A Short Proof Of The Riemann Hypothesis
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A short proof of the Riemann hypothesis Werner Raab We consider the meromorphic function u(s) =
1 (s − 1)ζ(s)
(<s > 0) and the Mellin transform u(3/2 − s) v(s) = π = sin(πs)
Z
∞
ts−1 w(t) dt
0
of the function w(t) =
=
∞ X
1 2πi
Z
1/2+i∞
t−s v(s) ds =
1/2−i∞
∞ X
Ress=−k t−s v(s)
k=0
u(3/2 + k)(−t)k =
k=0
∞ 1 X m t m ∆ u(3/2)( ) 1 + t m=0 1+t
with the differences m
∆ u(3/2) =
m µ ¶ X m k=0
k
(−1)k u(3/2 + k)
(|t| < 1 or −1/2 respectively). For positive real values of t we have Z ∞ 1 1 tiy w(t) = √ = dy → u(3/2) t 0 cosh(πy)y ζ(1 + iy) when t → 0 and 1 1 1 w( ) = √ t t t
Z 0
∞
1 t−iy = dy → u(1/2) cosh(πy)y ζ(1 + iy)
when t → 0. The properties w(t) = O(1) when t → 0 and w(t) = O(1/t) when t → ∞ of the Mellin inverse w(t) imply that the Mellin transform v(s) is holomorphic within the complex strip: 0 < <s < 1, as Riemann conjectured.
1
1 (s − 1)ζ(s)
(<s > 0) and the Mellin transform u(3/2 − s) v(s) = π = sin(πs)
Z
∞
ts−1 w(t) dt
0
of the function w(t) =
=
∞ X
1 2πi
Z
1/2+i∞
t−s v(s) ds =
1/2−i∞
∞ X
Ress=−k t−s v(s)
k=0
u(3/2 + k)(−t)k =
k=0
∞ 1 X m t m ∆ u(3/2)( ) 1 + t m=0 1+t
with the differences m
∆ u(3/2) =
m µ ¶ X m k=0
k
(−1)k u(3/2 + k)
(|t| < 1 or
Z 0
∞
1 t−iy = dy → u(1/2) cosh(πy)y ζ(1 + iy)
when t → 0. The properties w(t) = O(1) when t → 0 and w(t) = O(1/t) when t → ∞ of the Mellin inverse w(t) imply that the Mellin transform v(s) is holomorphic within the complex strip: 0 < <s < 1, as Riemann conjectured.
1
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