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

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) Dec 4, 2018 at 16:48
• 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) Dec 4, 2018 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. Dec 4, 2018 at 17:05
• @Winther Hi Hans. Happy Holidays. Thank you for the comments and embedded references! Much appreciated. -Mark Dec 4, 2018 at 19:03
• 3Blue1Brown recently posted a video on this very thing Nov 16, 2022 at 6:02

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, $$J_7\approx \frac\pi2-2.31\cdot10^{-11}.$$