Oscillatory integrals

1 Oscillatory integrals

Suppose that $\mathrm{\Phi }\in C^{\mathrm{\infty }}({ℝ}^d)$, $\psi \in 𝒟({ℝ}^d)$, and that $\mathrm{\Phi }$ is real-valued. Define $I:(0,\mathrm{\infty })\to ℂ$ by

$$I(\lambda )={\int }_{{ℝ}^d}{e}^{i\lambda \mathrm{\Phi }(x)}\psi (x)𝑑x,\lambda >0.$$

We call $\mathrm{\Phi }$ a phase and $\psi$ an amplitude, and $I(\lambda )$ an oscillatory integral.

The following proof follows Stein and Shakarchi.

Theorem 1.

If there is some $c>0$ such that $|(\nabla \mathrm{\Phi })(x)|\ge c$ for all $x\in \mathrm{supp}\psi$, then for each nonnegative integer $N$ there is some $c_N\ge 0$ such that

$$|I(\lambda )|\le c_N{\lambda }^{-N},\lambda >0.$$

Proof.

There is some $h\in 𝒟({ℝ}^d)$, $h\ge 0$, such that $h(x)=1$ for $x\in \mathrm{supp}\psi$. Define $a:{ℝ}^d\to {ℝ}^d$ by

$$a=h\frac{\nabla \mathrm{\Phi }}{{|\nabla \mathrm{\Phi }|}^2},$$

whose entries each belong to $𝒟({ℝ}^d)$, and define $L:C^{\mathrm{\infty }}({ℝ}^d)\to 𝒟({ℝ}^d)$ by

$$Lf=\frac{1}{i\lambda }\sum _{k=1}^d{a}_k{\partial }_{k}f=\frac{1}{i\lambda }(a\cdot \nabla )f.$$

$L$ satisfies, doing integration by parts and using the fact that $a$ has compact support,

$${\int }_{{ℝ}^d}(Lf)g𝑑x=\frac{1}{i\lambda }\sum _{k=1}^d{\int }_{{ℝ}^d}{a}_k({\partial }_{k}f)g𝑑x=\frac{1}{i\lambda }\sum _{k=1}^d-{\int }_{{ℝ}^d}f{\partial }_k(ga)dx.$$

Thus the transpose of $L$ is

$$L^{t}g=-\frac{1}{i\lambda }\sum _{k=1}^d{\partial }_k(ga)=-\frac{1}{i\lambda }\nabla \cdot (ga).$$

Furthermore, in $\mathrm{supp}\psi$,

$L(e^{i\lambda \mathrm{\Phi }})$ $=$ $e^{i\lambda \mathrm{\Phi }}{\displaystyle \sum _{k=1}^d}{a}_k({\partial }_k\mathrm{\Phi })$

$=$ $e^{i\lambda \mathrm{\Phi }}{\displaystyle \sum _{k=1}^d}{\displaystyle \frac{{\partial }_k\mathrm{\Phi }}{{|\nabla \mathrm{\Phi }|}^2}}{\partial }_k\mathrm{\Phi }$

$=$ $e^{i\lambda \mathrm{\Phi }}.$

Thus for any positive integer $N$ and for $x\in \mathrm{supp}\psi$, $L(e^{i\lambda \mathrm{\Phi }})(x)=e^{i\lambda \mathrm{\Phi }(x)}$, hence

$$I(\lambda )={\int }_{{ℝ}^d}{L}^N(e^{i\lambda \mathrm{\Phi }})\psi 𝑑x={\int }_{{ℝ}^d}{e}^{i\lambda \mathrm{\Phi }}{(L^t)}^N\psi 𝑑x.$$

But

$${\int }_{{ℝ}^d}|{(L^t)}^N\psi |𝑑x={\int }_{{ℝ}^d}|{\lambda }^{-N}{A}_N|𝑑x,$$

where $A_1=\nabla \cdot (\psi a)$ and $A_n=\nabla \cdot (A_{n-1}a)$. With

we obtain

$$|I(\lambda )|=\left|{\int }_{{ℝ}^d}{e}^{i\lambda \mathrm{\Phi }}{(L^t)}^N\psi 𝑑x\right|\le {\int }_{{ℝ}^d}|{(L^t)}^N\psi |𝑑x=c_N{\lambda }^{-N},$$

completing the proof. ∎

The following is an estimate for a one-dimensional oscillatory integral without an amplitude term.

