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I don't know why, but this equation seems to be true for several functions that I've tested. "a" is a real value and "f" is some function.

$$\lim_{n \to \infty} \int_0^a f(t, \sin{nt})\,\mathrm dx = \lim_{n \to \infty} \int_0^a f(t, \cos{nt})\,\mathrm dx$$

ex. $$\lim_{n \to \infty} \int_0^{\pi/2} \frac{\sin^2{nt}}{1+t} \,\mathrm dt = \lim_{n \to \infty} \int_0^{\pi/2} \frac{\cos^2{nt}}{1+t}\,\mathrm dt = \frac{1}{2}\log(1+\frac{\pi}{2})$$

Graph Image Here is a graph that graphs the equation above. https://www.desmos.com/calculator/20m87pszz8

My intuition tells me that each pair of neighboring hills made by sin and cos will eventually be indistinguishable and therefore the area will be the same. However, I don't know a way to prove this.

Does anyone know a proof or a theorem relating to this problem?

I appreciate your help.

Mittens
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Ichi Hama
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1 Answers1

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Your integral, at least the one in your posting, is of the form $$\int_I f(t)g(nt)\,dt$$ where $g$ is bounded and $T$-periodic ($g(x+T)=g(x)$ for all $x$), $I$ is an interval (finite or not), and $f$ is a nice function. The interpretation is that as $n\rightarrow\infty$, the periodic function $averages out$.

  • A result by Féjer, states that under if $f$ is integrable over $I$, then $$\lim_m\int_I f(t)g(nt)dt\xrightarrow{n\rightarrow\infty}\frac{1}{T}\int^T_0g(t)\,dt \cdot \int_I f(t)\,dt$$ Notice that $\sin^2(t)$ and $\cos^2(t)$ are both $\pi$ periodic and $\int^\pi_0\sin^2t\,dt=\int^\pi_0\cos^2t\,dt$

There a several postings on MSE about this with different levels of difficulty. Here are a few (1, 2, 3, 4).

  • On can also use the identities $\sin^2t=\frac{1-\cos 2t}{2}$ and $\cos^2t=\frac{1+\cos2t}{2}$ an use a well known result: Riemann-Lebesgue lemma, which states that for any nice function $f$ $$\lim_n\int_I f(t)\sin nt\,dt=0=\lim_n\int_I f(t)\cos nt\,dt$$ The Riemann-Lebesgue lemma can be obtained from the result of Féjer,
Mittens
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