# A Van der Corput style inequality for highly oscillatory integrals

Suppose $f$ has at least two continuous derivatives, $f'$ is monotonically increasing, and $f' \geq \lambda$ for some $\lambda > 0$. How might one find the upper bound $|\int_a^b \cos(f(x))| \leq 2/\lambda$?

I've tried a number of basic approaches and none have worked. For instance, I've tried rewriting $\cos(f(x))$ as $(\sin(f(x)))' / f'(x)$, which seemed promising until I realized that I cannot justify the inequality $|\int_a^b (\sin(f(x)))' / f'(x)| \leq (1/\lambda) |\int_a^b (\sin(f(x)))'|$. Of course, were that true, then the result would follow quickly by evaluating the new integral, applying the triangle inequality, and finally noting that $|\sin(x)| \leq 1$.

I've tried a similar approach using the mean value theorem for integrals but was only able to show that there is some $\theta$, $a < \theta < b$, so that

$|\int_a^b \cos(f(x))| \leq \frac{1}{\lambda} (|f(\theta) - f(a)| + |f(b) - f(\theta)|)$,

which is not strong enough either.

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The change of variables $t = f(x)$ transforms your integral to $$J(a,b) = \int_{f(a)}^{f(b)} \frac{\cos(t)}{f'(f^{-1}(t))}\, dt$$ The integrand is positive for $(n-1/2) \pi < t < (n+1/2)\pi$ if $n$ is even and negative if $n$ is odd. Let $$J_n = \int_{(n-1/2) \pi}^{(n+1/2) \pi} \frac{\cos(t)}{f'(f^{-1}(t))}\ dt$$

Thus $J_n$ alternate in sign and decrease in absolute value, and it's easy to see that $$|J(a,b)| \le |J_n| < \frac{1}{\lambda} \int_{(n-1/2)\pi}^{(n+1/2)\pi} \cos(t)\ dt = \frac{2}{\lambda}$$ where $f(a) > (n-1/2) \pi$.

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I think you might get a slight error due to the integral starting and ending midperiod. I just answered with the normal proof of the Van der Corput lemma where you get ${\displaystyle {3 \over \lambda}}$, giving a worse constant due to these issues. –  Zarrax Sep 18 '11 at 22:14
I was assuming the conditions on $f$ apply to the whole real line. If not, extend it so $f'$ is constant outside $[a,b]$. If $(n-1/2) \pi < f(a) < (n+1/2) \pi$, you can increase the integral by moving $a$ in one direction and decrease it by moving in the other direction, similarly for $b$. So for a local maximum of the absolute value you have full periods. –  Robert Israel Sep 19 '11 at 0:17
What if instead of $f'>0$, one had $|f'|>0$ as in the original van der Corput lemma? –  Cantor Apr 12 '13 at 9:47

If you allow ${\displaystyle {3 \over \lambda}}$ the standard proof of the Van der Corput lemma applies: As you observed, your integral can be rewritten as $$\int_a^b {(\sin f(x))' \over f'(x)}\,dx$$ Integrate this by parts, integrating $(\sin f(x))'$ and differentiating ${1 \over f'(x)}$. You obtain $${\sin(f(b)) \over f'(b)} - {\sin(f(a)) \over f'(a)} + \int_a^b {\sin(f(x)) f''(x) \over (f'(x))^2}\,dx$$ This is bounded in absolute value by $$\bigg|{\sin(f(b)) \over f'(b)} - {\sin(f(a)) \over f'(a)}\bigg| + \int_a^b {\big|\sin(f(x)) f''(x)\big| \over (f'(x))^2}\,dx$$ $|\sin(f(x))| \leq 1$ and $f''(x) \geq 0$, so the above is bounded by $$\bigg|{\sin(f(b)) \over f'(b)} - {\sin(f(a)) \over f'(a)}\bigg| + \int_a^b { f''(x) \over (f'(x))^2}\,dx$$ The integral on the right evaluates to ${1 \over f'(a)} - {1 \over f'(b)}$, so the above is bounded by $$\bigg|{\sin(f(b)) \over f'(b)} - {\sin(f(a)) \over f'(a)}\bigg| + {1 \over f'(a)} - {1 \over f'(b)}$$ Since $f'(x)$ is monotone, you have $${1 \over f'(a)} - {1 \over f'(b)} < {1 \over f'(a)}$$ $$< {1 \over \lambda}$$ Furthermore, $$\bigg|{\sin(f(b)) \over f'(b)} - {\sin(f(a)) \over f'(a)}\bigg| \leq \bigg|{1 \over f'(a)}\bigg| + \bigg|{1 \over f'(b)}\bigg|$$ $$\leq {2 \over \lambda}$$ So adding together, your integral is bounded by ${\displaystyle {3 \over \lambda}}$.

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Well..., Van derCorput gives you $|\int_a^b e^{if(x)}dx|\leq 2/{\lambda}$, but $|\int_a^b e^{if(x)}dx| = |\int_a^b \cos(f(x))dx + i \int_a^b \sin(f(x))dx| \geq |\int_a^b \cos(f(x))dx|$ (the real part of a complex number is smaller than the magnitude of the complex numebr). So, there is no need to go to $3/{\lambda}$ as an upper bound.

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