Computing $\lim_{n \rightarrow\infty} \int_{a}^{b}\left ( f(x)\left | \sin(nx) \right | \right )$ with $f$ continuous on $[a,b]$

Let $a,b \in \mathbb{R}$ and $\textit{f} :[a,b] \rightarrow \mathbb{R}$ continuous on $[a,b]$.

Does the sequence $\left (\int_{a}^{b} f(x)\left |\sin(nx) \right |dx \right )$ converge? If it does, what is its limit ?

I know how to solve this for $\left(\int_{a}^{b} f(x)\sin(nx)dx \right )$ with integration by parts when $f$ is a class $C^1$ function.

Here, I don't know how to deal with the absolute value and the non-differentiability of $\textit{f}$.

Any help is appreciated, thanks in advance.

I got it:

• Let $I_n=\int_{a}^{b} f(x)\left |\sin(nx) \right |dx$

• Let $p_n$ be the root of $sin(nx)$ that is closest to $a$ and $q_{n}$ the root of $sin(nx)$ that is closest to $b$ with $a<p_n$ and $q_n<b$, then $p_n$ and $q_n$ are both multiples of $\frac{\pi}{n}$

• Let's define a partition of $[a,b]$ $a=t_0<t_1=p_n<\ldots<t_{B_n-1}=q_n<t_{B_n}=b$ where the other $t_i$ are all the multiples of $\frac{\pi}{n}$ $\in ]p_n,q_n[$

then $$I_n=\int_{a}^{p_n} f(x)\left |\sin(nx) \right |dx + \int_{p_n}^{q_n} f(x)\left |\sin(nx) \right |dx+\int_{q_n}^{b} f(x)\left |\sin(nx) \right |dx$$

• Let's deal with $\int_{p_n}^{q_n} f(x)\left |\sin(nx) \right |dx$

It can be rewritten as $$\sum _{ k=1 }^{ B_n -1 }{ \int _{ t_k }^{ t_{k+1} }{ f(x)\left |\sin(nx) \right |dx } }$$

According to the mean value theorem, For every $[t_k,t_{k+1}]$, there exists an $x_k$ such that $\int _{ t_k }^{ t_{k+1} }{ f(x)\left |\sin(nx) \right |dx } = f(x_k)\int _{ t_k }^{ t_{k+1} }{\left |\sin(nx) \right |dx }$ It can be easily proven that $\int _{ t_k }^{ t_{k+1} }{\left |\sin(nx) \right |dx } = 2/n$ Then $$\int_{p_n}^{q_n} f(x)\left |\sin(nx) \right |dx = \sum _{ k=1 }^{ B_n -1 }{ f(x_k)2/n} = \sum _{ k=1 }^{ B_n -1 }{ (t_{k+1}-t_k)f(x_k)\frac{2}{\pi}}$$

$\sum _{ k=0 }^{ B_n }{ (t_{k+1}-t_k)f(x_k)}$ is a Riemann sum which converges to $\int_{a}^{b} f(x) dx$ since the norm of the partition is $\pi/n$

Then $$\sum _{ k=1 }^{ B_n -1 }{ (t_{k+1}-t_k)f(x_k)}$$ converges to $\int_{a}^{b} f(x) dx$ as n goes to infinity

Hence $$\frac{\pi}{2}\int_{p_n}^{q_n} f(x)\left |\sin(nx) \right |dx$$ converges to $\int_{a}^{b} f(x) dx$

• Let's deal with the remaining integrals.

Since $f$ is continuous on $[a,b]$, it is bounded by say $M$.

$| \int_{a}^{p_n} f(x)\left |\sin(nx) \right |dx | \leq M|\int_{a}^{p_n} \left |\sin(nx) \right |dx \leq M(p_n -a) \leq Mπ/n$

Therefore both remaining terms go to 0.

• Conclusion: $I_n$ converges to $$\frac{2}{\pi}\int_{a}^{b} f(x) dx$$
• This is the approach I would have taken so (+1). Just curious, but you posted the question and answered it. I didn't think that one could award oneself with the "best vote." How is that possible? -Mark – Mark Viola Jan 19 '17 at 20:44
• There's actually an article in the Help Center about that. There's a 48 hours delay before you can accept your own answer. – Gabriel Romon Jan 19 '17 at 20:51
• I've posted a few questions and provided answers that I believe to be solid. But I am reticent to award myself the "best answer" vote. Do you receive 15 points for doing so? – Mark Viola Jan 19 '17 at 20:56
• No, you don't receive additional points. Moreover, your answer need not float to the top, it will be ranked like any other answer on the page. – Gabriel Romon Jan 19 '17 at 21:00
• See my answer here for example. It stands at the bottom of the page with respect to votes. – Gabriel Romon Jan 19 '17 at 21:03

We should perhaps try to explain that $f(x)$ and $\sin(nx)$ are "asymptotically independent" as $n \to \infty$, so $$\lim_{n\to\infty}\frac{1}{b-a}\int_a^b f(x)\,|\sin(nx)|\,dx = \frac{1}{b-a}\int_a^b f(x)\,dx \cdot \frac{1}{\pi}\int_0^\pi \sin(x)\, dx$$ which agrees with Gabriel's answer.

A hint: For a given $n\gg 1$ consider partitions $$a=t_0<t_1<\ldots<t_N=b$$ of $[a,b]$ such that the $t_k$ with $1\leq k\leq N-1$ are the zeros of $x\mapsto \sin (n x)$ on $[a,b]$. Consider besides $f$ a step function $\tilde f$ assuming in each subinterval $[t_{k-1},t_k]$ a suitable constant value, see the edit below. Compare the integrals, and in the end use the uniform continuity of $f$ on $[a,b]$.

Edit. You may use the following version of the mean value theorem of integral calculus: When $f$ is continuous on $[u,v]$ and $g(x)\geq0$ on $[u,v]$ then there is a $\xi\in[u,v]$ such that $$\int_u^v f(x)g(x)\ dx= f(\xi) \int_u^v g(x)\ dx\ .$$

• I don't get how to choose the values of the step function...thanks for your help – Gabriel Romon Mar 12 '13 at 18:51