# 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 $\phi:\mathbb{R}\to\mathbb{C}$ belongs to $L^{1}([0,T])$, that $\sigma>-1$, and that $g(t)=t^{-\sigma}\phi(t)$ is $C^{\infty}$ on some neighborhood of $0$. Then $F:(0,\infty)\to\mathbb{C}$ defined by

 $F(\lambda)=\int_{0}^{T}e^{-\lambda t}\phi(t)dt$

satisfies

 $F(\lambda)\sim\sum_{n=0}^{\infty}\frac{g^{(n)}(0)\Gamma(\sigma+n+1)}{n!\lambda% ^{\sigma+n+1}},\qquad\lambda\to\infty.$
###### Proof.

Take $g$ to be $C^{\infty}$ on some interval with left endpoint $<0$ and right endpoint $s$, $0. For $p$ a nonnegative integer and $\lambda>1$, define

 $F_{p}(\lambda)=\int_{0}^{s}e^{-\lambda t}t^{\sigma+p}dt,$

which satisfies, doing the change of variable $\tau=\lambda t$,

 $\displaystyle F_{p}(\lambda)$ $\displaystyle=$ $\displaystyle\int_{0}^{\infty}e^{-\lambda t}t^{\sigma+p}dt-\int_{s}^{\infty}e^% {-\lambda t}t^{\sigma+p}dt$ $\displaystyle=$ $\displaystyle\lambda^{-(\sigma+p+1)}\int_{0}^{\infty}e^{-\tau}\tau^{\sigma+p}d% \tau-\int_{s}^{\infty}e^{-\lambda t}t^{\sigma+p}dt$ $\displaystyle=$ $\displaystyle\lambda^{-(\sigma+p+1)}\Gamma(\sigma+p+1)-\int_{s}^{\infty}e^{-% \lambda t}t^{\sigma+p}dt.$

Using the Cauchy-Schwarz inequality,

 $\displaystyle\int_{s}^{\infty}e^{-\lambda t}t^{\sigma+p}dt$ $\displaystyle=$ $\displaystyle\int_{s}^{\infty}e^{-\lambda t/2}e^{-\lambda t/2}t^{\sigma+p}dt$ $\displaystyle\leq$ $\displaystyle\left(\int_{s}^{\infty}e^{-\lambda t}dt\right)^{1/2}\left(\int_{s% }^{\infty}e^{-\lambda t}t^{2\sigma+2p}dt\right)^{1/2}$ $\displaystyle=$ $\displaystyle e^{-\lambda s/2}\lambda^{-1/2}\left(\int_{s}^{\infty}e^{-\lambda t% }t^{2\sigma+2p}dt\right)^{1/2}$ $\displaystyle<$ $\displaystyle e^{-\lambda s/2}\left(\int_{0}^{\infty}e^{-t}t^{2\sigma+2p}dt% \right)^{1/2}$ $\displaystyle=$ $\displaystyle e^{-\lambda s/2}\Gamma(2\sigma+2p+1)^{1/2}.$

For any nonnegative integer $m$ we have $e^{-\lambda s/2}=o_{m}(\lambda^{-(\sigma+m+1)})$ as $\lambda\to\infty$, hence, dealing with $\Gamma(2\sigma+2p+1)$ merely as a constant depending on $p$,

 $F_{p}(\lambda)=\lambda^{-(\sigma+p+1)}\Gamma(\sigma+p+1)+o_{m,p}(\lambda^{-(% \sigma+m+1)})$ (1)

as $\lambda\to\infty$.

Write

 $F(\lambda)=\int_{0}^{s}e^{-\lambda t}\phi(t)dt+\int_{s}^{T}e^{-\lambda t}\phi(% t)dt.$

One the one hand,

 $\left|\int_{s}^{T}e^{-\lambda t}\phi(t)dt\right|\leq\int_{s}^{T}e^{-\lambda t}% |\phi(t)|dt\leq e^{-\lambda s}\int_{s}^{T}|\phi(t)|dt\leq e^{-\lambda s}\left% \|\phi\right\|_{L^{1}},$

which shows that for any nonnegative integer $n$,

 $\int_{s}^{T}e^{-\lambda t}\phi(t)dt=o_{n}(\lambda^{-(\sigma+n+1)})$

as $\lambda\to\infty$.

