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The Riemann hypothesis

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Has anyone ever tried to attack the Riemann hypothesis themselves?

If you take the Fourier transform (the unitary, ordinary frequency variety) of 1/|x|^s, you will note that it is equal to 1/|x|^(1-s) up to a multiplicative constant, the exact same constant found in the functional equation Zeta[s] = ... Zeta[1-s]. s =1/2 is the only value for which the Fourier transform of 1/|x|^s is equal to itself. Coincidence?

This is not the complete proof of the Riemann hypothesis however I believe it is the right direction to look in.
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>>9148861
>Coincidence?
Yup.
>>
>>9148861
>If you take the Fourier transform (the unitary, ordinary frequency variety) of 1/|x|^s
how does this converge?
>>
>>9148902
How would you know?

>>9148924
It converges for 0 < s < 1 (the critical strip). Now if we were to assume that the same relation holds outside the critical strip, this is where "1 + 2 + 3 ... = -1/12" comes from.
>>
>>9148861
>Has anyone ever tried to attack the Riemann hypothesis themselves?
There have been many, MANY people who have attempted to tackle the problem. There have even been some people (though these people are pretty much all cranks, unfortunately) that claim to have solved it. I would totally recommend looking into what others have found while looking into this problem, if you are interested.
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>>9148973
Can you point me to other research that has thought of it in similar terms as what I described?
>>
>>9148861
Something as intriguing is the fact that the limit of the function 1/x as x approaches 2 from both negative and positive sides is exactly 1/2, which is the value of the zeta function for s = 0.
Coincidence? I think not
>>
>>9148991
The Fourier transform of the 1/|x|^s, the sums of which the Zeta function comes from is interesting as it is used in quantum physics.
>>
>>9148964
>It converges for 0 < s < 1 (the critical strip). Now if we were to assume that the same relation holds outside the critical strip, this is where "1 + 2 + 3 ... = -1/12" comes from.
no it converges for Re(s) > 1. therefore its useless.
>>
Ugh, obviously no one here knows what they are talking about. Screw it
>>
>>9149028
>Ugh, obviously no one here knows what they are talking about. Screw it
no your first assumption is already wrong. The fourier transform of 1/|x|^s doesnt converge for s < 1. (because the fouriertransform of 1/|x| does already not exist and 1/|x|^s > 1/|x| for 0 < s < 1 and |x| > 1) Therefore s = 1/2 isnt in the convergence radius.
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