Let $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and periodic with period $\displaystyle T > 0.$ For any real numbers $\displaystyle a<b$ evaluate $\displaystyle \lim_{n\to\infty} \int_a^b f(nx) \ dx.$

Printable View

- May 22nd 2009, 07:55 AMNonCommAlgTechniques of integration (6)
Let $\displaystyle f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and periodic with period $\displaystyle T > 0.$ For any real numbers $\displaystyle a<b$ evaluate $\displaystyle \lim_{n\to\infty} \int_a^b f(nx) \ dx.$

- May 22nd 2009, 10:54 AMTheAbstractionist
Hi

**NonCommAlg**.

A continuous function is integrable, and the integral over a bounded interval of an integrable periodic function is bounded. Hence, using the substitution $\displaystyle u=nx,$

$\displaystyle \lim_{n\,\to\,\infty}\int_a^bf(nx)\,dx$

$\displaystyle =\ \ \lim_{n\,\to\,\infty}\frac1n\int_{na}^{nb}f(u)\,du$

$\displaystyle =\ \ 0$ since the integral is bounded

Something tells me I may have done something wrong, because, well, surely it can’t be that simple … (Thinking) - May 22nd 2009, 01:12 PMOpalg
I'd agree with that approach up to that point, but I wouldn't go on to say that the limit is 0, because the length of the interval (in the u-integral) is getting unboundedly long.

Let $\displaystyle f_{\text{av}} = \frac1T\int_0^Tf(x)\,dx$ be the mean value of f over one period. Then the interval [na,nb] consists of $\displaystyle \frac{n(b-a)}T$ subintervals of length T (not counting odd bits at the ends, which we can dispose of with epsilons). So $\displaystyle \frac1n\int_{na}^{nb}f(u)\,du\approx (b-a)f_{\text{av}}$. That's my candidate for the limit. - May 22nd 2009, 02:15 PMTheAbstractionist
- May 22nd 2009, 05:54 PMNonCommAlg
Opalg's candidate, which is $\displaystyle \frac{b-a}{T} \int_0^T f(x) \ dx,$ is the correct answer. here's a more detailed solution:

$\displaystyle \int_a^b f(nx) \ dx = \frac{1}{n} \int_{na}^{nb} f(u) \ du= \frac{1}{n} \left[\sum_{k=0}^{m-1} \int_{na+kT}^{na+(k+1)T} f(u) \ du + \int_{na+mT}^{nb} f(u) \ du \right],$ where $\displaystyle na+mT < b \leq na+(m+1)T. \ \ \ \ \ (1)$

but $\displaystyle \int_{na+kT}^{na+(k+1)T} f(u) \ du =\int_0^T f(u) \ du,$ since $\displaystyle T$ is the period of $\displaystyle f.$ thus: $\displaystyle \int_a^b f(nx) \ dx = \frac{m}{n} \int_0^T f(u) \ du + \frac{1}{n}\int_{na+mT}^{nb} f(u) \ du. \ \ \ (2)$

we also from (1) have that $\displaystyle \lim_{n\to\infty}\frac{m}{n}=\frac{b-a}{T},$ which completes the proof because $\displaystyle f$ is bounded, say by $\displaystyle K,$ and thus:

$\displaystyle \left| \frac{1}{n} \int_{na+mT}^{nb} f(u) \ du \right| \leq(nb-na-mT)\frac{K}{n}=\left(\frac{b-a}{T}-\frac{m}{n} \right) \frac{K}{T}.$ hence: $\displaystyle \lim_{n\to\infty} \frac{1}{n} \int_{na+mT}^{nb} f(u) \ du =0,$ and the result follows from (2).