Si dimostra che
π
2
{\displaystyle \pi ^{2}}
Sia
n
{\displaystyle n}
intero positivo, definiamo
f
:
R
→
R
{\displaystyle f\colon \mathbb {R} \rightarrow \mathbb {R} }
f
(
x
)
:=
x
n
(
1
−
x
)
n
n
!
=
1
n
!
∑
k
=
0
n
(
n
k
)
(
−
1
)
k
x
n
+
k
,
{\displaystyle f(x):={\frac {x^{n}(1-x)^{n}}{n!}}={\frac {1}{n!}}\sum _{k=0}^{n}{n \choose k}(-1)^{k}x^{n+k},}
dove l'ultimo membro segue dal teorema binomiale . Poiché
f
(
x
)
{\displaystyle f(x)}
2
n
{\displaystyle 2n}
x
{\displaystyle x}
f
(
m
)
(
x
)
=
0
{\displaystyle f^{(m)}(x)=0}
x
∈
R
{\displaystyle x\in \mathbb {R} }
m
>
2
n
.
{\displaystyle m>2n.}
f
(
m
)
(
0
)
=
0
{\displaystyle f^{(m)}(0)=0}
m
<
n
{\displaystyle m<n}
x
{\displaystyle x}
f
(
x
)
{\displaystyle f(x)}
n
.
{\displaystyle n.}
Se
m
∈
{
n
,
n
+
1
,
…
,
2
n
}
{\displaystyle m\in \{n,n+1,\dots ,2n\}}
f
(
m
)
(
x
)
=
1
n
!
∑
k
=
m
−
n
n
(
n
k
)
(
n
+
k
)
!
(
n
+
k
−
m
)
!
(
−
1
)
k
x
n
+
k
−
m
,
{\displaystyle f^{(m)}(x)={\frac {1}{n!}}\sum _{k=m-n}^{n}{n \choose k}{\frac {(n+k)!}{(n+k-m)!}}(-1)^{k}x^{n+k-m},}
per cui per ogni
m
∈
{
n
,
n
+
1
,
…
,
2
n
}
{\displaystyle m\in \{n,n+1,\dots ,2n\}}
f
(
m
)
(
0
)
=
1
n
!
(
n
m
−
n
)
m
!
(
−
1
)
m
−
n
∈
Z
.
{\displaystyle f^{(m)}(0)={\frac {1}{n!}}{{n} \choose {m-n}}m!(-1)^{m-n}\in \mathbb {Z} .}
Queste considerazioni mostrano che
f
(
m
)
(
0
)
∈
Z
{\displaystyle f^{(m)}(0)\in \mathbb {Z} }
m
∈
N
.
{\displaystyle m\in \mathbb {N} .}
f
(
1
−
x
)
=
f
(
x
)
,
{\displaystyle f(1-x)=f(x),}
f
(
m
)
(
1
)
∈
Z
.
{\displaystyle f^{(m)}(1)\in \mathbb {Z} .}
Supponiamo ora, per assurdo, che esistano due interi positivi
a
{\displaystyle a}
b
{\displaystyle b}
π
2
=
a
b
.
{\displaystyle \pi ^{2}={\frac {a}{b}}.}
F
n
:
R
→
R
{\displaystyle F_{n}\colon \mathbb {R} \rightarrow \mathbb {R} }
F
n
(
x
)
:=
b
n
∑
k
=
0
n
(
−
1
)
k
π
2
(
n
−
k
)
f
(
2
k
)
(
x
)
.
{\displaystyle F_{n}(x):=b^{n}\sum _{k=0}^{n}(-1)^{k}\pi ^{2(n-k)}f^{(2k)}(x).}
Per quanto detto prima
F
n
(
0
)
{\displaystyle F_{n}(0)}
F
n
(
1
)
{\displaystyle F_{n}(1)}
f
(
2
n
+
2
)
(
x
)
=
0
,
{\displaystyle f^{(2n+2)}(x)=0,}
F
n
(
2
)
(
x
)
+
π
2
F
n
(
x
)
=
b
n
[
∑
k
=
0
n
(
−
1
)
k
π
2
(
n
−
k
)
f
(
2
k
+
2
)
(
x
)
+
π
2
∑
k
=
0
n
(
−
1
)
k
π
2
(
n
−
k
)
f
(
2
k
)
(
x
)
]
=
{\displaystyle F_{n}^{(2)}(x)+\pi ^{2}F_{n}(x)=b^{n}\left[\sum _{k=0}^{n}(-1)^{k}\pi ^{2(n-k)}f^{(2k+2)}(x)+\pi ^{2}\sum _{k=0}^{n}(-1)^{k}\pi ^{2(n-k)}f^{(2k)}(x)\right]=}
=
b
n
[
−
π
2
∑
h
=
1
n
+
1
(
−
1
)
h
π
2
(
n
−
h
)
f
(
2
h
)
(
x
)
+
π
2
∑
h
=
0
n
(
−
1
)
h
π
2
(
n
−
h
)
f
(
2
h
)
(
x
)
]
=
b
n
π
2
(
n
+
1
)
f
(
x
)
=
a
n
π
2
f
(
x
)
.
