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

I'm currently trying to prove the Riemann-Lebesgue lemma using lower Darboux-sums and an approximation of any integrable function $f: [0,1] \to \mathbb{R}$ defined as $$t(x) := \begin{cases} m_i & \text{for } x_i \leq x < x_{i+1},\; i < n\\ m_n & \text{for } x= 1 \end{cases},$$ where $m_i := \inf\nolimits_{\xi \in [x_i, x_{i+1}]} f(\xi)$ and $P = \{x_i\}_{1 \leq i \leq n}$ is a partition of the interval $[0,1]$. Now I want to show that $$\int_0^1 t(x) \cos(\lambda x) \; \mathrm dx \leq \varepsilon$$ for some $\varepsilon > 0$.

Our assistant professor told us that it was possible to actually integrate $t(x) \cos(\lambda x)$ but I just don't see how to. I guess I would have to make use of the product rule twice (as $\int t(x) \; \mathrm dx = s(f,P)$), but the problem is that I don't know the integral of the lower Darboux-sum (or whether it even exists) and also I don't know the derivative of $t(x)$ (or whether it exists).

How am I supposed to proceed? Is there a different trick to integrate $t(x) \cos(\lambda x)$?

Thanks for any answers in advance.

share|cite|improve this question
Try to do the integral by splitting $[0,1]$ into $[x_0,x_1], [x_1,x_2],\cdots$. This way, on each interval $t(x)$ is a constant, and $\cos$ is something you know how to integrate. – Soarer Apr 14 '11 at 20:55
$t(x)$ is piecewise constant – yoyo Apr 14 '11 at 20:56
Integrating piecewise, I arrive at $\frac{1}{\lambda} \sum_{l=1}^{n-1} m_l \cdot (\sin(\lambda x_{l+1}) - \sin(\lambda x_l))$. Can I now simply estimate $\sum_l m_l \leq n \sup f(x)$ and then for all $\lambda \geq \lambda_0 = 2 n \sup f(x)/\varepsilon$ the integral is less or equal to $\varepsilon$? I ask this, because in the solution the integral is basically the same as mine but contains $(x_{i+1} - x_i)$ instead of $m_l$, where of course $\sum_i x_{i+1} - x_i = 1$ which would be lots nicer. – Huy Apr 14 '11 at 21:17
let $\lambda\to\infty$ in $(1/\lambda)\sum_l(\sin(\lambda x_{l+1})-\sin(\lambda x_l))$. dont get hung up on details; the point is that this gets small as $\lambda$ gets big – yoyo Apr 15 '11 at 1:22
up vote 0 down vote accepted

We can integrate piecewise:

$\int_0^1 t(x) \cos(\lambda x) \; \mathrm dx = \sum_{i=1}^{n-1} \int_{x_i}^{x_{i+1}} m_i \cos(\lambda x) \; \mathrm dx = \sum_{i=1}^{n-1} m_i \cdot \int_{x_i}^{x_{i+1}} \cos(\lambda x) \; \mathrm dx$

$ = \sum_{i=1}^{n-1} m_i \cdot \left. \left( \frac{1}{\lambda}\cdot \sin(\lambda x) \right) \right|_{x_i}^{x_{i+1}}$

$ = \frac{1}{\lambda} \cdot \sum_{i=1}^{n-1} m_i \cdot (\sin(\lambda x_{i+1}) - \sin(\lambda x_i)).$

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.