# Oscillatory integrals

Jordan Bell
August 4, 2014

## 1 Oscillatory integrals

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

 $I(\lambda)=\int_{\mathbb{R}^{d}}e^{i\lambda\Phi(x)}\psi(x)dx,\qquad\lambda>0.$

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

The following proof follows Stein and Shakarchi.11 1 Elias M. Stein and Rami Shakarchi, Functional Analysis, p. 325, Proposition 2.1.

###### Theorem 1.

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

 $|I(\lambda)|\leq c_{N}\lambda^{-N},\qquad\lambda>0.$
###### Proof.

There is some $h\in\mathscr{D}(\mathbb{R}^{d})$, $h\geq 0$, such that $h(x)=1$ for $x\in\mathrm{supp}\,\psi$.22 2 Walter Rudin, Functional Analysis, second ed., p. 162, Theorem 6.20. Define $a:\mathbb{R}^{d}\to\mathbb{R}^{d}$ by

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

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

 $\int_{\mathbb{R}^{d}}(Lf)gdx=\frac{1}{i\lambda}\sum_{k=1}^{d}\int_{\mathbb{R}^% {d}}a_{k}(\partial_{k}f)gdx=\frac{1}{i\lambda}\sum_{k=1}^{d}-\int_{\mathbb{R}^% {d}}f\partial_{k}(ga)dx.$

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

 $\displaystyle L(e^{i\lambda\Phi})$ $\displaystyle=$ $\displaystyle e^{i\lambda\Phi}\sum_{k=1}^{d}a_{k}(\partial_{k}\Phi)$ $\displaystyle=$ $\displaystyle e^{i\lambda\Phi}\sum_{k=1}^{d}\frac{\partial_{k}\Phi}{|\nabla% \Phi|^{2}}\partial_{k}\Phi$ $\displaystyle=$ $\displaystyle e^{i\lambda\Phi}.$

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

 $I(\lambda)=\int_{\mathbb{R}^{d}}L^{N}(e^{i\lambda\Phi})\psi dx=\int_{\mathbb{R% }^{d}}e^{i\lambda\Phi}(L^{t})^{N}\psi dx.$

But

 $\int_{\mathbb{R}^{d}}|(L^{t})^{N}\psi|dx=\int_{\mathbb{R}^{d}}|\lambda^{-N}A_{% N}|dx,$

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

 $c_{N}=\int_{\mathbb{R}^{d}}|A_{N}|dx<\infty,$

we obtain

 $|I(\lambda)|=\left|\int_{\mathbb{R}^{d}}e^{i\lambda\Phi}(L^{t})^{N}\psi dx% \right|\leq\int_{\mathbb{R}^{d}}|(L^{t})^{N}\psi|dx=c_{N}\lambda^{-N},$

completing the proof. ∎

The following is an estimate for a one-dimensional oscillatory integral without an amplitude term.33 3 Elias M. Stein and Rami Shakarchi, Functional Analysis, p. 326, Proposition 2.2.

###### Lemma 2.

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

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}dx\right|\leq 3\lambda^{-1},\qquad\lambda% >0.$
###### Proof.

Write

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

which satisfies

 $\int_{a}^{b}(Lf)gdx=\int_{a}^{b}\frac{1}{i\lambda\Phi^{\prime}}f^{\prime}gdx=% \frac{1}{i\lambda\Phi^{\prime}}fg\bigg{|}_{a}^{b}-\int_{a}^{b}f\left(\frac{g}{% i\lambda\Phi^{\prime}}\right)^{\prime}dx.$

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

 $\displaystyle\int_{a}^{b}e^{i\lambda\Phi}dx$ $\displaystyle=$ $\displaystyle\frac{e^{i\lambda\Phi}}{i\lambda\Phi^{\prime}}\bigg{|}_{a}^{b}-% \int_{a}^{b}e^{i\lambda\Phi}\left(\frac{1}{i\lambda\Phi^{\prime}}\right)^{% \prime}dx$ $\displaystyle=$ $\displaystyle\frac{e^{i\lambda\Phi}}{i\lambda\Phi^{\prime}}\bigg{|}_{a}^{b}+% \frac{1}{i\lambda}\int_{a}^{b}e^{i\lambda\Phi}(\Phi^{\prime})^{-2}\Phi^{\prime% \prime}dx.$

