I tried to prove that $$\int_{a}^{b}\frac{|\cos (x)|}{x}dx\leq \frac{2}{\pi}\log\left(\frac{b}{a}\right)+O(1).$$ Of course $O(1)$ as a function of $b$, i.e. a bounded function of $b$. $a$ is considered constant with respect to $b$. I found this inequality in a paper. There, it is written that we can be prove it by splitting the integral in integrals of length $2\pi$ and by using the basic equality $$\frac{1}{2\pi}\int_0^{2\pi}|\cos(x)|dx=\frac{2}{\pi}.$$ I tried to follow the aforementioned hint, but I got nowhere. I suppose that the term $\log(b/a)$ pops out from the integral $\int_{a}^{b}x^{-1}dx$, but I could not make this integral appear and obtain the desired inequality at the same time.

Do you have any hint? I suspect that it must be easy, but I am probably missing something now. I hope it is not a duplicate. Thanks in advance for your help!

  • $\begingroup$ I suspect you do a Taylor expansion for this problem. $\endgroup$ – user23793 Nov 28 '18 at 2:09
  • $\begingroup$ @user23793 And what do you do after that actually? $\endgroup$ – Sachpazis Stelios Nov 28 '18 at 2:11
  • $\begingroup$ What are the conditions on $a$ and $b$? When you say "prove it by splitting the integral in integrals of length $2\pi$", the length of $[a,b]$ may be less than $2\pi$, in which case you may have to consider some cases. Also, could you clarify what you mean in the sentence I quoted? $\endgroup$ – user23793 Nov 28 '18 at 2:34
  • $\begingroup$ @user23793 Usually the length of the interval is large, a lot greater than $2\pi.$ There are no other conditions on $a$ and $b$. We just have to know that both sides are treated as functions of $b$, while $a$ is considered to be constant in terms of $a$. I also edited the question about this clarification. $\endgroup$ – Sachpazis Stelios Nov 28 '18 at 2:39
  • $\begingroup$ The reason I ask is that there might be a singularity at $0$ if $0\in [a,b]$. It seems that there is some missing information. $\endgroup$ – user23793 Nov 28 '18 at 2:44

We assume that $a\geq 4\pi$ and $b>a$.

One way to prove the inequality is to partition the range $[a,b]$ into the sets $I_n = [ 2 \pi n , 2\pi (n+1)] $. In particular, we have that $$\int_{a}^{b}\frac{|\cos (x)|}{x}dx = \sum_{n=n_0}^{n_1} \int_{I_n \cap [a,b]} \frac{|\cos (x)|}{x}dx$$ with $n_0 = \lfloor a/2\pi \rfloor$ and $n_1 = \lceil b/2\pi \rceil$.

Now, we have that $$\int_{a}^{b}\frac{|\cos (x)|}{x}dx \leq \sum_{n=n_0}^{n_1} \int_{I_n} \frac{|\cos (x)|}{x}dx \leq \sum_{n=n_0}^{n_1} \int_{I_n} \frac{|\cos (x)|}{2\pi n}dx = \frac{2}{\pi} \sum_{n=n_0}^{n_1}\frac{1}{2\pi n} \leq \frac{2}{\pi} \int_{\lfloor a/2\pi \rfloor -1}^{\lceil b/2\pi \rceil} \frac{dn}{2\pi n} \\= \frac{2}{\pi} \log\left(\frac{b}a\right) + O(1)\,.$$

  • $\begingroup$ Ahh, I knew that it had to be something simple. That time I was stuck. Anyway, thanks for the answer! $\endgroup$ – Sachpazis Stelios Nov 29 '18 at 1:32
  • $\begingroup$ Just a comment. You don't have the $2\pi$ in the denominators after using the integral equality. However, the rest remains valid, because the $2\pi$'s will cancel in the logarithm at the end. $\endgroup$ – Sachpazis Stelios Nov 29 '18 at 1:39

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.