1. Techniques of integration (6)

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

2. Originally Posted by NonCommAlg
Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and periodic with period $T > 0.$ For any real numbers $a evaluate $\lim_{n\to\infty} \int_a^b f(nx) \ dx.$
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 $u=nx,$

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

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

$=\ \ 0$ since the integral is bounded

Something tells me I may have done something wrong, because, well, surely it can’t be that simple …

3. Originally Posted by TheAbstractionist
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 $u=nx,$
$\lim_{n\,\to\,\infty}\int_a^bf(nx)\,dx$
$=\ \ \lim_{n\,\to\,\infty}\frac1n\int_{na}^{nb}f(u)\,du$
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 $f_{\text{av}} = \frac1T\int_0^Tf(x)\,dx$ be the mean value of f over one period. Then the interval [na,nb] consists of $\frac{n(b-a)}T$ subintervals of length T (not counting odd bits at the ends, which we can dispose of with epsilons). So $\frac1n\int_{na}^{nb}f(u)\,du\approx (b-a)f_{\text{av}}$. That's my candidate for the limit.

4. Originally Posted by Opalg
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.
Aha, that was where I went wrong. I knew I had made a mistake somewhere but just couldn’t see where.

Thanks, Opalg.

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

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

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

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

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