# An integral trigonometric inequality

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.

• the first question looks strange: we integrate the odd function against $[-1,1]$ – Fedor Petrov Oct 23 '18 at 19:06
• @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). – Lviv Scottish Book Oct 24 '18 at 5:50
• Maybe she was thinking about the integral from 0 to 1? – Fedor Petrov Oct 24 '18 at 6:58
• 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. – Lviv Scottish Book Nov 23 '18 at 10:15
• You have inverted $a_\nu$ and $b_\nu$ in the two inequalities. And even better with "dt". :) – Wolfgang Nov 23 '18 at 10:18