Lemma 2.

Let , and suppose that $\mathrm{\Phi }\in C^2(ℝ)$ is real-valued, that either ${\mathrm{\Phi }}^{\prime \prime }(x)\ge 0$ for all $x\in [a,b]$ or ${\mathrm{\Phi }}^{\prime \prime }(x)\le 0$ for all $x\in [a,b]$, and that ${\mathrm{\Phi }}^{\prime }(x)\ge 1$ for all $x\in [a,b]$. Then

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 3{\lambda }^{-1},\lambda >0.$$

Proof.

Write

$$L=\frac{1}{i\lambda {\mathrm{\Phi }}^{\prime }}\frac{d}{dx},$$

which satisfies

$${\int }_a^b(Lf)g𝑑x={\int }_a^b\frac{1}{i\lambda {\mathrm{\Phi }}^{\prime }}{f}^{\prime }g𝑑x={\frac{1}{i\lambda {\mathrm{\Phi }}^{\prime }}fg|}_a^b-{\int }_a^{b}f{\left(\frac{g}{i\lambda {\mathrm{\Phi }}^{\prime }}\right)}^{\prime }𝑑x.$$

With $f=e^{i\lambda \mathrm{\Phi }}$ and $g=1$, we have $Lf=e^{i\lambda \mathrm{\Phi }}$ and hence

${\displaystyle {\int }_a^b}{e}^{i\lambda \mathrm{\Phi }}𝑑x$ $=$ ${{\displaystyle \frac{e^{i\lambda \mathrm{\Phi }}}{i\lambda {\mathrm{\Phi }}^{\prime }}}|}_a^b-{\displaystyle {\int }_a^b}{e}^{i\lambda \mathrm{\Phi }}{\left({\displaystyle \frac{1}{i\lambda {\mathrm{\Phi }}^{\prime }}}\right)}^{\prime }𝑑x$

$=$ ${{\displaystyle \frac{e^{i\lambda \mathrm{\Phi }}}{i\lambda {\mathrm{\Phi }}^{\prime }}}|}_a^b+{\displaystyle \frac{1}{i\lambda }}{\displaystyle {\int }_a^b}{e}^{i\lambda \mathrm{\Phi }}{({\mathrm{\Phi }}^{\prime })}^{-2}{\mathrm{\Phi }}^{\prime \prime }𝑑x.$

For $\lambda >0$, using that ${\mathrm{\Phi }}^{\prime }(x)\ge 1$ for all $x\in [a,b]$ the boundary terms have absolute value

$$\left|\frac{e^{i\lambda \mathrm{\Phi }(b)}}{i\lambda {\mathrm{\Phi }}^{\prime }(b)}-\frac{e^{i\lambda \mathrm{\Phi }(a)}}{i\lambda {\mathrm{\Phi }}^{\prime }(a)}\right|\le \frac{1}{\lambda |{\mathrm{\Phi }}^{\prime }(b)|}+\frac{1}{\lambda |{\mathrm{\Phi }}^{\prime }(a)|}\le \frac{2}{\lambda }.$$

Because ${\mathrm{\Phi }}^{\prime \prime }\ge 0$ or ${\mathrm{\Phi }}^{\prime \prime }\le 0$ on $[a,b]$,

