We know that polylogarithms of complex argument sometimes have simple real and imaginary parts, e.g.


Is there a closed form (free of polylogs and imaginary numbers) for the imaginary part of

$\text{Li}_3\left(\frac{2}{3}-i \frac{2\sqrt{2}}{3}\right)$

  • $\begingroup$ Are you asking how to accurately calculate this? Putting the "closed-form" tag on a question is a poor substitute for spelling out what you want. When you edit your Question, please include more context, such as why the value of interest or what difficulty you encountered in solving for it. $\endgroup$ – hardmath Jan 6 '16 at 23:09
  • $\begingroup$ Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, make your question clear. Do you want a closed-form exact answer, a real approximation, or something else? $\endgroup$ – Rory Daulton Jan 6 '16 at 23:19
  • $\begingroup$ I hope the question is more clear now $\endgroup$ – user12588 Jan 7 '16 at 20:47

Inspired by this answer and by the comments below it, we could express it in terms of a generalized hypergeometric function as the following:

$$ \Im\left[\text{Li}_3\left(\frac{2}{3}-\frac{2\sqrt{2}}{3}i\right)\right] = \frac{1}{3}\arcsin^3\left(\frac{\sqrt3}{3}\right) - \frac{2\sqrt3}{3}{_4F_3}\!\left(\begin{array}c \tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32, \tfrac32,\tfrac32\end{array}\middle|\,\frac13\right). $$

  • 2
    $\begingroup$ Edifying though this expression may be, it's hard to regard it as anything else but a closed form for the hypergeometric term in terms of the trilogarithm, not the other way around. ;) $\endgroup$ – David H Jan 10 '16 at 15:22
  • $\begingroup$ @DavidH "free of polylogs and imaginary numbers". That was the best of me. It would be nice to look behind the hypergeom expression. $\endgroup$ – user153012 Jan 10 '16 at 15:35
  • $\begingroup$ This has a very nice symmetry to my answer here, $$\Im\left[\operatorname{Li}_3\big(1+i\big)\right] =-\frac13\arcsin^3\left(\frac{\sqrt2}{2}\right)+\sqrt2\;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)$$ $\endgroup$ – Tito Piezas III Jun 24 at 16:00

Inspired in turn by user 153012's answer which is similar to my answer in this post, then more generally, for any real $k>1$,

$$\Im\left[\operatorname{Li}_3\left(\frac2k\,\big(1\pm\sqrt{1-k}\big)\right)\right] =\color{red}\mp\frac13\arcsin^3\left(\frac1{\sqrt k}\right)\pm\frac2{\sqrt k}\;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\frac1k\right)$$

where the OP's case was just $k=3$.

Edit: Courtesy of Oussama Boussif in his answer here, there is also a broad identity for $\rm{Li}_2(x)$ but for the real part,

$$\Re\left[\rm{Li}_{2}\left(\frac{1}{2}+iq\right)\right]=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{\left(\frac{1+4q^2}{4}\right)}}^{2}-\frac{{\arctan{(2q)}}^{2}}{2} $$


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