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.

  • 4
    $\begingroup$ $\cos^n{x} \rightarrow 0$ almost everywhere. So use the dominated convergence theorem. $\endgroup$
    – Aphelli
    Jul 5, 2019 at 9:13
  • $\begingroup$ @Mindlack There it is! I didn't aware of the fact that $\cos ^n x\to 0$ a.e. Thank you! $\endgroup$
    – Bach
    Jul 5, 2019 at 9:16
  • $\begingroup$ If you know what the dominated convergence theorem is, you should certainly know that $\cos^n x \to 0$ a.e....... $\endgroup$ Jul 5, 2019 at 9:36

1 Answer 1


We can use the dominated convergence theorem.

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

$$ \lim_{n\to\infty}\int_{-\infty}^\infty (\cos x)^nf(x)dx = 0$$

  • $\begingroup$ $\cos(x)^n\to 0$ does not hold for all $x\in\Bbb R$, but only for almost all $x\in\Bbb R$. This is enough, though. $\endgroup$ Jul 5, 2019 at 13:48
  • $\begingroup$ @MarsPlastic You're right. Thank you! $\endgroup$
    – J.A.G
    Jul 5, 2019 at 13:50

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