# 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) – Mark Viola Dec 4 '18 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) – Winther Dec 4 '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 4 '18 at 17:05
• @Winther Hi Hans. Happy Holidays. Thank you for the comments and embedded references! Much appreciated. -Mark – Mark Viola Dec 4 '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 4 '18 at 19:39