{\displaystyle =b^{n}\left[-\pi ^{2}\sum _{h=1}^{n+1}(-1)^{h}\pi ^{2(n-h)}f^{(2h)}(x)+\pi ^{2}\sum _{h=0}^{n}(-1)^{h}\pi ^{2(n-h)}f^{(2h)}(x)\right]=b^{n}\pi ^{2(n+1)}f(x)=a^{n}\pi ^{2}f(x).}
Da questi calcoli segue che:
d
d
x
[
F
n
(
1
)
(
x
)
sin
(
π
x
)
−
π
F
n
(
x
)
cos
(
π
x
)
]
=
π
2
a
n
f
(
x
)
sin
(
π
x
)
.
{\displaystyle {\frac {d}{dx}}\left[F_{n}^{(1)}(x)\sin(\pi x)-\pi F_{n}(x)\cos(\pi x)\right]=\pi ^{2}a^{n}f(x)\sin(\pi x).}
Si ha quindi:
π
a
n
∫
0
1
f
(
x
)
sin
(
π
x
)
d
x
=
1
π
[
F
n
(
1
)
(
x
)
sin
(
π
x
)
−
π
F
n
(
x
)
cos
(
π
x
)
]
0
1
=
F
n
(
1
)
+
F
n
(
0
)
∈
Z
.
{\displaystyle \pi a^{n}\int _{0}^{1}f(x)\sin(\pi x)dx={\frac {1}{\pi }}\left[F_{n}^{(1)}(x)\sin(\pi x)-\pi F_{n}(x)\cos(\pi x)\right]_{0}^{1}=F_{n}(1)+F_{n}(0)\in \mathbb {Z} .}
Poiché per ogni
x
∈
(
0
,
1
)
{\displaystyle x\in (0,1)}
x
n
(
1
−
x
)
n
∈
(
0
,
1
)
,
{\displaystyle x^{n}(1-x)^{n}\in (0,1),}
0
<
f
(
x
)
<
1
n
!
.
{\displaystyle 0<f(x)<{\frac {1}{n!}}.}
Quindi
π
a
n
∫
0
1
f
(
x
)
sin
(
π
x
)
d
x
∈
(
0
,
π
a
n
n
!
)
∩
Z
.
{\displaystyle \pi a^{n}\int _{0}^{1}f(x)\sin(\pi x)dx\in \left(0,{\frac {\pi a^{n}}{n!}}\right)\cap \mathbb {Z} .}
D'altra parte,
lim
n
→
+
∞
π
a
n
n
!
=
0
,
{\displaystyle \lim _{n\rightarrow +\infty }{\frac {\pi a^{n}}{n!}}=0,}
quindi, per
n
{\displaystyle n}
(
0
,
π
a
n
n
!
)
⊂
(
0
,
1
)
.
{\displaystyle \left(0,{\frac {\pi a^{n}}{n!}}\right)\subset (0,1).}
Abbiamo quindi trovato che
π
a
n
∫
0
1
f
(
x
)
sin
(
π
x
)
d
x
∈
(
0
,
1
)
∩
Z
.
{\displaystyle \pi a^{n}\int _{0}^{1}f(x)\sin(\pi x)dx\in (0,1)\cap \mathbb {Z} .}
Ma non esistono interi nell'intervallo
(
0
,
1
)
{\displaystyle (0,1)}
π
2
{\displaystyle \pi ^{2}}
π
{\displaystyle \pi }
![{\displaystyle F_{n}^{(2)}(x)+\pi ^{2}F_{n}(x)=b^{n}\left[\sum _{k=0}^{n}(-1)^{k}\pi ^{2(n-k)}f^{(2k+2)}(x)+\pi ^{2}\sum _{k=0}^{n}(-1)^{k}\pi ^{2(n-k)}f^{(2k)}(x)\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/426415bf982e3a6e69e1ce1ad3d83b0698adc087)
![{\displaystyle =b^{n}\left[-\pi ^{2}\sum _{h=1}^{n+1}(-1)^{h}\pi ^{2(n-h)}f^{(2h)}(x)+\pi ^{2}\sum _{h=0}^{n}(-1)^{h}\pi ^{2(n-h)}f^{(2h)}(x)\right]=b^{n}\pi ^{2(n+1)}f(x)=a^{n}\pi ^{2}f(x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a439eca3f0106c408dd96a49932537ee3ee6f05a)
![{\displaystyle {\frac {d}{dx}}\left[F_{n}^{(1)}(x)\sin(\pi x)-\pi F_{n}(x)\cos(\pi x)\right]=\pi ^{2}a^{n}f(x)\sin(\pi x).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/592b744485a2bbf3c759d92b7117cac174ecaa15)
![{\displaystyle \pi a^{n}\int _{0}^{1}f(x)\sin(\pi x)dx={\frac {1}{\pi }}\left[F_{n}^{(1)}(x)\sin(\pi x)-\pi F_{n}(x)\cos(\pi x)\right]_{0}^{1}=F_{n}(1)+F_{n}(0)\in \mathbb {Z} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59b771135caffa396b3e67948df141ba01150943)
