Timeline for An integral involving the argument of the Gamma function and the Riemann Hypothesis
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 18, 2018 at 21:06 | vote | accept | OneTwoOne | ||
| Dec 18, 2018 at 18:58 | comment | added | user64494 | The question is unclearly formulated: usually $\arg z$ stands for the main value of the argument of a complex number $z$ and the Arg[z] command realizes it. | |
| Dec 18, 2018 at 18:53 | comment | added | juan | @user64494 Plot it from 0 to 20, the problem is that the argument of gamma can be defined as a continuous function that sometimes cross pi, 3pi and so on. Of course you can consider the integral with the discontinuous argument. But usually when one speaks about the argument of Gamma one uses the continuous one. | |
| Dec 18, 2018 at 18:49 | comment | added | user64494 | Can you elaborate your comment? The plots Plot[Arg[Gamma[1/4 + I * t/2]], {t, 0, 10}] and Plot[Im[LogGamma[1/4 + I * t/2]], {t, 0, 10}] are identical | |
| Dec 18, 2018 at 18:39 | comment | added | juan | @user64494 You can not use here Arg[Gamma[]] because this is not the continuous argument. use instead Im[LogGamma[ and you will get my value. | |
| Dec 18, 2018 at 18:30 | comment | added | user64494 | Unfortunately, the output of the Mathematica's code N[-Pi/4*(EulerGamma + Log[4])] equals $-1.54214$. This is not in accordance with the numeric value NIntegrate[t * Arg[Gamma[1/4 + I*t/2]]/(1/4 + t^2)^2, {t, 0, Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15], i. e. $ -1.62060929754175$. | |
| Dec 18, 2018 at 17:37 | history | answered | juan | CC BY-SA 4.0 |