In mathematics, a Borwein integral is an integral involving products of sinc(ax), where the sinc function is given by sinc(x) = sin(x)/x for x not equal to 0, and sinc(0) = 1.
These integrals are remarkable for exhibiting apparent patterns which, however, eventually break down. An example is as follows,
∫
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
This pattern continues up to
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
13
)
x
/
13
d
x
=
π
/
2
.
Nevertheless, 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
.
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.
An example for a longer series,
∫
0
∞
2
cos
(
x
)
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
111
)
x
/
111
d
x
=
π
/
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
,
is shown in together with an intuitive mathematical explanation of the reason why the original and the extended series break down. In this case, 1/3 + 1/5 + … + 1/111 < 2, but 1/3 + 1/5 + … + 1/113 > 2.
Given a sequence of real numbers,
a
0
,
a
1
,
a
2
,
.
.
.
, a general formula for the integral
∫
0
∞
∏
k
=
0
n
sin
(
a
k
x
)
a
k
x
d
x
can be given. To state the formula, one will need to consider sums involving the
a
k
. In particular, if
γ
=
(
γ
1
,
γ
2
,
.
.
.
,
γ
n
)
∈
{
±
1
}
n
is an
n
-tuple where each entry is
±
1
, then we write
b
γ
=
a
0
+
γ
1
a
1
+
γ
2
a
2
+
⋯
+
γ
n
a
n
, which is a kind of alternating sum of the first few
a
k
, and we set
ϵ
γ
=
γ
1
γ
2
⋯
γ
n
, which is either
±
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
where
C
n
=
1
2
n
n
!
∏
k
=
1
n
a
k
∑
γ
∈
{
±
1
}
n
ϵ
γ
b
γ
n
sgn
(
b
γ
)
In the case when
a
0
>
|
a
1
|
+
|
a
2
|
+
⋯
+
|
a
n
|
, we have
C
n
=
1
.
Furthermore, if there is an
n
so that for each
k
=
0
,
.
.
.
,
n
−
1
we have
0
<
a
n
<
2
a
k
and
a
1
+
a
2
+
⋯
+
a
n
−
1
<
a
0
<
a
1
+
a
2
+
⋯
+
a
n
−
1
+
a
n
, 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
,
.
.
.
,
n
−
1
but
C
n
=
1
−
(
a
1
+
a
2
+
⋯
+
a
n
−
a
0
)
n
2
n
−
1
n
!
∏
k
=
1
n
a
k
The first example is the case when
a
k
=
1
2
k
+
1
. Note that if
n
=
7
then
a
7
=
1
15
and
1
3
+
1
5
+
1
7
+
1
9
+
1
11
+
1
13
≈
0.955
but
1
3
+
1
5
+
1
7
+
1
9
+
1
11
+
1
13
+
1
15
≈
1.02
, so because
a
0
=
1
, we get that
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
⋯
sin
(
x
/
13
)
x
/
13
d
x
=
π
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
)
)
which is equal to the value given previously.
