6
$\begingroup$

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.

  • 1
    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

1 Answers 1

3

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.

  • 0
    I had just realized this error, but was still trying to understand the inequality. Thank you Siminore, that was really helpfull!!2012-08-08