${\displaystyle \frac{1}{\lambda }}\left|{\displaystyle {\int }_a^b}{e}^{i\lambda \mathrm{\Phi }}{({\mathrm{\Phi }}^{\prime })}^{-2}{\mathrm{\Phi }}^{\prime \prime }𝑑x\right|$ $\le$ ${\displaystyle \frac{1}{\lambda }}{\displaystyle {\int }_a^b}|{({\mathrm{\Phi }}^{\prime })}^{-2}{\mathrm{\Phi }}^{\prime \prime }|𝑑x$

$=$ ${\displaystyle \frac{1}{\lambda }}\left|{\displaystyle {\int }_a^b}{({\mathrm{\Phi }}^{\prime })}^{-2}{\mathrm{\Phi }}^{\prime \prime }𝑑x\right|$

$=$ ${\displaystyle \frac{1}{\lambda }}\left|{\displaystyle \frac{1}{{\mathrm{\Phi }}^{\prime }(a)}}-{\displaystyle \frac{1}{{\mathrm{\Phi }}^{\prime }(b)}}\right|$

$\le$ ${\displaystyle \frac{1}{\lambda }};$

the final inequality uses the fact that the two terms inside the absolute value are both $\ge 1$, and thus the absolute value can be bounded by the larger of them. Putting together the two inequalities,

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }}𝑑x\right|\le \frac{2}{\lambda }+\frac{3}{\lambda }=3{\lambda }^{-1},\lambda >0,$$

proving the claim. ∎

Lemma 3.

Let , and suppose that $\mathrm{\Phi }\in C^2(ℝ)$ is real-valued, that either ${\mathrm{\Phi }}^{\prime \prime }(x)\ge 0$ for all $x\in [a,b]$ or ${\mathrm{\Phi }}^{\prime \prime }(x)\le 0$ for all $x\in [a,b]$, and that there is some $\mu >0$ such that $|{\mathrm{\Phi }}^{\prime }(x)|\ge \mu$ for all $x\in [a,b]$. Then

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 3{\mu }^{-1}{\lambda }^{-1},\lambda >0.$$

Proof.

${\mathrm{\Phi }}^{\prime }$ is continuous on $[a,b]$, so, by the intermediate value theorem, either ${\mathrm{\Phi }}^{\prime }(x)\ge \mu$ for all $x\in [a,b]$ or ${\mathrm{\Phi }}^{\prime }(x)\le -\mu$ for all $x\in [a,b]$. Let $ϵ=1$ in the first case and $ϵ=-1$ in the second case, and define ${\mathrm{\Phi }}_0=ϵ\frac{\mathrm{\Phi }}{\mu }$. Then applying Lemma 2, for $\lambda >0$ we have, writing ${\lambda }_0=\mu \lambda$,

$$\left|{\int }_a^b{e}^{i{\lambda }_0{\mathrm{\Phi }}_0(x)}𝑑x\right|\le 3{\lambda }_0^{-1},$$

i.e.

$$\left|{\int }_a^b{e}^{iϵ\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 3{(\mu \lambda )}^{-1}.$$

If $ϵ=1$ this is the claim. If $ϵ=-1$, then the above integral is the complex conjugate of the integral in the claim, and these have the same absolute values. ∎

Theorem 4.

Let , and suppose that $\mathrm{\Phi }\in C^2(ℝ)$ is real-valued, that either ${\mathrm{\Phi }}^{\prime \prime }(x)\ge 0$ for all $x\in [a,b]$ or ${\mathrm{\Phi }}^{\prime \prime }(x)\le 0$ for all $x\in [a,b]$, and there is some $\mu >0$ such that $|{\mathrm{\Phi }}^{\prime }(x)|\ge \mu$ for all $x\in [a,b]$. Suppose also that $\psi \in C^1(ℝ)$. Then with

$$c_{\psi }=3\left(|\psi (b)|+{\int }_a^b|{\psi }^{\prime }(x)|𝑑x\right),$$

we have

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}\psi (x)𝑑x\right|\le c_{\psi }{\mu }^{-1}{\lambda }^{-1}.$$

Proof.

