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May 19, 2017 at 17:09 answer added user64494 timeline score: 0
May 19, 2017 at 9:48 answer added Zurab Silagadze timeline score: 3
May 17, 2017 at 20:12 comment added Carlo Beenakker and the imaginary part is $-i\frac{1}{12}\pi^3-i\pi\,{\rm Li}_2(-2)$, with the definition $\ln(-x)=i\pi+\ln x $ for $x<0$, which indeed confirms the missing $-i\pi\ln^2 3$.
May 17, 2017 at 17:58 comment added Robert Israel The real part appears to be $-(7/6) \zeta(3)$.
May 17, 2017 at 9:27 history edited J. M. isn't a mathematician
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May 17, 2017 at 9:25 history edited Zurab Silagadze CC BY-SA 3.0
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May 17, 2017 at 9:17 history asked Zurab Silagadze CC BY-SA 3.0