Van der Corput lemma

Remarks

The point is that the bound does not depend on $a$, $b$ or $\phi$. For $k = 1$, write $e^{i\lambda\phi} = (i\lambda\phi')^{-1} \frac{d}{dx} e^{i\lambda\phi}$ and integrate by parts; monotonicity of $\phi'$ makes the integral of $\abs{(1/\phi')'}$ telescope, and one gets $c_1 = 3$. For $k \ge 2$, use induction on $k$. On the set where $\abs{\phi^{(k-1)}} \ge \lambda^{-1/k}$, apply the case $k - 1$ after rescaling. This set misses at most an interval of length about $\lambda^{-1/k}$, where the integrand is bounded by $1$.

Example: for $\phi(x) = x^2/2$ and $k = 2$, the lemma gives $\abs{\int_a^b e^{i\lambda x^2/2} \, dx} \le c_2 \lambda^{-1/2}$ uniformly in $a, b$; the Fresnel integral $\int_{-\infty}^{\infty} e^{i\lambda x^2/2} \, dx = \sqrt{2\pi/\lambda} \, e^{i\pi/4}$ shows that the power $\lambda^{-1/2}$ cannot be improved.

For $k = 1$ the monotonicity of $\phi'$ is needed. Without it, no bound $c_1\lambda^{-1}$ holds uniformly in $\phi$ and in the length of the interval.

A corollary adds an amplitude: if $\psi$ is smooth on $[a, b]$, then $\abs{\int_a^b e^{i\lambda\phi}\psi} \le c_k\lambda^{-1/k} \big(\abs{\psi(b)} + \int_a^b \abs{\psi'}\big)$.

Going deeper

For a student

This is the basic one-dimensional tool for oscillatory integrals. It gives decay rates for Fourier transforms of measures on curves, such as $\abs{\hat\sigma(\xi)} \lesssim \abs{\xi}^{-1/2}$ for arc length on a curve with nonzero curvature. In higher dimensions there is no equally clean analogue, and one uses stationary phase or sublevel set estimates instead.