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Question

Let $f:[1,\infty)\to \mathbb{R}$ s.t:

  • $f\in C^{1}$
  • $f(x)\ge 0$
  • $\forall x<y:f(y)\le f(x)$ , $x\ne y \in [1,\infty) $
  • $\lim_{x\to \infty} xf(x)=L\in (0,\infty)$

Prove $\int_{1}^{\infty}f(x) \cos{x} \ dx < \infty $ and $\int_{1}^{\infty} \left| f(x) \cos{x} \right| \ dx $ diverges.

I want to show $\lim_{x\to \infty }f(x)=0$ because then I can prove $\int_{1}^{\infty}f(x) \cos{x} \ dx < \infty $ by Dirichlet's test for convergence of improper integral. As for $\int_{1}^{\infty} \left| f(x) \cos{x} \right| \ dx $ I have not idea where to begin. I would appreciate guidance/answer.

Thank you in advance!

EDIT

  • I see now why $\lim_{x\to \infty }f(x)=0$.
Lior
  • 623
  • Your conditions imply that $0 \le f(x) < \frac{2L}{x}$ for sufficiently large $x$. – Martin R Jul 10 '23 at 13:08
  • @MartinR You're right. Thanks!! – Lior Jul 10 '23 at 13:12
  • @MartinR I edited the post and added what else needed to be proven (I forgot to mention it in the first place). – Lior Jul 10 '23 at 13:18
  • For the second integral apply the fact that $\int\limits_1^\infty{|\cos x|\over x},dx$ is divergent. – Ryszard Szwarc Jul 10 '23 at 14:30
  • @RyszardSzwarc In order to use that you need to show $\frac{|\cos x|}{x} \le \left| f(x) \cos{x} \right| $. How can I deduce that?

    I thought using $\lim_{x\to \infty} xf(x)=L\in (0,\infty)$ and say $0<\frac{L}{2}<xf(x)$ for sufficiently large $x$. Then multiply it by $|\cos{x}|$ and devide by $x$.

     – Lior Jul 10 '23 at 15:17
  • That's exactly what I had in mind: $f(x)|\cos x|\ge {L\over 2x}|\cos x|$ for large $x.$ – Ryszard Szwarc Jul 10 '23 at 16:13
  • @RyszardSzwarc Cool. Thanks :) – Lior Jul 10 '23 at 16:18
  • You came up with the same idea :) – Ryszard Szwarc Jul 10 '23 at 16:19
  • It follows from Dirichlet's test for convergence of improper integrals that $\int_{1}^{\infty}f(x) \cos{x} \ dx < \infty $ -see e.g. https://math.stackexchange.com/questions/141048/dirichlets-test-for-convergence-of-improper-integrals – Botnakov N. Jul 10 '23 at 22:09
  • @BotnakovN. I know, I said it in the post. – Lior Jul 11 '23 at 06:46

0 Answers0