# Extract imaginary part of $\text{Li}_3\left(\frac{2}{3}-i \frac{2\sqrt{2}}{3}\right)$ in closed form

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

$\mathrm{Re}[\text{Li}_2(i)]=-\frac{\pi^2}{48}$

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

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

• 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. Commented Jan 6, 2016 at 23:09
• 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? Commented Jan 6, 2016 at 23:19
• I hope the question is more clear now Commented Jan 7, 2016 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:

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

• 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. ;) Commented Jan 10, 2016 at 15:22
• @DavidH "free of polylogs and imaginary numbers". That was the best of me. It would be nice to look behind the hypergeom expression. Commented Jan 10, 2016 at 15:35
• This has a very nice symmetry to my answer here, $$\Im\left[\operatorname{Li}_3\big(1+i\big)\right] =-\frac13\arcsin^3\left(\frac{\sqrt2}{2}\right)+\sqrt2\;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)$$ Commented Jun 24, 2019 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,

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