# Let $f$ be a Lebesgue integrable function on $\mathbb R$. Prove that $\lim_{n\to\infty}\int_{-\infty}^\infty (\cos x)^nf(x)dx=0$

Let $$f$$ be a Lebesgue integrable function on $$\mathbb R$$. Prove that $$\lim_{n\to\infty}\int_{-\infty}^\infty (\cos x)^nf(x)dx=0$$

My attempt:

Firstly, I know the statement is true when the integrand is $$\cos(nx)f(x)$$ which is implied by the Riemann-Lebesgue lemma. (See this post.)

Now I want to apply the same method, i.e., using simple functions to deduce the limit of the integral of $$(\cos x)^n$$ over any interval is zero. However, I think there is not a nice formula for the integral of $$(\cos x)^n$$. Though we know the recurrence formula: $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx,$$ indeed, this can lead us to the conclusion by noticing the fact that the first term above on the RHS tends to zero as $$n\to\infty$$ and also $$\lim_{n\to\infty}\frac{(n-1)!!}{n!!}=0.$$

I think this should not be the best way for us to deal with this kind of problems since this relies on the recurrence formula of $$\int\cos^n x dx$$. Can someone provide other approaches? Thank you.

• $\cos^n{x} \rightarrow 0$ almost everywhere. So use the dominated convergence theorem. Jul 5, 2019 at 9:13
• @Mindlack There it is! I didn't aware of the fact that $\cos ^n x\to 0$ a.e. Thank you!
– Bach
Jul 5, 2019 at 9:16
• If you know what the dominated convergence theorem is, you should certainly know that $\cos^n x \to 0$ a.e....... Jul 5, 2019 at 9:36

Consider $$g_{n}(x) = (cos(x))^{n}f(x)$$ measurable. Then $$|g_{n}(x)| = |(cos(x))^{n}f(x)| = |(cos(x))^{n}||f(x)| = |cos(x)|^{n}|f(x)| \leqslant |f(x)|$$. We have that |f| is integrable because f is integrable. Thus,
$$\lim_{n\to\infty}\int_{-\infty}^\infty (\cos x)^nf(x)dx = \int_{-\infty}^\infty \lim_{n \to \infty}(\cos x)^nf(x)dx$$
Now, since $$\lim_{n \to \infty}(cos(x))^{n}f(x) = 0$$, because $$\lim_{n \to \infty}(cos(x))^{n} = 0$$ a. e, we have
$$\lim_{n\to\infty}\int_{-\infty}^\infty (\cos x)^nf(x)dx = 0$$
• $\cos(x)^n\to 0$ does not hold for all $x\in\Bbb R$, but only for almost all $x\in\Bbb R$. This is enough, though. Jul 5, 2019 at 13:48