Timeline for $\operatorname{Li}_{2} \left(\frac12 \right)$ vs $\operatorname{Li}_{2} \left(-\frac12 \right)$ : some long summation expressions
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| 2 days ago | comment | added | Username_qs | According to the numerical calculations, there won't be rational numbers $a_n$ such that $\operatorname{Li}_2(-\frac12)=a_1\zeta(2)+a_2\beta(2)+a_3\pi\ln2+a_4\ln^22$. And the similarity of the Direct Sum Conjecture on multiple zeta value is speculated, so I think there's no closed form for $\operatorname{Li}_2(-\frac12)$. | |
| Jul 5 at 4:20 | comment | added | Srini | Agreed, but I knew that from dilogarithm property relating dilog($z^2$) to dilog(z)+dilog(-z). But the wishful closed form in my mind should not have another dilog term in the result. | |
| Jul 5 at 4:12 | comment | added | FDP | $\operatorname{Li}_{2}(-\frac12)$ is equal to a rational combination of $\operatorname{Li}_{2}(\frac14)$, $\zeta(2)$,$\ln^2 2$. | |
| Jul 5 at 4:04 | comment | added | FDP | As i said, there is argument (not proof) that dilog(-1/2) is not equal to $a\ln 2+b\zeta(2)$ with $a,b$ rational numbers. | |
| Jul 5 at 4:01 | comment | added | Srini | Not expecting, but wishful thinking. Something like $\operatorname{Li}_{2}(\frac12)$ | |
| Jul 5 at 3:58 | comment | added | FDP | What kind of closed-form are you expecting for $\text{Li}_2\left(-\frac{1}{2}\right)$? Empirically one knows this is not a rational combination of $\zeta(2)$ and $\ln 2$. | |
| Jul 5 at 3:54 | comment | added | FDP | There is a more simpler series for $\text{Li}_2\left(-\frac{1}{2}\right)$ since $\displaystyle \text{Li}_2\left(-\frac{1}{2}\right)=\frac{1}{2}\int_0^1\frac{\ln x}{1+\frac{x}{2}}dx$ | |
| S Jul 5 at 1:18 | history | suggested | David Galea | CC BY-SA 4.0 |
Minor Latex Edits (formatting of Li)
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| Jul 5 at 0:57 | review | Suggested edits | |||
| S Jul 5 at 1:18 | |||||
| Jul 2 at 13:39 | history | edited | Srini |
Changed tags to add polylog
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| Jul 2 at 6:25 | history | asked | Srini | CC BY-SA 4.0 |