Sequel for my proof for Riemann's hypothesis

[imag94]

Sequel to my Proof for Riemann Hypothesis.

Riemann conjured that the function
\xi = \int \frac {1}{\ln(x)}
has a root at \frac{1}{2} when s=2

Let
f ^-1 (x)=\frac{1}{\ln (x)}
f(x)= e^ \frac {1}{ln (x)}
Taking log on both sides, \frac {1}{\ln(x)} = \ln e^\frac{1}{1/ln (x)}
Integrating both sides,
\int \frac {1}{ln(x)}= \int \ln e^{1}{ln x}
\int \ln(e^\\frac {1}{ln (x)} = e^ \ln (e^\frac{1}{ln (x)}

Value of \xi function is got by taking the integral between 2 and that number.
Taking limits of r.h.s from \frac{1}{2} to 1 and from 1 to 2 and adding them up for finding the value of the \xi function at \frac {1}{2} due to Cauchy’s principal numbers method.
We get,
e^\ln e^ \infty – e^ \ln e^\frac{1){0.5} + e^ \ln e^0.5 – e^ \ln e^\infty
(\infty - 0.2363) + (4.2321 - \infty)
Here we see that small numbers are added to uncountable infinity which preserve infinity value and so the equation becomes,
[Tes]\infty - \infty = 0

thus proving Riemann hypothesis that \xi has a root at \frac {1}{2}

Mathew Cherian

[\Tex]

[imag94]

I posted using Tex code exactly as given in the tutorial, and it seems not functioning in both my attempts. Please advice.

[g_edgar]

I posted using Tex code exactly as given in the tutorial, and it seems not functioning in both my attempts. Please advice.

do not capitalize ... tex not Tex

This has tex ... \frac{1}{2}
This is the same thing with Tex ... \frac{1}{2}
As you can probably see, it does not produce the intended result.

[D H]

Riemann conjured that the function
\xi = \int \frac {1}{\ln(x)}
has a root at \frac{1}{2} when s=2

Mathew, your function is not the Riemann zeta function and your hypothesis is not the Riemann hypothesis. The Riemann hypothesis claims that the only non-trivial zeroes the Riemann zeta function (i.e., values with non-zero imaginary part that are zeroes of the Riemann zeta function) have real part equal to 1/2.


Here we see that small numbers are added to uncountable infinity which preserve infinity value and so the equation becomes,
\infty - \infty = 0

thus proving Riemann hypothesis ...

Mathew, any proof based on infinity - infinity being zero is not a valid proof.

[ramsey2879]

Sequel to my Proof for Riemann Hypothesis.

Riemann conjured that the function
\xi = \int \frac {1}{\ln(x)}
has a root at \frac{1}{2} when s=2

Let
f ^-1 (x)=\frac{1}{\ln (x)}
f(x)= e^ \frac {1}{ln (x)}
Taking log on both sides, \frac {1}{\ln(x)} = \ln e^\frac{1}{1/ln (x)}
Integrating both sides,
\int \frac {1}{ln(x)}= \int \ln e^{1}{ln x}
\int \ln(e^\\frac {1}{ln (x)} = e^ \ln (e^\frac{1}{ln (x)}

Value of \xi function is got by taking the integral between 2 and that number.
Taking limits of r.h.s from \frac{1}{2} to 1 and from 1 to 2 and adding them up for finding the value of the \xi function at \frac {1}{2} due to Cauchy’s principal numbers method.
We get,
e^\ln e^ \infty – e^ \ln e^\frac{1){0.5} + e^ \ln e^0.5 – e^ \ln e^\infty
(\infty - 0.2363) + (4.2321 - \infty)
Here we see that small numbers are added to uncountable infinity which preserve infinity value and so the equation becomes,
\infty - \infty = 0

thus proving Riemann hypothesis that \xi has a root at \frac {1}{2}

Mathew Cherian



I tried correcting from Tex to tex. Otherwise no changes made. I believe you need to put {} around a few of the argments as needed. The subject is outside my scope, but it seems to me that you cannot ignore the 0.2363 or 4.2321 unless you are dealing with subtracting or adding fractions with infinity as the denominator of the fractions.