Why do the Borwein integrals stop being $\frac{\pi}{2}$?

Question

I just received the book "single digits - In praise of Small Numbers" by Marc Chamberland.

In this book, he gives an interesting integral

$$\displaystyle \int_0^\infty \dfrac{\sin x}{x} = \dfrac{\pi}{2}$$

$$\displaystyle \int_0^\infty \dfrac{\sin(x)}{x}\dfrac{\sin(x/3)}{x/3} = \dfrac{\pi}{2}$$

$$\displaystyle \int_0^\infty \dfrac{\sin(x)}{x}\dfrac{\sin(x/3)}{x/3}\dfrac{\sin(x/5)}{x/5} = \dfrac{\pi}{2}$$

$$\displaystyle \int_0^\infty \dfrac{\sin(x)}{x}\dfrac{\sin(x/3)}{x/3}\dfrac{\sin(x/5)}{x/5}\dfrac{\sin(x/7)}{x/7} = \dfrac{\pi}{2}$$

$$\displaystyle \int_0^\infty \dfrac{\sin(x)}{x}\dfrac{\sin(x/3)}{x/3}\dfrac{\sin(x/5)}{x/5}\dfrac{\sin(x/7)}{x/7} \dfrac{\sin(x/9)}{x/9}= \dfrac{\pi}{2}$$

$$\displaystyle \int_0^\infty \dfrac{\sin(x)}{x}\dfrac{\sin(x/3)}{x/3}\dfrac{\sin(x/5)}{x/5}\dfrac{\sin(x/7)}{x/7} \dfrac{\sin(x/9)}{x/9}\dfrac{\sin(x/11)}{x/11}= \dfrac{\pi}{2}$$

$$\displaystyle \int_0^\infty \dfrac{\sin(x)}{x}\dfrac{\sin(x/3)}{x/3}\dfrac{\sin(x/5)}{x/5}\dfrac{\sin(x/7)}{x/7} \dfrac{\sin(x/9)}{x/9}\dfrac{\sin(x/11)}{x/11}\dfrac{\sin(x/13)}{x/13} = \dfrac{\pi}{2}$$

At this point, it is tempting to speculate that this pattern goes on forever, but we run into problems and this is another example of jumping to conclusions too soon.

$$\displaystyle \int_0^\infty \dfrac{\sin(x)}{x}\dfrac{\sin(x/3)}{x/3}\dfrac{\sin(x/5)}{x/5}\dfrac{\sin(x/7)}{x/7} \dfrac{\sin(x/9)}{x/9}\dfrac{\sin(x/11)}{x/11}\dfrac{\sin(x/13)}{x/13}\dfrac{\sin(x/15)}{x/15} = \dfrac{467807924713440738696537864469 \pi }{935615849440640907310521750000}$$

I calculated the next several and they are nice approximations to the results above, but not that result

$$\dfrac{17708695183056190642497315530628422295569865119 \pi }{35417390788301195294898352987527510935040000000}$$
$$\dfrac{8096799621940897567828686854312535486311061114550605367511653 \pi }{16193600755941299921751838065715269433640150152124763150000000}$$
$$\dfrac{2051563935160591194337436768610392837217226815379395891838337765936509 \pi }{4103129007448718822870650414175026723860506854636748901313920000000000}$$
$$\dfrac{37193167701690492344448194533283488902041049236760438302965167901187323851384840067287863 \pi }{74386376780038719358535506076609218130495936637120586884474907521986965251324791250000000}$$

He states "The explanation for this change is a bit technical, but the critical reason is that $\dfrac{1}{3} + \dfrac{1}{5} + \ldots + \dfrac{1}{13} \lt 1$, whereas, adding the next term $\frac{1}{15}$ pushes the sum over $1$, making a difference in the value of the integral."

He does not mention the researcher, but I'd like to know what is a "bit technical" explanation or if there is a more analytical or mathematical rationale or a reference to the research?

