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
sinc
(
a
x
)
{\displaystyle \operatorname {sinc} (ax)}
, where the sinc function is given by
sinc
(
x
)
=
sin
(
x
)
/
x
{\displaystyle \operatorname {sinc} (x)=\sin(x)/x}
for
x
{\displaystyle x}
not equal to 0, and
sinc
(
0
)
=
1
{\displaystyle \operatorname {sinc} (0)=1}
.[1] [2]
These integrals are remarkable for exhibiting apparent patterns that eventually break down. The following is an example.
∫
0
∞
sin
(
x
)
x
d
x
=
π
2
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
d
x
=
π
2
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
sin
(
x
/
5
)
x
/
5
d
x
=
π
2
{\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
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
13
)
x
/
13
d
x
=
π
2
.
{\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,
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
15
)
x
/
15
d
x
=
467807924713440738696537864469
935615849440640907310521750000
π
=
π
2
−
6879714958723010531
935615849440640907310521750000
π
≈
π
2
−
2.31
×
10
−
11
.
{\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
2
cos
(
x
)
{\displaystyle 2\cos(x)}
, the pattern holds up over a longer series,[3]
∫
0
∞
2
cos
(
x
)
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
111
)
x
/
111
d
x
=
π
2
,
{\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
∫
0
∞
2
cos
(
x
)
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
111
)
x
/
111
sin
(
x
/
113
)
x
/
113
d
x
<
π
2
.
{\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<{\frac {\pi }{2}}.}
In this case, 1 / 3 + 1 / 5 + … + 1 / 111 < 2, but 1 / 3 + 1 / 5 + … + 1 / 113 > 2.
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,
a
0
,
a
1
,
a
2
,
…
{\displaystyle a_{0},a_{1},a_{2},\ldots }
, a general formula for the integral
∫
0
∞
∏
k
=
0
n
sin
(
a
k
x
)
a
k
x
d
x
{\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
a
k
{\displaystyle a_{k}}
. In particular, if
γ
=
(
γ
1
,
γ
2
,
…
,
γ
n
)
∈
{
±
1
}
n
{\displaystyle \gamma =(\gamma _{1},\gamma _{2},\ldots ,\gamma _{n})\in \{\pm 1\}^{n}}
is an
n
{\displaystyle n}
-tuple where each entry is
±
1
{\displaystyle \pm 1}
, then we write
b
γ
=
a
0
+
γ
1
a
1
+
γ
2
a
2
+
⋯
+
γ
n
a
n
{\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
a
k
{\displaystyle a_{k}}
, and we set
ε
γ
=
γ
1
γ
2
⋯
γ
n
{\displaystyle \varepsilon _{\gamma }=\gamma _{1}\gamma _{2}\cdots \gamma _{n}}
, which is either
±
1
{\displaystyle \pm 1}
. With this notation, the value for the above integral is
∫
0
∞
∏
k
=
0
n
sin
(
a
k
x
)
a
k
x
d
x
=
π
2
a
0
C
n
{\displaystyle \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
=
1
2
n
n
!
∏
k
=
1
n
a
k
∑
γ
∈
{
±
1
}
n
ε
γ
b
γ
n
sgn
(
b
γ
)
{\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
a
0
>
|
a
1
|
+
|
a
2
|
+
⋯
+
|
a
n
|
{\displaystyle a_{0}>|a_{1}|+|a_{2}|+\cdots +|a_{n}|}
, we have
C
n
=
1
{\displaystyle C_{n}=1}
.
Furthermore, if there is an
n
{\displaystyle n}
such that for each
k
=
0
,
…
,
n
−
1
{\displaystyle k=0,\ldots ,n-1}
we have
0
<
a
n
<
2
a
k
{\displaystyle 0<a_{n}<2a_{k}}
and
a
1
+
a
2
+
⋯
+
a
n
−
1
<
a
0
<
a
1
+
a
2
+
⋯
+
a
n
−
1
+
a
n
{\displaystyle a_{1}+a_{2}+\cdots +a_{n-1}<a_{0}<a_{1}+a_{2}+\cdots +a_{n-1}+a_{n}}
, which means that
n
{\displaystyle n}
is the first value when the partial sum of the first
n
{\displaystyle n}
elements of the sequence exceed
a
0
{\displaystyle a_{0}}
, then
C
k
=
1
{\displaystyle C_{k}=1}
for each
k
=
0
,
…
,
n
−
1
{\displaystyle k=0,\ldots ,n-1}
but
C
n
=
1
−
(
a
1
+
a
2
+
⋯
+
a
n
−
a
0
)
n
2
n
−
1
n
!
∏
k
=
1
n
a
k
{\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
a
k
=
1
2
k
+
1
{\displaystyle a_{k}={\frac {1}{2k+1}}}
.
Note that if
n
=
7
{\displaystyle n=7}
then
a
7
=
1
15
{\displaystyle a_{7}={\frac {1}{15}}}
and
1
3
+
1
5
+
1
7
+
1
9
+
1
11
+
1
13
≈
0.955
{\displaystyle {\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}+{\frac {1}{13}}\approx 0.955}
but
1
3
+
1
5
+
1
7
+
1
9
+
1
11
+
1
13
+
1
15
≈
1.02
{\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
a
0
=
1
{\displaystyle a_{0}=1}
, we get that
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
13
)
x
/
13
d
x
=
π
2
{\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
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
15
)
x
/
15
d
x
=
π
2
(
1
−
(
3
−
1
+
5
−
1
+
7
−
1
+
9
−
1
+
11
−
1
+
13
−
1
+
15
−
1
−
1
)
7
2
6
⋅
7
!
⋅
(
1
/
3
⋅
1
/
5
⋅
1
/
7
⋅
1
/
9
⋅
1
/
11
⋅
1
/
13
⋅
1
/
15
)
)
,
{\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.
References
^ 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
^ Baillie, Robert (2011). "Fun With Very Large Numbers". arXiv :1105.3943 [math.NT ].
^ Hill, Heather M. (September 2019). Random walkers illuminate a math problem (Volume 72, number 9 ed.). American Institute of Physics. pp. 18–19.
^ 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
^ Baez, John (September 20, 2018). "Patterns That Eventually Fail" . Azimuth . Archived from the original on 2019-05-21.
^ 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
External links