Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 6 down vote accepted

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 $f(a) > (n-1/2) \pi$.

share|cite|improve this answer
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 ${\displaystyle {3 \over \lambda}}$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.