One the other hand, for each nonnegative integer $N$, Taylor’s theorem tells us that the function $r_{N}:(r,s)\to\mathbb{C}$ defined by

 $r_{N}(t)=g(t)-\sum_{n=0}^{N}\frac{g^{(n)}(0)}{n!}t^{n},\qquad t\in(r,s),$

satisfies

 $|r_{N}(t)|\leq\sup|g^{(N+1)}(\tau)|\cdot\frac{|t|^{N+1}}{(N+1)!},$

where the supremum is over those $\tau$ strictly between $0$ and $t$. Then for $t\in(0,s)$,

 $|r_{N}(t)|\leq\sup_{0<\tau

Using the definition of $r_{N}$,

 $\displaystyle\int_{0}^{s}e^{-\lambda t}\phi(t)dt$ $\displaystyle=$ $\displaystyle\int_{0}^{s}e^{-\lambda t}t^{\sigma}\sum_{n=0}^{N}\frac{g^{(n)}(0% )}{n!}t^{n}dt+\int_{0}^{s}e^{-\lambda t}t^{\sigma}r_{N}(t)dt$ $\displaystyle=$ $\displaystyle\sum_{n=0}^{N}\frac{g^{(n)}(0)}{n!}F_{n}(\lambda)+\int_{0}^{s}e^{% -\lambda t}t^{\sigma}r_{N}(t)dt$

and using the inequality for $r_{N}(t)$,

 $\displaystyle\left|\int_{0}^{s}e^{-\lambda t}t^{\sigma}r_{N}(t)dt\right|$ $\displaystyle\leq$ $\displaystyle\int_{0}^{s}e^{-\lambda t}t^{\sigma}|r_{N}(t)|dt$ $\displaystyle\leq$ $\displaystyle\sup_{0<\tau $\displaystyle=$ $\displaystyle\sup_{0<\tau

Using this and (1),

 $\int_{0}^{s}e^{-\lambda t}t^{\sigma}r_{N}(t)dt=O_{N}(\lambda^{-(\sigma+N+2)}).$

Putting together what we have shown, for any nonnegative integer $N$, as $\lambda\to\infty$,

 $\displaystyle F(\lambda)$ $\displaystyle=$ $\displaystyle\sum_{n=0}^{N}\frac{g^{(n)}(0)}{n!}F_{n}(\lambda)+O_{N}(\lambda^{% -(\sigma+N+2)})+O_{N}(\lambda^{-(\sigma+N+2)})$ $\displaystyle=$ $\displaystyle\sum_{n=0}^{N}\frac{g^{(n)}(0)}{n!}\lambda^{-(\sigma+n+1)}\Gamma(% \sigma+n+1)+\sum_{n=0}^{N}\frac{g^{(n)}(0)}{n!}\cdot o_{N,n}(\lambda^{-(\sigma% +N+1)})$ $\displaystyle+O_{N}(\lambda^{-(\sigma+N+2)})$ $\displaystyle=$ $\displaystyle\sum_{n=0}^{N}\frac{g^{(n)}(0)}{n!}\lambda^{-(\sigma+n+1)}\Gamma(% \sigma+n+1)+o_{N}(\lambda^{-(\sigma+N+1)}),$

which proves the claim. ∎

## 2 Laplace’s method for an interval

###### Theorem 2.

Suppose that $a, that $f\in C^{2}([a,b],\mathbb{R})$, and that there is a unique $x_{0}\in[a,b]$ at which $f$ is equal to its supremum over $[a,b]$. Suppose also that $a and that $f^{\prime\prime}(x_{0})<0$. Then

 $\int_{a}^{b}e^{Mf(x)}dx\sim e^{Mf(x_{0})}\sqrt{\frac{2\pi}{-Mf^{\prime\prime}(% x_{0})}}.$

as $M\to\infty$.

###### Proof.

We remark first that $f^{\prime}(x_{0})=0$ 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 $M\to\infty$. We shall prove that the liminf and the limsup of this ratio are both 1, which will prove the claim. Let $\epsilon>0$. Because $f^{\prime\prime}:[a,b]\to\mathbb{R}$ is continuous, there is some $\delta>0$ such that $|x-x_{0}|<\delta$ implies $f^{\prime\prime}(x)\geq f^{\prime\prime}(x_{0})-\epsilon$; we take $\delta$ small enough that $(x_{0}-\delta,x_{0}+\delta)\subset[a,b]$. Writing

 $f(x)=f(x_{0})+f^{\prime}(x_{0})(x-x_{0})+R_{1}(x)=f(x_{0})+R_{1}(x),\qquad x% \in[a,b],$

Taylor’s theorem tells us that for each $x\in[a,b]$ there is some $\xi_{x}$ strictly between $x_{0}$ and $x$ such that

