Timeline for An integral involving the argument of the Gamma function and the Riemann Hypothesis
Current License: CC BY-SA 4.0
27 events
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| Dec 23, 2018 at 6:55 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 21, 2018 at 22:10 | review | Close votes | |||
| Dec 26, 2018 at 3:01 | |||||
| Dec 20, 2018 at 20:33 | comment | added | juan | @OneTwoOne Your reasoning is not correct. The difference between $\arg\zeta(1/2+it)$ and the other expression you equals it is not bounded. Some argument of a complex number and another one may differ in any multiple of $2\pi$. We know this is not bounded. | |
| Dec 20, 2018 at 18:51 | comment | added | OneTwoOne | @juan, the singularities at the zeros $x_i$ can be treated/circumvented by considering $\lim_{\delta \rightarrow 0^+} (x_{i} + \delta)$. See the addendum. | |
| Dec 20, 2018 at 18:23 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 20, 2018 at 18:03 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 20, 2018 at 13:06 | comment | added | GH from MO | Consider the principal branch of the logarithm on $\mathbb{C}\setminus(-\infty,0]$. Let $\rho$ be a third root of unity. Check that $\log\rho^2$ is not equal to $2\log\rho$. Think about the reasons. | |
| Dec 20, 2018 at 12:43 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 20, 2018 at 12:09 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 20, 2018 at 11:47 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 20, 2018 at 11:42 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 18, 2018 at 22:27 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 18, 2018 at 22:25 | comment | added | juan | ... then you are making the $\arg\Gamma(1/4+it/2)$ discontinuous at certain points, but not the zeros of zeta. Nevertheless the two functions will be equal modulo $\pi$. | |
| Dec 18, 2018 at 22:23 | comment | added | juan | @OneTwoOne $\frac{x}{2}\log\pi-\arg\Gamma(1/4+it/2)$ is a very simple function, continuous and indefinitely differentiable. $\arg\zeta(1/2+ix)$ is continuous except at zeros of zeta (to simplify I assume the Riemann Hypothesis here) the function has jumps at the zeros of zeta. The equality you write is only true $\mod \pi$. But if you consider the function of user 64494 | |
| Dec 18, 2018 at 22:19 | comment | added | juan | @OneTwoOne What you have done have nothing to do with the Riemann Hypothesis. $\arg\zeta(1/2+it)$ contains many information about the zeros. $\Gamma(1/4+it/2)$ almost nothing. | |
| Dec 18, 2018 at 22:16 | comment | added | juan | @OneTwoOne This discontinuous function that uses user 64494 is not the same as you consider. RH is much more difficult than you think. | |
| Dec 18, 2018 at 22:13 | comment | added | juan | @OneTwoOne It is not so easy to explain. It is defined in the book by Titchmarsh Section 9.3. $\arg\zeta(1/2+iT)$ is obtained by continuous variation along the straight lines joining 2, $2+iT$, $1/2+iT$, starting with the value $0$. It is a discontinuous function. While $\arg\Gamma(1/4+it/2)$ is usually meaning as the continuous argument. So the relation you writes between then is not true. | |
| Dec 18, 2018 at 22:12 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 18, 2018 at 22:00 | comment | added | juan | @OneTwoOne No. This not disproof the Riemann hypothesis. when one uses $\arg\zeta(1/2+ix)$ one refers to the "discontinuous" arg defined in a certain specific way. Your formula for $\arg\zeta(1/2+it)$ is not correct. It is only an equality $\bmod \pi$ | |
| Dec 18, 2018 at 21:11 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 18, 2018 at 21:06 | vote | accept | OneTwoOne | ||
| Dec 18, 2018 at 21:06 | history | edited | OneTwoOne | CC BY-SA 4.0 |
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| Dec 18, 2018 at 18:34 | comment | added | user64494 | Its numeric value equals $-1.6206092975417 $. Isn't it enough? | |
| Dec 18, 2018 at 17:37 | answer | added | juan | timeline score: 16 | |
| Dec 18, 2018 at 16:35 | comment | added | მამუკა ჯიბლაძე | Did you try to use that $\arg$ is the imaginary part of $\log$? | |
| Dec 18, 2018 at 14:55 | review | First posts | |||
| Dec 18, 2018 at 15:38 | |||||
| Dec 18, 2018 at 14:51 | history | asked | OneTwoOne | CC BY-SA 4.0 |