波尔文积分 (英語:Borwein integral )是一种由波尔文父子发现的性质特殊的积分 ,常用于作为看似存在的数学规律最终失效的例子。2001年,大卫·波尔文 和乔纳森·波尔文 共同发表了这个涉及sinc函数 的积分[ 1] 。
常见的例子为:
∫
0
∞
sin
(
x
)
x
d
x
=
π
2
∫
0
∞
sin
(
x
)
x
sin
(
x
/
3
)
x
/
3
d
x
=
π
2
∫
0
∞
sin
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x
)
x
sin
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x
/
3
)
x
/
3
sin
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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}}}
这种规律一直到
∫
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}}.}
都是成立的。
但是到了下一个数,这个规律就突然失效了:
∫
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]&\simeq {\frac {\pi }{2}}-2.31\times 10^{-11}.\end{aligned}}}
对于给定的一系列非零实数,即
a
0
,
a
1
,
a
2
⋯
{\displaystyle a_{0},a_{1},a_{2}\cdots }
,可以给出
∫
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}}\mathrm {d} x}
的封闭公式形式。为了计算这个公式,其中需要做的就是计算含有
a
k
{\displaystyle a_{k}}
相关的量之和。特别的,设
γ
=
(
γ
1
,
γ
2
,
⋯
,
γ
n
)
∈
{
±
1
}
n
{\displaystyle \gamma =(\gamma _{1},\gamma _{2},\cdots ,\gamma _{n})\in \{\pm 1\}^{n}}
即由
±
1
{\displaystyle \pm 1}
构成的
n
{\displaystyle n}
元组,于是可以写成
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}}
即有关
a
k
{\displaystyle a_{k}}
的各种加减形式的总和,并且令
ε
γ
=
γ
1
γ
2
⋯
γ
n
{\displaystyle \varepsilon _{\gamma }=\gamma _{1}\gamma _{2}\cdots \gamma _{n}}
(其结果为
±
1
{\displaystyle \pm 1}
)。基于上述定义,可以得到该积分的值为:
∫
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}}\mathrm {d} x={\frac {\pi }{2a_{0}}}\cdot C_{n}}
其中:
C
n
=
1
2
n
⋅
n
!
∏
k
=
1
n
a
k
⋅
∑
γ
∈
{
±
1
}
n
ε
γ
b
γ
n
sgn
(
b
n
)
{\displaystyle C_{n}={\frac {1}{2^{n}\cdot n!\prod _{k=1}^{n}a_{k}}}\cdot \sum _{\gamma \in \{\pm 1\}^{n}}\varepsilon _{\gamma }b_{\gamma }^{n}\operatorname {sgn} (b_{n})}
在这里如果
a
0
>
|
a
1
|
+
|
a
2
|
+
⋯
+
|
a
n
|
{\displaystyle a_{0}>|a_{1}|+|a_{2}|+\cdots +|a_{n}|}
,那么有
C
n
=
1
{\displaystyle C_{n}=1}
。
进一步地,如果存在一个
n
{\displaystyle n}
对于每个
k
=
0
,
⋯
,
n
−
1
{\displaystyle k=0,\cdots ,n-1}
总有
0
<
a
n
<
2
a
k
{\displaystyle 0<a_{n}<2a_{k}}
成立,并且有
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}}
,即
n
{\displaystyle n}
为首次超过
a
0
{\displaystyle a_{0}}
的前几项之和时的元素数量,即当
k
=
0
,
⋯
,
n
−
1
{\displaystyle k=0,\cdots ,n-1}
时有
C
k
=
1
{\displaystyle C_{k}=1}
,但在其他情况时:
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}\cdot n!\prod _{k=1}^{n}a_{k}}}}
在这里令
a
k
=
1
2
k
+
1
{\displaystyle a_{k}={\frac {1}{2k+1}}}
,即当
n
=
7
{\displaystyle n=7}
时
a
7
=
1
15
{\displaystyle a_{7}={\frac {1}{15}}}
,此时
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}
但是
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}
,又由于
a
0
=
1
{\displaystyle a_{0}=1}
,于是该公式成立(并且移去其中任何因子也成立):
∫
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}}\cdot {\frac {\sin {\frac {x}{3}}}{\frac {x}{3}}}\cdots {\frac {\sin {\frac {x}{13}}}{\frac {x}{13}}}\mathrm {d} x={\frac {\pi }{2}}}
但在另一方面,则有:
∫
0
∞
sin
x
x
⋅
sin
x
3
x
3
⋯
sin
x
15
x
15
d
x
=
π
2
[
1
−
(
1
3
+
1
5
+
1
7
+
1
9
+
1
11
+
1
13
+
1
15
−
1
)
7
2
6
⋅
7
!
⋅
(
3
⋅
5
⋅
7
⋅
9
⋅
11
⋅
13
⋅
15
)
−
1
]
{\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x}}\cdot {\frac {\sin {\frac {x}{3}}}{\frac {x}{3}}}\cdots {\frac {\sin {\frac {x}{15}}}{\frac {x}{15}}}\mathrm {d} x={\frac {\pi }{2}}\left[1-{\frac {\left({\frac {1}{3}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{9}}+{\frac {1}{11}}+{\frac {1}{13}}+{\frac {1}{15}}-1\right)^{7}}{2^{6}\cdot 7!\cdot (3\cdot 5\cdot 7\cdot 9\cdot 11\cdot 13\cdot 15)^{-1}}}\right]}
即与前面给出的公式的结果相同。