Define $J:[a,b]\to ℂ$ by

$$J(x)={\int }_a^x{e}^{i\lambda \mathrm{\Phi }(u)}𝑑u,$$

which satisfies $J^{\prime }(x)=e^{i\lambda \mathrm{\Phi }(x)}$. Integrating by parts,

$${\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}\psi (x)𝑑x={\int }_a^b{J}^{\prime }(x)\psi (x)𝑑x={J(x)\psi (x)|}_a^b-{\int }_a^{b}J(x){\psi }^{\prime }(x)𝑑x,$$

and as $J(a)=0$ this is equal to

$$J(b)\psi (b)-{\int }_a^{b}J(x){\psi }^{\prime }(x)𝑑x.$$

Lemma 3 tells us that $|J(x)|\le 3{\mu }^{-1}{\lambda }^{-1}$ for all $x\in [a,b]$, so

$$\left|J(b)\psi (b)-{\int }_a^{b}J(x){\psi }^{\prime }(x)𝑑x\right|\le 3{\mu }^{-1}{\lambda }^{-1}|\psi (b)|+3{\mu }^{-1}{\lambda }^{-1}{\int }_a^b|{\psi }^{\prime }(x)|𝑑x,$$

proving the claim. ∎

The following is van der Corput’s lemma.

Lemma 5 (van der Corput’s lemma).

Let and suppose that $\mathrm{\Phi }\in C^2(ℝ)$ is real-valued and satisfies ${\mathrm{\Phi }}^{\prime \prime }(x)\ge 1$ for all $x\in [a,b]$. Then

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 8{\lambda }^{-1/2},\lambda >0.$$

Proof.

Because ${\mathrm{\Phi }}^{\prime }$ is strictly increasing on $[a,b]$, ${\mathrm{\Phi }}^{\prime }$ has at most one zero in this interval. If ${\mathrm{\Phi }}^{\prime }(x_0)=0$, then for $x\ge x_0+{\lambda }^{-1/2}$ we have ${\mathrm{\Phi }}^{\prime }(x)\ge {\lambda }^{-1/2}$, and applying Lemma 3 with $\mu ={\lambda }^{-1/2}$,

$$\left|{\int }_{[x_0+{\lambda }^{-1/2},b]}{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 3{\mu }^{-1}{\lambda }^{-1}=3{\lambda }^{-1/2}.$$

For $x\le x_0-{\lambda }^{-1/2}$ we have ${\mathrm{\Phi }}^{\prime }(x)\le -{\lambda }^{-1/2}$, and applying Lemma 3 with $\mu ={\lambda }^{-1/2}$,

$$\left|{\int }_{[a,x_0-{\lambda }^{-1/2}]}{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 3{\mu }^{-1}{\lambda }^{-1}=3{\lambda }^{-1/2}.$$

But

$$\left|{\int }_{[x_0-{\lambda }^{-1/2},x_0+{\lambda }^{-1/2}]\cap [a,b]}{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le {\int }_{[x_0-{\lambda }^{-1/2},x_0+{\lambda }^{-1/2}]\cap [a,b]}𝑑x\le 2{\lambda }^{-1/2},$$

and

$${\int }_a^b={\int }_{[a,x_0-{\lambda }^{-1/2}]}+{\int }_{[x_0-{\lambda }^{-1/2},x_0+{\lambda }^{-1/2}]\cap [a,b]}+{\int }_{[x_0+{\lambda }^{-1/2},b]},$$

so

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 3{\lambda }^{-1/2}+2{\lambda }^{-1/2}+3{\lambda }^{-1/2}=8{\lambda }^{-1/2}.$$

If there is no $x_0\in [a,b]$ such that ${\mathrm{\Phi }}^{\prime }(x_0)=0$, then either ${\mathrm{\Phi }}^{\prime }>0$ on $[a,b]$ or on $[a,b]$. In the first case, because ${\mathrm{\Phi }}^{\prime }$ is strictly increasing on $[a,b]$, ${\mathrm{\Phi }}^{\prime }(x)>{\lambda }^{-1/2}$ for $x\in [a+{\lambda }^{-1/2},b]$, and applying Lemma 3 with $\mu ={\lambda }^{-1/2}$ gives

