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?

  • 7
    $\begingroup$ This is an interesting example of "jumping to a conclusion." (+1) $\endgroup$
    – Mark Viola
    Dec 4, 2018 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, 2018 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, 2018 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, 2018 at 19:03
  • 3
    $\begingroup$ 3Blue1Brown recently posted a video on this very thing $\endgroup$
    – user170231
    Nov 16, 2022 at 6:02

1 Answer 1


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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .