Accurate Proof and varification for Riemann Hypothesis

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Accurate Proof verification of Riemann’s Hypothesis

Riemann Hypothesis states that $$\int \frac{1}{ln (x)}$$ has a root at $$\frac{1}{2}$$ when s=2

The time series expansion of Log function is,

[tex]\ln(x) = \frac {[x-1}{[x-2}+ \frac{1){3} \frac{x-3}{x-4} + \frac{1}{5}\frac{x-5}{x-6}+……. [\tex]<br /> Let it be equal to [tex]mx + c [\tex] because of the Linear nature of Log function.<br /> <br /> Now,<br /> \[tex]int \frac{1}{ln(x)}=\int\frac{1}{mx+c}[\tex]<br /> If we take x=2 from Log function we can deduce m=0.35 and c= 0.0007<br /> Riemann stipulates that for any value x \int \frac{1}{ln (x) will have to be taken between limits 2 and x<br /> So,<br /> [tex]\int^2 _1/2 \frac{1}{ln(x)} [\tex] is done using Cauchys principal number taken between 2 to 1 and from 1 to ½<br /> Which is, <br /> [tex]{\ln(035x1+0.0007)-\\ln(0.5x0.35+0.0007)} + {\ln(0.35x2 +0.0007)-\ln(0.35x1+0.0007)}[\tex]<br /> Taking the proper order of integrations and signs we get<br /> $$[-1.04782 + 0.35568]+[-1.04782+1.73897]=-0.69+0.69 =0[\tex]<br /> <br /> Which proves that the root of Riemann’s ξ function is 1/2 when s=2<br /> <br /> <br /> Mathew Cherian$$[/tex][/tex][/tex][/tex][/tex]

 

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Mathew, some words of advice:

1. Learn to use the [ tex ] feature properly. It is tex with a lowercase t, not Tex.

2. Learn to use the Preview button.

3. Learn what the Riemann zeta function is. Your integral is the logarithmic integral, not the Riemann zeta function.

4. Learn what the Riemann says. It says the all non-trivial (complex) zeros of the Riemann zeta hypothesis have real part = 1/2.

5. Learn how to do math.

6. Learn that this site has rules against posts of this sort.