An equation involving Non-Trivial Zeros of the Riemann Zeta function
$begingroup$
$rho$ is a Non-Trivial Zero of the Riemann Zeta function if and only if
$$displaystyleint_1^{+infty} lfloor xrfloor x^{-2-rho} dx =int_1^{+infty} lfloor xrfloor { x }x^{-2-rho} dx $$
where $lfloor x rfloor$ is the floor function and ${ x} = x-lfloor x rfloor$.
Note that the left member is just $frac{zeta(rho +1)}{rho+1}$.
The above equation let us explore the Riemann Zeta function in the plane $Re(s)>1$ for finding the zeros in the critical strip.
If in some manner we can show that $rho$ is a zero iff $2Re(rho)-1+rho$ is a zero using the above equation, the Riemann Hypothesis will be true.
My ask however is only to prove the above equation.
integration complex-analysis equivalence-relations riemann-zeta riemann-hypothesis
asked Dec 7 '18 at 13:18
Saverio PicozziSaverio Picozzi
12
$endgroup$
add a comment |
$begingroup$
$rho$ is a Non-Trivial Zero of the Riemann Zeta function if and only if
$$displaystyleint_1^{+infty} lfloor xrfloor x^{-2-rho} dx =int_1^{+infty} lfloor xrfloor { x }x^{-2-rho} dx $$
where $lfloor x rfloor$ is the floor function and ${ x} = x-lfloor x rfloor$.
Note that the left member is just $frac{zeta(rho +1)}{rho+1}$.
The above equation let us explore the Riemann Zeta function in the plane $Re(s)>1$ for finding the zeros in the critical strip.
If in some manner we can show that $rho$ is a zero iff $2Re(rho)-1+rho$ is a zero using the above equation, the Riemann Hypothesis will be true.
My ask however is only to prove the above equation.
integration complex-analysis equivalence-relations riemann-zeta riemann-hypothesis
asked Dec 7 '18 at 13:18
Saverio PicozziSaverio Picozzi
12
$endgroup$
1
$begingroup$
Why making things complicated ? $rho $ is a non-trivial zero iff $int_0^{infty} (lfloor xrfloor-x) x^{-1-rho} dx =0$. The difference between the values of $int_1^{infty} { x} f(x)x^{-s-1} dx $ and $int_1^{infty} frac12 f(x) x^{-s-1} dx$ is what makes Dirichlet series complicated
$endgroup$
– reuns
Dec 7 '18 at 13:29
$begingroup$
Your integral representation is just $zeta (rho) / rho$. I have posted a relation involving $zeta(rho +1)$ when $rho$ is a non-trivial zero of the function.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:04
1
$begingroup$
You are using that $sint_1^infty lfloor xrfloor^2 x^{-s-1}dx = sum_n (2n-1) n^{-s}$, hiding it is making things complicated
$endgroup$
– reuns
Dec 7 '18 at 14:10
$begingroup$
Exactly, that's just what i was hiding.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:12
add a comment |
$begingroup$
$rho$ is a Non-Trivial Zero of the Riemann Zeta function if and only if
$$displaystyleint_1^{+infty} lfloor xrfloor x^{-2-rho} dx =int_1^{+infty} lfloor xrfloor { x }x^{-2-rho} dx $$
where $lfloor x rfloor$ is the floor function and ${ x} = x-lfloor x rfloor$.
Note that the left member is just $frac{zeta(rho +1)}{rho+1}$.
The above equation let us explore the Riemann Zeta function in the plane $Re(s)>1$ for finding the zeros in the critical strip.
If in some manner we can show that $rho$ is a zero iff $2Re(rho)-1+rho$ is a zero using the above equation, the Riemann Hypothesis will be true.
My ask however is only to prove the above equation.
integration complex-analysis equivalence-relations riemann-zeta riemann-hypothesis
asked Dec 7 '18 at 13:18
Saverio PicozziSaverio Picozzi
12
$endgroup$
$rho$ is a Non-Trivial Zero of the Riemann Zeta function if and only if
$$displaystyleint_1^{+infty} lfloor xrfloor x^{-2-rho} dx =int_1^{+infty} lfloor xrfloor { x }x^{-2-rho} dx $$
where $lfloor x rfloor$ is the floor function and ${ x} = x-lfloor x rfloor$.
Note that the left member is just $frac{zeta(rho +1)}{rho+1}$.
The above equation let us explore the Riemann Zeta function in the plane $Re(s)>1$ for finding the zeros in the critical strip.
If in some manner we can show that $rho$ is a zero iff $2Re(rho)-1+rho$ is a zero using the above equation, the Riemann Hypothesis will be true.
