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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}`.2012-08-08
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    I have reformatted your integrand. Please make sure I've done so correctly.2012-08-08
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    Yes, that's correct. Thank you both!2012-08-08
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    The purported inequality [does not hold](http://www.wolframalpha.com/input/?i=integrate%28sin%28x%29%2F%282*sin%28x%2F5%29%29%2C{x%2C0%2CPi}%29+%3C+Pi) 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$2012-08-08
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    Sasha is right. My mathematica command had an unfortunate error as well, and the inequality does not hold, so what i wrote in my question is wrong. I am so sorry. But if this is wrong i wonder what Giaquinta meant, because... if this does not hold, i have a bigger problem than before.2012-08-08
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    The given integral is greater than $n$ $$\begin{equation*} \int_{0}^{\pi }\frac{\sin x}{2\sin \left(\frac{x}{2n+1}\right)}dx\geq \int_{0}^{\pi } \frac{\sin x}{\frac{2x}{2n+1}}dx=\frac{1}{2}\left( 2n+1\right) \int_{0}^{\pi }\frac{\sin x}{x}\geq n, \end{equation*}$$ because for $0\leq x\leq \pi $ $$\begin{equation*} \sin \left(\frac{x}{2n+1}\right)\leq \frac{x}{2n+1}. \end{equation*}$$2012-08-08
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    Thanks Americo, I was observing this fact empirically with mathematica. Nevertheless Siminore found the mistake in the textbook where this whole misunderstanding came from.2012-08-08

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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!!2012-08-08