Techniques of integration (6)

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May 22nd 2009, 07:55 AM
NonCommAlg

Techniques of integration (6)

Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and periodic with period $T > 0.$ For any real numbers $a<b$ evaluate $\lim_{n\to\infty} \int_a^b f(nx) \ dx.$
May 22nd 2009, 10:54 AM
TheAbstractionist

Quote:

Originally Posted by NonCommAlg

Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be continuous and periodic with period $T > 0.$ For any real numbers $a<b$ 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 … (Thinking)
May 22nd 2009, 01:12 PM
Opalg

Quote:

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.
May 22nd 2009, 02:15 PM
TheAbstractionist

Quote:

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. (Happy)
May 22nd 2009, 05:54 PM
NonCommAlg

Quote:

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).