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Problem 1. Suppose that $\xi>0$ and $\sin(2\xi)<0$. Let $$b_\nu=(N-v+1)\tfrac{\pi}{\xi}\quad\mbox{for}\quad\nu=1,\dots,N:=\big[\tfrac{\xi}{\pi}\big].$$ Prove that $$\mathrm{sgn}(\sin \xi)\int_{-1}^1t\prod_{\nu=1}^n(t^2-b_\nu^2)\sin(\xi t)\,dt>0$$ for every $n=1,2,\dots, N$.

Problem 2. Suppose that $\xi>0$ and $\sin(2\xi)>0$. Let $$a_\nu=(N-v+\tfrac12)\tfrac{\pi}{\xi}\quad\mbox{for}\quad\nu=1,\dots,N:=\big[\tfrac{\xi}{\pi}\big].$$ Prove that $$\mathrm{sgn}(\sin \xi)\int_{-1}^1\prod_{\nu=1}^n(t^2-a_\nu^2)\cos(\xi t)\,dt>0$$ for every $n=1,2,\dots, N$.

(The problem is posed 29.10.2016 by Marija Stanić on page 24 of Volume 1 of the Lviv Scottish Book).

More information on these problems can be found in this paper of G.Milovanovic, A.Cvetkovic, and M.Stanic.

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    $\begingroup$ the first question looks strange: we integrate the odd function against $[-1,1]$ $\endgroup$ – Fedor Petrov Oct 23 '18 at 19:06
  • $\begingroup$ @FedorPetrov Thank you for the comment. You are absolutely right. The first integral is equal to zero and Problem 1 has a trivial negative answer. So, I removed the first question but left the second one (which does not seem to have the obvious answer). $\endgroup$ – Lviv Scottish Book Oct 24 '18 at 5:50
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    $\begingroup$ Maybe she was thinking about the integral from 0 to 1? $\endgroup$ – Fedor Petrov Oct 24 '18 at 6:58
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    $\begingroup$ Marija Stanic has informed me that in the formulation of Problem 1 the variable $t$ was missing, which made the corresponding function odd. Now it is corrected. $\endgroup$ – Lviv Scottish Book Nov 23 '18 at 10:15
  • $\begingroup$ You have inverted $a_\nu$ and $b_\nu$ in the two inequalities. And even better with "dt". :) $\endgroup$ – Wolfgang Nov 23 '18 at 10:18

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