Changing variables, keeping in mind that f′′(x0)<0,



abeMf(x)𝑑xeMf(x0)(-Mf′′(x0)2)-1/2 (2)

is lower bounded by


so we get that the liminf of (2) as M is lower bounded by


But this is true for all ϵ>0 and (2) and its liminf do not depend on ϵ, so the liminf of (2) as M is lower bounded by π. In other words,

lim infMabeMf(x)𝑑xeMf(x0)(-Mf′′(x0)2π)-1/21.

Let ϵ>0 with f′′(x0)+ϵ<0; this is possible because f′′(x0)<0. Because f′′:[a,b] is continuous there is some δ>0 such that |x-x0|<δ implies that f′′(x)f′′(x0)+ϵ; we take (x0-δ,x0+δ)[a,b]. Taylor’s theorem tells us that for any x[a,b] there is some ξx strictly between x0 and x such that


Therefore, as |x-x0|<δ implies that |ξx-x0|<δ,

f(x)f(x0)+f′′(x0)+ϵ2(x-x0)2. (3)

Furthermore, f:[a,b] is continuous, so it makes sense to define


Because x0 is not in this union of intervals, by hypothesis we know that C<f(x0), and we define η=f(x0)-C>0. This means that for all x[a,x0-δ][x0+δ,b], f(x)f(x0)-η. Then

abeMf(x)𝑑x = ax0-δeMf(x)𝑑x+x0-δx0+δeMf(x)𝑑x+x0+δbeMf(x)𝑑x
= (b-a-2δ)eMC+x0-δx0+δeMf(x)𝑑x
< (b-a)eMC+x0-δx0+δeMf(x)𝑑x.

For the integral over (x0-δ,x0+δ),

x0-δx0+δeMf(x)𝑑x x0-δx0+δeM(f(x0)+f′′(x0)+ϵ2(x-x0)2)𝑑x
= eMf(x0)x0-δx0+δeMf′′(x0)+ϵ2(x-x0)2𝑑x
< eMf(x0)-eMf′′(x0)+ϵ2(x-x0)2𝑑x.

Changing variables, and keeping in mind that f′′(x0)+ϵ<0,

-eMf′′(x0)+ϵ2(x-x0)2𝑑x = -e-y2(-M2(f′′(x0)+ϵ))-1/2𝑑y
= (-M2π(f′′(x0)+ϵ))-1/2.



which we rearrange as


As M the first term on the right-hand side tends to 0, because η>0. Therefore,

lim supMabeMf(x)𝑑xeMf(x0)(-M2π(f′′(x0)+ϵ))-1/21.

This is true for all ϵ>0, so it holds that

lim supMabeMf(x)𝑑xeMf(x0)(-Mf′′(x0)2π)-1/21,

completing the proof. ∎