 $R_{1}(x)=\frac{f^{\prime\prime}(\xi_{x})}{2}(x-x_{0})^{2}.$

Thus for $|x-x_{0}|<\delta$ we have $|\xi_{x}-x_{0}|<\delta$, so

 $f(x)\geq f(x_{0})+\frac{f^{\prime\prime}(x_{0})-\epsilon}{2}(x-x_{0})^{2}.$

Using this inequality, which applies for any $x\in(x_{0}-\delta,x_{0}+\delta)$, and because the integrand in the following integral is positive,

 $\displaystyle\int_{a}^{b}e^{Mf(x)}dx$ $\displaystyle\geq$ $\displaystyle\int_{x_{0}-\delta}^{x_{0}+\delta}e^{Mf(x)}dx$ $\displaystyle\geq$ $\displaystyle\int_{x_{0}-\delta}^{x_{0}+\delta}e^{M\left(f(x_{0})+\frac{f^{% \prime\prime}(x_{0})-\epsilon}{2}(x-x_{0})^{2}\right)}dx$ $\displaystyle=$ $\displaystyle e^{Mf(x_{0})}\int_{x_{0}-\delta}^{x_{0}+\delta}e^{-M\frac{-f^{% \prime\prime}(x_{0})+\epsilon}{2}(x-x_{0})^{2}}dx.$

Changing variables, keeping in mind that $f^{\prime\prime}(x_{0})<0$,

 $\int_{x_{0}-\delta}^{x_{0}+\delta}e^{-M\frac{-f^{\prime\prime}(x_{0})+\epsilon% }{2}(x-x_{0})^{2}}dx=\int_{-\delta\sqrt{M\frac{-f^{\prime\prime}(x_{0})+% \epsilon}{2}}}^{\delta\sqrt{M\frac{-f^{\prime\prime}(x_{0})+\epsilon}{2}}}e^{-% y^{2}}\left(M\frac{-f^{\prime\prime}(x_{0})+\epsilon}{2}\right)^{-1/2}dy.$

Thus

 $\frac{\int_{a}^{b}e^{Mf(x)}dx}{e^{Mf(x_{0})}\left(\frac{-Mf^{\prime\prime}(x_{% 0})}{2}\right)^{-1/2}}$ (2)

is lower bounded by

 $\left(\frac{-f^{\prime\prime}(x_{0})+\epsilon}{-f^{\prime\prime}(x_{0})}\right% )^{-1/2}\int_{-\delta\sqrt{M\frac{-f^{\prime\prime}(x_{0})+\epsilon}{2}}}^{% \delta\sqrt{M\frac{-f^{\prime\prime}(x_{0})+\epsilon}{2}}}e^{-y^{2}}dy,$

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

 $\left(\frac{-f^{\prime\prime}(x_{0})+\epsilon}{-f^{\prime\prime}(x_{0})}\right% )^{-1/2}\sqrt{\pi}.$

But this is true for all $\epsilon>0$ and (2) and its liminf do not depend on $\epsilon$, so the liminf of (2) as $M\to\infty$ is lower bounded by $\sqrt{\pi}$. In other words,

 $\liminf_{M\to\infty}\frac{\int_{a}^{b}e^{Mf(x)}dx}{e^{Mf(x_{0})}\left(\frac{-% Mf^{\prime\prime}(x_{0})}{2\pi}\right)^{-1/2}}\geq 1.$

Let $\epsilon>0$ with $f^{\prime\prime}(x_{0})+\epsilon<0$; this is possible because $f^{\prime\prime}(x_{0})<0$. Because $f^{\prime\prime}:[a,b]\to\mathbb{R}$ is continuous there is some $\delta>0$ such that $|x-x_{0}|<\delta$ implies that $f^{\prime\prime}(x)\leq f^{\prime\prime}(x_{0})+\epsilon$; we take $(x_{0}-\delta,x_{0}+\delta)\subset[a,b]$. Taylor’s theorem tells us that for any $x\in[a,b]$ there is some $\xi_{x}$ strictly between $x_{0}$ and $x$ such that

 $f(x)=f(x_{0})+\frac{f^{\prime\prime}(\xi_{x})}{2}(x-x_{0})^{2}.$

Therefore, as $|x-x_{0}|<\delta$ implies that $|\xi_{x}-x_{0}|<\delta$,

 $f(x)\leq f(x_{0})+\frac{f^{\prime\prime}(x_{0})+\epsilon}{2}(x-x_{0})^{2}.$ (3)

