Accurate Proof verification of Riemann’s Hypothesis(adsbygoogle = window.adsbygoogle || []).push({});

Riemann Hypothesis states that [tex] \int \frac{1}{ln (x)}[/tex] has a root at [tex]\frac{1}{2}[/tex] 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]

Let it be equal to [tex] mx + c [\tex] because of the Linear nature of Log function.

Now,

\[tex]int \frac{1}{ln(x)}=\int\frac{1}{mx+c}[\tex]

If we take x=2 from Log function we can deduce m=0.35 and c= 0.0007

Riemann stipulates that for any value x \int \frac{1}{ln (x) will have to be taken between limits 2 and x

So,

[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

Which is,

[tex]{\ln(035x1+0.0007)-\\ln(0.5x0.35+0.0007)} + {\ln(0.35x2 +0.0007)-\ln(0.35x1+0.0007)}[\tex]

Taking the proper order of integrations and signs we get

[tex][-1.04782 + 0.35568]+[-1.04782+1.73897]=-0.69+0.69 =0[\tex]

Which proves that the root of Riemann’s ξ function is 1/2 when s=2

Mathew Cherian

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# Accurate Proof and varification for Riemann Hypothesis

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