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I'm having trouble verifying this inequality. It goes like this (appears in Giaquinta, Mathematical analysis, linear and metric structures, page 445): $$ \int_{0}^{\pi} \cfrac{\sin(x)}{\sin\left(\frac{x}{2n+1}\right)} dx \leq\frac{ 2(n+1)\pi}{2n+1} \leq 2\pi $$ Of course, the last inequality is obvious. The first one, however, I can't show. I've tried bounding $\sin(x)$ by $1$, and then calculating the integral with mathematica, but it comes out unbounded. When I put $n=1,2,3...$ or any finite number in mathematica, the result is numerically true, but I want to show this for any "$n$", and mathematica gives me a very complicated function (depending on $n$) with imaginary units and hypergeometric functions. I guess I'm missing out a very simple argument here. Any ideas?

Edit: I have edited so that the formula is identical to the one of the book.

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If you just write out function names like that, $\TeX$ interprets them as strings of variable names, which it italicizes. To get the right formatting, you need to use the predefined commands like \sin or, if you need a function for which there's no predefined command, \operatorname{name}. – joriki Aug 8 '12 at 14:52
I have reformatted your integrand. Please make sure I've done so correctly. – Cameron Buie Aug 8 '12 at 15:11
Yes, that's correct. Thank you both! – Pani Aug 8 '12 at 15:14
The purported inequality does not hold already for $n=2$. Indeed, $\int_0^\pi \frac{\sin(x)}{2 \sin(x/5)} \mathrm{d} x = \frac{\pi}{2} + \frac{5}{16} \sqrt{50+22\sqrt{5}} \approx 4.68317 > \pi$ – Sasha Aug 8 '12 at 15:49
Thanks Americo, I was observing this fact empirically with mathematica. Nevertheless Siminore found the mistake in the textbook where this whole misunderstanding came from. – Pani Aug 8 '12 at 17:05

Giaquinta and Modica are right, although they are wrong. Let me explain: the book suggests to compute $G_n \left( \frac{2 \pi}{2n+1} \right)$, where $$ G_n(x) = \int_0^x D_n(t)\, dt $$ and $D_n$ is the standard Dirichlet kernel. In their computation there is a very stupid mistake: they change variable and they forget to change the differential! Here is a corrected version of their formula on page 445: $$ \begin{align*} \|G_n\|_\infty &= G_n \left( \frac{2 \pi}{2n+1} \right) = \int_0^{2\pi/(2n+1)} \frac{\sin (n+1/2)s)}{\sin (s/2)}ds\\ &= \frac{2}{2n+1} \int_0^\pi \frac{\sin t}{\sin \frac{t}{2n+1}}dt \leq \frac{2}{2n+1} \cdot (2n+1)\pi = 2\pi. \end{align*} $$ We have used the fact that $t \mapsto \frac{\sin t}{\sin \frac{t}{2n+1}}$ is decreasing, so that it is always less that its limit as $t \to 0^+$.

I hope my answer will be useful.

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I had just realized this error, but was still trying to understand the inequality. Thank you Siminore, that was really helpfull!! – Pani Aug 8 '12 at 17:02

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