对黎曼猜想的证明

non-trivial zero points of ζ ( s) ;
z
ζ (ρ)
exists infinite non-trivial zero points lying on line
Re ( s )
=
1 2
.
Riemann further conjectured that all non-trivial zeros of
Γ ( s) here is the Euler gamma function
∏ 1
Γ(s)
=
s
∞ n=1
⎛⎜⎝1 +
s n
⎞ ⎟⎠
⎛⎜⎝1 +
1 n
⎞−s ⎟⎠
.
(4)
Moreover, this zeta function ζ ( s) has the following properties:
z ζ ( s) has an analytic continuation to the whole complex plane except for a simple pole at
s =1;
z ζ ( s) has zero points at s = −2, −4," , in half plane Re ( s) < 0 . These zero points are
+
Γ′ ⎛ 1− s Γ ⎜⎝ 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
=
0 .（15）
Since
∑ Γ′ ⎛ 1− s
Γ ⎜⎝ 2
⎞ ⎟⎠
+
Γ′ Γ
⎛1− s ⎜⎝ 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
=
−α
∞ n=o
Re ( s )
=
1 2
⇔
A(s) A(s ) =1.
Lemma 2 is true.
Lemma 3:
0 < Re ( s) < 1 ⇒ ζ (n) (s) = ζ (n) (1− s)
(21)
Proof: In region 0 < Re ( s) < 1, if we select any suitable simple closed contour C, which is
symmetry for real axis and line
Re ( s )
=
1 2
,
then
the
two
functions,
ζ (s)
and
ζ (1− s) , have
completely identical boundary values in total. According to Cauchy derivative formula, we have
⎡⎛ ⎢⎢⎣⎜⎝
n
+
1 4
+
⎛ ⎜⎝
n
+
1 4
⎞2 ⎟⎠
−
α2 4
−
β2 4
α 2
⎞2 ⎟⎠
+
β2 4
⎤ ⎥ ⎥⎦
⎡⎛ ⎢⎢⎣⎜⎝
n
+
1 4
−
α 2
⎞2 ⎟⎠
+
β2 4
⎤ ⎥ ⎥⎦
,
(16)
and
∑∞
⎛ ⎜⎝
n
+
1 4
⎞2 ⎟⎠
−
α2 4
−
β2 4
≠0,
n=o
⎡⎢⎢⎣⎛⎜⎝
n
+
1 4
+
α 2
⎞2 ⎟⎠
In this paper, a proof of the Riemann Hypothesis is given. We first obtain the ratio
ζ (s) ζ (1− s)
limited value of the Riemann zeta functions
to
at non-trivial
(23)
Consequently, 0 < Re ( s) < 1 ⇒ ζ (n) ( s) = ζ (n) (1− s) .
Lemma 3 is established.
Lemma 4:
Re(s) =
1 2
⇔
A(s) ≡ 1.
(24)
Proof: Let s = σ + it , some mathematicians [10] had proved that
zero points by L’Hospital Rule. And then we prove that all non-trivial zero points of
ζ (s)
1
have real part 2 . Thus, we provide evidence for the correctness of the
i.e., Proof 2:
so,
α
=
0
⇔
A(s)
A(s
)
=
A
⎛ ⎜⎝
1 2
+ iβ
⎞ ⎟⎠
A ⎛⎜⎝
1 2
− iβ
⎞ ⎟⎠
=1,
(13)
Re ( s )
=
1 2
⇔

⎛ ⎜⎝
s
=
1 2
+α
+ iβ
⎞ ⎟⎠
∨
⎛ ⎜⎝
s
=
1 2
+α
− iβ
⎞ ⎟⎠
⇒
A(s)
A(s
)
=πα
Γ
⎛ ⎜⎝
Γ
+
β2 4
⎤ ⎥ ⎥⎦
⎡⎢⎢⎣⎛⎜⎝
n
+
1 4
−
α 2
⎞2 ⎟⎠
+
β2 4
⎤ ⎥ ⎥⎦
(17)
then
Γ′ Γ
⎛1− s ⎜⎝ 2
⎞ ⎟⎠
+
Γ′ ⎛ 1− s Γ ⎜⎝ 2
⎞ ⎟⎠
−
Γ′ Γ
⎛ ⎜⎝
s 2
⎞ ⎟⎠
−
Γ′ ⎛ Γ ⎜⎝
s 2
⎞ ⎟⎠
=
0
⇔
α
=
0.
(18)
Combining (15) and (18), we obtain
Re(s) =
1 2
⇔
A(s) A(s
)
=1
(9)
Proof 1: Let
s = 1 +α + iβ 2
and
s = 1 +α − iβ 2
(where
α
<
1 2
is any of real constants,
β ≠ 0 is real variable), such that Re ( s) = 1 + α , Im ( s) ≠ 0 , then
5

3. Proof of RH
Let ρ = σ + iβ ( 0 < σ < 1) is anyone of non-trivial zero points of (n) order of ζ ( s) , then
the functions ζ ( s) and ζ (1− s) ) are both equal to zero at point ρ = σ + it . Namely,
s
−
1 n
⎞ ⎟⎠ ,
(7)
where γ is Euler constant, then
∑ ∑ Γ′
Γ
(s1 ) −
Γ′ Γ
(
s2
)
=
1 s2
+
∞ n=1
n
1 + s2
−
1 s1
−
∞ n=1
n
1 + s1
∑ =
∞⎛ ⎜
n=0 ⎝
n
1 + s2
−
n
1 + s1
⎞ ⎟ ⎠
(8)
Hence, Lemma 1 is established. Lemma 2:
called trivial zero points;
z ζ ( s) has infinite complex zero points ρ = σ + iβ in region 0 < σ < 1. These zero
points are called non-trivial zero points. Furthermore, ρ,1− ρ, ρ ,1− ρ are also
2. Lemmas
Lemma 1:
∑ Re ( s )
>
0
⇒
Γ′ Γ
(s1 )
−
Γ′ Γ
( s2
)
=
∞⎛ ⎜
n=0 ⎝
n
1 + s2
−
n
1 + s1
⎞ ⎟ ⎠
,
(6)
where
Γ′ Γ
(
s
)
=
Γ′ ( s ) Γ(s)
.
Proof: Since
∑ −
Γ′ (s)