Let $I_n=\int_{\pi}^{2\pi} \frac {\left|\sin(nx)\right|}{x}dx$

($a$) Prove $I_n \leq \ln(2)$

($b$) Prove $I_n \geq \frac {2}{\pi}(\frac 1{n+1}+ \frac 1{n+2}+...+\frac 1{2n})$

For ($a$)

We have

$$\left|\sin(nx)\right|\leq 1 \forall x\in\mathbb{R}$$

Let's divide by $x$ and integrate from $\pi$ to $2\pi$

We get:

$$\int_{\pi}^{2\pi}\frac{\left|\sin(nx)\right|}{x}dx\leq \int_{\pi}^{2\pi}\frac 1xdx$$

Leading to

$$I_n \leq \ln(2)$$

For ($b$) however I don't know how to start. Actually I tried to apply substitution $u=nx$ and then do a sum of integrals but doesn't really work.


Note that by letting $t=nx$ and then $s=t-(n\pi+ k\pi)$ we have that $\left|\sin(t)\right|=\left|\sin(s)\right|$, and $$\begin{align}I_n&=\int_{\pi}^{2\pi} \frac {\left|\sin(nx)\right|}{x}dx= \int_{n\pi}^{2n\pi} \frac {\left|\sin(t)\right|}{t}dt\\ & =\sum_{k=0}^{n-1}\int_{t=n\pi+k\pi}^{n\pi+(k+1)\pi}\frac {\left|\sin(t)\right|}{t}dt= \sum_{k=0}^{n-1}\int_{s=0}^{\pi}\frac {\left|\sin(s)\right|}{n\pi+k\pi+s}ds\\&\geq \sum_{k=0}^{n-1}\frac {1}{n\pi+k\pi+\pi}\int_{s=0}^{\pi}\left|\sin(s)\right|ds=\frac{2}{\pi}\left(\frac 1{n+1}+ \frac 1{n+2}+\dots+\frac 1{2n}\right).\end{align}$$

  • $\begingroup$ This is indeed what I was talking about, please be more specific. $\endgroup$ – C. Cristi May 4 '18 at 12:50
  • $\begingroup$ Also what do you mean by $s=0$ on the lower bound of the last integral $\endgroup$ – C. Cristi May 4 '18 at 12:50
  • $\begingroup$ Oh right, I see now $\endgroup$ – C. Cristi May 4 '18 at 12:51
  • $\begingroup$ Can you add more steps between the last 2? $\endgroup$ – C. Cristi May 4 '18 at 12:51
  • 1
    $\begingroup$ Since $\sin(t)=-\sin(t+\pi)$ then $|\sin(t)|=|\sin(t+n\pi)|$ for any integer $n$. $\endgroup$ – Robert Z May 4 '18 at 13:09

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