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This post is inspired in the Borwein integral, and in a problem proposed by Ovidiu Furdui in Crux Mathematicorum, that is the Problem 3707, in page 151.

I've considered integrals of the form $$\int_0^\infty\prod_{k=1}^n\frac{\sin^2\left(\frac{x}{k}\right)}{\pi^2-\left(\frac{x}{k}\right)^2}dx\tag{1}$$ for integers $n\geq 1$. My belief is that $(1)$ isn't exactly Ovidiu's integral.

Question. Can you show that the case $n=3$ the integral in $(1)$ is $\neq 0$? Is it possible to find $n>2$ such that the integral in $(1)$ is equals to $0$ again? Many thanks.

Please if the integral in $(1)$ is Ovidiu's integral, or it is in the literatute, add a comment. I think that isn't the same integral, using Wolfram Alpha online calculator an alternate form of the difference of integrands, for the second case is showed as using the input

sin^2(x)/(pi^2-x^2)sin^2(x/2)/(pi^2-(x/2)^2)- sin^4(x)/((pi^2-x^2)(4pi^2-x^2))

References:

[1] Borwein integral, from the encyclopedia Wikipedia.

[2] Problem 3707, Crux Mathematicorum, Volume 38, Number 4, April 2012.

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  • $\begingroup$ The web page of Crux Mathematicorum (I believe that is a journal of the Canadian Mathematical Society) is cms.math.ca/crux from which you can see its Digital Archive. $\endgroup$
    – user142929
    Sep 3, 2019 at 10:00

1 Answer 1

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$$I_3=\int_0^\infty \frac{ \sin ^2(x)\sin ^2\left(\frac{x}{2}\right)\sin ^2\left(\frac{x}{3}\right)}{\left(\pi ^2-x^2\right) \left(\pi ^2-\frac{x^2}{4}\right) \left(\pi ^2-\frac{x^2}{9}\right)}\,dx=-\frac{27 \sqrt{3}}{320 \pi ^4}=-0.00150029\cdots.$$

and $I_4=-\frac{9 \left(17 \sqrt{3}+10\right)}{2240 \pi ^6}$.

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  • $\begingroup$ What quick with this answer! Many thanks, I known just the approximation $\approx -0.00150029$ $\endgroup$
    – user142929
    Sep 3, 2019 at 11:05
  • $\begingroup$ Other problem from which I tried to do variants, but I've failed, is Exercise 4.76 from the book Claude George, Exercises in Integration, Problem Books in Mathematics, Springer-Verlag (1984). I add it, if you are interesting in this kind of questions. $\endgroup$
    – user142929
    Sep 4, 2019 at 8:10

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