Timeline for Conjectured closed form of $\int\limits_0^1 \frac{\ln y \operatorname{Li}_2 (-y)}{1-y^2} \, dy$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 17 at 15:58 | comment | added | Jorge Zuniga | @Iosif_Pinelis It seems that link does not work currently. Take a look at researchgate.net/publication/… | |
| Apr 17 at 2:26 | comment | added | Iosif Pinelis |
@JorgeZuniga : I have tried PacletInstall[ CloudObject[ "https://www.wolframcloud.com/obj/pisco125/MultipleZetaValues-1.2.0.\ paclet"]] but got $Failed. Can you please help with this?
|
|
| Mar 17 at 21:44 | comment | added | Jorge Zuniga | Actually not. As the references that I placed indicate it, this is a mix-up of methods of Zhao's paper (mainly) and Valean's books. Summation by parts to evaluate $I$ is a technique used by Valean and many others. In $S$ sum of odd terms equal to sum of monotone and alternating is a basic technique. I used a couple of known weight 4 sums that were taken from Zhao's tables. The proof was short, correct, complete, in-time and pretty clear. | |
| Mar 17 at 21:17 | comment | added | Martin.s | Copied solution of coronel loan valean? | |
| Mar 13 at 12:01 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
edited body
|
| Mar 12 at 22:38 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
added 16 characters in body
|
| Mar 12 at 22:17 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
added 3977 characters in body
|
| Mar 7 at 19:15 | comment | added | Martin.s | The links you provided are challenging for me as an undergraduate student to comprehend. I'm seeking a conjectured closed form instead | |
| Mar 7 at 19:11 | comment | added | Martin.s | Thank you, @Jorge Zuniga, for the response. Unfortunately, I don't have Mathematica, so I can't use the decompose command | |
| Mar 7 at 19:06 | comment | added | Jorge Zuniga | @Martin.s, Thanks, but I think (and many readers as well, I guess) that this result is exactly what you were asking for. Anyway Mathematica has a command to decompose computation into steps if you want to learn how this result was obtained. Also the links I left provide the details. I do not see how you want to go further the question you have made. | |
| Mar 7 at 18:59 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
deleted 1 character in body
|
| Mar 7 at 18:37 | comment | added | Martin.s | You didn't get it. I'm asking what I can learn from a response given by Mathematica software | |
| Mar 7 at 18:17 | comment | added | Steven Stadnicki | @Martin.s It's very confusing what you're looking for if not a closed form for the integral. | |
| Mar 7 at 17:58 | comment | added | Martin.s | This is closed form not an answer | |
| Mar 7 at 17:50 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
added 205 characters in body
|
| Mar 7 at 17:31 | history | edited | Jorge Zuniga | CC BY-SA 4.0 |
added 128 characters in body
|
| Mar 7 at 17:24 | history | answered | Jorge Zuniga | CC BY-SA 4.0 |