# 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. Borwein integrals involve products of $\operatorname {sinc} (ax)$ , where the sinc function is given by $\operatorname {sinc} (x)=\sin(x)/x$ for $x$ not equal to 0, and $\operatorname {sinc} (0)=1$ .

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

{\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

$\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/13)}{x/13}}\,dx={\frac {\pi }{2}}.$ At the next step the obvious pattern fails,

{\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 $2\cos(x)$ , the pattern holds up over a longer series,

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

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

${\frac {\pi }{2}}\left(1-{\frac {3\cdot 5\cdot \cdot \cdot 113\cdot (1/3+1/5+\dots +1/113-2)^{56}}{2^{55}\cdot 56!}}\right)$ The reason the original and the extended series break down has been demonstrated with an intuitive mathematical explanation. 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.

## General formula

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

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

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

$C_{n}={\frac {1}{2^{n}n!\prod _{k=1}^{n}a_{k}}}\sum _{\gamma \in \{\pm 1\}^{n}}\varepsilon _{\gamma }b_{\gamma }^{n}\operatorname {sgn}(b_{\gamma })$ In the case when $a_{0}>|a_{1}|+|a_{2}|+\cdots +|a_{n}|$ , we have $C_{n}=1$ .

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

$C_{n}=1-{\frac {(a_{1}+a_{2}+\cdots +a_{n}-a_{0})^{n}}{2^{n-1}n!\prod _{k=1}^{n}a_{k}}}$ The first example is the case when $a_{k}={\frac {1}{2k+1}}$ .

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

$\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/13)}{x/13}}\,dx={\frac {\pi }{2}}$ which remains true if we remove any of the products, but that

{\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

{\begin{aligned}\int _{0}^{\infty }\prod _{k=0}^{n}{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx\end{aligned}} becomes less than ${\frac {\pi }{2}}$ when $n$ exceeds 6, it never becomes much less, and in fact Borwein and Bailey have shown

{\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

$\int _{0}^{\infty }2\cos x\prod _{k=0}^{n}{\frac {\sin(x/(2n+1))}{x/(2n+1)}}\,dx$ becomes less than ${\frac {\pi }{2}}$ when $n$ exceeds 55, we have

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

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

$\prod _{n=0}^{\infty }{\frac {\sin(2x/(2n+1))}{2x/(2n+1)}}=\prod _{n=1}^{\infty }\cos \left({\frac {x}{n}}\right)$ and with a change of variables obtain

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

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