# Borwein integral

In mathematics, a Borwein integral is an integral whose unusual properties were first presented by mathematicians David Borwein and Jonathan Borwein in 2001.[1] Borwein integrals involve products of ${\displaystyle \operatorname {sinc} (ax)}$, where the sinc function is given by ${\displaystyle \operatorname {sinc} (x)=\sin(x)/x}$ for ${\displaystyle x}$ not equal to 0, and ${\displaystyle \operatorname {sinc} (0)=1}$.[1][2]

These integrals are remarkable for exhibiting apparent patterns that eventually break down. The following is an example.

{\displaystyle {\begin{aligned}&\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}\\[10pt]&\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\,dx={\frac {\pi }{2}}\\[10pt]&\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}{\frac {\sin(x/5)}{x/5}}\,dx={\frac {\pi }{2}}\end{aligned}}}

This pattern continues up to

${\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/13)}{x/13}}\,dx={\frac {\pi }{2}}.}$

At the next step the obvious pattern fails,

{\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/15)}{x/15}}\,dx&={\frac {467807924713440738696537864469}{935615849440640907310521750000}}~\pi \\[5pt]&={\frac {\pi }{2}}-{\frac {6879714958723010531}{935615849440640907310521750000}}~\pi \\[5pt]&\approx {\frac {\pi }{2}}-2.31\times 10^{-11}.\end{aligned}}}

In general, similar integrals have value π/2 whenever the numbers 3, 5, 7… are replaced by positive real numbers such that the sum of their reciprocals is less than 1.

In the example above, 1/3 + 1/5 + … + 1/13 < 1, but 1/3 + 1/5 + … + 1/15 > 1.

With the inclusion of the additional factor ${\displaystyle 2\cos(x)}$, the pattern holds up over a longer series,[3]

${\displaystyle \int _{0}^{\infty }2\cos(x){\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/111)}{x/111}}\,dx={\frac {\pi }{2}},}$

but

${\displaystyle \int _{0}^{\infty }2\cos(x){\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/111)}{x/111}}{\frac {\sin(x/113)}{x/113}}\,dx\approx {\frac {\pi }{2}}-2.3324\times 10^{-138}.}$

In this case, 1/3 + 1/5 + … + 1/111 < 2, but 1/3 + 1/5 + … + 1/113 > 2. The exact answer can be calculated using the general formula provided in the next section, and a representation of it is shown below. Fully expanded, this value turns into a fraction that involves two 2736 digit integers.

${\displaystyle {\frac {\pi }{2}}\left(1-{\frac {3\cdot 5\cdot \cdot \cdot 113\cdot (1/3+1/5+\dots +1/113-2)^{56}}{2^{55}\cdot 56!}}\right)}$

The reason the original and the extended series break down has been demonstrated with an intuitive mathematical explanation.[4][5] In particular, a random walk reformulation with a causality argument sheds light on the pattern breaking and opens the way for a number of generalizations.[6]

## General formula

Given a sequence of nonzero real numbers, ${\displaystyle a_{0},a_{1},a_{2},\ldots }$, a general formula for the integral

${\displaystyle \int _{0}^{\infty }\prod _{k=0}^{n}{\frac {\sin(a_{k}x)}{a_{k}x}}\,dx}$

can be given.[1] To state the formula, one will need to consider sums involving the ${\displaystyle a_{k}}$. In particular, if ${\displaystyle \gamma =(\gamma _{1},\gamma _{2},\ldots ,\gamma _{n})\in \{\pm 1\}^{n}}$ is an ${\displaystyle n}$-tuple where each entry is ${\displaystyle \pm 1}$, then we write ${\displaystyle b_{\gamma }=a_{0}+\gamma _{1}a_{1}+\gamma _{2}a_{2}+\cdots +\gamma _{n}a_{n}}$, which is a kind of alternating sum of the first few ${\displaystyle a_{k}}$, and we set ${\displaystyle \varepsilon _{\gamma }=\gamma _{1}\gamma _{2}\cdots \gamma _{n}}$, which is either ${\displaystyle \pm 1}$. With this notation, the value for the above integral is

${\displaystyle \int _{0}^{\infty }\prod _{k=0}^{n}{\frac {\sin(a_{k}x)}{a_{k}x}}\,dx={\frac {\pi }{2a_{0}}}C_{n}}$

where

${\displaystyle C_{n}={\frac {1}{2^{n}n!\prod _{k=1}^{n}a_{k}}}\sum _{\gamma \in \{\pm 1\}^{n}}\varepsilon _{\gamma }b_{\gamma }^{n}\operatorname {sgn}(b_{\gamma })}$

In the case when ${\displaystyle a_{0}>|a_{1}|+|a_{2}|+\cdots +|a_{n}|}$, we have ${\displaystyle C_{n}=1}$.

Furthermore, if there is an ${\displaystyle n}$ such that for each ${\displaystyle k=0,\ldots ,n-1}$ we have ${\displaystyle 0 and ${\displaystyle a_{1}+a_{2}+\cdots +a_{n-1}, which means that ${\displaystyle n}$ is the first value when the partial sum of the first ${\displaystyle n}$ elements of the sequence exceed ${\displaystyle a_{0}}$, then ${\displaystyle C_{k}=1}$ for each ${\displaystyle k=0,\ldots ,n-1}$ but

${\displaystyle C_{n}=1-{\frac {(a_{1}+a_{2}+\cdots +a_{n}-a_{0})^{n}}{2^{n-1}n!\prod _{k=1}^{n}a_{k}}}}$

The first example is the case when ${\displaystyle a_{k}={\frac {1}{2k+1}}}$.