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

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

Because $\Phi^{\prime\prime}\geq 0$ or $\Phi^{\prime\prime}\leq 0$ on $[a,b]$,

 $\displaystyle\frac{1}{\lambda}\left|\int_{a}^{b}e^{i\lambda\Phi}(\Phi^{\prime}% )^{-2}\Phi^{\prime\prime}dx\right|$ $\displaystyle\leq$ $\displaystyle\frac{1}{\lambda}\int_{a}^{b}|(\Phi^{\prime})^{-2}\Phi^{\prime% \prime}|dx$ $\displaystyle=$ $\displaystyle\frac{1}{\lambda}\left|\int_{a}^{b}(\Phi^{\prime})^{-2}\Phi^{% \prime\prime}dx\right|$ $\displaystyle=$ $\displaystyle\frac{1}{\lambda}\left|\frac{1}{\Phi^{\prime}(a)}-\frac{1}{\Phi^{% \prime}(b)}\right|$ $\displaystyle\leq$ $\displaystyle\frac{1}{\lambda};$

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

 $\left|\int_{a}^{b}e^{i\lambda\Phi}dx\right|\leq\frac{2}{\lambda}+\frac{3}{% \lambda}=3\lambda^{-1},\qquad\lambda>0,$

proving the claim. ∎

###### Lemma 3.

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

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}dx\right|\leq 3\mu^{-1}\lambda^{-1},% \qquad\lambda>0.$
###### Proof.

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

 $\left|\int_{a}^{b}e^{i\lambda_{0}\Phi_{0}(x)}dx\right|\leq 3\lambda_{0}^{-1},$

i.e.

 $\left|\int_{a}^{b}e^{i\epsilon\lambda\Phi(x)}dx\right|\leq 3(\mu\lambda)^{-1}.$

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

###### Theorem 4.

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

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

we have

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}\psi(x)dx\right|\leq c_{\psi}\mu^{-1}% \lambda^{-1}.$
###### Proof.

Define $J:[a,b]\to\mathbb{C}$ by

 $J(x)=\int_{a}^{x}e^{i\lambda\Phi(u)}du,$

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

 $\int_{a}^{b}e^{i\lambda\Phi(x)}\psi(x)dx=\int_{a}^{b}J^{\prime}(x)\psi(x)dx=J(% x)\psi(x)\bigg{|}_{a}^{b}-\int_{a}^{b}J(x)\psi^{\prime}(x)dx,$

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

 $\left|J(b)\psi(b)-\int_{a}^{b}J(x)\psi^{\prime}(x)dx\right|\leq 3\mu^{-1}% \lambda^{-1}|\psi(b)|+3\mu^{-1}\lambda^{-1}\int_{a}^{b}|\psi^{\prime}(x)|dx,$

proving the claim. ∎

The following is van der Corput’s lemma.44 4 Elias M. Stein and Rami Shakarchi, Functional Analysis, p. 328, Proposition 2.3.

###### Lemma 5 (van der Corput’s lemma).

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

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}dx\right|\leq 8\lambda^{-1/2},\qquad% \lambda>0.$
###### Proof.

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

 $\left|\int_{[x_{0}+\lambda^{-1/2},b]}e^{i\lambda\Phi(x)}dx\right|\leq 3\mu^{-1% }\lambda^{-1}=3\lambda^{-1/2}.$

For $x\leq x_{0}-\lambda^{-1/2}$ we have $\Phi^{\prime}(x)\leq-\lambda^{-1/2}$, and applying Lemma 3 with $\mu=\lambda^{-1/2}$,

 $\left|\int_{[a,x_{0}-\lambda^{-1/2}]}e^{i\lambda\Phi(x)}dx\right|\leq 3\mu^{-1% }\lambda^{-1}=3\lambda^{-1/2}.$

But

 $\left|\int_{[x_{0}-\lambda^{-1/2},x_{0}+\lambda^{-1/2}]\cap[a,b]}e^{i\lambda% \Phi(x)}dx\right|\leq\int_{[x_{0}-\lambda^{-1/2},x_{0}+\lambda^{-1/2}]\cap[a,b% ]}dx\leq 2\lambda^{-1/2},$

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

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}dx\right|\leq 3\lambda^{-1/2}+2\lambda^{-% 1/2}+3\lambda^{-1/2}=8\lambda^{-1/2}.$

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

 $\displaystyle\left|\int_{a}^{b}e^{i\lambda\Phi(x)}dx\right|$ $\displaystyle\leq$ $\displaystyle\left|\int_{[a,a+\lambda^{-1/2}]\cap[a,b]}e^{i\lambda\Phi(x)}dx% \right|+\left|\int_{[a+\lambda^{-1/2},b]}e^{i\lambda\Phi(x)}dx\right|$ $\displaystyle\leq$ $\displaystyle\lambda^{-1/2}+3\mu^{-1}\lambda^{-1}$ $\displaystyle=$ $\displaystyle 4\lambda^{-1/2}.$

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

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}dx\right|\leq 4\lambda^{-1/2}.$

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

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}dx\right|\leq 4\lambda^{-1/2}<8\lambda^{-% 1/2}.$

###### Lemma 6.

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

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}dx\right|\leq 8\mu^{-1/2}\lambda^{-1/2},% \qquad\lambda>0.$
###### Proof.

