Dyadic Decomposition For Oscillatory Integral In Terence Tao's notes on oscillatory integrals, Tao mentions that if $a(x)$ is a smooth, compactly supported phase, with $a(x) = 1$ in a neighborhood of the origin, then for any $N$,
$$ \int a(x) e^{\lambda i x^2}\; dx = e^{i\pi/4} \sqrt{\pi/\lambda} + O_{N,a}(\lambda^{-N}) $$
He mentions that this bound can be proven by 'a dyadic decomposition'. I've tried to come up with a dyadic decomposition strategy that yields this bound but something appears to be alluding me. What dyadic decomposition strategy might yield this bound? Below is the approach my intuition suggests, which might guide your answer, but if there is an obvious approach that I am missing, there is no need to read the approach below: 
If we define $\beta_0(x)$ to be a smooth function equal to $1$ on $[-1,1]$ and vanishing outside of $[-2,2]$, and then define $\beta_n(x) = \beta_0(x/2^n) - \beta_0(x/2^{n-1})$, then for each $x \in \mathbf{R}$, $\sum_{n = 0}^\infty \beta_n(x) = 1$. Thus $\{ \beta_n \}$ is a partition of unity, with $\beta_n$ supported on $|x| \sim 2^n$ for each $n$. I want to isolate the behaviour of the integral when $|x| \lesssim \lambda^{-1/2}$, where the phase is stationary, so I might want to decompose the integral as
$$ \int a(x) e^{\lambda i x^2} = \sum_{n = 0}^\infty \int a(x) \beta_n(x \cdot \lambda^{1/2}) e^{\lambda i x^2}. $$
However, for $n \geq 1$, I am unable to obtain any better bound than
$$ \left| \sum_{n = 1}^\infty \int \beta_n(x \cdot \lambda^{1/2}) e^{\lambda i x^2} \right| \lesssim \lambda^{-1/2}. $$
Even if I was able to obtain a bound $O(\lambda^{-N})$ for this term, it seems unclear how we might reintroduce the amplitude $a$ to obtain a bound on
$$ \left| \sum_{n = 1}^\infty \int a(x) \beta_n(x \cdot \lambda^{1/2}) e^{\lambda i x^2} \right|. $$
We can certainly sum up to $n \lesssim \log(\lambda^{1/2})$, where $\beta_n(x \cdot \lambda^{1/2}) a(x) = \beta_n(x \cdot \lambda^{1/2})$, but for $n \gtrsim \log(\lambda^{1/2})$ we obtain problems because $a$ modifies the behaviour of $\beta_n$.
 A: I think you are referring to these notes notes8.tao (p. 6). As stated in the notes we have $\int_{\mathbb{R}}e^{i\lambda x^2}\text{d}x=e^{\pi i/4}\sqrt{\frac{\pi}{\lambda}}$. Now we want to show that for a bump function $a$ with $a(x)=1$ for $x$ close to $0$ we have that
$$\int e^{i\lambda x^2}a(x)\text{d}x-e^{\pi i/4}\sqrt{\frac{\pi}{\lambda}}=\mathcal{O}_{N,a}(\lambda^{-N})$$
W.l.o.g. we can assume that $a(x)=1$ for $|x|\leq 1$ and $a(x)=0$ for $|x|\geq 2$ (otherwise we just scale and adjust the following proof accordingly). Then we set $\widetilde{a}(x)=a(x/2)-a(x)$ and $\beta_n(x)=\widetilde{a}(x/2^n)$. Note that $\widetilde{a}$ is a bump function which vanishes near $0$. One can also verify that $\text{supp}(\beta_n)\subset \left[2^n,2^{n+2}\right]$ and $a(x)+\sum_{n=0}^{\infty}\beta_n(x)=1$ for all $x\in\mathbb{R}$. Therefore
$$\int e^{i\lambda x^2}\text{d}x-\int e^{i\lambda x^2}a(x)\text{d}x=\sum_{n=0}^{\infty}\int e^{i\lambda x^2}\beta_n(x)\text{d}x$$
Since $\widetilde{a}$ vanishes near $0$ we have that (this is also stated in the notes)
$$\int e^{i\lambda x^2}\widetilde{a}(x/R)\text{d}x=\mathcal{O}_{N,a}(\lambda^{-N}R^{-N})$$.
Hence
$$\left|\sum_{n=0}^{\infty}\int e^{i\lambda x^2}\beta_n(x)\text{d}x\right|=\left|\sum_{n=0}^{\infty}\int e^{i\lambda x^2}\widetilde{a}(x/2^n)\text{d}x\right|\lesssim \lambda^{-N}\sum_{n=0}^{\infty}\left(\frac{1}{2^{N}}\right)^n\lesssim \lambda^{-N}$$
which finishes the proof.