Note that if ${\displaystyle n=7}$ then ${\displaystyle a_{7}={\frac {1}{15}}}$ and ${\displaystyle {\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}+{\frac {1}{13}}\approx 0.955}$ but ${\displaystyle {\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}+{\frac {1}{13}}+{\frac {1}{15}}\approx 1.02}$, so because ${\displaystyle a_{0}=1}$, we get that

${\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/13)}{x/13}}\,dx={\frac {\pi }{2}}}$

which remains true if we remove any of the products, but that

{\displaystyle {\begin{aligned}&\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/15)}{x/15}}\,dx\\[5pt]={}&{\frac {\pi }{2}}\left(1-{\frac {(3^{-1}+5^{-1}+7^{-1}+9^{-1}+11^{-1}+13^{-1}+15^{-1}-1)^{7}}{2^{6}\cdot 7!\cdot (1/3\cdot 1/5\cdot 1/7\cdot 1/9\cdot 1/11\cdot 1/13\cdot 1/15)}}\right),\end{aligned}}}

which is equal to the value given previously.

## Infinite products

While the integral

{\displaystyle {\begin{aligned}\int _{0}^{\infty }\prod _{k=0}^{n}{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx\end{aligned}}}

becomes less than ${\displaystyle {\frac {\pi }{2}}}$ when ${\displaystyle n}$ exceeds 6, it never becomes much less, and in fact Borwein and Bailey[7] have shown

{\displaystyle {\begin{aligned}\int _{0}^{\infty }\prod _{k=0}^{\infty }{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx&=\int _{0}^{\infty }\lim _{n\to \infty }\prod _{k=0}^{n}{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx\\[5pt]&=\lim _{n\to \infty }\int _{0}^{\infty }\prod _{k=0}^{n}{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx\\[5pt]&\approx {\frac {\pi }{2}}-0.0000352\end{aligned}}}

where we can pull the limit out of the integral thanks to the dominated convergence theorem. Similarly, while

${\displaystyle \int _{0}^{\infty }2\cos x\prod _{k=0}^{n}{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx}$

becomes less than ${\displaystyle {\frac {\pi }{2}}}$ when ${\displaystyle n}$ exceeds 55, we have

${\displaystyle \int _{0}^{\infty }2\cos x\prod _{k=0}^{n}{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx\approx {\frac {\pi }{2}}-2.9629\cdot 10^{-42}}$

Furthermore, using the Weierstrass factorizations

${\displaystyle {\frac {\sin x}{x}}=\prod _{n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)\qquad \cos x=\prod _{n=0}^{\infty }\left(1-{\frac {4x^{2}}{\pi ^{2}(2n+1)^{2}}}\right)}$

one can show

${\displaystyle \prod _{n=0}^{\infty }{\frac {\sin(2x/(2n+1))}{2x/(2n+1)}}=\prod _{n=1}^{\infty }\cos \left({\frac {x}{n}}\right)}$

and with a change of variables obtain[8]

${\displaystyle \int _{0}^{\infty }\prod _{n=1}^{\infty }\cos \left({\frac {x}{n}}\right)\,dx={\frac {1}{2}}\int _{0}^{\infty }\prod _{n=0}^{\infty }{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx\approx {\frac {\pi }{4}}-0.0000176}$

and[7][9]

${\displaystyle \int _{0}^{\infty }\cos(2x)\prod _{n=1}^{\infty }\cos \left({\frac {x}{n}}\right)\,dx={\frac {1}{2}}\int _{0}^{\infty }\cos(x)\prod _{n=0}^{\infty }{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx\approx {\frac {\pi }{8}}-7.4073\cdot 10^{-43}}$

## References

1. ^ a b c Borwein, David; Borwein, Jonathan M. (2001), "Some remarkable properties of sinc and related integrals", The Ramanujan Journal, 5 (1): 73–89, doi:10.1023/A:1011497229317, ISSN 1382-4090, MR 1829810, S2CID 6515110
2. ^ Baillie, Robert (2011). "Fun With Very Large Numbers". arXiv:1105.3943 [math.NT].
3. ^ Hill, Heather M. (September 2019). Random walkers illuminate a math problem (Volume 72, number 9 ed.). American Institute of Physics. pp. 18–19.
4. ^ Schmid, Hanspeter (2014), "Two curious integrals and a graphic proof" (PDF), Elemente der Mathematik, 69 (1): 11–17, doi:10.4171/EM/239, ISSN 0013-6018
5. ^ Baez, John (September 20, 2018). "Patterns That Eventually Fail". Azimuth. Archived from the original on 2019-05-21.
6. ^ Satya Majumdar; Emmanuel Trizac (2019), "When random walkers help solving intriguing integrals", Physical Review Letters, 123 (2): 020201, arXiv:1906.04545, Bibcode:2019arXiv190604545M, doi:10.1103/PhysRevLett.123.020201, ISSN 1079-7114, PMID 31386528, S2CID 184488105
7. ^ a b Borwein, J. M.; Bailey, D. H. (2003). Mathematics by experiment : plausible reasoning in the 21st century (1st ed.). Wellesley, MA: A K Peters.
8. ^ Borwein, Jonathan M. (2004). Experimentation in mathematics : computational paths to discovery. David H. Bailey, Roland Girgensohn. Natick, Mass.: AK Peters. ISBN 1-56881-136-5. OCLC 53021555.
9. ^ Bailey, David H.; Borwein, Jonathan M.; Kapoor, Vishaal; Weisstein, Eric W. (2006-06-01). "Ten Problems in Experimental Mathematics". The American Mathematical Monthly. 113 (6): 481. doi:10.2307/27641975. JSTOR 27641975.