From Wikibooks, open books for an open world
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Jump to navigation
Jump to search
Theorem 1
For all integers,
n
>
1
{\displaystyle n>1}
,
∫
[
0
,
1
]
n
1
1
−
∏
i
=
1
n
x
i
∏
i
=
1
n
d
x
i
=
ζ
(
n
)
{\displaystyle \int _{[0,1]^{n}}{\frac {1}{1-\prod _{i=1}^{n}x_{i}}}\prod _{i=1}^{n}\mathrm {d} x_{i}=\zeta (n)}
Proof
Using the power series of
(
1
−
x
)
−
1
{\displaystyle (1-x)^{-1}}
,
∫
[
0
,
1
]
n
1
1
−
∏
i
=
1
n
x
i
∏
i
=
1
n
d
x
i
=
∫
[
0
,
1
]
n
∑
j
=
0
∞
(
∏
i
=
1
n
x
i
)
j
∏
i
=
1
n
d
x
i
{\displaystyle \int _{[0,1]^{n}}{\frac {1}{1-\prod _{i=1}^{n}x_{i}}}\prod _{i=1}^{n}\mathrm {d} x_{i}=\int _{[0,1]^{n}}\sum _{j=0}^{\infty }\left(\prod _{i=1}^{n}x_{i}\right)^{j}\prod _{i=1}^{n}\mathrm {d} x_{i}}
Evaluating,
=
∫
[
0
,
1
]
n
∑
j
=
0
∞
∏
i
=
1
n
x
i
j
d
x
i
{\displaystyle =\int _{[0,1]^{n}}\sum _{j=0}^{\infty }\prod _{i=1}^{n}x_{i}^{j}\mathrm {d} x_{i}}
=
∑
j
=
0
∞
∏
i
=
1
n
∫
0
1
x
i
j
d
x
i
{\displaystyle =\sum _{j=0}^{\infty }\prod _{i=1}^{n}\int _{0}^{1}x_{i}^{j}\mathrm {d} x_{i}}
Evaluating the integral,
=
∑
j
=
0
∞
∏
i
=
1
n
1
j
+
1
{\displaystyle =\sum _{j=0}^{\infty }\prod _{i=1}^{n}{\frac {1}{j+1}}}
Evaluating the product,
=
∑
j
=
1
∞
∏
i
=
1
n
1
j
{\displaystyle =\sum _{j=1}^{\infty }\prod _{i=1}^{n}{\frac {1}{j}}}
=
∑
j
=
1
∞
1
j
n
{\displaystyle =\sum _{j=1}^{\infty }{\frac {1}{j^{n}}}}
Using the definition of the zeta function that holds only for
ℜ
n
>
1
{\displaystyle \Re n>1}
,
∫
[
0
,
1
]
n
1
1
−
∏
i
=
1
n
x
i
∏
i
=
1
n
d
x
i
=
∑
j
=
1
∞
1
j
n
=
ζ
(
n
)
{\displaystyle \int _{[0,1]^{n}}{\frac {1}{1-\prod _{i=1}^{n}x_{i}}}\prod _{i=1}^{n}\mathrm {d} x_{i}=\sum _{j=1}^{\infty }{\frac {1}{j^{n}}}=\zeta (n)}
for all integers
n
>
1
{\displaystyle n>1}
◼
{\displaystyle \blacksquare }
Note
It can be noted that,
∫
0
1
1
1
−
x
d
x
{\displaystyle \int _{0}^{1}{\frac {1}{1-x}}\mathrm {d} x}
fails to converge, as
−
lim
x
→
1
log
(
1
−
x
)
=
∞
{\displaystyle -\lim _{x\to 1}\log(1-x)=\infty }
.