Let $\phi (x) = x^2$, $k,\lambda\in \mathbb{R}$, and $[a,b]\subset\mathbb{R}$ and define the integral $$I(\lambda;k)=\int_{a}^{b}e^{i(\lambda\phi(x)-kx)}dx$$ I am trying to prove that $|I(\lambda;k)|\leq c_\phi|\lambda|^{-\frac{1}{2}}$ such that $c_{\phi}$ is a constant depending only on $\phi$.
Van der Corput Lemma states that for $\phi:[a,b]\rightarrow \mathbb{R}$, a smooth function such that $|\phi^{(n)}(x)|\geq 1$ $\forall x \in (a,b)$ and $\psi$ a complex valued differentiable function we have, $$\Bigg|\int_{a}^{b}e^{i\lambda \phi(x)} \psi(x)dx \Bigg| \leq c_n|\lambda|^{-\frac{1}{n}}\Big[|\psi(b)|+\int_{a}^{b}|\psi'(x)|dx\Big]$$ where $c_n$ is an absolute constant.
I have tried to use this by taking $n=2$ and $\psi(x)=e^{-kx}$ but on the RHS I am left with $c_2|\lambda|^{-\frac{1}{2}}(1+|k|(b-a))$ so the bound is clearly not dependent only on $\phi$.
This is actually a special case of Lemma 2.7 (page 42) in https://www.jstor.org/stable/24896258?seq=1#metadata_info_tab_contents
Any help will be very much appreciated.
I would complete the square, $$I(\lambda;k)= \int_a^b e^{i[\lambda(x-\frac{k}{2\lambda})^2-\frac{k^2}{4\lambda}]}dx = e^{-i\frac{k^2}{4\lambda}}\int_{a-\frac{k}{2\lambda}}^{b-\frac{k}{2\lambda}}e^{i\lambda y^2}dy.$$ Then $$|I(\lambda;k)| \leq \sup_{\alpha<\beta}\bigg\vert\int_\alpha^\beta e^{i\lambda y^2}dy\bigg\vert,$$ and the supremum can be bounded using Van der Corput's lemma.