My ask however is only to prove the above equation.
integration complex-analysis equivalence-relations riemann-zeta riemann-hypothesis
integration complex-analysis equivalence-relations riemann-zeta riemann-hypothesis
asked Dec 7 '18 at 13:18
Saverio PicozziSaverio Picozzi
12
asked Dec 7 '18 at 13:18
Saverio PicozziSaverio Picozzi
12
edited Dec 7 '18 at 13:26
Saverio Picozzi
asked Dec 7 '18 at 13:18
Saverio PicozziSaverio Picozzi
12
asked Dec 7 '18 at 13:18
Saverio PicozziSaverio Picozzi
12
asked Dec 7 '18 at 13:18
Saverio PicozziSaverio Picozzi
12
12
1
$begingroup$
Why making things complicated ? $rho $ is a non-trivial zero iff $int_0^{infty} (lfloor xrfloor-x) x^{-1-rho} dx =0$. The difference between the values of $int_1^{infty} { x} f(x)x^{-s-1} dx $ and $int_1^{infty} frac12 f(x) x^{-s-1} dx$ is what makes Dirichlet series complicated
$endgroup$
– reuns
Dec 7 '18 at 13:29
$begingroup$
Your integral representation is just $zeta (rho) / rho$. I have posted a relation involving $zeta(rho +1)$ when $rho$ is a non-trivial zero of the function.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:04
1
$begingroup$
You are using that $sint_1^infty lfloor xrfloor^2 x^{-s-1}dx = sum_n (2n-1) n^{-s}$, hiding it is making things complicated
$endgroup$
– reuns
Dec 7 '18 at 14:10
$begingroup$
Exactly, that's just what i was hiding.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:12
add a comment |
1
$begingroup$
Why making things complicated ? $rho $ is a non-trivial zero iff $int_0^{infty} (lfloor xrfloor-x) x^{-1-rho} dx =0$. The difference between the values of $int_1^{infty} { x} f(x)x^{-s-1} dx $ and $int_1^{infty} frac12 f(x) x^{-s-1} dx$ is what makes Dirichlet series complicated
$endgroup$
– reuns
Dec 7 '18 at 13:29
$begingroup$
Your integral representation is just $zeta (rho) / rho$. I have posted a relation involving $zeta(rho +1)$ when $rho$ is a non-trivial zero of the function.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:04
1
$begingroup$
You are using that $sint_1^infty lfloor xrfloor^2 x^{-s-1}dx = sum_n (2n-1) n^{-s}$, hiding it is making things complicated
$endgroup$
– reuns
Dec 7 '18 at 14:10
$begingroup$
Exactly, that's just what i was hiding.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:12
1
1
$begingroup$
Why making things complicated ? $rho $ is a non-trivial zero iff $int_0^{infty} (lfloor xrfloor-x) x^{-1-rho} dx =0$. The difference between the values of $int_1^{infty} { x} f(x)x^{-s-1} dx $ and $int_1^{infty} frac12 f(x) x^{-s-1} dx$ is what makes Dirichlet series complicated
$endgroup$
– reuns
Dec 7 '18 at 13:29
$begingroup$
Why making things complicated ? $rho $ is a non-trivial zero iff $int_0^{infty} (lfloor xrfloor-x) x^{-1-rho} dx =0$. The difference between the values of $int_1^{infty} { x} f(x)x^{-s-1} dx $ and $int_1^{infty} frac12 f(x) x^{-s-1} dx$ is what makes Dirichlet series complicated
$endgroup$
– reuns
Dec 7 '18 at 13:29
$begingroup$
Your integral representation is just $zeta (rho) / rho$. I have posted a relation involving $zeta(rho +1)$ when $rho$ is a non-trivial zero of the function.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:04
$begingroup$
Your integral representation is just $zeta (rho) / rho$. I have posted a relation involving $zeta(rho +1)$ when $rho$ is a non-trivial zero of the function.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:04
1
1
$begingroup$
You are using that $sint_1^infty lfloor xrfloor^2 x^{-s-1}dx = sum_n (2n-1) n^{-s}$, hiding it is making things complicated
$endgroup$
– reuns
Dec 7 '18 at 14:10
$begingroup$
You are using that $sint_1^infty lfloor xrfloor^2 x^{-s-1}dx = sum_n (2n-1) n^{-s}$, hiding it is making things complicated
$endgroup$
– reuns
Dec 7 '18 at 14:10
$begingroup$
Exactly, that's just what i was hiding.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:12
$begingroup$
Exactly, that's just what i was hiding.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:12
add a comment |
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$begingroup$
Why making things complicated ? $rho $ is a non-trivial zero iff $int_0^{infty} (lfloor xrfloor-x) x^{-1-rho} dx =0$. The difference between the values of $int_1^{infty} { x} f(x)x^{-s-1} dx $ and $int_1^{infty} frac12 f(x) x^{-s-1} dx$ is what makes Dirichlet series complicated
$endgroup$
– reuns
Dec 7 '18 at 13:29
$begingroup$
Your integral representation is just $zeta (rho) / rho$. I have posted a relation involving $zeta(rho +1)$ when $rho$ is a non-trivial zero of the function.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:04
1
$begingroup$
You are using that $sint_1^infty lfloor xrfloor^2 x^{-s-1}dx = sum_n (2n-1) n^{-s}$, hiding it is making things complicated
$endgroup$
– reuns
Dec 7 '18 at 14:10
$begingroup$
Exactly, that's just what i was hiding.
$endgroup$
– Saverio Picozzi
Dec 7 '18 at 14:12