By the Riemann-Lebesgue lemma, I have shown that for any finite interval measurable set $I$ of finite measure, any $h \in \mathbb{R}$, $$\lim_{n \to\infty}\int_I \cos (n(x+h)) \mathop{dx} = 0.$$

I also have an arbitrary real sequence $\{a_n\}$, and I would like to show $$\lim _{n \to \infty} \int_I \cos\left(n\left(x+\frac{a_n}{n}\right)\right)\mathop{dx} = 0$$

How can we see this? I tried by noting that for fixed $k$, we have $$\lim _{n \to \infty} \int_I \cos\left(n\left(x+\frac{a_k}{k}\right)\right)\mathop{dx} = 0$$ by the first equation above. I tried to do take this and do a "diagonalization argument," but I convinced myself that it would not work (if a countable number of sequences converge to a common limit, its "diagonal" sequence does not necessarily converge to that limit, if at all). I guess I haven't exploited the periodicity of cosine yet... any suggestions? Thanks!


Notice that your function can be easily integrated, so we don't need the Riemann-Lebesgue lemma.

Say, your interval is $I=[a,b]$. Then

$$\int_a^b \cos(nx+a_n) dx=\left.\frac{1}{n}\sin(nx+a_n)\right|_a^b$$

which converges to $0$ as $n\rightarrow\infty$.

More generally:

Claim: Let $f\in L^1(I)$ and $(a_n)_n$ any sequence of real numbers. Then $$\lim_{n\rightarrow\infty}\int_I \cos(nx+a_n) f(x) dx=0$$

The proof is the same as for the Riemann-Lebesgue lemma:

  • Assume without loss of generality that $f$ is smooth (smooth functions are dense in $L^1(I)$).
  • Do an integration by parts, integrating the $\cos$-term and giving a factor $1/n$.
  • Notice that this converges to $0$.
  • $\begingroup$ What happens if we generalize to measurable sets $I$ of finite measure? $\endgroup$
    – angryavian
    Feb 28 '14 at 16:53
  • $\begingroup$ The same, we also don't need finite measure. In fact this holds also on all of $\mathbb{R}$. $\endgroup$
    – J.R.
    Feb 28 '14 at 16:56
  • $\begingroup$ The point of Riemann-Lebesgue and the oscillatory integrals in general is to integrate by parts. That really is the whole story. $\endgroup$
    – J.R.
    Feb 28 '14 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.