The value of $\displaystyle\int_0^{\infty } \left(\prod_{i=1}^n \frac{\sin \left(\frac{x}{2 i-1}\right)}{\frac{x}{2 i-1}}\right) \, \mathbb{d}x$

Asked Sep 29 '15 at 08:13

Active Nov 05 '22 at 10:30

Viewed 169 times

I came across this problem on Facebook the other day, so probably you guys have already seen it.

Define the following integral: $$I(n):=\int_0^{\infty } \left(\prod _{i=1}^n \frac{\sin \left(\frac{x}{2 i-1}\right)}{\frac{x}{2 i-1}}\right) \, \mathbb{d}x$$

Using Mathematica, I found the symbolic solutions are:

$$I(1)=I(2)=\dots=I(7)=\frac\pi2$$

However,

$$I(8)=\frac{467807924713440738696537864469 }{935615849440640907310521750000}\pi\approx0.499999999993\pi$$ $$I(9)=\frac{17708695183056190642497315530628422295569865119}{35417390788301195294898352987527510935040000000}\pi\approx0.499999994\pi$$

I didn't go further. My question is, should $I(n)=\frac\pi2$ for all positive integers $n$, and the above results are just some roundoff errors in symbolic calculation?

Edit 5 Nov 2022

3Blue1Brown offers an incredible explaination in here. It is much easier to see the reason with Fourier transform!

edited Nov 05 '22 at 10:30

Rodrigo de Azevedo

asked Sep 29 '15 at 08:13

zytsang

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I think this has appeared in some question on unexpected results or something like that. I can't find the link now, though. It is a funny property, but not so strange after it is explained. – mickep Sep 29 '15 at 08:21
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It's explained here: http://mathoverflow.net/a/11607/4678 – Hans Lundmark Sep 29 '15 at 09:15
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http://mathworld.wolfram.com/BorweinIntegrals.html – Jack D'Aurizio Sep 29 '15 at 12:28
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See also Borwein integral. – Lucian Sep 29 '15 at 14:19

The value of $\displaystyle\int_0^{\infty } \left(\prod_{i=1}^n \frac{\sin \left(\frac{x}{2 i-1}\right)}{\frac{x}{2 i-1}}\right) \, \mathbb{d}x$

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