Furthermore, $f:[a,b]\to\mathbb{R}$ is continuous, so it makes sense to define

 $C=\sup_{x\in[a,x_{0}-\delta]\cup[x_{0}+\delta,b]}f(x).$

Because $x_{0}$ is not in this union of intervals, by hypothesis we know that $C, and we define $\eta=f(x_{0})-C>0$. This means that for all $x\in[a,x_{0}-\delta]\cup[x_{0}+\delta,b]$, $f(x)\leq f(x_{0})-\eta$. Then

 $\displaystyle\int_{a}^{b}e^{Mf(x)}dx$ $\displaystyle=$ $\displaystyle\int_{a}^{x_{0}-\delta}e^{Mf(x)}dx+\int_{x_{0}-\delta}^{x_{0}+% \delta}e^{Mf(x)}dx+\int_{x_{0}+\delta}^{b}e^{Mf(x)}dx$ $\displaystyle\leq$ $\displaystyle\int_{a}^{x_{0}-\delta}e^{MC}dx+\int_{x_{0}-\delta}^{x_{0}+\delta% }e^{Mf(x)}dx+\int_{x_{0}+\delta}^{b}e^{MC}dx$ $\displaystyle=$ $\displaystyle(b-a-2\delta)e^{MC}+\int_{x_{0}-\delta}^{x_{0}+\delta}e^{Mf(x)}dx$ $\displaystyle<$ $\displaystyle(b-a)e^{MC}+\int_{x_{0}-\delta}^{x_{0}+\delta}e^{Mf(x)}dx.$

For the integral over $(x_{0}-\delta,x_{0}+\delta)$,

 $\displaystyle\int_{x_{0}-\delta}^{x_{0}+\delta}e^{Mf(x)}dx$ $\displaystyle\leq$ $\displaystyle\int_{x_{0}-\delta}^{x_{0}+\delta}e^{M\left(f(x_{0})+\frac{f^{% \prime\prime}(x_{0})+\epsilon}{2}(x-x_{0})^{2}\right)}dx$ $\displaystyle=$ $\displaystyle e^{Mf(x_{0})}\int_{x_{0}-\delta}^{x_{0}+\delta}e^{M\frac{f^{% \prime\prime}(x_{0})+\epsilon}{2}(x-x_{0})^{2}}dx$ $\displaystyle<$ $\displaystyle e^{Mf(x_{0})}\int_{-\infty}^{\infty}e^{M\frac{f^{\prime\prime}(x% _{0})+\epsilon}{2}(x-x_{0})^{2}}dx.$

Changing variables, and keeping in mind that $f^{\prime\prime}(x_{0})+\epsilon<0$,

 $\displaystyle\int_{-\infty}^{\infty}e^{M\frac{f^{\prime\prime}(x_{0})+\epsilon% }{2}(x-x_{0})^{2}}dx$ $\displaystyle=$ $\displaystyle\int_{-\infty}^{\infty}e^{-y^{2}}\left(-\frac{M}{2}(f^{\prime% \prime}(x_{0})+\epsilon)\right)^{-1/2}dy$ $\displaystyle=$ $\displaystyle\left(-\frac{M}{2\pi}(f^{\prime\prime}(x_{0})+\epsilon)\right)^{-% 1/2}.$

Therefore

 $\int_{a}^{b}e^{Mf(x)}dx<(b-a)e^{MC}+e^{Mf(x_{0})}\left(-\frac{M}{2\pi}(f^{% \prime\prime}(x_{0})+\epsilon)\right)^{-1/2},$

which we rearrange as

 $\frac{\int_{a}^{b}e^{Mf(x)}dx}{e^{Mf(x_{0})}\left(-\frac{M}{2\pi}(f^{\prime% \prime}(x_{0})+\epsilon)\right)^{-1/2}}<(b-a)e^{-M\eta}\left(-\frac{M}{2\pi}(f% ^{\prime\prime}(x_{0})+\epsilon)\right)^{1/2}+1.$

As $M\to\infty$ the first term on the right-hand side tends to $0$, because $\eta>0$. Therefore,

 $\limsup_{M\to\infty}\frac{\int_{a}^{b}e^{Mf(x)}dx}{e^{Mf(x_{0})}\left(-\frac{M% }{2\pi}(f^{\prime\prime}(x_{0})+\epsilon)\right)^{-1/2}}\leq 1.$

This is true for all $\epsilon>0$, so it holds that

 $\limsup_{M\to\infty}\frac{\int_{a}^{b}e^{Mf(x)}dx}{e^{Mf(x_{0})}\left(\frac{-% Mf^{\prime\prime}(x_{0})}{2\pi}\right)^{-1/2}}\leq 1,$

completing the proof. ∎