0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 '19 at 10:00
1
$\begingroup$

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

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

-->
$\endgroup$
2
  • $\begingroup$ What quick with this answer! Many thanks, I known just the approximation $\approx -0.00150029$ $\endgroup$
    – user142929
    Sep 3 '19 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 '19 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.