## Proof of Riemann Hypothesis

**Authors:** Toshiro Takami

Let a be real number of 0< a <1.
(1)= cos[x*ln1]/1^a – cos[x*ln2]/2^a + cos[x*ln3]/3^a – cos[x*ln4]/4^a + cos[x*ln5]/5^a............
(2)= sin[x*ln1]/1^a – sin[x*ln2]/2^a + sin[x*ln3]/3^a – sin[x*ln4]/4^a + sin[x*ln5]/5^a.............
Then, at this time,
The imaginary solution of the equation (1)^2+(2)^2=0 exists only when a = 0.5.
x is an infinite non-trivial zero.
At the same time satisfying (1) and (2) is x, that is, an infinitely present non-tribial zero.
(1) is
\sum_{n=1}^{infty} [cos(x*ln(2n-1))/(2n-1)^0.5)- cos(x*ln(2n)/(2n)^0.5)] =0
(2) is
\sum_{n=1}^{infty} [sin(x*ln(2n-1))/(2n-1)^0.5) - sin(x*ln(2n)/(2n)^0.5)] =0

**Comments:** 9 Pages. serios mistake consist, re-load-up

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### Submission history

[v1] 2019-02-28 04:36:53 (removed)

[v2] 2019-02-28 17:46:37 (removed)

[v3] 2019-03-04 13:56:54 (removed)

[v4] 2019-03-08 02:53:41

[v5] 2019-03-08 17:19:49

[v6] 2019-03-10 05:08:41

[v7] 2019-03-11 02:08:38

[v8] 2019-03-12 19:14:08

[v9] 2019-03-13 17:36:31

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