$\left|{\displaystyle {\int }_a^b}{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|$ $\le$ $\left|{\displaystyle {\int }_{[a,a+{\lambda }^{-1/2}]\cap [a,b]}}{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|+\left|{\displaystyle {\int }_{[a+{\lambda }^{-1/2},b]}}{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|$

$\le$ ${\lambda }^{-1/2}+3{\mu }^{-1}{\lambda }^{-1}$

$=$ $4{\lambda }^{-1/2}.$

In the second case, for $x\in [a,b-{\lambda }^{-1/2}]$, and applying Lemma 3 with $\mu ={\lambda }^{-1/2}$ also gives

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 4{\lambda }^{-1/2}.$$

Therefore, if ${\mathrm{\Phi }}^{\prime }$ does not have a zero on $[a,b]$ then

∎

Lemma 6.

Let and suppose that $\mathrm{\Phi }\in C^2(ℝ)$ is real-valued and that there is some $\mu >0$ such that $|{\mathrm{\Phi }}^{\prime \prime }(x)|\ge \mu$ for all $x\in [a,b]$. Then

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 8{\mu }^{-1/2}{\lambda }^{-1/2},\lambda >0.$$

Proof.

${\mathrm{\Phi }}^{\prime \prime }$ is continuous on $[a,b]$, so by the intermediate value theorem either ${\mathrm{\Phi }}^{\prime \prime }(x)\ge \mu$ for all $x\in [a,b]$ or ${\mathrm{\Phi }}^{\prime \prime }(x)\le -\mu$ for all $x\in [a,b]$. Let $ϵ=1$ in the first case and $ϵ=-1$ in the second case, and define ${\mathrm{\Phi }}_0=ϵ\frac{\mathrm{\Phi }}{\mu }$. Then ${\mathrm{\Phi }}_0^{\prime \prime }(x)\ge 1$ for all $x\in [a,b]$, and applying Lemma 5,

$$\left|{\int }_a^b{e}^{i\mu \lambda {\mathrm{\Phi }}_0(x)}𝑑x\right|\le 8{(\mu \lambda )}^{-1/2},\lambda >0,$$

i.e.

$$\left|{\int }_a^b{e}^{iϵ\lambda \mathrm{\Phi }(x)}𝑑x\right|\le 8{(\mu \lambda )}^{-1/2},\lambda >0.$$

If $ϵ=1$ this is the inequality in the claim. If $ϵ=-1$, then the above integral is the complex conjugate of the integral in the claim, and these have the same absolute values. ∎

We use the above to prove the following estimate which involves an amplitude.

Theorem 7.

Let and suppose that $\mathrm{\Phi }\in C^2(ℝ)$ is real-valued and that there is some $\mu >0$ such that $|{\mathrm{\Phi }}^{\prime \prime }(x)|\ge \mu$ for all $x\in [a,b]$. Suppose also that $\psi \in C^1(ℝ)$. Then with

$$c_{\psi }=8\left(|\psi (b)|+{\int }_a^b|{\psi }^{\prime }(x)|𝑑x\right),$$

we have

$$\left|{\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}\psi (x)𝑑x\right|\le c_{\psi }{\mu }^{-1/2}{\lambda }^{-1/2},\lambda >0.$$

Proof.