$\Phi^{\prime\prime}$ is continuous on $[a,b]$, so by the intermediate value theorem either $\Phi^{\prime\prime}(x)\geq\mu$ for all $x\in[a,b]$ or $\Phi^{\prime\prime}(x)\leq-\mu$ for all $x\in[a,b]$. Let $\epsilon=1$ in the first case and $\epsilon=-1$ in the second case, and define $\Phi_{0}=\epsilon\frac{\Phi}{\mu}$. Then $\Phi_{0}^{\prime\prime}(x)\geq 1$ for all $x\in[a,b]$, and applying Lemma 5,

 $\left|\int_{a}^{b}e^{i\mu\lambda\Phi_{0}(x)}dx\right|\leq 8(\mu\lambda)^{-1/2}% ,\qquad\lambda>0,$

i.e.

 $\left|\int_{a}^{b}e^{i\epsilon\lambda\Phi(x)}dx\right|\leq 8(\mu\lambda)^{-1/2% },\qquad\lambda>0.$

If $\epsilon=1$ this is the inequality in the claim. If $\epsilon=-1$, 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.55 5 Elias M. Stein and Rami Shakarchi, Functional Analysis, p. 328, Corollary 2.4.

###### Theorem 7.

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

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

we have

 $\left|\int_{a}^{b}e^{i\lambda\Phi(x)}\psi(x)dx\right|\leq c_{\psi}\mu^{-1/2}% \lambda^{-1/2},\qquad\lambda>0.$
###### Proof.

Define $J:[a,b]\to\mathbb{C}$ by

 $J(x)=\int_{a}^{x}e^{i\lambda\Phi(u)}du,$

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

 $\int_{a}^{b}e^{i\lambda\Phi(x)}\psi(x)dx=\int_{a}^{b}J^{\prime}(x)\psi(x)dx=J(% x)\psi(x)\bigg{|}_{a}^{b}-\int_{a}^{b}J(x)\psi^{\prime}(x)dx.$

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 $|J(x)|\leq 8\mu^{-1/2}\lambda^{-1/2}$, so

 $\left|J(b)\psi(b)-\int_{a}^{b}J(x)\psi^{\prime}(x)dx\right|\leq 8\mu^{-1/2}% \lambda^{-1/2}|\psi(b)|+8\mu^{-1/2}\lambda^{-1/2}\int_{a}^{b}|\psi^{\prime}(x)% |dx,$

completing the proof. ∎

## 2 Bessel functions

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

 $J_{n}(\lambda)=\frac{1}{2\pi}\int_{0}^{2\pi}e^{i\lambda\sin x}e^{-inx}dx,% \qquad\lambda\in\mathbb{R}.$

Let

 $I_{1}=\left[0,\frac{\pi}{4}\right],\quad I_{2}=\left[\frac{3\pi}{4},\pi\right]% ,\quad I_{3}=\left[\pi,\frac{5\pi}{4}\right],\quad I_{4}=\left[\frac{7\pi}{4},% 2\pi\right],$

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

 $I_{5}=\left[\frac{\pi}{4},\frac{3\pi}{4}\right],\quad I_{6}=\left[\frac{5\pi}{% 4},\frac{7\pi}{4}\right],$

on which $|\sin x|\geq\frac{1}{\sqrt{2}}$. Write $\Phi(x)=\sin x$ and $\psi(x)=e^{-inx}$. $\Phi^{\prime}(x)=\cos(x)$ and $\Phi^{\prime\prime}(x)=-\sin(x)$, 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

 $\left|\int_{I_{k}}e^{i\lambda\Phi(x)}\psi(x)dx\right|\leq c_{\psi}\mu^{-1}% \lambda^{-1}=3\left(1+\frac{\pi n}{4}\right)\cdot\sqrt{2}\cdot\lambda^{-1}.$

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

 $\left|\int_{I_{k}}e^{i\lambda\Phi(x)}\psi(x)dx\right|\leq c_{\psi}\mu^{-1/2}% \lambda^{-1/2}=8\left(1+\frac{\pi n}{2}\right)\cdot 2^{1/4}\cdot\lambda^{-1/2}.$

Therefore

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

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

 $J_{n}(\lambda)=O_{n}(\lambda^{-1/2})$

as $\lambda\to\infty$.