This is an interesting example of "jumping to a conclusion." (+1) — Mark Viola, Dec 04 '18 at 16:48
@MarkViola: Thanks. There is a post somewhere on the site that has a bunch of examples that are similar in nature, but I cannot find it at the moment, — Moo, Dec 04 '18 at 16:50
A derivation of the integral is given on Wiki::Borwein integral which explains the result (i.e. why it suddenly fails to hold true once the sum of some series gets large enough) — Winther, Dec 04 '18 at 17:02
A paper in the same spirit (formulas that holds for the first $N$ integers and then suddenly fails) might also be of interest "Fun with large numbers" by R. Baillie. — Winther, Dec 04 '18 at 17:05
@Winther Hi Hans. Happy Holidays. Thank you for the comments and embedded references! Much appreciated. -Mark — Mark Viola, Dec 04 '18 at 19:03
There's a variant of this identity that holds until 15,341,178,777,673,149,429,167,740,440,969,249,338,310,888 but fails at the next numbers :) see here. — Arnaud D., Dec 04 '18 at 19:39
@ArnaudD.: Excellent, thank you - I will have to play around with that one. — Moo, Dec 04 '18 at 20:43
Does this answer your question? Calculating $\int_0^\infty \frac{\sin(x)}{x} \frac{\sin(x / 3)}{x / 3} \frac{\sin(x / 5)}{x / 5} \cdots \frac{\sin(x / 15)}{x / 15} \ dx$ — Integrand, Oct 28 '21 at 11:46
@user170231: Very nice video - thanks for that link! I wonder what tool is used to do all those beautiful animations and to create a video like that. — Moo, Nov 16 '22 at 09:05

score 8 · Answer 1 · answered Nov 24 '21 at 20:20

Sort of an answer.

These integrals are known as the Borwein Integrals, found by David and Johnathan Borwein in Some remarkable properties of sinc and related integrals (2001).

According to wikipedia, if we have a sequence of nonzero reals $a_0,a_1,...,a_n$, we may evaluate $$\int_0^\infty \prod_{k=0}^{n}\frac{\sin a_k x}{a_kx}dx=\frac{\pi}{2a_0}C_n,$$ where $$C_n=\frac{1}{2^nn!\prod_{k=1}^{n}a_k}\sum_{\gamma\in\{\pm1\}^n}\varepsilon_\gamma b_\gamma^n\text{ sgn}(b_\gamma),$$ $$\gamma=(\gamma_1,\gamma_2,...,\gamma_n)\in\{-1,1\}^n,\qquad \varepsilon_\gamma=\gamma_1\gamma_2\cdots\gamma_n,$$ and $$b_\gamma=a_0+a_1\gamma_1+a_2\gamma_2+\dots+a_n\gamma_n.$$ Then, apparently, when $a_0>|a_1|+|a_2|+\dots+|a_n|,$ we have $C_n=1$. I am not sure how this follows from the explicit evaluation, but I'll update when I find out.

Anyway, taking $a_k=\frac1{2k+1}$ and $J_n=\int_0^\infty\prod_{k=0}^{n}\sin(a_kx)/(a_kx)\, dx$, we get $$J_1=\frac\pi2\cdot\frac{1}{2\cdot\tfrac13}\left(\frac43-\frac23\right)=\frac\pi2,$$ $$J_2=\frac\pi2\cdot\frac1{2^2\cdot2\cdot\tfrac13\tfrac15}\left(\left(\frac{23}{15}\right)^2-\left(\frac{17}{15}\right)^2-\left(\frac{13}{15}\right)^2+\left(\frac{7}{15}\right)^2\right)=\frac\pi2,$$ and so on. As you can see, there are $2^n$ terms for $J_n$, and I don't wanna have to write out anymore of them.

Finally, when $n=7$, we have $$\sum_{k=1}^{7}|a_k|=\sum_{k=1}^{7}\frac1{2k+1}=\frac{46207}{45045}\approx 1.0218>a_0=1,$$ and indeed, $$J_7\approx \frac\pi2-2.31\cdot10^{-11}.$$

Why do the Borwein integrals stop being $\frac{\pi}{2}$?

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