Define $J:[a,b]\to ℂ$ by

$$J(x)={\int }_a^x{e}^{i\lambda \mathrm{\Phi }(u)}𝑑u,$$

which satisfies $J^{\prime }(x)=e^{i\lambda \mathrm{\Phi }(x)}$. Integrating by parts,

$${\int }_a^b{e}^{i\lambda \mathrm{\Phi }(x)}\psi (x)𝑑x={\int }_a^b{J}^{\prime }(x)\psi (x)𝑑x={J(x)\psi (x)|}_a^b-{\int }_a^{b}J(x){\psi }^{\prime }(x)𝑑x.$$

and as $J(a)=0$ this is equal to

$$J(b)\psi (b)-{\int }_a^{b}J(x){\psi }^{\prime }(x)𝑑x.$$

But for each $x\in [a,b]$ we have by Lemma 6 that $|J(x)|\le 8{\mu }^{-1/2}{\lambda }^{-1/2}$, so

$$\left|J(b)\psi (b)-{\int }_a^{b}J(x){\psi }^{\prime }(x)𝑑x\right|\le 8{\mu }^{-1/2}{\lambda }^{-1/2}|\psi (b)|+8{\mu }^{-1/2}{\lambda }^{-1/2}{\int }_a^b|{\psi }^{\prime }(x)|𝑑x,$$

completing the proof. ∎

2 Bessel functions

For $n\in \mathbb{Z}$, the $n$th Bessel function of the first kind $J_n:ℝ\to ℝ$ is

$$J_n(\lambda )=\frac{1}{2\pi }{\int }_0^{2\pi }{e}^{i\lambda \sin x}{e}^{-inx}𝑑x,\lambda \in ℝ.$$

Let

$$I_1=[0,\frac{\pi }{4}],I_2=[\frac{3\pi }{4},\pi ],I_3=[\pi ,\frac{5\pi }{4}],I_4=[\frac{7\pi }{4},2\pi ],$$

on which $|\cos x|\ge \frac{1}{\sqrt{2}}$, and

$$I_5=[\frac{\pi }{4},\frac{3\pi }{4}],I_6=[\frac{5\pi }{4},\frac{7\pi }{4}],$$

on which $|\sin x|\ge \frac{1}{\sqrt{2}}$. Write $\mathrm{\Phi }(x)=\sin x$ and $\psi (x)=e^{-inx}$. ${\mathrm{\Phi }}^{\prime }(x)=\cos (x)$ and ${\mathrm{\Phi }}^{\prime \prime }(x)=-\sin (x)$, and for $I_1,I_2,I_3,I_4$ we apply Theorem 4 with $\mu =\frac{1}{\sqrt{2}}$. For each of $I_1,I_2,I_3,I_4$ we compute $c_{\psi }=3\left(1+\frac{\pi n}{4}\right)$, which gives us

$$\left|{\int }_{I_k}{e}^{i\lambda \mathrm{\Phi }(x)}\psi (x)𝑑x\right|\le c_{\psi }{\mu }^{-1}{\lambda }^{-1}=3\left(1+\frac{\pi n}{4}\right)\cdot \sqrt{2}\cdot {\lambda }^{-1}.$$

For $I_5$ and $I_6$, we apply Theorem 7 with $\mu =\frac{1}{\sqrt{2}}$. For each of $I_5$ and $I_6$ we compute $c_{\psi }=8\left(1+\frac{\pi n}{2}\right)$, which gives us

$$\left|{\int }_{I_k}{e}^{i\lambda \mathrm{\Phi }(x)}\psi (x)𝑑x\right|\le c_{\psi }{\mu }^{-1/2}{\lambda }^{-1/2}=8\left(1+\frac{\pi n}{2}\right)\cdot 2^{1/4}\cdot {\lambda }^{-1/2}.$$

Therefore

$$|J_n(\lambda )|\le 4\cdot \frac{1}{2\pi }\cdot 3\left(1+\frac{\pi n}{4}\right)\cdot \sqrt{2}\cdot {\lambda }^{-1}+2\cdot \frac{1}{2\pi }\cdot 8\left(1+\frac{\pi n}{2}\right)\cdot 2^{1/4}\cdot {\lambda }^{-1/2},$$

which shows that for each $n\in \mathbb{Z}$,

$$J_n(\lambda )=O_n({\lambda }^{-1/2})$$

as $\lambda \to \mathrm{\infty }$.