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?

edited Dec 4 '18 at 16:43

Xander Henderson

14.2k103554

asked Dec 4 '18 at 16:22

Moo

5,53131020

5

$begingroup$
This is an interesting example of "jumping to a conclusion." (+1)
$endgroup$
– Mark Viola
Dec 4 '18 at 16:48

1

$begingroup$
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)
$endgroup$
– Winther
Dec 4 '18 at 17:02

1

$begingroup$
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.
$endgroup$
– Winther
Dec 4 '18 at 17:05

2

$begingroup$
@Winther Hi Hans. Happy Holidays. Thank you for the comments and embedded references! Much appreciated. -Mark
$endgroup$
– Mark Viola
Dec 4 '18 at 19:03

1

$begingroup$
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.
$endgroup$
– Arnaud D.
Dec 4 '18 at 19:39

|
show 3 more comments