# Playing with Integrals: Limits and Integrals 4

This note wasn't a planned one, but yesterday I tried to solve a very beautiful problem proposed to me by Gabriel Stuart Romon. The problem asked to find the following limit

$\lim_{n\to\infty}\int^b_a f(x)|\sin(nx)|\,dx, \forall f(x)\subset C^0 [a,b].$

The detailed solution to this problem, written by Gabriel may be found here. However, I want to discuss another problem, which I found on math.stackexchange. The key idea of it helped me to elaborate the solution to Gabriel's problem. I hope you'll find it beautiful and interesting.

Problem. If $$g(x):\mathbb{R}\rightarrow\mathbb{R}$$ is a continuous periodic function with period 1 and $$f(x):[0,1]\rightarrow\mathbb{R}$$ is a continuous function, prove that $\lim_{n\to\infty}\int^1_0 f(x)g(nx)\,dx=\left(\int^1_0 f(x)\,dx\right)\left(\int^1_0 g(x)\,dx\right).$

Solution. First of all we'll break our single integral into the sum of integrals over the shorter intervals.

$\int^1_0 f(x)g(nx)\,dx=\sum^{n-1}_{i=0}\int^{\frac{i+1}{n}}_{\frac{i}{n}} f(x)g(nx)\,dx=$ $=\frac{1}{n}\left(\sum^{n-1}_{i=0}\int^{i+1}_i f\left(\frac{u}{n}\right)g(u)\,du\right)$

By applying mean-value theorem we'll transform our sum into something similar to a Riemann sum and we'll use the fact that $$g(x)$$ is periodic in our favor.

$\frac{1}{n}\left(\sum^{n-1}_{i=0}\int^{i+1}_{i} f\left(\frac{u}{n}\right)g(u)\,du\right)=\frac{1}{n}\sum^{n-1}_{i=0}\left(f(\xi_i)\int^{i+1}_{i} g(u)\,du\right)=$ $=\frac{1}{n}\left(\sum^{n-1}_{i=0}f(\xi_i)\right)\left(\int^1_0 g(u)\,du\right)$

Because every $$\xi_i\in\left[\dfrac{i}{n},\dfrac{i+1}{n}\right]$$, we have that the limit of our sum is $$\int^1_0 f(x)\, dx$$, so

$\lim_{n\to\infty}\int^1_0 f(x)g(nx)\,dx=\lim_{n\to\infty}\left(\frac{1}{n}\sum^{n-1}_{i=0}f(\xi_i)\right)\left(\int^1_0 g(x)\,dx\right)=$ $=\left(\int^1_0 f(x)\,dx\right)\left(\int^1_0 g(x)\,dx\right).$

Note by Nicolae Sapoval
5 years ago

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Hello Nicolae! Great post as always. :)

Can you please explain the step where you take $$f(\xi_i)$$ out of the integral? I fail to understand why it is a valid step.

Many thanks!

- 5 years ago

Yep, I apply the Mean Value Theorem, so it allows me to pull out the $$f(\xi_i)$$.

- 5 years ago

Thanks for the quick response!

I have never seen mean value theorem for integration, I am sorry if I ask stupid questions. From what I see at wiki, it states that this theorem is applicable if $$\varphi$$ ($$g(x)$$ in our case) does not change sign on $$(a,b)$$ (or $$i/n,(i+1)/n)$$ in our case). How do you know that this condition is followed here?

- 5 years ago

Yes, you're right my friend! thanks for pointing that out. Here is the missing part.

In case if $$g(x)$$ isn't positive for all reals (so it changes sign) we may find a constant $$M$$, such that $$g(x)+M> 0$$. The existence of the constant is caused by the afct that the function $$g(x)$$ is integrable. By propostion we proved

$\lim_{n\to\infty}\int^1_0f(x)(g(nx)+M)\,dx=\left(\int^1_0 (g(nx)+M)\,dx\right)\left(\int^1_0f(x)\,dx\right)\Leftrightarrow$ $\Leftrightarrow \lim_{n\to\infty}\int^1_0f(x)g(nx)\,dx=\left(\int^1_0 g(nx)\,dx\right)\left(\int^1_0f(x)\,dx\right).$

Comment. Those parts with $$M$$ are not affected by the limit and theya re equal on the $$RHS$$ and the $$LHS$$, so the problem is done.

- 5 years ago